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https://ocw.mit.edu/courses/7-012-introduction-to-biology-fall-2004/7.012-fall-2004.zip
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It comes acquainted with different antigens. And recall that what we were talking about was the following, that there were several kinds of phagocytic cells. Phagocytic cells are cells that chew up other things, both macrophages and even more frequently, dendritic cells, many of which hang around lymph nodes by the way. They process antigens into oligopeptides. The oligopeptides get presented on the surface of these cells. Let's say, here's a macrophage, in the form of through the class 2 MHC molecules which are displayed on the surfaces of the cells. And, here's a typical oligopeptide that has been chewed up from one of the antigens that was previously internalized, eaten up by the macrophage, a dendritic cell, and then presented on the surface. Recall, then, we have an itinerant macrophage or dendritic cell. Could we turn up the sound just a little, just a notch? Thank you. And this dendritic cell or macrophage, I'll just call it a macrophage for the moment, is then moving largely through the lymph nodes, but wherever it moves, it's carrying along this oligopeptide. And recall that we liken this voyage of it to a Middle Eastern market where there's a lot of bazaar stalls on either side of the road, and where instead of the usual male merchants, there's a lot of female merchants hanging out. And these females are called T helper cells. They're a kind of T lymphocyte or T cell. TH refers to their function. And here, all of these T helper cells, I'll indicate each of them here as a pink circle. And these T helper cells display on their surface to blow them up to a large size a T cell receptor that is organized much the way immunoglobulin molecule and antibody molecule's organized. That is to say, it has variable and constant regions. It's generated through the rearrangement of antibody-like genes. But it only functions, this T cell receptor or as it's called in the trade the TCR, only functions to sense the presence of antigens in the extracellular space. In fact, it senses antigens in the context of the MHC class 2. So here's an MHC class 2 molecule. We can think of the MHC class 2 as being a hand, which is presenting this oligopeptide. I haven't drawn a hand, but you can pretend it's a hand. And this MHC class 2 is being presented by either a macrophage or a dendritic cell. And recall, we talked about the voyage of this macrophage or dendritic cell through this street here, and all these T helper cells are kind of lazily waiting along on the sidelines looking at what this phagocytic cell is hocking. Most of the T helper cells are totally uninterested in what he's hocking. But one of them is struck. It's love at first sight, this one over here, let's say, because her T cell receptor precisely recognizes this oligopeptide in the context of the MHC class 2 molecule. And, obviously I would like to draw thousands of these T helper cells here, each of which bears a different T cell receptor on her surface. I'm only showing one. And recall that after they make this encounter, the T helper cell gets all excited because she says, oh, I can't believe it, you have exactly the oligopeptide that's recognized by my receptor. And so, she gets all excited, and what she does is she proliferates because that's about all that excited cells can do. Well, they can do other things, but again, we don't want to talk about it. Anyhow, so this particular T helper cell undergoes a clonal expansion, and is now activated, i.e. activated not only psychologically but physiologically by having encountered the antigen-presenting cell. The MHC class 2, the macrophage is called an antigen-presenting cell. It's using its MHC class 2 molecules to do so. Macrophages and dendritic cells are pretty much similar in this respect. Macrophages go all over the body in the tissues. Dendritic cells tend to inhabit the lymph nodes. But from the point of view of our discussion today, we can imagine that they're functionally essentially equivalent. And having said that, now these T helper cells go looking for a congenial B cell. So, now we have a third actor in the drama, and the congenial B cell looks like this. And, the B cell has the following thing. The B cell has also on its surface, MHC class 2 molecules, which can be used in antigen presentation. But the B cell has an addition as we said last time, already, on its surface, IGM molecules. An IGM is a brand of antibody. Keep in mind that really one of the paradoxes that we haven't really fully settled on is the following question or the following issue. How is it that when a B cell sees a cognate antigen, why does it get stimulated? In other words, what is it that induces the B cell to start proliferating? Here's another version of what I showed you last time, where a B cell which makes the right antibody gets stimulated, but a B cell that doesn't does not get stimulated. And that's the issue we're wrestling with right now. So here on the surface of this B cell is an IGM molecule. Keep in mind an IGM molecule is an antibody molecule. Immunoglobulin means IG. It's the earliest form of antibody that's made. Once again, it is antigen specific. Its variable region has been rearranged through the fusion of different VDJ segments and somatic hypermutation. And, this B cell has a very interesting property. This is a naïve cell, and what this B cell does is as follows. It moves around the body, and if this B cell happens to find an antigen, which is recognized by its antibody, the IGM antibody. Then the B cell will bind this antigen using its IGM molecule to do so. More importantly, it will then internalize this antigen and chew it up into little pieces, and then present it on it surface via the MHC class 2 molecule. So, let's just review what we've been saying. Before, we were talking about macrophages and dendritic cells, which gobbled up whatever they could, processed whatever they gobbled up, and put it on the surface again as MHC class 2, and therefore the macrophages and the dendritic cells are really like sewer rates. They'll just chew on anything and they'll put it on their surface. They're totally promiscuous in what they present on their surface. But here, the B cell is doing something rather similar. But the B cell isn't presenting whatever it happens to stumble across on its surface. The B cell is extraordinarily selective at what it presents on its surface. It only presents on its surface those antigens which are recognized by its IGM molecule. So here, it uses its IGM molecule to grab hold of this antigen. It internalizes the antigen, and then presents it back on the surface as an MHC class 2 molecule. So, there's a profound contrast in the behavior of these two kinds of antigen-presenting cells. The macrophages and dendritic cells, they just gobble up everything, and whatever they find, they put on their surface. They don't care with MHC class 2. The B cell is extraordinarily selective and specific. It will pull in not through regular phagocytosis. It will pull in using its IGM receptor, specific antigens that are recognized by its IGM molecule, and then externalize it using its MHC class 2 molecule to do so. Let's go back to the drama of the activated T helper cell, and here's the activated T helper cell. We'll draw her in pink. The T helper cell has just had an encounter with a macrophage, or dendritic cell. And, she's just left this marketplace. And now, she's very excited. So here is her T cell receptor. And she's very excited. She's putting out all kinds of growth factors and multiplying all over the place. And she starts looking again now for a B cell with which he can react. Now, most of the B cells in the body will not have an epitope that she recognizes. Most of the B cells in the body will have picked up other kinds of things that happen to recognize by their T cell receptor, and will present it on the surface. A rare B cell will happen to internalize an antigen, and put on the surface, which is the same antigen that was recognized previously in the previous encounter. And so, now this T helper cell goes around looking for an attractive male. What's an attractive male for her? An attractive male for her is one whose MHC class 2 molecule is recognized directly in the context, and this oligopeptide is recognized by her T cell receptor. And so, she'll come over here and she'll say excitedly to the B cell, you can't believe what just happened. She will say, I was just there. I just went through the market. I was sitting there in the marketplace, and along came a macrophage, and presented me with an oligopeptide that exactly fit in my T cell receptor. And now, she says excitedly, here I find a B cell has exactly the same oligopeptide presenting it to me. Isn't that a coincidence? And the B cell says, come on lady, get to the point. And she says, I just had an encounter with a macrophage dendritic cell. I recognized the same oligopeptide in the macrophage dendritic cell that you have. And, the B cell says, well, I guess this must be some kind of meaningful encounter, and so these two cells get together. And what happens now is that the T cell, having recognized the oligopeptide, presented on the surface of the B cell now begins to send out signals to stimulate the B cell to proliferate. And this B cell now begins to proliferate. And this B cell now begins to proliferate, as is indicated on this overhead, and eventually it starts making IGM molecules. It makes more of them, and then through the class switching that we talked about last time, it'll make eventually IGG secreted antibody gamma globulin molecules. So you see here the three essential cell types that participate in this. And why is it so complicated? Because it's extraordinarily important that the immune system doesn't inadvertently make antibodies that are inappropriate to express, because as we said before, if a certain of those antibodies and indeed possibly many of them could be autoreactive. And what do I mean by autoreactive? I mean reactive with self. They could be antibodies that react with one's own tissues. And in so doing, they could create series kinds of autoimmune diseases. So, we have this sequence of failsafe reactions. So, when finally the decision for the B cell to get activated depends on a previous encounter with the same oligopeptide by a macrophage or dendritic cell, the T helper cell acting as an intermediary and now activating the B cell, once the T cell tells the B cell that the T cell has had a previous encounter with exactly the same oligopeptide, on that occasion being presented by a macrophage or dendritic cell. And that is actually the mechanism by which we get this clonal expansion of B cells in the immune system, and ultimately how we get the productive antibody molecules. I mean, this is the image I showed you earlier, but I never really explained to you what the biology behind that is. I just said that antigen encounter on the part of the B cell causes that B cell to enjoy clonal expansion. And now, we've gone through the detail of deciding how three different cell types interact, collaborate with one another to create the antibody response because this B cell then goes on to produce IGM as it already is doing, and then eventually IGG, and possibly a series of other immunoglobulins, IGE and IGA which have other purposes. Now, all of this actually is an important prelude to our main topic of discussion today, which is the disease of AIDS. And, let me just add one other detail to this because the ability of a T helper cell to recognize MHC class 2 molecules depends on another cell surface molecule expressed by the T helper cell. And this other T cell surface molecule is called CD4. CD4, we don't have to worry of what it stands for. CD4 is not an antigen specific receptor. CD4, instead, only recognizes MHC class 2 molecules no matter what they're carrying. So, there are MHC class 2 molecules which I've implied to you can carry thousands of different oligopeptides. CD4 doesn't care what's being carried by the MHC class 2. It just binds to MHC class 2 molecules, thereby telling the T helper cell that an encounter has been made with an antigen-presenting cell. So, the ligand for CD4 is part of the MHC class 2 molecule. Now, that all leaves us in a very nice segue to the whole disease of AIDS. Let's just remember how the disease of AIDS was discovered. In 1981, there were a group of five young men who were all subsequently determined to be gay, be homosexual, who were discovered in San Francisco to have a very unusual kind of immunodeficiency. They all had night sweats. They got different kinds of otherwise unusual diseases. For example one of the things they got was a disease called Kaposi's sarcoma, which was otherwise known only in old southern Italian and Jewish men, Kaposi's sarcoma. But these were young men, and they were neither southern Italian nor Jewish. They got pneumocystis carinii, which is a microbial infection of the lung. And, in fact, they got all kinds of herpes virus infections. And they were all seen in a cluster by an alert physician who saw something very unusual, and therefore said, perhaps correctly, that they had acquired immunodeficiency. Now, this acquired immunodeficiency is a syndrome. A syndrome, by the way, for your information, is a whole collection of symptoms that appear together. That's what a syndrome means. So, this term AIDS came from the fact that they had a whole series of symptoms. And, this was an acquired immunodeficiency rather than a congenital immunodeficiency rather than a congenital immunodeficiency because given the complexity of the immune system, you can imagine correctly that there are a lot of people in the world who were born with congenitally defective immune systems that are immunodeficient from birth because there's so many different proteins involved in regulating all of these different immune responses. But this was really different. It was an acquired immunodeficiency. It was seen in a very special subgroup, and so the race was on over the next two years to figure out what was going on. Now, by coincidence, starting in 1970-'71, retrovirus research had begun. And as it turned out, retrovirus search, President Nixon's war on cancer, retrovirus research was motivated largely by the notion that human cancers are caused by retrovirus infections. And, that led to the war on cancer. And the notion behind the war on cancer was totally wrong because it turns out that only a minute fraction of human cancers have anything to do with retrovirus infections, although as we've said earlier, retroviruses proved to be very important tools experimentally for discovering proto-oncogenes and oncogenes in the genome. But if you ask, what fraction of human cancers are actually due to a human being infected by a retrovirus? It's almost zero. It's a small fraction of a percent. Nonetheless, in the 1970s, there was an enormous effort made in trying to figure out all of the biology of retroviruses. And by the late 1970s, people concluded this was very interesting science. Indeed, proto-oncogenes and oncogenes were discovered, but that it was pretty irrelevant to understanding directly how human cancer arose, which human cancers could be explained rather by somatic mutations in the genome. In 1981, the AIDS infection arose. And, what happened subsequently is that there was a race on to try to find out what the infectious agent was because it seemed to be an infectious agent. It was being spread from one gay young man to another that was used to induce this. And within two years, by 1983, the culprit retrovirus had been found. I'm telling you this long song and dance to give you the following insight. If there had not been a decade of earlier retrovirus research, it could have taken the scientific community many, many years to figure out what was causing AIDS. But through happenstance, through a sheer stroke of luck, by the time the first individuals suffering from AIDS were encountered in '81, there was already a backlog of a decade's worth of detailed retrovirus research, which made it possible to discover, to discern almost within months what was causing it. And, the agent that was causing it was a retrovirus. The retrovirus here has indicated, very schematically, these artists' drawings never have any resemblance of what things really look like. And if they do, it's only by coincidence. Let me borrow your laser pointer here for a second. So here, and this is what a retrovirus looks like just to give you a feeling. In the center, there are two single stranded RNA molecules. The virus is diploid. There's two copies of the genome for reasons we still don't understand. Surrounding it is a so-called nucleocapsid, which is responsible for protecting the RNA molecules. These two pink dots are reverse-transcriptase molecules because as you'll recall, when retroviruses infect a cell, they carry the enzyme with them into the cell. You could say, why don't they make it after they get into the cell? And it's not totally obvious why, but this is what they do. There's another shell of proteins out here. And then, beyond that is a lipid bilayer. And this lipid bilayer is, as you may recall, stolen from cell from which the virus is protruding because if you look at retrovirus-infected cells, here's the plasma membrane of a retrovirus-infected cell. Here you can see a nucleo-capsid forming with the RNA molecules. And this shoves its way, protrudes its way through the plasma membrane, stealing a patch of plasma membrane from the infected cell, and at the same time this part of the plasma membrane carries with it viral glycoproteins. And viral glycoproteins, they're obviously glycosylated, as are many other extracellular proteins. And in this case, they're indicated with these yellow ovals, and these viral glycoproteins are used to attach to subsequently infected cells so what happens is that when the retrovirus gets out of the cell, I'll draw it again schematically here, it has this glycoproteins coat on it with the plasma membrane, and it uses these glycoproteins spikes. I just won't put the yellow ovals on them, to attach to cells which need to be infected. So, here's a target cell that needs to be infected. So, the target cell, and how does this virus know how to attach to this cell and not to other cells? Because on the surface of the target cell are certain cellular proteins, which are used for normal cell physiology, which are there, and which the virus has opportunistically developed an affinity for. So here on the surface of a target cell might be a normal cellular protein to which the viral glycoprotein combined. Or, if you want to get technical, this enables the virus particle to adsorb, to attach to. Notice the D here rather than the B, to adsorb to the surface of the target cell. Importantly, what's the cell surface protein of the target cell to which HIV virus adsorbs? It's our old friend CD4. I.e. the HIV particle likes to adsorb, preferentially adsorbs to the surface of cells that express the CD4 molecule on the surface. Note, by the way, that just five minutes ago, we described a totally different function of CD4. CD4 over here was said to represent the means by which the T helper cell can recognize MHC molecules being displayed on the surface of either of these dendritic cells or B cells. But here we see CD4 in a totally different context. Here, the CD4 represents the docking site to which the viral glycoprotein can attach, enabling the virus, which came to be called human immunodeficiency virus. In fact, the virus was discovered by two groups simultaneously, one of them called HDLV-3, the other called lymphadenopathy virus. The first group was American. The second group was French, and they allowed Herald Varmus, one of the co-discoverers of the proto-oncogene to act as sort of the judge to see what it would be called because there was great political tension. Would it get the American or the French name depending on which of the two warring scientists, and they were warring, could claim discovery? So, he had a Solomonic decision. He decided to name it human immunodeficiency virus. That was a compromise. And by the way, some people in less than charitable mood say, well, of course he named it human immunodeficiency virus because those are almost his own initials. But, I think that's unfair. These were his initials. Here's human immunodeficiency virus. He named it for a perfectly good reason. Anyhow, that broke the Franco-American diplomatic tension, and now one began to realize that HIV or human immunodeficiency virus attacked T helper cells and preferentially infected T helper cells by virtue of the ability of the virus particle to dock itself to the CD4 molecules presented on the surface of these cells. By the way, what happens afterwards, after the virus becomes adsorbed to the surface of an infectable cell such as a T helper cell. So, here's the virus particle. Here's the surface of a T helper cell. What happens then is these two lipid bilayers fuse with one another so that now they became one, and now the internal contents, the nucleocapsid which contains the RNA and the reverse transcriptase now has direct topological access into the cytoplasm of the cell. In fact, the glycoprotein, the yellow ovals there of human immunodeficiency virus actually had two functions. First, it specifically recognizes the CD4 molecules to which it then anchors or adsorbs the virus particle. And secondly, it also has fusing functions, i.e. it's capable of causing the lipid bilayer of the virion, or the virus particle, to fuse with that of the plasma membrane of the infected target cell. And, once it's in there, then the virus can begin to do its replication. Now, the replication of the HIV virus was already pretty well understood by the time that HIV was discovered in 1982-'83 because of this backlog of retrovirus research. And just to review for you how retroviruses replicate, RNA is put into the cell, single stranded RNA. It's called plus strand RNA because it is of the same polarity of the same strandedness as messenger RNA. If it were complementary to messenger RNA, then it would be called minus strand RNA. This is reverse transcribed by the reverse transcriptase, which is carried into the cell. RT stands for reverse transcriptase. And now, one gets a double stranded DNA molecule, a copy of the virus. And this DNA copy is sometimes called a provirus. And just to review, this provirus is then subsequently integrated into the chromosomal DNA of the cell. So, here's the provirus. Here's the chromosomal DNA. And then, this integrated provirus then serves as a template for making progeny plus-stranded RNA. And this progeny plus-stranded RNA, which is, by the way, forward transcribed by RNA polymerase too, which does the bulk of the heavy lifting in terms of making RNA in the nucleus, this plus-stranded RNA can have two functions recall. One, it can serve as a template on ribosomes for making viral proteins such as the viral capsid proteins. And two, the plus stranded RNA can in turn be encapsidated, i.e. it can become packaged. Encapsidate equals, it can become packaged by the viral proteins to make progeny virus particles, which can then bud, as I've indicated here, from the surface of the infected cell. And in fact, we can imagine three classes of viral proteins that are required for replication. First is the reverse transcriptase, which is encoded by the viral RNA. Second are the capsid proteins which carry the RNA, and third are the viral glycoproteins up here which these glycoprotein spikes which are trans-membrane proteins that protrude from the virion, and allow the virion to adsorb to the surface of infected cells. It turns out that this virus has become an extremely difficult virus to deal with. For most viruses that we have encountered over the last 100 years, one has had great success in making vaccines against these viruses including, as we discussed in great detail, poliovirus. In fact, for smallpox, another virus, the vaccine effort was so successful that about 20 years ago, the last case of smallpox finally occurred in Eritrea in northeast African when some herdsmen had the last documented case. And since that time, there have been no documented cases of smallpox in the wild, and there's only two or three stocks of smallpox virus surviving. One of them is in some type of research center is Moscow, and the other is probably in the Communicable Disease Center in Atlanta, Georgia. And, there's been great debate, by the way. Should one get rid of those surviving stalks, or should one keep them for research? By now, you guys aren't vaccinated against smallpox because nobody gets it anymore, and there's a certain risk of getting a small pox vaccine. I am, so I'm not worried, but maybe you should be because starting about 20-25 years ago, one stopped vaccinating people against smallpox because it just doesn't seem to be necessary. Why give them the risk of having some disease which happens in one out of a million vaccinees (sic) instead of just leaving them unvaccinated? Well, I digress. Back to HIV, the fact is we've had enormous lack of success in making a good kind of vaccine against HIV, and why is that? Well, one of the critical things is that HIV is attacking and replicating in the T helper cells, and the T helper cells it turns out are the lynch pains of the immune response. Keep in mind that the T helper cells that I've shown you in this diagram over here represent these critical cellular messengers between the dendritic cells and the macrophages on the one hand, and the B cells on the other. You wipe them out, and the ability to make new antibodies is totally compromised. It turns out the T helper cells can also help to make another class of cells which are called cytotoxic T cells, another kind of T cell, cytotoxic, and these cytotoxic T cells have on their surfaces T cell receptors, which they can use to recognize infected cells, and kill infected cells. So, the cytotoxic T cells aren't involved in making antibody responses at all. The cytotoxic T cells are involved in recognizing cells that are expressing unusual or foreign antigens on their surface, and killing those cells. That's the function of the cytotoxic T cells. Obviously, it's quite different from the helper T cells. But once again, the activation of the cytotoxic T cells, and empowering them to make these attacks on abnormal cells depends on the helper T cells. Once again, the helper T cells represent the lynch pins, the keystones, of the immune response. But, because of this tropism, and when I use the word tropism, I mean because of the desire of the virus to phase towards and infect a certain subset of cells in the body, this tropism of HIV for infecting and killing helper T cells, the production of antibodies is strongly compromised on the one hand, and the production of cytotoxic T cells is compromised on the other. There's another aspect of HIV infection which is also very insidious, and that's the following. It turns out that the body can initially make an immune response against an infecting HIV particle. And here's kind of what things look like. I hope this shows up here. Who could lend me a laser pointer again? Excellent, thank you. OK, so here's what happens. And, there's two graphs here. On one hand are the cytotoxic T cells, and their level is shown on the solid line here. So, look at the course of infection. It's plotted here in weeks and years. If you see what happens in a primary HIV infection, and here on the right on this ordinate here is the viral titer indicated on a log scale. So, this is a semi-log graph for the viral titer. And what you see over here is that when you initially infected it, there's an enormous burst of viral titer. It goes up by four or five orders of magnitude, and then it falls down dramatically by two or three orders of magnitude. And it goes on, and it remains depressed by two or three orders of magnitude below its initial height for a number of years. What's going on then? The immune system has come to grips with the presence of the HIV virus, and begins successfully to try to eliminate it. How does the immune system eliminate HIV? By two mechanisms. First of all, the immune system makes neutralizing antibodies of the sort that float through the serum, and are able to glom onto the virus particle, attach to the virus particle, and thereby prevent it from being infectious. And secondly, as I mentioned last time, the immune system also can recognize virus-infected cells and kill them. And by killing a virus-infected cell, the immune system prevents that cell from continuing to function as a factory for putting out new virus particles. So, there's two ways by which virus particles are eliminated, but note here that the virus infected cell which is critically important among all the cell types in the body are the T helper cells. So, certain components of the immune system are killing the T helper cells that are involved in harboring and producing HIV virus. So, there's an auto destruction on the part of the immune system. Look, at the same time, at the titer, at the number of CD4 cells, and they're indicated here on the left ordinate, in this case, cells per microliter. CD4 cells originally started up here. They go down by a factor of two or three for the first weeks, and then over a period of years there's this ongoing struggle between the HIV particle and the immune system as the number of CD4 cells, and the CD4 cells we've said before, the CD4 positive cells are these T helper cells as the number of these cells per microliter of blood progressively declines further and further and further. And finally, the number of CD4 positive cells, i.e. T helper cells, gets so low that the body is totally overwhelmed, and the patient then dies of an opportunistic infection. What do I mean by an opportunistic infection? Well, what I mean is that we are surrounded all the time by all kinds of microbes, which given the chance will kill us within a couple days. Keep in mind, I told you that there are in your gut, as many bacterial cells as there are the rest of your body. And some of these bacteria are really nasty. I remember, my grandfather got kicked in the belly by a horse, and three days later he was dead. Why? Because some of the bacteria got out of his gut, got into his peritoneal fluid cavity, and gone. This was the pre-antibiotic era, by the way, never new him very well because he died in 1916. I'm just telling you that your gut is full of all kinds of nasty thing on your skin. Not just on my skin, but on your skin there are billions of staph aureus bacteria. They're just waiting to cause you problems. Don't look. It's OK. Don't look too closely. They're just waiting to cause a nasty infection as well. Everyday we breathe in all kinds of awful microbes, fungi, and all these kinds of things including pneumocystis. And, rarely do we get sick because of the extraordinary competence of the immune system to respond to such a diversity of infectious agents, and to hold them at bay. In the 20th century, the percentage of people who die of infectious diseases has plummeted both because of the immune system and because we're eating healthier up to a point, and because of antibiotics and antifungals. But if the immune system is defective, all the antibiotics in the world, and all the antifungal agents can't save a patient if their CD4 cells get down very, very low because these antifungals always worked as collaborators with the immune system. They get rid of a bulk of the infection, but the immune system has to wipe out the residue. And what you see here is a struggle going on for a period of three, four, five, six years where the viral titer is successfully held low, and then all of a sudden as the immune system weakens the viral titer goes up to high levels, wipes out the residual T helper cells, and death invariably ensues. Now, I've given you one reason why the immune system can't deal with this virus, because virtually all other viruses attack various tissues throughout our body, but they don't attack the immune system itself. Here, we're having a virus which is attacking the defense of the body, that is to say, the immune system. So, one reason is the continuing depletion of the T helper cells. They can regenerate themselves for a period of time, very impressively long period of time, four, five, six, seven years. But ultimately, they get worn out, they die. Another reason is this is antigenic variation. Now, if you look at the retrovirus particle, what you see is on the surface the glycoprotein. It's right here. And, the glycoprotein is used by antibodies to recognize and bind to the virus particle and neutralize it, same as with poliovirus. But let's imagine, as happens to be the case, that the virus is highly error prone in replicating its genome. When I say error prone, I mean that instead of the host cell, polymerase, which makes ultimately one mistake out of a billion, the viral replication machinery makes mistakes all the time. It's quite defective in the fidelity and the faithfulness with which it replicates nucleic acid. That means that after each cycle of replication, there are in effect mutant viruses that have been produced, mutant progeny viruses, and where the mutation rate instead of being one in 10-9 might be one in 10-2 or one in 10-3. And that means that there are continually novel variants of HIV being produced in a person's body. Let's imagine that that person has developed antibodies against the viral glycoprotein of the virus that initially infected him or her. Let's imagine that. And those antibodies are successful in eliminating most of the virus particles of the sort that initially infected that individual. But now we can imagine the possibility that in the imperative weeks or months, a new strain of HIV will arise within that individual's body a mutant strain in which the sequences that code the viral glycoprotein had been changed slightly. And now, the viral glycoprotein has changed slightly its epitopes. And the initially developed neutralizing antibody that recognized the initial cohort of virus coming into the cell in the individual no longer works because the virus has undertaken a strategy of immune evasion, sometimes it's called immunoevasion, in which now the viral glycoprotein, although it's still competent to affect a replication cycle to adsorb and fuse to the surface of an affected cell. Many of the epitopes, many of the oligopeptide antigens on the surface of the glycoprotein have been changed slightly through amino acid substitutions, through point mutations. And hence, the initially developed antibody, which previously was successful in glomming on and neutralizing this virus particle is rendered ineffective. And now, this second wave, this new strain of HIV will grow up and expand in that individual, and once again provoke a new immune response. And, the same cycle will repeat it. The second strain will now soon be eliminated, but while the elimination is going on, there's strong Darwinian selective pressure favoring the outgrowth of yet another mutant strain which is not recognized by either of the two initial antibody responses. And so, over this period of many years, what's happening is that the virus in the immune system are playing continual cat and mouse games with one another. The immune system goes after the virus; the virus moves over here; the immune system goes after that; and so you have a succession of antigenic variance. Here's one variant. Here's another variant. Here's another variant, and so forth. By the time the immune system succeeds in getting rid of the first variant, a new variant has appeared, and then the immune system ramps up its defenses and tries to get rid of that. And, it succeeds almost. But by the time that has happened, yet a third variant has appeared. And so, there are these continual clonal successions. A clonal succession represents a time where one clonal virus explodes, expands. It's soon eliminated, collapses, and then another clone comes up and expands. And this goes on. This works OK for about four, five, six years. But ultimately, the ability of the T helper cells to replenish themselves and to continue to mount an effective immune response fails. There's yet another aspect of HIV infection which is so insidious, and that's the following. Let's look at the viral life cycle right here, and the provirus, remember the provirus is this thing right here, which is integrated into the chromosomal DNA. And, we can assume that this provirus is transcribed by RNA polymerase too, and I will tell you that the promoter of the provirus is carried in by the proviral DNA, and actually depends on transcription factors that are present in the T helper cell. In fact, you'll recall that the T helper cell gets excited sometimes, and other times it's not excited. And she gets excited if I can attach gender to a T helper cell, when she encounters macrophages and dendritic cells, and/or when she encounters B cells. Other times, the T helper cell is kind of quiet and unactivated (sic). And, what happens when a T helper cell gets activated through these encounters? The T helper cell starts making her own transcription factors, which are used in order to facilitate these complex biological interactions with both antigen-presenting cells, macrophages and dendritic cells, and later on, B cells. And those same transcription factors that the T helper cell uses to turn on its own expression program are used by the provirus to transcribe its own DNA, i.e. HIV has evolved a proviral promoter sequence here which takes advantage of transcription factors that are present uniquely in an activated T cell. And when those transcription factors are available, not only does the T cell become activated, but the provirus becomes transcribed because these transcription factors now enable RNA polymerase 2 of the host to transcribe the provirus. But let's imagine now, if we follow that scenario to its conclusion, what happens when the T cell is not activated? When the T cell is quiescent, when it's quiet, these transcription factors are unavailable to attach to the promoter of the integrated provirus, and as a consequence the provirus will not be transcribed. It won't make RNA, and in that situation, how will anybody know that there's an integrated provirus in there? Well, the provirus is not being transcribed. The transcripts aren't being used to make viral protein. So in effect, the only evidence for the existence of HIV in the cell is this segment of DNA. In other words, this provirus can hide out in an unactivated quiescent T cell indefinitely. And, the immune system can't know that there's a provirus hiding out in this T helper cell because it's not being transcribed. And therefore, one can have a quiescent T helper cell, and several other cell types in the body, macrophages also, which aren't transcribing their proviruses. Well, you'll say, so what? It doesn't make any difference. If it's not being transcribed, it's not going to hurt the individual. But, keep in mind that the idea of getting rid of a viral infection is to eliminate all traces of a viral genome from an infected individual, and that's what happens with smallpox and with poliovirus, and with measles and virtually all the other infections we have. But here, we have a situation where the viral genome can hide out in a latent or inapparent (sic) configuration. There's no way to know it's there, and it may reemerge days, weeks, months, even years later because this previously transcriptionally silent provirus may suddenly be present in a cell which suddenly becomes activated. And now, an individual who "thought" quote unquote that he or she had gotten rid of HIV infection all of a sudden realizes there are viral genomes still hiding out in the body. And what that means is that in effect, it's absolutely impossible to rid the body of HIV infection ever. Once an individual is infected, in fact, that individual is infected for life. There's no way on Earth what we have at present of getting rid of the viral infection because the viral genome is always hiding out here or there in different interstices of the immune system, hiding out in transcriptionally silent state. Of course, we have very effective drugs against HIV now. Some drugs inhibit the reverse transcriptase. Others inhibit the processing of the capsid proteins, that is, the proteins which these capsid proteins happen to be cleaved from a large, high molecular-weight protein precursor into individual proteins, and there's an inhibitor of the protease that cleaves these proteins to the mature size. And, those drugs together hold the viral infection at bay for maybe 10, 15, 20 years. But keep in mind that even though the viral infection is being stopped by these drugs, first of all, the virus is always hiding out in the bodies of such individuals in this latent, hidden form, and secondly, there may be a slow depletion of their T helper cells in spite of the effectiveness of these drugs. On that cheerful note, I wish you a good day.
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https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/8.04-spring-2016.zip
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PROFESSOR: We have to ask what happens here? This series for h of u doesn't seem to stop. You go a 0, a 2, a 4. Well, it could go on forever. And what would happen if it goes on forever? So if it goes on forever, let's calculate what this aj plus 2 over aj as j goes to infinity. Let's see how the coefficients vary as you go higher and higher up in the polynomial. That should be an interesting thing. So I pass the aj that is on the right side and divide it, and now on the right-hand side there's just this product of factors. And as j goes to infinity, it's much larger than 1 or e, whatever it is, and the 2 and the 1 in the denominator. So this goes like 2j over j squared. And this goes roughly like 2 over j. So as you go higher and higher up, by the time j is a billion, the next term is 2 divided by a billion. And they are decaying, which is good, but they're not decaying fast enough. That's a problem. So let's try to figure out if we know of a function that decays in a similar way. So you could do it some other way. I'll do it this way. e to the u squared-- let's look at this function-- is this sum from n equals 0 to infinity 1 over n u squared to the n. So it's u to the 2n-- 1 over n factorial, sorry. So now, since we have j's and they jump by twos, these exponents also here jump by two. So that's about right. So let's think of 2n being j, and therefore this becomes a sum where j equals 0, 2, 4, and all that of 1 over-- so even j's. 2n is equal to j-- so j over 2 factorial over 1, and then you have u to the j. So if I think of this as some coefficient c sub j times u to the j, we've learned that c sub j is equal to 1 over j divided by two factorial. In which case, if that is true, let's try to see what this cj plus 2 over cj-- the ratio of two consecutive coefficients in this series. Well, cj plus 2 would be j plus 2 over 2 factorial, like this. That's the numerator, because of that formula. And the denominator would have just j over 2 factorial. Now, these factorials make sense. You don't have to worry that they are factorials of halves, because j is even. And therefore, the numerators are even-- divided by 2. These are integers. These are ordinary factorials. There are factorials of fractional numbers. You've seen them probably in statistical physics and other fields, but we don't have those here. This is another thing. So this cancels. If you have a number and the number plus 1, which is here, you get j plus 2 over 2, which is 2 over j plus 2. And that's when j is largest, just 2 over j, which is exactly what we have here. So this supposedly nice, innocent function, polynomial here-- if it doesn't truncate, if this recursive relation keeps producing more and more and more terms forever-- will diverge. And it will diverge like so. If the series does not truncate, h of u will diverge like e to the u squared. Needless to say, that's a disaster. Because, first, it's kind of interesting to see that here, yes, you have a safety factor, e to the minus u squared over 2. But if h of u diverges like e to the u squared, you're still in trouble. e to the u squared minus u squared over 2 is e to the plus u squared over 2. And it actually coincides with what we learned before, that any solution goes like either plus or minus u squared over 2. So if h of u doesn't truncate and doesn't become a polynomial, it will diverge like e to the u squared, and this solution will diverge like e to the plus u squared over 2, which was a possibility. And it will not be normalizable. So that's basically the gist of the argument. This differential equation-- whenever you work with arbitrary energies, there's no reason why the series will stop. Because e there will have to be equal to 2j plus 1, which is an integer. So unless je is an integer, it will not stop, and then you'll have a divergent-- well, not divergent; unbounded-- far of u that is impossible to normalize. So the requirement that the solution be normalizable quantizes the energy. It's a very nice effect of a differential equation. It's very nice that you can see it without doing numerical experiments, that what's going on here is an absolute requirement that this series terminates. So here, phi of u would go like e to the u squared over 2, what we mentioned there, and it's not a solution. So if the series must terminate, the numerator on that box equation must be 0 for some value of j, and therefore there must exist a j such that 2j plus 1 is equal to the energy. So basically, what this means is that these unit-free energies must be an odd integer. So in this case, this can be true for j equals 0, 1, 2, 3. In each case, it will terminate the series. With j equals 0, 1, 2, or 3 there, you get some values of e that the series will terminate. And when this series terminates, aj plus 2 is equal to 0. Because look at your box equation. aj, you got her number, and then suddenly you get this 2j plus 1 minus e. And if that's 0, the next one is zero. So, yes, you get something interesting even for j equals 0. Because in that case, you can have a0, but you will have no a2, just the constant. So I will write it. So if aj plus 2 is equal to zero, h of u will be aj u to the j plus aj minus 2u to the j minus 2. And it goes down. The last coefficient that exists is aj, and then you go down by two's. So let's use the typical notation. We call j equals n, and then the energy is 2n plus 1. The h is an u to the n plus an minus 2. You do the n minus 2, and it goes on. If n is even, it's an even solution. If n is odd, it's an odd solution. And the energy e, remember, was h omega over 2 times e-- so 2n plus 1. So we'll move the 2 in, and e will be equal to h omega n plus 1/2. And n in all these solutions goes from 0, 1, 2, 3. We can call this the energy en. So here you see another well-known, famous fact that energy levels are all evenly spaced, h omega over 2, one by one by one-- except that there's even an offset for n equals 0, which is supposed to be the lowest energy state of the oscillator. You still have a 1/2 h bar omega. This is just saying that if you have the potential, the ground state is already a little bit up. You would expect that-- you know there's no solutions with energy below the lowest point of the potential. But the first solution has to be a little bit up. So it's here and then they're all evenly spaced. And this begins with E0; for n equals 0, e1. And there's a little bit of notational issues. We used to call the ground state energy sometimes e1, e2, e3, going up, but this time it is very natural to call it E0 because it corresponds to n equals 0. Sorry. Those things happen. No, it's not an approximation. It's really, in a sense, the following statement. Let me remind everybody of that statement. When you have even or odd solutions, you can produce a solution that you may say it's a superposition, but it will not be an energy eigenstate anymore. Because the even solution that stops, say, at u to the 6 has some energy, and the odd solution has a different energy. So these are different energy eigenstates. So the energy eigenstates, we prove for one-dimensional potentials, are not chosen to be even or odd for bound states. They are either even or odd. You see, a superposition-- how do we say like that? Here we have it. If this coefficient is even, the energy sum value-- if this coefficient is odd, the energy will be different. And two energy eigenstates with different energies, the sum is not an energy eigenstate. You can construct the general solution by superimposing, but that would be general solutions of the full time-dependent Schrodinger equation, not of the energy eigenstates. The equation we're aiming to solve there is a solution for energy eigenstates. And although this concept I can see now from the questions where you're getting, it's a subtle statement. Our statement was, from quantum mechanics, that when we would solve a symmetric potential, the bound states would turn out to be either even or odd. It's not an approximation. It's not a choice. It's something forced on you. Each time you find the bound state, it's either even or it's odd, and this turned out to be this case. You would have said the general solution is a superposition, but that's not true. Because if you put a superposition, the energy will truncate one of them but will not truncate the other series. So one will be bad. It will do nothing. So if this point is not completely clear, please insist later, insist in recitation. Come back to me office hours. This point should be eventually clear. Good. So what are the names of these things? These are called Hermite polynomials. And so back to the differential equation, let's look at the differential equations when e is equal to 2n plus 1. Go back to the differential equation, and we'll write d second du squared Hn of u. That will be called the Hermite polynomial, n minus 2udHn du plus e minus 1. But e is 2n plus 1 minus 1 is 2n Hn of u is equal to 0. This is the Hermite's differential equation. And the Hn's are Hermite polynomials, which, conventionally, for purposes of doing your algebra nicely, people figured out that Hn of u is convenient if-- it begins with u to the n and then it continues down u to the n minus 2 and all these ones here. But here people like it when it's 2 to the n, u to the n-- a normalization. So we know the leading term must be u to the n. If you truncate with j, you've got u to the j. You truncate with n, you get u to the n. Since this is a linear differential equation, the coefficient in front is your choice. And people's choice has been that one and has been followed. A few Hermite polynomials, just a list. H0 is just 1. H1 is 2u. H2 is 4u squared minus 2. H3 is our last one, 8u cubed minus 12u, I think. I have a little typo here. Maybe it's wrong. So you want to generate more Hermite polynomials, here is a neat way that is used sometimes. And these, too, are generating functional. It's very nice actually. You will have in some homework a little discussion. Look, you put the variable z over there. What is z having to do with anything? u we know, but z, why? Well, z is that formal variable for what is called the generating function. So it's equal to the sum from n equals 0 to infinity. And you expand it kind of like an exponential, zn over n factorial. But there will be functions of u all over there. If you expand this exponential, you have an infinite series, and then you have to collect terms by powers of z. And if you have a z to the 8, you might have gotten from this to the fourth, but you might have gotten it from this to the 3 and then two factors of this term squared or a cross-product. So after all here, there will be some function of u, and that function is called the Hermite polynomial. So if you expand this with Mathematica, say, and collect in terms of u, you will generate the Hermite polynomials. With this formula, it's kind of not that difficult to see that the Hermite polynomial begins in this way. And how do you check this is true? Well, you would have to show that such polynomials satisfy that differential equation, and that's easier than what it seems. It might seem difficult, but it's just a few lines. Now, I want you to feel comfortable enough with this, so let me wrap it up, the solutions, and remind you, well, you had always u but you cared about x. So u was x over a. So let's look at our wave functions. Our wave functions phi n of x will be the Hermite polynomial n of u, which is of x over a, times e to the minus u squared over 2, which is minus x squared over 2a squared. And you should remember that a squared is h bar over m omega. So all kinds of funny factors-- in particular, this exponential is e to the minus x squared m omega over h squared over 2. I think so-- m omega over 2h bar. Let me write it differently-- m omega over 2h bar x squared. That's that exponential, and those are the coefficients. And here there should be a normalization constant, which I will not write. It's a little messy. And those are the solutions. And the energies en were h bar omega over 2 n plus 1/2, so E0 is equal to h bar omega over 2. E1 is 3/2 of h bar omega, and it just goes on like that.
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https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/8.05-fall-2013.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality, educational resources for free. To make a donation or to view additional materials, from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: All right. So, this homework that is due on Friday contains some questions on the harmonic oscillator. And the harmonic oscillator is awfully important. I gave you notes on that. And I want to use about half of the lecture, perhaps a little less, to go over some of those points in the notes concerning the harmonic oscillator. After that, we're going to begin, essentially, our study of dynamics. And we will give the revision, today, of the Schrodinger equation. It's the way Dirac, in his textbook on quantum mechanics, presents the Schrodinger equation. I think it's actually, extremely insightful. It's probably not the way you should see it the first time in your life. But it's a good way to think about it. And it will give you a nice feeling that this Schrodinger equation is something so fundamental and so basic that it would be very hard to change or do anything to it and tinker with it. It's a rather complete theory and quite beautiful [? idea. ?] So we begin with the harmonic oscillator. And this will be a bit quick. I won't go over every detail. You have the notes. I think that's pretty much all you need to know. So we'll leave it at that. So the harmonic oscillator is a quantum system. And as quantum systems go, they're inspired by classical systems. And the classical system is very famous here. It's the system in which, for example, you have a mass and a spring. And it does an oscillation for which the energy is written as p squared over 2m plus 1/2 m, omega squared, x squared. And m omega squared is sometimes called k squared, the spring constant. And you are supposed to do quantum mechanics with this. So nobody can tell you this is what the harmonic oscillators in quantum mechanics. You have to define it. But since there's only one logical way to define the quantum system, everybody agrees on what the harmonic oscillator quantum system is. Basically, you use the inspiration of the classical system and declare, well, energy will be the Hamiltonian operator. p will be the momentum operator. And x will be the position operator. And given that these are operators, will have a basic commutation relation between x and p being equal to i h-bar. And that's it. This is your quantum system. Hamiltonian is-- the set of operators that are relevant for this are the x the p, and the energy operator that will control the dynamics. You know also you should specify a vector space, the vector space where this acts. And this will be complex functions on the real line. So this will act in wave functions that define the vector space, sometimes called Hilbert space. It will be the set of integrable functions on the real line, so complex functions on the real line. These are your wave functions, a set of states of the theory. All these complex functions on the real line work. I won't try to be more precise. You could say they're square integrable. That for sure is necessary. And we'll leave it at that. Now you have to solve this problem. And in 804, we discussed this by using the differential equation and then through the creation annihilation operators. And we're going do it, today, just through creation and annihilation operators. But we want to emphasize something about this Hamiltonian and something very general, which is that you can right the Hamiltonian as say 1/2m, omega squared, x squared. And then you have plus p squared, m squared, omega squared. And a great solution to the problem of solving the Hamiltonian-- and it's the best you could ever hope-- is what is called the factorization of the Hamiltonian, in which you would manage to write this Hamiltonian as some operator times the dagger operator. So this is the ideal situation. It's just wonderful, as you will see, if you can manage to do that. If you could manage to do this factorization, you would know immediately what is the ground state energy, how low can it go, something about the Hamiltonian. You're way on your way of solving the problem. If you could just factorize it. Yes? AUDIENCE: [INAUDIBLE] if you could just factorize it in terms of v and v instead of v dagger and v? PROFESSOR: You want to factorize in which way instead of that? AUDIENCE: Would it be helpful, if it were possible, to factor it in terms of v times v instead of v dagger? PROFESSOR: No, no, I want, really, v dagger. I don't want v v. That that's not so good. I want that this factorization has a v dagger there. It will make things much, much better. So how can you achieve that? Well, it almost looks possible. If you have something like this, like a squared plus b squared, you write it as a minus ib times a plus ib. And that works out. So you try here, 1/2 m, omega squared, x minus ip over m omega, x plus ip over m omega. And beware that's not quite right. Because here, you have cross terms that cancel. You have aib b and minus iba. And they would only cancel if a and b commute. And here they don't commute. So it's almost perfect. But if you expand this out, you get the x squared for sure. You get this term. But then you get an extra term coming from the cross terms. And please calculate it. Happily, it's just a number, because the commutator of x and b is just a number. So the answer for this thing is that you get, here, x squared plus this is equal to this, plus h-bar over m omega, times the unit operator. So here is what you could call v dagger. And this is what we'd call v. So what is your Hamiltonian? Your Hamiltonian has become 1/2 m, omega squared, v dagger v, plus, if you multiply out, H omega times the identity. So we basically succeeded. And it's as good as what we could hope or want, actually. I multiply this out, so h-bar omega was the only thing that was left. And there's your Hamiltonian. Now, in order to see what this tells you, just sandwich it between any two states. Well, this is 1/2 m, omega squared, psi, v dagger, v, psi, plus 1/2 half h, omega. And assume it's a normalized state, so it just gives you that. So this thing is the norm of the state, v psi. You'd think it's dagger and it's this. So this is the norm squared of v psi. And therefore that's positive. So H, between any normalized state, is greater than or equal to 1/2 h-bar omega. In particular, if psi is an energy eigenstate, so that H psi is equal to E psi. If psi is an energy eigenstate, then you have this. And back here, you get that the energy must be greater than or equal to 1/2 h omega, because H and psi gives you an E. The E goes out. And you're left with psi, psi, which is 1. So you already know that the energy is at least greater than or equal to 1/2 h omega. So this factorization has been very powerful. It has taught you something extremely nontrivial about the spectrum of the Hamiltonian. All energy eigenstates must be greater than or equal to 1/2 h omega. In fact, this is so good that people try to do this for almost any problem. Any Hamiltonian, probably the first thing you can try is to establish a factorization of this kind. For the hydrogen atom, that factorization is also possible. There will be some homework sometime later on. It's less well known and doesn't lead to useful creation and annihilation operators. But you can get the ground state energy in a proof that you kind of go below that energy very quickly. So a few things are done now to clean up this system. And basically, here I have the definition of v and v dagger. Then you define a to be square root of m omega over 2 h-bar, v. And a dagger must be m omega over 2 h-bar v dagger. And I have not written for you the commutator of v and v dagger. We might as well do the commutator of a and a dagger. And that commutator turns out to be extremely simple. a with a dagger is just equal to 1. Now things that are useful, relations that are useful is-- just write what v is in here so that you have a formula for a and a dagger in terms of x and p. So I will not bother writing it. But it's here already. Maybe I'll do the first one. m omega over 2 h-bar. v is here would be x, plus ip over m omega. And you can write the other one there. So you have an expression for a and a dagger in terms of x and p. And that can be inverted as well. And it's pretty useful. And it's an example of formulas that you don't need to know by heart. And they would be in any formula sheet. And the units and all those constants make it hard to remember. But here they are. So you should know that x is a plus a dagger up to a constant. And p is a dagger minus a. Now p is Hermitian, that's why there is an i here. So that this, this anti-Hermitian, the i becomes a Hermitian operator. x is manifestly Hermitian, because a plus a dagger is. Finally, you want to write the Hamiltonian. And the Hamiltonian is given by the following formula. You know you just have to put the v and v dagger, what they are in terms of the creation, annihilation operators. So v dagger, you substitute a dagger. v, you go back here and just calculate it. And these calculations really should be done. It's something that is good practice and make sure you don't make silly mistakes. So this operator is so important it has been given a name. It's called the number operator, N. And its eigenvalues are numbers, 0, 1, 2, 3, all these things. And the good thing about it is that, once you are with a's and a daggers, all this m omega, h-bar are all gone. This is all that is happening here. The basic energy is h-bar omega. Ground state energies, what we'll see is 1/2 h-bar omega. And this is the number operator. So this is written as h-bar omega, number operator-- probably with a hat-- like that. So when you're talking about eigenvalues, as we will talk soon, or states for which these thing's are numbers, saying that you have a state that is an eigenstate of the Hamiltonian is exactly the same thing as saying that it's an eigenstate of the number operator. Because that's the only thing that is an operator here. There's this plus this number. So this number causes no problem. Any state multiplied by a number is proportional to itself. But it's not true that every state multiplied by a dagger a is proportional to itself. So being an eigenstate of N means that acting on a state, N, gives you a number. But then H is just N times the number. So H is also an eigenstate. So eigenstates of N or eigenstates of H are exactly the same thing. Now there's a couple more properties that maybe need to be mentioned. So I wanted to talk in terms of eigenvalues. I would just simply write the energy eigenvalue is therefore equal h-bar omega, the number eigenvalue-- so the operator is with a hat-- plus 1/2. So in terms of eigenvalues, you have that. From here, the energy is greater than 1/2 h omega. So the number must be greater or equal than 0 on any state. And that's also clear from the definition of this operator. On any state, the expectation value of this operator has to be positive. And therefore, you have this. So two more properties that are crucial here are that the Hamiltonian commuted with a is equal to minus h omega a and that the Hamiltonian committed with a dagger is plus h omega a dagger. Now there is a reasonably precise way of going through the whole spectrum of the harmonic oscillator without solving differential equations, almost to any degree, and trying to be just very logical about it. It's possible to deduce the properties of the spectrum. So I will do that right now. And we begin with the following statement. We assume there is some energy eigenstate. So assume there is a state E such that the Hamiltonian-- for some reason in the notes apparently I put hats on the Hamiltonian, so I'll start putting hats here-- so that the states are labeled by the energy. And this begins a tiny bit of confusion about the notation. Many times you want to label the states by the energy. We'll end up labeling them with the number operator. And then, I said, it will turn out, when the number operator is 0, we'll put a 0 in here. And that doesn't mean 0 energy. It means energy equal 1/2 h-bar omega. So if you assume there is an energy eigenstate, that's the first step in the construction. You assume there is one. And what does that mean? It means that this is a good state. So it may be normalized. It may not be normalized. In any case, it should be positive. I put first the equal, but I shouldn't put the equal. Because we know in a complex vector space, if a state has 0 norm, it's 0. And I want to say that there's really some state that is non-0, that has this energy. If the state would be 0, this would become a triviality. So this state is good. It's all good. Now with this state, you can define, now, two other states, acting with the creation, annihilation operators. I didn't mention that name. But a dagger is going to be called the creation operator. And this is the destruction or annihilation operator. And we built two states, E plus is a dagger acting on E. And E minus is a acting on E. Now you could fairly ask a this moment and say, well, how do you know these states are good? How do you know they even exist? How do you know that if you act with this, don't you get an inconsistent state? How do you know this makes sense? And these are perfectly good questions. And in fact, this is exactly what you have to understand. This procedure can give some funny things. And we want to discuss algebraically why some things are safe and why some things may not quite be safe. And adding an a dagger, we will see it's safe. While adding a's to the state could be fairly unsafe. So what can be bad about the state? It could be a 0 state, or it could be an inconsistent state. And what this an inconsistent state? Well, all our states are represented by wave functions. And they should be normalizable. And therefore they have norms that are positive, norms squared that are positive. Well you may find, here, that you have states that have norms that are negative, norm squareds that are negative. So this thing that should be positive, algebraically you may show that actually you can get into trouble. And trouble, of course, is very interesting. So I want to skip this calculation and state something that you probably checked in 804, several times, that this state has more energy than E and, in fact, has as much energy as E plus h-bar omega. Because a dagger, the creation operator, adds energy, h-bar omega. And this subtracts energy, h-bar omega. This state has an energy, E plus, which is equal to E plus h-bar omega. And E minus is equal to E minus h-bar omega. Now how do you check that? You're supposed to act with a Hamiltonian on this, use the commutation relation that we wrote up there, and prove that those are the energy eigenvalues. So at this moment, you can do the following. So these states have energies, they have number operators, they have number eigenvalues. So we can test, if these states are good, by computing their norms. So let's compute the norm, a dagger on E, a dagger on E for the first one. And we'll compute a E, a E. We'll do this computation. We just want to see what this is. Now remember how you do this. An operator acting here goes with a dagger into the other side. So this is equal to E a, a dagger, E. Now a, a dagger is not quite perfect. It differs from the one that we know is an eigenvalue for this state, which is the number operator. So what is a, a dagger in terms of N? Well, a, a dagger-- it's something you will use many, many times-- is equal to a commutator with a dagger plus a dagger a. So that's 1 plus the number operator. So this thing is E 1 plus the number operator acting on the state E. Well, the 1 is clear what it is. And the number operate is clear. If this has some energy E, well, I can now what is the eigenvalue of the number operator because the energy on the number eigenvalues are related that way. So I will simply call it the number of E and leave it at that. Times EE. So in here, the computation is easier because it's just E a dagger a E. That's the number, so that's just NE times EE. OK, so these are the key equations we're going to be using to understand the spectrum quickly. And let me say a couple of things about them. So I'll repeat what we have there, a dagger E a dagger E is equal to 1 plus NE EE. On the other hand, 888 aE aE is equal to NE EE. OK, so here it goes. Here is the main thing that you have to think about. Suppose this state was good, which means this state has a good norm here. And moreover, we've already learned that the energy is greater than some value. So the number operator of this state could be 0-- could take eigenvalue 0. But it could be bigger than 0, so that's all good. Now, at this stage, we have that-- for example, this state, a dagger E has number one higher than this one, than the state E because it has an extra factor of the a dagger which adds an energy of h omega. Which means that it adds number of 1, So if this state has some number, this state has a number which is bigger. So suppose you keep adding. Now, look at the norm of this state. The norm of this state is pretty good because this is positive and this is positive. If you keep adding a daggers here, you always have that this state, the state with two a daggers, you could use that to find its norm. You could use this formula, put in the states with one a dagger here. But the states with one a dagger already has a good norm. So this state with two a daggers would have also good norm. So you can go on step by step using this equation to show that as long as you keep adding a daggers, all these states will have positive norms. And they have positive norms because their number eigenvalue is bigger and bigger. And therefore, the recursion says that when you add one a dagger, you don't change the sign of this norm because this is positive and this is positive, and this keeps happening. On the other hand, this is an equation that's a lot more dangerous. Because this says that in this equation, a lowers the number. So if this has some number, NE, this has NE minus 1. And if you added another a here, you would use this equation again and try to find, what is the norm of things with two a's here? And put in the one with one a here and the number of that state. But eventually, the number can turn into a negative number. And as soon as the number turns negative, you run into trouble. So this is the equation that is problematic and the equation that you need to understand. So let me do it in two stages. Here are the numbers. And here is 5 4, 3, 2, 1, 0. Possibly minus 1, minus 2, and all these numbers. Now, suppose you start with a number that is an integer. Well, you go with this equation. This has number 4. Well, you put an a. Now it's a state with number 3, but its norm is given 4 times that. So it's good. Now you go down another 1, you have a state with number 3, with number 2, with number 1, with number 0. And then if you keep lowering, you will get minus 1, which is not so good. We'll see what happens. Well, here you go on and you start producing the states-- the state with number 4, state with number 3, state with number 2, state with number 1. And state here, let's call it has an energy E prime. And it has number equal 0. Number of E prime equals 0. So you look at this equation and it says aE prime times aE prime is equal N E prime times E prime E prime. Well, you obtain this state at E prime, and it was a good state because it came from a state that was good before. And therefore, when you did the last step, you had the state at 1 here, with n equals to 1, and then that was the norm of this state. So this E E prime is a fine number positive. But the number E prime is 0. So this equation says that aE prime aE prime is equal to 0. And if that's equal to 0, the state aE prime must be equal to 0. And 0 doesn't mean the vacuum state or anything. It's just not there. There's no such state. You can't create it. You see, aE prime would be a state here with number minus 1. And everything suggests to us that that's not possible. It's an inconsistent state. The number must be less than 1. And we avoided the inconsistency because this procedure said that as you go ahead and do these things, you eventually run into this state E prime at 0 number. But then, you get that the next state is 0 and there's no inconsistency. Now, that's one possibility. The other possibility that could happen is that there are energy eigenstates that have numbers which are not-- well, I'll put it here. That are not integer. So maybe you have a state here with some number E which is not an integer. It doesn't belong to the integers. OK, so what happens now? Well, this number is positive. So you can lower it and you can put another state with number 1 less. Also, not integer and it has good norm. And this thing has number 2.5, say. Well, if I use the equation again, I put the 2.5 state with its number 2.5 and now I get the state with number 1.5 and it still has positive norm. Do it again, you find the state with 0.5 number and still positive norm. And looking at this, you start with a state with 0.5, with 0.5 here. And oops, you get a state that minus 0.5. And it seems to be good, positive norm. But then, if this is possible, you could also build another state acting with another a. And this state is now very bad because the N for this state was minus 1/2. And therefore, if you put that state, that state at the minus 1/2, you get the norm of the next one that has one less. And this state now is inconsistent. So you run into a difficulty. So what are the ways in which this difficulty could be avoided? What are the escape hatches? There are two possibilities. Well, the simplest one would be that the assumption is bad. There's no state with fractional number because it leads to inconsistent states. You can build them and they should be good, but they're bad. The other possibility is that just like this one sort of terminated, and when you hit 0-- boom, the state became 0. Maybe this one with a fractional one, before you run into trouble you hit a 0 and the state becomes 0. So basically, what you really need to know now on the algebraic method cannot tell you is how many states are killed by a. If maybe the state of 1/2 is also killed by a, then we would have trouble. Now, as we will see now, that's a simple problem. And it's the only place where it's interesting to solve some equation. So the equation that we want to solve is the equation a on some state is equal to 0. Now, that equation already says that this possibility is not going to happen. Why? Because from this equation, you can put an a dagger on this. And therefore, you get that NE is equal to 0. This is the number operator, so the eigenvalue of the number operator, we call it NE. So in order to be killed by a, you have to have NE equals 0. So in the fractional case, no state will be killed and you would arrive to an inconsistency. So the only possibility is that there's no fractional states. So it's still interesting to figure out this differential equation, what it gives you. And why do we call it a differential equation? Because a is this operator over there. It has x and ip. So the equation is x a E equals 0, which is square root of m omega over 2 h bar x x plus ip over m omega on E equals 0. And you've translated these kind of things. The first term is an x multiplying the wave function. We can call it psi E of x. The next term, the coefficient in front is something you don't have to worry, of course. It's just multiplying everything, so it's just irrelevant. So have i over m omega. And p, as you remember, is h bar over i d dx of psi E of x zero. So it's so simple differential equation, x plus h bar over m omega d dx on psi E of x is equal to 0. Just one solution up to a constant is the Gaussian that you know represents a simple harmonic oscillator. So that's pretty much the end of it. This ground state wave function is a number times the exponential of minus m omega over 2 h bar x squared. And that's that. This is called the ground state. It has N equals 0 represented as a state. We say this number is N equals 0. So this state is the thing that represents this psi E. In other words, psi E of x is x with 0. And that 0 is a little confusing. Some people think it's the 0 vector. That's not good. This is not the 0 vector. The 0 vector is not a state. It's not in the Hilbert space. This is the ground state. Then, the worst confusion is to think it's the 0 vector. The next confusion is to think it's 0 energy. That's not 0 energy, it's number equals 0. The energy is, therefore, 1/2 h bar omega. And now, given our discussion, we can start building states with more oscillators. So we build a state with number equal 1, which is constructed by an a dagger on the vacuum. This has energy 1 h bar omega more. It has number equal to 1. And that's sometimes useful to just make sure you understand why N on a dagger on the vacuum is a dagger a a dagger on the vacuum. Now, a kills the vacuum, so this can be replaced by the commutator, which is 1. And therefore, you're left with a dagger on the vacuum. And that means that the eigenvalue of n hat is 1 for this state. Moreover, this state is where normalized 1 with 1 actually gives you a good normalization if 0 is well-normalized. So we'll take 0 with 0 to be 1, the number 1. And that requires fixing that N0 over here. Now, these are things that you've mostly seen, so I don't want to say much more about them. I'd rather go through the Schrodinger thing that we have later. So let me conclude by just listing the general states, and then leaving for you to read what is left there in the notes so that you can just get an appreciation of how you use it. And with the practice problems, you'll be done. So here it is. Here is the answer. The n state is given by 1 over square root of n factorial a dagger to the n acting on the vacuum. And these n states are such that m with n is delta mn. So here we're using all kinds of things. First, you should check this is well normalized, or read it and do the calculations. And these are, in fact, orthogonal unless they have the same number of creation operators are the same number. Now, that had to be expected. These are eigenstates of a Hermitian operator. The N operator is Hermitian. Eigenstates of a Hermitian operator with different eigenvalues are always orthogonal to each other. If you have eigenstates of a Hermitian operator with the same eigenvalue, if you have a degeneracy, you can always arrange them to make them orthogonal. But if the eigenvalues are different, they are orthogonal. And there's no degeneracies in this spectrum whatsoever. You will, in fact, argue that because there's no degeneracy in the ground state, there cannot be degeneracy anywhere else. So this result, this orthonormality is really a consequence of all the theorems we've proven. And you could check it by doing the algebra and you would start moving a and a daggers. And you would be left with either some a's or some a daggers. If you're left with some a's, they would kill the thing on the right. If you're left with some a daggers, it would kill the thing on the left. So this can be proven. But this is just a consequence that these are eigenstates of the Hermitian operator n that have different eigenvalues. And therefore, you've succeeded in constructing a full decomposition of the state space of the harmonic oscillator. We spoke about the Hilbert space. Are now very precisely, see we can say this is u0 plus u1 plus u2 where uk is the states of the form alpha k, where N on k-- maybe I should put n here. It looks nicer. n. Where N n equal n n. So every one-dimensional subspace is spanned by that state of number n. So you have the states of number 0, states of number 1, states of number 2. These are all orthogonal subspaces. They add up to form everything. It's a nice description. So the general state in this system is a complex number times the state with number 0 plus the complex number states of number 1, complex number, and that. Things couldn't have been easier in a sense. The other thing that you already know from 804 is that if you try to compute expectation values, most of the times you want to use a's and a daggers. So the typical thing that one wants to compete is on the state n, what is the uncertainty in x on the state n? How much is it? What is the uncertainty of momentum on the energy eigenstate of number n? These are relatively straightforward calculations. If you have to do the integrals, each one-- by the time you organize all your constants-- half an hour, maybe 20 minutes. If you do it with a and a daggers, this computation should be five minutes, or something like that. We'll see that done on the notes. You can also do them yourselves. You probably have played with them a bit. So this was a brief review and discussion of them spectrum. It was a little detailed. We had to argue things carefully to make sure we don't assume things. And this is the way we'll do also with angular momentum in a few weeks from now. But now I want to leave that, so I'm going to take questions. If there are any questions on this logic, please ask. Yes. AUDIENCE: [INAUDIBLE] for how you got a dagger, a, a dagger, 0, 2 dagger, 0? PROFESSOR: Yes, that calculation. So let me do at the step that I did in words. So at this place-- so the question was, how did I do this computation? Here I just copied what N is. So I just copied that. Then, the next step was to say, since a kills this, this is equal to a dagger times a a dagger minus a dagger a. Because a kills it. And I can add this, it doesn't cost me anything. Now, I added something that is convenient, so that this is a dagger commutator of a with a dagger on 0. This is 1, so you get that. It's a little more interesting when you have, for example, the state 2, which is 1 over square root of 2 a dagger a dagger on 0. I advise you to try to calculate n on that. And in general, convince yourselves that n is a number operator, which means counts the number of a daggers. You'll have to use that property if you have N with AB. It's N with A B and then A N with B. The derivative property of the bracket has to be used all the time. So Schrodinger dynamics, let's spend the last 20 minutes of our lecture on this. So basically, it's a postulate of how evolution occurs in quantum mechanics. So we'll state it as follows. What is time in quantum mechanics? Well, you have a state space. And you see the state space, you've seen it in the harmonic oscillator is this sum of vectors. And these vectors were wave functions, if you wish. There's no time anywhere there. There's no time on this vector space. This vector space is an abstract vector space of functions or states, but time comes because you have clocks. And then you can ask, where is my state? And that's that vector on that state space. And you ask the question a littler later and the state has moved. It's another vector. So these are vectors and the vectors change in time. And that's all the dynamics is in quantum mechanics. The time is sort of auxiliary to all this. So we must have a picture of that. And the way we do this is to imagine that we have a vector space H. And here is a vector. And that H is for Hilbert space. We used to call it in our math part of the course V, the complex vector space. And this state is the state of the system. And we sometimes put the time here to indicate that's what it is. At time t0, that's it. Well, at time t, some arbitrary later time, it could be here. And the state moves. But one thing is clear. If it's a state of a system, if we normalize it, it should be of unit length. And we can think of a sphere in which this unit sphere is the set of all the tips of the vectors that have unit norm. And this vector will move here in time, trace a trajectory, and reach this one. And it should do it preserving the length of the vector. And in fact, if you don't use a normalized vector, it has a norm of 3. Well, it should preserve that 3 because you'd normalize the state once and forever. So we proved in our math part of the subject that an operator that always preserves the length of all vectors is a unitary operator. So this is the fundamental thing that we want. And the idea of quantum mechanics is that psi at time t is obtained by the action of a unitary operator from the state psi at time t0. And this is for all t and t0. And this being unitary. Now, I want to make sure this is clear. It can be misinterpreted, this equation. Here, psi at t0 is an arbitrary state. If you had another state, psi prime of t0, it would also evolve with this formula. And this U is the same. So the postulate of unitary time evolution is that there is this magical U operator that can evolve any state. Any state that you give me at time equal 0, any possible state in the Hilbert space, you plug it in here. And by acting with this unitary operator, you get the state at the later time. Now, you've slipped an extraordinary amount of physics into that statement. If you've bought it, you've bought the Schrodinger equation already. That is going to come out by just doing a little calculation from this. So the Schrodinger equation is really fundamentally, at the end of the day, the statement that this unitary time evolution, which is to mean there's a unitary operator that evolves any physical state. So let's try to discuss this. Are there any questions? Yes. AUDIENCE: So you mentioned at first that in the current formulation [INAUDIBLE]? PROFESSOR: A little louder. We do what in our current formulation? AUDIENCE: So if you don't include time [INAUDIBLE]. PROFESSOR: That's right. There's no start of the vector space. AUDIENCE: Right. So is it possible to consider a vector space with time? PROFESSOR: Unclear. I don't think so. It's just nowhere there. What would it mean, even, to add time to the vector space? I think you would have a hard time even imagining what it means. Now, people try to change quantum mechanics in all kinds of ways. Nobody has succeeded in changing quantum mechanics. That should not be a deterrent for you to try, but should give you a little caution that is not likely to be easy. So we'll not try to do that. Now, let me follow on this and see what it gives us. Well, a few things. This operator is unique. If it exists, it's unique. If there's another operator that evolves states the same way, it must be the same as that one. Easy to prove. Two operators that attack the same way on every state are the same, so that's it. Unitary, what does it mean that u t, t0 dagger times u t, t0 is equal to 1? Now, here these parentheses are a little cumbersome. This is very clear, you take this operator and you dagger it. But it's cumbersome, so we write it like this. This means the dagger of the whole operator. So this is just the same thing. OK, what else? u of t0, t0, it's the unit operator. If the times are the same, you get the unit operator for all t0 because you're getting psi of t0 here and psi of t0 here. And the only operator that leaves all states the same is the unit operator. So this unitary operator must become the unit operator, in fact, for the two arguments being equal. Composition. If you have psi t2, that can be obtained as U of t2, t1 times the psi of t1. And it can be obtained as u of t2, t1, u of t1, t0, psi of t0. So what do we learn from here? That this state itself is u of t2, t0 on the original state. So u of t2, t0 is u of t2, t1 times u of t1, t0. It's like time composition is like matrix multiplication. You go from t0 to t1, then from t1 to t2. It's like the second index of this matrix. In the first index of this matrix, you are multiplying them and you get this thing. So that's composition. And then, you have inverses as well. And here are the inverses. In that equation, you take t2 equal to t0. So the left-hand side becomes 1. And t1 equal to t, so you get u of t0, t be times u of t, t0 is equal to 1, which makes sense. You propagate from t0 to t. And then from t to t0, you get nothing. Or if it's to say that the inverse of an operator-- the inverse of this operator is this one. So to take the inverse of a u, you flip the arguments. So I'll write it like that, the inverse minus 1 of t, t0. You just flip the arguments. It's u of t0, t. And since the operator is Hermitian, the dagger is equal to the inverse. So the inverse of an operator is equal to the dagger. so t, t0 as well. So this one we got here. And Hermiticity says that the dagger is equal to the inverse. Inverse and dagger are the same. So basically, you can delete the word "inverse" by flipping the order of the arguments. And since dagger is the same as inverse, you can delete the dagger by flipping the order of the arguments. All right, so let's try to find the Schrodinger equation. So how c we c the Schrodinger equation? Well, we try obtaining the differential equation using that time evolution over there. So the time evolution is over there. Let's try to find what is d dt of psi t. So d dt of psi of t is just the d dt of this operator u of t, t0 psi of t0. And I should only differentiate that operate. Now, I want an equation for psi of t. So I have here psi of t0. So I can write this as du of t, t0 dt. And now put a psi at t. And then, I could put a u from t to t0. Now, this u of t and t0 just brings it back to time t0. And this is all good now, I have this complicated operator here. But there's nothing too complicated about it. Especially if I reverse the order here, I'll have du dt of t, t0 and u dagger of t, t0. And I reverse the order there in order that this operator is the same as that, the one that is being [INAUDIBLE] that has the same order of arguments, t and t0. So I've got something now. And I'll call this lambda of t and t0. So what have I learned? That d dt of psi and t is equal to lambda of t, t0 psi of t. Questions? I don't want to loose you in their derivation. Look at it. Anything-- you got lost, notation, anything. It's a good time to ask. Yes. AUDIENCE: Just to make sure when you differentiated the state by t, the reason that you don't put that in the derivative because it doesn't have a time [INAUDIBLE] necessarily, or because-- oh, because you're using the value at t0. PROFESSOR: Right. Here I looked at that equation and the only part that has anything to do with time t is the operator, not the state. Any other comments or questions? OK, so what have we learned? We want to know some important things about this operator lambda because somehow, it's almost looking like a Schrodinger equation. So we want to see a couple of things about it. So the first thing that I will show to you is that lambda is, in fact, anti-Hermitian. Here is lambda. I could figure out, what is lambda dagger? Well, lambda dagger is you take the dagger of this. You have to think when you take the dagger of this thing. It looks a little worrisome, but this is an operator. This is another operator, which is a time derivative. So you take the dagger by doing the reverse operators and daggers. So the first factor is clearly u of t, t0. And then the dagger of this. Now, dagger doesn't interfere at all with time derivatives. Think of the time derivative-- operator at one time, operator at another slightly different time. Subtract it. You take the dagger and the dagger goes through the derivative. So this is d u dagger t, t0 dt. So I wrote here what lambda dagger is. You have here what lambda is. And the claim is that one is minus the other one. It doesn't look obvious because it's supposed to be anti-Hermitian. But you can show it is true by doing the following-- u of t, t0 u dagger of t, t0 is a unitary operator. So this is 1. And now you differentiate with respect to t. If you differentiate with respect to t, you get du dt of t, t0 u dagger of t, t0 plus u of t, t0 du dagger of t, t0 equals 0 because the right-hand side is 1. And this term is lambda. And the second term is lambda dagger. And they add up to 0, so lambda dagger is minus lambda. Lambda is, therefore, anti-Hermitian as claimed. Now, look. This is starting to look pretty good. This lambda depends on t and t0. That's a little nasty though. Why? Here is t. What is t0 doing here? It better not be there. So what I want to show to you is that even though this looks like it has a t0 in there, there's no t0. So we want to show this operator is actually independent of t0. So I will show that if you have lambda of t, t0, it's actually equal to lambda of t, t1 for any t1. We'll show that. Sorry. [LAUGHTER] PROFESSOR: So this will show that you could take t1 to be t0 plus epsilon. And take the limit and say the derivative of this with respect of t0 is 0. Or take this to mean that it's just absolutely independent of t0 and t0 is really not there. So if you take t1 equal t dot plus epsilon, you could just conclude from these that this lambda with respect to t0 is 0. No dependence on t0. So how do we do that? Let's go a little quick. This is du t, t0 dt times u dagger of t, t0. Complete set of states said add something. We want to put the t1 here. So let's add something that will help us do that. So let's add t, t0 and put here a u of t0, t1 and a u dagger of t0, t1. This thing is 1, and I've put the u dagger of t, t0 here. OK, look at this. T0 and t1 here and t dot t1 there like that. So actually, we'll do it the following way. Think of this whole thing, this d dt is acting just on this factor. But since it's time, it might as well be acting on all of this factor because this has no time. So this is d dt on u t, t0 u t0, t1. And this thing is u of t1m t0. The dagger can be compensated by this. And this dagger is u of t0, t. This at a t and that's a comma. t0, t. Yes. OK, so should I go there? Yes. We're almost there. You see that the first derivative is already d dt of u of t, t1. And the second operator by compensation is u of t1, t, which is the same as u dagger of t, t1. And then, du of t, t1 u dagger of t, t1 is lambda of t, t1. So it's a little sneaky, the proof, but it's totally rigorous. And I don't think there's any step you should be worried there. They're all very logical and reasonable. So we have two things. First of all, that this quantity, even though it looks like it depends on t0, we finally realized that it does not depend on t0. So I will rewrite this equation as lambda of t. And lambda of t is anti-Hermitian, so we will multiply by an i to make it Hermitian. And in fact, lambda has units of 1 over time. Unitary operators have no units. They're like numbers, like 1 or e to the i phi, or something like that-- have no units. So this has units of 1 over time. So if I take i h bar lambda of t, this goes from lambda being anti-Hermitian-- this operator is now Hermitian. This goes from lambda having units of 1 over time to this thing having units of energy. So this is a Hermitian operator with units of energy. Well, I guess not much more needs to be said. If that's a Hermitian operator with units of energy, we will give it a name called H, or Hamiltonian. i h bar lambda of t. Take this equation and multiply by i h bar to get i h bar d dt of psi is equal to this i h bar lambda, which is h of t psi of t. Schrodinger equation. So we really got it. That's the Schrodinger equation. That's the question that must be satisfied by any system governed by unitary time evolution. There's not more information in the Schrodinger equation than unitary time evolution. But it allows you to turn the problem around. You see, when you went to invent a quantum system, you don't quite know how to find this operator u. If you knew u, you know how to evolve anything. And you don't have any more questions. All your questions in life have been answered by that. You know how to find the future. You can invest in the stock market. You can do anything now. Anyway, but the unitary operator then gives you the Hamiltonian. So if somebody tells you, here's my unitary operator. And they ask you, what is the Hamiltonian? You go here and calculate I h bar lambda, where lambda is this derivative. And that's the Hamiltonian. And we conversely, if you are lucky-- and that's what we're going to do next time. If you have a Hamiltonian, you try to find the unitary time evolution. That's all you want to know. But that's a harder problem because you have a differential equation. You have h, which is here , and you are to find u. So it's a first-order matrix differential equation. So it's not a simple problem. But why do we like Hamiltonians? Because Hamiltonians have to do with energy. And we can get inspired and write quantum systems because we know the energy functional of systems. So we invent a Hamiltonian and typically try to find the unitary time operator. But logically speaking, there's not more and no less in the Schrodinger equation than the postulate of unitary time evolution. All right, we'll see you next week. In fact-- [APPLAUSE] Thank you.
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https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/8.06-spring-2018.zip
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PROFESSOR: Today, we continue with scattering. And we begin by reviewing what were the main ideas that have already been explored. And here they are. I've summarized the main results on the blackboard. And we begin with an expansion of our solutions for the case of central potentials. This psi effect represents a solution away from the scattering center. The scattering center is this region where there is a potential that the particles fill. And this solution that we've written on top is a solution written for spherically symmetric potentials in the case that the potential is 0. That is, that solution is valid away from the potential. We constructed it with the free particle solutions-- the jl and the nl of kr-- of the spherical Schrodinger or radial Schrodinger equation. In particular, we identified these objects called phase shifts that we discussed in the process of scattering with spherical waves. And the result is that the way we introduce the phase shifts, the tangent of the phase shift is determined by the ratio of these two coefficients, Bl over Al. So there's a phase shift for every value of l beginning from l equals 0. Then we have that given that relation, the solution above, when you use asymptotic expansions for jl and nl, takes that form with a sine of kr minus l pi over 2 plus a phase shift, which is another way where you can read the phase shift of your solution. We're going to focus today for a little while in understanding the physics of the phase shifts and how you find the phase shifts. But supposing you find the phase shifts and that was the derivation that was very non-trivial that we did last time, we find what this scattering amplitude is in terms of the phase shift. So this was our main hard work establishing that formula. Once you have that formula, that f of k in preliminary work was shown to represent the scattering solution far away from the source. Far away from the source we have the incoming wave and the outgoing spherical wave modulated by fk of theta. Now, in general, you can have an fk of theta and phi, but then the phase shift method is not quite suitable for that. For that, we will discuss some other methods today that can allow us to do that. So once you have this f of k, you know the wave far away. You know how it contributes to the cross-section, and you know how the phase shifts. Each phase shift contributes to a total cross-section, which is given here. So these were our main results. So let's leave this results here and discuss your general computation of a phase shift. You're given a problem, maybe like the problems in the p set, and you need to find the phase shift. So where do you begin? There's nothing in there that seems to say, OK, this is the way you're going to find the phase shit. So let's do this general computation of the phase shift. Computation of the phase shifts. So I think you all know-- and there's clear intuition-- that if you want to figure out the phase shift, at some point you have to look at this region, r less than a, where the potential is. Because after all, the phase shift depends on the potential. So that's unavoidable. You have to get dirty with that potential and solve it. So what is the equation that you have to solve? It's the radial Schrodinger equation. So you have to solve-- you set energy in the Schrodinger equation equal h squared k squared over 2m because we're using k from the beginning to represent the energy of the state. So energy is going to be that. And the radial Schrodinger equation is minus h squared over 2m, d second dr squared plus the potential. Finally, the potential shows up. You have to solve that. Plus the potential centrifugal barrier, h squared l times l plus 1 over 2m r squared. All that acting on a solution u sub l-- I will put the k as well of r-- equal the energy, which is h squared k squared over 2m ulk of r. So this step cannot be avoided. You have to solve it. You have to solve this equation. So you solve this for r less than a. This for r less than a. You have some hints of what you're going to encounter because you know from solving radial equations that ul behaves like r to the l plus 1 as r goes to 0. That's a boundary condition for the l-th wave. And well, with that, you're supposed to solve for this ul. OK. So you solve for this ul. And so what? Where are the phase shifts? You're done with it. And you need to know where are the phase shifts. Well, remember our notation is that we have a radial function, l of k of r. In the radial equation you know that the radial part of the solution of the Schrodinger equation is-- the full solution is a radial part. So ylm. In this case, yl0. But this radial thing is ulk of r over r. That is presumably something you still remember, that the radial solution was u over r. OK. So you've solved it. You've got this quantity. And you say, all right, what do I do next? Well, what you have to do next is to match to the solutions that lie outside. There is these solutions that hold for r greater than a, where the potential is 0. So those have to be matched to this solution. So let me draw a little diagram. We have diagram for r. And here is a. In this region, the solution is Rlk of r. That's the solution. And this solution here, this is for the l-th partial wave. The solution outside is the l-th partial wave that I've written there. So the solution here is some Al Jl of kr plus Bl nl of kr. So this one you've determined already. You did solve the Schrodinger equation. We cannot help you. Now it depends on what potential you have. You have to solve it each time. But you now have the solution up to A. But you know the general form of the solution for r greater than A, and these two have to match here. There might be cases in which there is a delta function precisely here, like in the p set, in which case you know that with a delta function, the derivatives have to match in a funny way. There's a discontinuity in the derivatives, and the wave functions do have to match. So let's assume, in general, that there's no delta function there. And let's do that case. If you have a delta function you could do a similar case. And the way to do this is actually to match the wave functions and the derivatives. So matching at r equal a. So you must have Rlk of a is equal to Al Jl of ka plus Bl nl of ka. The derivatives must also match. Let me use primes to indicate derivatives with respect to the argument. Most people use that for primes. So the derivative of this quantity Rlk a prime-- so that means the derivative of this function of r evaluated at a. And you differentiate with respect to a in order to get the units not to change. You can multiply by an a. You have to multiply then by an a on the right-hand side when I take the derivative. Now, when I take the derivative here with respect to r, I take the derivative of J with respect to the argument, and then the relative of the argument with respect to r. That takes out the k out. Since I put an a, I get a ka times Al J prime l at ka. The derivative of J with respect to the argument evaluated at ka plus Bl n prime l at ka. So here it is. I've matched the function and the derivatives. So I'm giving you an algorithm to do it any time and all times. At this point, the right thing to do is to divide the equations because that gets rid of a lot of things that we are not interested in. We're not interested in anything except the phase shift. That's all we want, the phase shift. So how are we going to get the phase shift? Well, the phase shift comes from Bl over Al. So let's form the ratio here. So I'll form the ratio aRlk prime of a over Rlk at a. And then I have ka. That's still there. And then let me divide-- I'm going to divide these two things, but then I can divide the numerator and the denominator by Al. So here I'll have the ratio of Jl prime ka plus Bl over Al n prime l of ka over-- my equation came out a little unbalanced, but it's not so bad. ka. Let's lower this. OK. Numerator. And here you get Jl of ka plus Bl over Al nl of ka. But this is nothing but ka and the ratio Jl prime at ka minus tan delta k-- no, delta l. That's what the ratio Bl over Al is. n prime l of ka over Jl of ka minus tan delta l eta. Not eta. It's n, actually. n of ka. That's it. Actually, we've solved the problem. Why? Because this number is supposed to be known because you've solved the equation. You did solve the equation, the Schrodinger equation, from r equals 0 to r equal a. It's completely solved. So that number is known. The Bessel functions, spherical Bessels, are known. And all you need to find is tangent delta, and that equation determines it. Should I write the formula for tangent delta? OK. Let's write. Let's write it. Yeah. OK. One minute to write it. It's just solving from here. Tan delta of l is equal J prime l at ka minus Rl prime at a over kRl of k of a Jl of ka over nl prime at ka minus Rl prime of a over kRl of a, that same constant nl of ka. So that's solving for tangent delta from this side equal to this side. This shows that, in principle, once you have the solution from 0 to a, you have solved the problem of the phase shift. You know the phase shift. It's a matter of just calculating them. Might be a little messy. But if your calculation is important and this is a problem you really want to solve, you will be willing to look at the spherical Bessel functions, which, in fact, are trigonometric functions times powers. They're not as difficult as the ordinary Bessel functions. So these are easy to work with. And these things are, many times, done numerically. For a given potential, you calculate the first 10 phase shifts, and you get a solution that is very accurate. So there's a lot of power in this formulation. So you have a solution. You have the phase shifts.
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https://ocw.mit.edu/courses/5-111-principles-of-chemical-science-fall-2008/5.111-fall-2008.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Please settle down and take a look at this question. OK, let's take 10 seconds. I think that it's a simple math mistake is between one and two at least. So, the trick here is you know the p h and the p k a and you want to find the ratio so you can subtract and do the log. So maybe we'll have this question later or something similar and we can try this one again. So we're going to talk about buffers again today. I just feel the need to take a moment and reflect on the historic events of the last 24 hours, and talk about how it will affect chemistry. So some of you may have voted for the first time. Some of you may have worked on a campaign for the first time. Some of you may have been very active in a campaign for the first time, either for Obama or McCain, that you got involved. And I thought just to put this election in a little bit of the historic perspective in terms of about being an undergraduate student or a student and working on a political campaign or being part of a political movement. So, my father was very active as a political student activist. But the difference between some of you and my father was that he was a political activist at the University of Hamberg in Germany in the 1930's in Hitler's Germany. So he was the leader of the left wing student organization. That was something that put one's life at risk to take on that role at that time. So, things were heating up a little bit and the gestapo were discussing some of the activities with the left wing student leaders at college campuses in Germany. And some of them, after the discussions, no one knew where they went, they seemed to disappear. Now my father was very concerned about this and he decided to lay low for a while, and so he thought I'll do a semester at another university. And he told his parents that if the gestapo came looking for him, that they should send him a telegram saying "Your Aunt Millie is sick." Since he did not have an Aunt Milly, he knew that that would mean get out now. So he went to another university and he was doing a semester there, and someone he knew told him you really need to go into hiding. But he didn't really trust this person, so we packed a bag with a few clothes and some toiletries, but he didn't actually leave. Then the next day he came home and there was a telegram under his door. So, you can guess what the telegram said. He grabbed the bag that was already packed and headed down the stairs. The gestapo was coming up the stairs. My father's name was Heinz Leopold Lushinski and the gestapo said to him, "Do you know Herr Lushinski? And my father said, "Yes, of course, he lives on the top floor." The gestapo went up, my father went down, and he didn't go back to Germany for 30 years. So he came to the United States as a political refugee and became a citizen. He voted in every election, every possibility, he was very, very active. My family was very, very active in politics. He gave money every year to the American Civil Liberties Union to protect civil liberties, and he also gave money to the American Rifle Association. He always liked to have a plan b. So, it was sometimes a little humbling to be the only child of this man. He was in his 50's when I was born, and I thought how can I live up to something like this? Am I ever going to risk my life for what I believe in. If given that choice would I do the right thing? And I don't know if I'll ever get an answer to that question, but I talked to my father about this and he said all I need to do is work hard, find something that I love doing, some way that I can contribute, and that's what's really important -- contributing is really important. So I was drawn to teaching, and I love teaching here at MIT because you all are so talented and smart, and it is really an honor and a privilege to be involved in your education. But I feel that in the last 24 hours, we have all received an additional call to service. That president elect Obama said in the campaign that his top priorities are going to be scientific research, coming up with clean energy technologies, and improving healthcare. He called to scientists and engineers. And last night the American people said yes, we like that vision, and they elected him president. So we have been called, you have been called, he has reached out to students and said, students of science and engineering, you need to contribute. And it's been a while since any president has really called to action, scientists and engineers. And last time that happened, a man went on the moon. So let's see what we can do this time. The next challenge is clean energy, healthcare. It's going to be really important for sciences and engineers to get involved, and at the core an energy technologies, and at the core of medicine is chemistry. So you are in the right place right now. You are going to be the generation that needs to solve these problems, because if you don't solve the energy problem and don't come up with clean alternatives, there isn't going to be much of a planet left for another generation to try to solve those problems. So it's going to be your job and your job is starting right now with the education that you can get at MIT. So, it's actually somewhat interesting that today, the day after this election, we are going to talk about one of the units that students in this class have had the most difficulty with over the years, acid based titrations. This has been the undoing of some chemistry individuals. It has been the undoing of some grades of A. It has been the undoing, perhaps, of some interest in chemistry. But I would like to say today, at this moment, it will not be your undoing, it will be your triumph. Every year I challenge students to do the best job on acid based titration ever, and people have been doing well. This might be the last time I teach in the fall. You have actually had the highest grades so far in this class, in the history of the class that I know of, and so this is the challenge. So right after this election, your challenge is to conquer chemistry starting one acid and one base at a time. So, ready to do some acid based titrations? Who are the naysayers in this crowd? Just a few people up there. All right. I have to tell you that what I'm going to tell you about acid based titrations will seem like it makes pretty good sense as I'm saying it. But often, people inform me that when they actually go to work the problems on the test, it's a little less clear on what they're supposed to be doing. So the key to acid based titrations is really to work problems. And so we have, for your benefit, assigned problems for the problem-set due Friday. And so after today, you should be set to do all of the problems on the problem-set. And in terms of acid based titration, you will need a lot of this knowledge again in organic chemistry, biochemistry, if you go to medical school -- I used to TA medical students, they didn't know how to do this. And I said "Who taught you freshmen chemistry?" So it's good to learn to this now here today, work problems, take the next test, and guaranteed it'll be on the final again. So you'll learn it now, you'll get lots of points, both on the final and the third exam. All right, so acid based titrations, they're not that hard, but there are not a lot of equations to use, and I think that people in chemistry are used to what equation do I use. No, it's really about thinking about what's going on in the problem, and as the problem proceeds, as more, say, strong base is added, the problem changes. So it's figuring out where you are in the titration and knowing what sort of steps to apply. So here are some titration curves, and one thing you may be asked to do is draw a titration curve, so you should be familiar with what they look like. So we talked last time about strong acids and strong bases. So if you have a strong base, you're going to have a basic p h, and then as you add the strong acid, you will go to the equivalence point, equivalence point when you've added the same amount of moles of acid as there is base or base as there is acid, equal number of moles. And when you mix a strong acid in a strong base, you form a salt, and the salt is neutral in p h, because the conjugate of a strong acid or a strong base, is ineffectual, it doesn't affect the p h, it's neutral. So we have p h 7, and then you continue to add, in this case, a more strong acid, and the p h goes down. So for the other titration it's pretty much the same, except you start at acidic p h's, go up to neutral p h, and then go basic. So we talked about these last time and we worked a couple of problems, but now we're going to move into the slightly more difficult type of problem, which has to do with when you have a weak acid or a weak base being titrated. So let's look at the difference of the curve to start off with. So here we have the strong acid and the strong base, and here we have a weak acid and a strong base. One thing you may notice right off is that the equivalence point has a different p h. So, a strong acid and strong base again, mix, you form a salt that's neutral, p h 7. But if you're titrating a weak acid in a strong base, the conjugate of the strong base will be ineffective, but the conjugate of the weak acid will act as a base. So the p h then, at the equivalence point, when you've added equal number of moles of your strong base as you had weak acid, then you'll have the conjugate base around, and the p h will be greater than 7. So in working the problems, if you get an answer with this type of titration problem that's different than that for p h at the equivalence point, you're going to know that you did something wrong, you need to go back and check your math. Another big difference has to do with the curve shape down here, and so you notice a difference over here than over there. And in a titration that involves a weak acid in a strong base, you have a part of the curve that's known as a buffering region, and the p h is fairly flat in this buffering region as shown down here. So that's in contrast, there's no such buffering region on this side. So here the p h will go up, it'll level off, and then go up again. And this, for some of you, is probably the frustration in doing acid based titrations in lab, because you're adding and nothing's happening and nothing's happening and nothing's happening, and you're in this region, then all of a sudden you add just a little more and you're up here. So notice how steep that is over here. So sometimes when you're in the buffering region, it seems like you're never going to reach the end of the titration and then it'll happen all too quickly. So buffering region, remember a buffer is something that has a conjugate, weak acid and weak base pair, and then in a buffering region, the p h pretty much stays fairly constant in that region. It acts as a buffer, neutralizing the p h, maintaining the p h by being a source or sink of protons, and so here the p h then is staying constant in that buffering region. So those are some of the differences between the type of curves. Another point that I will mention or term I will mention that has to do with weak acid in strong base or a weak base in strong acid is this 1/2 equivalence point concept. So 1/2 equivalence point you've added 1/2 of the amount of strong base that you need to get to the equivalence point, and that's right in the middle of that buffering region. So that's another point where you'll be asked to calculate the p h. So now let's look at different points in a titration. So, first let's walk through and just think about what is happening. So when we start in this titration of a weak acid in a strong base, before we've added any of the strong base, all we have is a weak acid. So it is a weak acid in water type problem. And so here I've drawn our acid, and the acid has its proton, which is going to give up when you start doing the titration. So that's what we have at zero volume. Then we start adding our strong base, and the strong base is going to react with the acid, one-to-one stoichiometry, it's a strong base. It'll pull off protons off the same number of moles of the strong acid as the number of moles of the strong base that were added. And so then, you'll start to have a mixture of your conjugates, your weak acid and your conjugate base. So the base is a minus here. And so if you have a mixture of a weak acid in its conjugate base, that's a buffer, and so you'll move in to the buffering region here. So that's at any volume that is greater than zero and less than the equivalence point is going to be around in that region. Then we have a special category of the buffering region, which is when you've added the volume to get to the 1/2 equivalence point. And when you've done that, you will have converted 1/2 of the weak acid it its conjugate base, so you'll have equal number of moles of your weak acid as moles of the conjugate base -- 1/2 has been converted. And so that's a special category right there. Then you get to the equivalence point. At the equivalence point, you've added the same number of moles of strong base as the number of moles of weak acid you have, so you've converted all of your weak acid to it's conjugate base. So all you have is conjugate base now, and so that's controlling the p h, so the p h should be greater than 7. So that's a weak base in water problem. And if you keep going, then you're going to end up with a strong base in water problem. The weak base will still be around, but it will be negligibly affecting the p h compared to the fact that you're dumping strong acid into your titration. And so that's this part of the curve. So you see that in one type of problem, one titration problem, you actually have a lot of sub problems, or sub types of problems, you'll have weak acid buffer, special category of buffer, a conjugate base or a salt issue, and then a strong base. And this is one of the things that people have trouble with in the titrations, because we may not ask you to do all the points, we may just sort of jump in somewhere, and say okay, what is the p h at the equivalenced point, and you need to think about what's happened to get to the equivalence point. Or we may jump in and ask you about a region that would be in the buffering region, and you have to remember that at that point you should have some of the weak acid and also some of the conjugate bases being formed. So, it seems like there are a lot of different things, but there are only five types of problems. But in a titration curve, you run into a lot of those different types at different points in the problem. So now let's go the other direction and consider titration of a weak base with a strong acid. So here's what that curve would look like. You're going to start basic, of course, because you're starting with a weak base, you haven't added any strong acid yet. As you add strong acid, the p h will decrease. Because it is a weak base, you will be forming some of its conjugate as you add the strong acid, and so you'll go through a buffering region again where the curve would be flat, where the p h will be pretty much the same for region of time. Then the curve will drop again and you'll get to the equivalence point. At the equivalence point, you've added the same amount of moles of strong acid as you had weak base, so all of your weak base is converted to its conjugate acid, and so you should be acidic at the equivalence point, and then the curve goes down. So again, we can think about this in terms of what is happening. In the beginning it's just a weak base in water problem, but as you add strong acid, you were pronating some of your base and forming its conjugate acid here, and you're in the going to be in the buffering region. Then at the 1/2 equivalence point, you've added enough moles of strong acid to convert 1/2 of the weak base to its conjugate, so those are going to be equal to each other -- the number of moles of the weak base and the number of moles of its conjugate acid. At the equivalence point, you've converted all of the weak base you started with to its conjugate acid, so it'll be a weak acid in water problem, and then at the end it's strong acid. So the trick is to recognizing what type of problem you're being asked to do, and a lot of times if people get a question and they just write down OK, at this point in the titration curve, it's going to be a weak base in water problem. And just writing that down, most of the time if you get that far, you do the rest of the problem correctly. So just identifying the type, there are only 5, of problems gets you a long way to getting the right answer. So let's do an example. We're going to titrate a weak acid with a strong base. We have 25 mils of 0.1 molar acid with 0.15 moles of a strong base, n a o h, we're given the k a for the acid. First we start with 0 mils of the strong base added. So what type of problem is this? It's a weak acid problem. So we know how to write the equation for a weak acid or for an acid in water. We have the acid in water going to hydronium ions and a conjugate base. So weak acid. For weak acid, we're going to use our k a, and we're going to set up our equilibrium expression. So here we have 0.1 molar of our acid. We're going to have some of that go away in the equilibrium, forming hydronium ion and some conjugate base, and so we know we have expressions for the concentrations at equilibrium. And we can use our k a, k a for acid, it's a weak acid problem, and we can look at products over reactants. So, see, now we're doing a titration problem, but you already know how to do this problem because we've seen a weak acid in water problem before. So we have x squared over 0.10 minus x here. We can assume x is small, and get rid of this minus x, and then later go back and check it, so that just makes the math a little bit easier. And we can solve for x and then we can check -- we can take this value, 0.00421 over 0.1 and see whether that's less than 5%, it's close but it is. So that assumption is OK. If it wasn't, what would we have to do? Quadratic equation. All right, so now, here's a sig fig question. Tell me how many sig figs this p h actually has. OK, 10 seconds. So, in the first part of the problem we had a concentration that had 2 significant figures, the 0.10 molar. Sometimes later, people have extra significant figures that they're carrying along, but we had those 2, and so we're going to have 2 after the decimal point then in the answer of the p h. So again, the number of significant figures that are limiting are going to be the number after the decimal point. All right, so we have one p h value, and now we're going to move on. So let me just put our one p h value down. We have volume of strong base, and p h over here, and we're starting here with zero moles added. We have a p h of 2 . 38. It's a weak acid, so it should be an acidic p h, which it is. All right, so now let's move into the titration problem, and now 5 . 0 mils of the strong base have been added, and we need to find what the p h is now. So it's a strong base, so it's going to react almost completely, that's our assumption. If it's strong, it goes completely. And so, the number of moles of the strong base that we add will convert all of the same number of moles of our acid over to its conjugate. So we can just do a subtraction then. So first, we need to know the initial moles of the acid that we had. We had 25 mils, 0.10 molar. We calculate the number of moles for the hydroxide added, we added 5 mils, it was 0.15 molar, and so we can calculate the number of moles of the strong base that were added. So the strong base will react completely with the same number of moles of the weak acid. And we're going to do then -- we have the moles of the weak acid here, minus the number of moles of the strong base we've added, and so we're going to have 1 . 75 times 10 to the minus 3 moles of the weak acid left. So, then how many moles of the conjugate base will be formed by this reaction? What do you think? Same number. So 0.75 times 10 to the minus 3. So always remember that in these titration problems, nothing has been added yet, you're at zero mils added. Some amount of some subtractions are going to have to occur because something has happened. You've converted something, things are different than when you started. All right, so now we have weak acid and we have moles of its conjugate, what type of problem is this? If you have a weak acid and its conjugate base -- buffer, right. So we're going to do a buffer problem and we need to know the molarity first. So we have moles over volume -- again, the volume, you had 25 mils to begin with, you added 5 more. So you have to have the total volume 30 mils, and we can calculate then the concentrations of both. Now we can set up our equilibrium table, and this looks like a buffer problem because it is, and by looking like a buffer problem you something over here, you have your weak acid over here, but you have something over here now, it's not zero now, we're starting with some conjugate base. So we have 0.0583 minus x on one side, and we 0.025 molar plus x on the other side. We can use k a again. This is set up as an acid in water going to hydronium ions and conjugate base, so we can use our k a, set things up, and we can always say let's see if x is small, make an assumption, check it later. That'll simplify the math. So we get rid of the plus x and the minus x. Again, we're saying that if x is small, the initial concentrations are going to be more or less the same as the concentrations after the equilibration occurs. And we can calculate 4 . 13 times 10 to the minus 4, as x, that is a pretty small number. And we have to check it, and yup, it's small enough, it's under 5%, so that's OK. So now we can plug this in. X is our hydronium ion concentration minus log of the hydronium ion concentration is p h, and we can calculate p h to 3 . 38 -- again, we're limited by two significant figures in the concentration. So now we've added 5 mils down here, and our p h has gone up a little bit, it's now at 3 . 38 over here. There's another option for a buffer problem. What's the one equation in this unit? Our friend Henderson Hasselbalch. And yes, you can use that here too, assuming that you check the assumption and it's OK. Most people will prefer to do this because it is a bit easier. So, you weren't given, though, the p k a in this problem, you were given the k a, so pretty easy to calculate -- minus log of the k a is the p k a. So you can calculate that, put that in. You have your concentrations and it should be concentrations, but you may notice that if you actually had moles the volume would cancel here. So here are the concentrations, but with the same volume, the volume term does cancel. It makes this a little faster and it gives the same answer, which is great. To use Henderson Hasselbalch you also need the 5% rule to be true, because Henderson Hasselbalch is assuming that x is small. It's assuming that the initial concentrations and the concentrations after equilibrium are about the same. So we can check the assumption. We can back-calculate the hydronium ion concentration, which would be x, and see if it's small, we already know it is, so it's OK. So there are 2 options for buffer problems, but do not use the Henderson Hasselbalch equation when it isn't in the buffering region, it doesn't hold then. So again, you check the assumption, and if it's OK, it's fine. If not, you need to use option one and you need to use the quadratic equation. All right, so buffering region. Now we're at the special kind of problem in the buffering region, the 1/2 equivalence point. So here you've added 1/2 the number of moles of the strong base to convert 1/2 the moles of the weak acid to its conjugate. So at this point, the concentration of h a equals the concentration of a minus -- equal number of moles in the same volume, those are equal. You can use Henderson Hasselbalch here, and find that if they're equal, you're talking about minus log of 1, so the p h is going to equal the p k a. And you're done with this type of problem. I have been known to put 1/2 equivalence problems on an exam, because exams are often long, you have only 50 minutes, there's lots of different type of problems, and this problem should not take you a long amount of time. You do not have to prove to me that this is true. All you need to remember, 1/2 equivalence point, p h equals p k a, and if you calculate the p k a, you're done. So this is a short type of problem. If you remember the definition of 1/2 equivalence point, it's easy to do. So now we have another number, so 3 . 75, and we're working on our curve. Now let's move to the equivalence point. At the equivalence point, you've added the same number of moles of your strong base as you had weak acid. So you've converted all of your weak acid to its conjugate base. So the p h should be greater than 7. Now all you have is conjugate base, that's basic, p h should be greater than 7. So when you are doing this titration, you have your weak acid and your strong base. You're going to be forming a salt here, and a salt problem, you can tell me about salts. And so, just remind me, what does the n a plus contribute to the p h here. It's going to be neutral. And what about this guy down here? Yeah, so it's going to be basic. So, the sodium, anything group 1, group 2, no effect on p h, they're neutral. But if you have a conjugate base of a weak acid, that's going to be basic. Salt problems, really just part of what you already know about. So always check your work. If your p h doesn't make sense from what you know, you might have made a math mistake. So let's calculate the actual p h at the equivalence point. We know that it should be basic, but what is it going to be? So first, we need to know how much of the strong base we had to add, because we need to know about all the moles. So how much of this did we need to add. So we needed to add enough of the strong base that you converted all of the moles of the weak acid to its conjugate. So we had 2 . 5 times 10 to the minus 3 moles of our weak acid. So that's all going to be converted to the moles of the conjugate base, and so that's going to be equal to the number of moles we needed to do it. So we needed 2 . 5 times 10 to the minus 3 moles of our strong base to do that complete conversion. We know the concentration of the base was 0.15 . So we would have needed 1 . 67 times 10 to the minus 2 liters of this concentration added to reach the equivalence point. So then the total volume that we're going to have at the equivalence point is the 25 mils that we had to begin with, plus this 16 . 7 mils to make this final, total volume. And remember, you always need to think, what is the total volume, how much has been added to get to this point in the titration curve. Then we can calculate molarity, so we know how many moles of conjugate base have been formed, and we know the new volume, so we can calculate the concentration of the conjugate base. So now, you can help me solve this problem. Set up an equation for me to solve it. Let's take 10 seconds. That's the best score we've had today. Yup. So now we're talking about a conjugate base. So we have converted all of the weak acid to the conjugate base, and so it's a weak base in water problem, so we're going to talk about a k b. If you were only given the k a for this problem, how would you find k b -- what interconnects k a and k b? K w, right. So you can calculate, here it's given to you, but you could calculate it if you had a calculator, and you would find that this is true. Now it's a weak base in water problem. We're not in the buffering region anymore. We've converted all of our weak acid to the conjugate. So it's a weak base in water problem. So we have x squared, 0.06, that was the concentration we calculated, minus x. So again, think about what type of problem it is. So again, weak base in water problem -- x squared over 0.06 minus x. And we can assume that x is small, and calculate a value for x, which is 0.83 times 10 to the minus 6, and then we're going to calculate p o h, because now x is the hydroxide ion concentration. Because in a weak base in water problem, here in this type of problem, the base, and here is your acid -- the conjugate of this acid is the base, hydroxide, and the conjugate of this weak base is its conjugate acid over here, so now when we are solving for x, we're solving for hydroxide ion in concentration, so we're calculating a p o h, which then we can calculate a p h from. So we can take 14 minus 5 . 74 and get our value. And it's bigger than neutral, it's 8, it's basic, and that makes sense, it is a weak base in water problem. So, let's see, it's 8 . 26, so now we're up here in our curve, and we're at 8 . 26, and that's going to be greater than 7 for this type of problem. So that makes sense, it's good. Greater than 7 is what we want to see. So now, you've gone too far -- you've passed the equivalence point, and you keep adding your strong base in. Now you still have some of the weak conjugate base around. So you still have this around, but you only have 1 . 83 times 10 to the minus 6 molar of it. So very little amount -- x is small. So your p h is going to be dictated by the amount of extra strong base you're adding. So this is similar, then, to a strong acid or strong base in water problem. So if you're 5 mils past the equivalence point, 5 mils times your concentration of a strong base, so you have extra, 7 . 5 times 10 to the minus 4 moles extra. So then you need to calculate a concentration of that, and so you remember the whole volume -- you're 5 mils past, you had 25 miles to start with, and you had to add 16 . 7 mils to get to the equivalence point. And you have, that's your total volume, you get a concentration, that's your concentration of hydroxide, it reacts completely, you don't have to do any equilibrium table here. It's going complete, it's a strong base. You could try adding that value of your other weak base to this, but remember, that's times 10 to the minus 6, so it's not going to be significant with significant figures. So you can just use this value -- plug it in to p o h, calculate it, and then calculate p h. And so now we're somewhere up here at p h 12 . 21, 5 mils past. And there we've worked a titration problem. So let's review what we saw. In the beginning, zero mils of the strong base, we have a weak acid in water problem. We moved into the buffering region where we had our weak acid and the conjugate base of that weak acid. At the equivalence point, we've converted all of the weak acid to the conjugate base, so it's a weak base problem. And then beyond the equivalence point, it's a strong base problem. That's what we've just worked. So, we can check these all off now. You know how to do all of these types of problems. And there are not that many, you just need to figure out where to apply what. And if you can do that, you're all set, this unit will be easy for you, and you can go through and make me very happy on the exam. There's nothing -- well, there are few things in life as beautiful to me as a perfectly worked titration problem. It really, it brings me joy, and I've had people write on the exam sometimes, "I hope that my solution to this brings you joy." And I will often write, "Yes, it does," and put a smiley face. Because it really is nice to see these beautifully worked. I know, I'm a little nerdy and geeky, but after yesterday, being smart and a nerd and a geek is cool again. All right, so let me just tell you where we're going. We have five more minutes, and actually that's perfect, because I can get through some rules in those 5 minutes. So let's do 5 minutes of rules. Oxidation reduction doesn't have a lot of rules, so five minutes is actually all we need to do that. Oxidation reduction involves equilibrium, it involves thermodynamics. I like it because it's really important for reactions occurring in the body, and acid bases as well -- p k a's are really important to that. And so, between acid base and oxidation reduction, you cover the way a lot of enzymes work. So let me give you five minutes of rules, and that will serve you well in this unit. Some of these are pretty simple. For free elements, each atom has an oxidation number of 0, so this would be 0. So, oxidation number of 0 in a free element. For ions that are composed of one atom, the oxidation number is equal to the charge of the atom, so lithium plus 1 ions would have an oxidation number of plus 1. Again, pretty straightforward. Group one and group two make your lives easy. They seem to have a lot of consistent rules. Group one metals in the periodic table have oxidation numbers of 1. Group two metals have oxidation numbers of plus 2. Aluminum is plus 3 in all its compounds. Pretty simple. Now we get to things that are a little more complicated but still useful, oxygen. Oxygen is mostly minus 2, but there are exceptions to that, such as in peroxides where it can have an oxidation number of minus 1, and if it's with a group one metal, it can be minus 1. Remember, group one, and actually group two here, that's plus 1, always plus 1, always plus 1, always plus 2, and so hydrogen has to accommodate that. So usually plus 1, except when it's in a binary complex with these particular metals that are in group one or group two. Fluorine, almost always minus 1, or always minus 1 -- other halogens, a chloride, bromide, iodide, also usually negatives, but if they're with oxygen, then it changes. So, here is an example. And in neutral molecules, the sum of the oxidation numbers must be 0. When the molecule has a charge, the sum of the oxidation numbers must be equal to that charge. So, let's do a quick example. Hydrogen, in this case, is going to be what? Plus 1, so it's not with a group one, group two metal here. So what does that leave for nitrogen? And that makes the sum, plus 1, which is equal to the sum of that molecule, so that works. So we might not have known nitrogen, but we can figure it out if we know the rules for hydrogen and we know what it all has to equal up to. And so, this unit is sometimes a relief after oxidation reduction, because it's all about simple adding and subtracting, it's not so bad. OK, oxidation numbers do not have to be integers. Example here, you have superoxide, what would its oxidation number be? Minus 1/2. And those are the rules, and then on Friday, we'll come back and we'll look at some examples.
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https://ocw.mit.edu/courses/5-08j-biological-chemistry-ii-spring-2016/5.08j-spring-2016.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit Mit OpenCourseWare at ocw.mit.edu. ELIZABETH NOLAN: So where we left off last time, we were talking about using antibiotics as tools to study the ribosome. And recall that antibiotics have many different structures, can bind to the ribosome at different places. And we closed with talking about this antibiotic, puromycin, that can bind to the A-site and cause chain termination, and also molecules that are derivatives of puromycin, such as that more elaborate one with a C75 there. And so the example of a system where puromycin has been employed, and this is just one of many, many examples, but also gives us a little new information about players in translation, involves studies of elongation factor peak, so EFP. And if you recall, where I closed last time was with the comment that this EFP over the years was implicated in a variety of cellular processes. But its precise function remained unclear. And so Rodnina and co-workers conducted a series of experiments to ask, what is the effect of EFP on peptide bond formation when different dipeptides are in the P-site? OK? And their experiments were motivated by the fact that there was some preliminary work out there suggesting that EFP accelerates peptide bond formation, but really, the details were unclear. So we're going to look at the experiment, their initial experiment they did, which led to some new understanding about how EFP affects the translation process. So what is it that they want to do in this experiment effectively? Imagine we have our ribosome, and we have our three sites, OK? And so what they do in this experiment is they have a dipeptide loaded in the P-site, OK, where x is some amino acid. OK, and then what they want to do is have puromycin in the A-site and then effectively monitor for peptide bond formation with or without EFP added such that the product is effectively a tripeptide, where we have fMat, the amino acid and puromycin, OK? And keep in mind, if this is what's being monitored, there needs to be a step to hydrolyze this tripeptide off the tRNA that's in the P-site, OK? And throughout this work, how they monitored this is that they have a radio label on the formal methionine. So you can imagine that you can somehow separate and see the dipeptide as well as this tripeptide-like molecule with the puromycin attached. So how to set up an experiment to test this? So they do a stop-flow experiment, so you heard some more about that method in recitation last week. And so in thinking about this, we need to think about what will be mixed. So what are the components of each syringe? How will this reaction be quenched? And so beginning to think about that, the question is, how do we even get the ribosome we need to start with in order to see the reaction? Right? So imagine that the goal is to have a pre-translocation ribosome, so effectively that dipeptide is in the P-site, and the A-site's empty. And then that assembled post-translocation ribosome needs to be mixed with puromycin such that puromycin can enter the A-site and peptide bond formation can occur. OK? So there's quite a bit of work that needs to happen to even get this experiment set up, because somehow that post-translocational ribosome needs to be made. OK, so if we think about this from the standpoint of the experiment and using the stop-flow to rapidly mix, we have syringe 1, and we have syringe 2, and we have our mixer. OK, so what are we going to put in syringe 1? OK, so here, we're going to have the post-translocation ribosome. The A-site is empty, and the P-site holds the dipeptide attached to the tRNA. And then in syringe 2, we're going to have puromycin here. OK, so before we get to EFP, I'm thinking about how we're going to look at that in this reaction and what it does. How are we going to get here? So what needs to be done to get this post-translocational ribosome? Is it in the sigma catalog? Bio rad? No way! And even if it were, you would be broke needing to purchase enough to do this experiment, right? You talked about needing high concentrations in recitation last week for these types of experiments. So where does this come from? What you need to do before even getting this into your syringe here, to do a rapid mixing experiment? AUDIENCE: You have to isolate it from cells? ELIZABETH NOLAN: OK, so what is the likelihood of isolating-- well, what's it, what do you need to isolate from cells? AUDIENCE: Well, you're going to need to modify it afterwards because there'll be all sorts of other things. ELIZABETH NOLAN: Right, but what's it? AUDIENCE: The ribosome. ELIZABETH NOLAN: OK, so we need a ribosome. Right? What else do we need? So we need the ribosome, and we need to get this into the P-site. So how are we going to get that dipeptidyl tRNA into the P-site? AUDIENCE: You need an mRNA. ELIZABETH NOLAN: We need an mRNA, and we're going to design that mRNA based on what amino acids we're interested in. So we need to come up with an mRNA. What else do we need? So think back to the whole cycle. AUDIENCE: You need EF-Tu, GTP. You need everything necessary to form the fMat to x-peptide. ELIZABETH NOLAN: Yeah. So what does that mean first? And when does that bond form? That's the next thing, right? So can we deliver this species to the P-site, based on what we understand about translation in the past four or five lectures? No. Right? So first, the initiation complex needs to be prepared in lab, which means you need initiation factors, a ribosome, mRNA, the initiator tRNA. And then that initiation complex needs to be purified, which is done by a type of sucrose gradient centrifugation. OK, and then what? Once that initiation complex is formed, there needs to be a round of elongation, where the ternary complex of EF-Tu. The amino acid and GTP comes in to deliver that x-tRNA x to this A-site, and then have peptide bond formation occur. OK? And then, we also need the help of EFG to move that to the P-site, right? So that whole cycle we've talked about from a fundamental perspective needs to be done at the bench in order to get here. So there's a lot of factors that need to be purified and obtained, quite a bit of effort to just even set this experiment up. OK? So always think about where these things come from. So we have this. We have puromycin, right? And then we want to look at the effective EFP. So the idea is, are there differences in peptide bond formation? Is it accelerated in the presence of EFP, as how some of this preliminary data indicated? And if so, is that for all amino acids? Or is it specific for certain amino acids, right? So we need to include EFP. And in these experiments, it was either omitted or included in each syringe. And something just to think about when thinking about these rapid mixing experiments is what happens in the mixer, right? If you're having the same volume, which is the case coming from syringe 1 and 2, you're going to have a dilution in here of all of the components. Right? So these are going to be rapidly mixed in the absence or presence of EFP. There'll be some time to allow for reaction to occur. And then, in this case, the reaction is going to be quenched. So it's the quench flow-type setup that came up in the recitation notes from last week. So in this case, we're going to have a syringe 3 with a quencher. And in this particular work, they used base, so sometimes it's acid. Sometimes it's base. And this was a solution of KOH. OK, so then after some time, OK, we can have the reaction quench. OK, and then there'll be some sort of workup and product analysis. OK? So in this case, they chose to hydrolyse the peptidyl tRNA's and look at the peptide fragments. So you can imagine you need a method that's going to separate fMet x, whatever amino acid x is, from that product there. And then the radio label on the fMat is used for quantification. So what happens here? And I'll just give a summary, and then we'll look at it in more detail. So what they did in these experiments-- and recall that JoAnne talked about in recitation last week, when doing these kinetic experiments, you have to tweak them quite a bit to get the exact good conditions to observe what you want to see. So imagine that happened. We have our k observed, and I'm going to show these on a log scale. So always keep in mind, paying attention to what type of scale the axes are in. And so what we're going to look at is the k observed for formation of this tripeptide, depending on amino acid. And I'm going to generalize a bunch of the data here, and then we'll look at all the individual cases. OK? So here, we have x does not equal proline OK, and here, not colored in, is no EFP. And shaded is k observed for the reactions conducted in the presence of EFP. OK? So what was observed in these studies, looking at having many different amino acids here? With that, many of these amino acids showed negligible difference, whether or not EFP was included in the reaction. OK? And we can look at that data in more detail from the paper on the slide. What was very striking about these initial experiments was what happened in this case, when x equals proline here. So effectively, what they observed in this case was about 90-fold rate acceleration. Effectively, if we compare the k observed for peptide bond formation in the absence of EFP, we see it's significantly diminished for proline if EFP isn't there. And along those lines, it was known before that proline attached to its tRNA is a poorly reactive tRNA. So different aminoacyl tRNA's react differently in the ribosome. So there's that layer of complexity we haven't really talked about in this class yet here. So if we take a look at all these different examples, this one is the outlier. OK? So what these data indicated is that EFP has some special role in accelerating peptide bond formation for peptide bonds that contain a C-terminal proline residue here for that. And so these experiments were just a starting point for many additional experiments that ended up showing EFP is really critical for helping the ribosome translate sequences that have consecutive prolines in a row. So either three prolines or maybe a PPG sequence here. And in the absence of EFP, what can happen is that the ribosome stalls. So these aminoacyl tRNA's are not very reactive, and the ribosome just gets kind of stuck. And you can imagine that's not good for the cell. And then if we bring these observations back around to some of these early works that were suggesting EFP has a role in a diversity of different cellular processes, what might we ask? We might ask, well, where do the sequences of multiple prolines come up? So what types of proteins have three prolines in a row some place in their sequence? Or something like PPG. And so they took a look at that. And if we think about E. coli, there's about 4,000 different proteins, and there's a subset of around 270 that have these types of sequences in them. So not hugely common, but they exist. And so then ask, what do these proteins do? Right? Provided a function is known. And so what we see is within that subset of about 270 proteins, there's examples of proteins that are involved in regulation, in metabolism, you know, important cellular processes. So you can begin to understand why it might be that this protein got implicated in all these different types of phenomena, right? But in terms of the details, it's really back here in terms of how this translation factor is helping the ribosome make a certain subset of peptide bonds there. So if you're curious about this, the paper's really wonderful. There's a number of additional interesting experiments that are done and additional methods to these kinetics there. I'm happy to point you in that direction. So yes? AUDIENCE: Does this rate of the reaction affect upon ribosome folding? ELIZABETH NOLAN: It could. I mean, basically, you're talking about what happens as the polypeptide extrudes from the ribosome, right? And if you're stalled and have some piece of this nascent polypeptide on the outside. Ribosomes stalling, yeah, what does that do in terms of how trigger factor, for instance, interacts. That's something we'll talk about in the next module, and we'll be getting there on Wednesday, I hope, if not Friday. So with that, we're going to close discussions of module 1 in the ribosome with looking at some biotechnology and thinking about how we can use this fundamental understanding of the ribosome to do some new things. And so we're going to talk about re-engineering translation and ways to use this machinery to incorporate unnatural amino acids. And so to begin thinking about this, we can just consider some questions. And so many of us in this room are chemists or chemistry majors. We can think about organic chemistry, so 5.12, 5.13, and all of the different organic transformations that are presented. So if we think about all these organic transformations and how they're available to synthetic chemists, we see a lot of versatility. And we can simply ask ourselves, can such versatility be achieved for protein modification? What is the toolkit? How can that toolkit be expanded? And then thinking about this further, can we use the translation machinery? So is it possible to modify the translation machinery to allow us to make peptides or proteins that have unnatural amino acids? So amino acids are moieties that are not the canonical ones. And can we do this in cells? Can we do this in a test tube? And if we can, what does that provide us with in terms of possibilities? So the answer is yes, and we're going to focus on the how and strengths and limitations in terms of our discussions of this machinery here. I also note-- I believe, JoAnne, this will come up. Will you be talking about this in the nucleotide parts, too? JOANNE STUBBE: If we get that far. ELIZABETH NOLAN: If we get that far. So in addition to here, this may come up again towards the end of the course, as a tool. So hopefully we'll get that far, because that's exciting. So let's think about re-engineering translation. And we can think about two things. We can think about the genetic code here, and we can think about the ribosome. And so I'll just present you with the questions. If we consider the genetic code, what can be done to this genetic code to change an amino acid in a protein? And if we think about the ribosome, what can be done to the ribosome to change an amino acid in a protein? And effectively, can we expand the genetic code to encode something other than what it's supposed to encode? So can this code allow us to encode an unnatural amino acid? And from the standpoint of the ribosome, is it possible to design new ribosomes? So can we make a new ribosome that can incorporate unnatural amino acids into proteins? So these are separate but related, and we're going to first discuss basically reassigning-- is it possible to reassign a codon? So why would we want to do this? And let's think about that for a minute. And what do I mean by expanding the genetic code? So if we think about the genetic code, we all know that it encodes these 20 amino acids building blocks, there's the start codons and the stop codon. And effectively, the codons are all used up, right? There aren't extra codons floating around that we could poach and assign to something else here. So can we overcome this? And why would we want to do that? Just broadly, if we think about being able to put something other than a natural amino acid in a protein at a specific location-- so exactly where we want it-- that opens up many possibilities for experiments. And we can think about those experiments both happening within a cell or outside of a cell. And these are experiments that just wouldn't be so easy or feasible otherwise. So maybe we'd like to study protein structure. What could we do? So fluorine is used in NMR quite a bit. Imagine if you could site-specifically label an unnatural amino acid that has a CF3 group, for example, and use that in your NMR studies. So that's something you'll get to think about in the context of problem set two. Ways to study protein function, protein localization. So for instance, instead of attaching GFP, which is big, to a protein of interest, maybe it's possible to incorporate a fluorescent amino acid that lets you see that protein in the cell. Protein-protein interactions. And maybe we'd like to make a new protein that has some desired characteristic. So there's a lot of possibilities to such technology. Just to keep in mind, what do many of us do? Many of us are familiar with site-directed mutagenesis, where we can change an amino acid in a protein. And we learn many, many things from this, but it is limited to naturally occurring amino acids. Right? So we'd like something more versatile. If we think about strategies also just a little bit, backing up here. OK, the first thing I'll just point out is that how I'm going to divide this, just in case this wasn't clear, is considering the native ribosome and then considering engineered ribosomes. And this is where we're going to focus today. And if we consider strategies, other strategies to incorporate unnatural amino acids, and I guess I'll call these standard, we can imagine chemical and biosynthetic. And I'm not going to go over a plethora of examples for either route. There'll be some slides included in the posted lecture notes that gives examples and pros and cons. But one example I will give here is just thinking from the standpoint of a chemical modification, what's an example and why we might want to do better. OK, so this is independent of something like site-directed mutagenesis, where you're having an organism do the work. So if we just consider an example of a chemical modification, there's certain amino acid side chains that are amenable to modification. So imagine you purify a protein, and you want to somehow tag that or label it, right? One option is to modify cysteine residues. And so iodoacetamide and related reagents are commonly employed, so imagine that you have some cysteine. You can react this with iodoacetamide that has some R group, right? What happens? Here, OK. You can get a covalent modification, and maybe this is a fluorophore or something else, right? So this is terrific, but what are some potential problems? AUDIENCE: Sorry, would this be a way to modify the amino acid before it's incorporated into the protein? Or would this be something you would do to modify the cysteine in an assembled protein? ELIZABETH NOLAN: Yeah, this would be after the fact. So imagine you have some protein. You've isolated your protein, and you have some cysteine. Right, and you'd like-- for some reason, you'd like to modify this protein. So maybe a fluorophore to see it. Maybe you know, a CF3 group for NMR here, which then gets to the point, what are possible problems with this? AUDIENCE: Do you have to use a mild base to be deprotonated, or is it maybe deprotonated based on where it is in the protein? ELIZABETH NOLAN: Yeah, so that gets to an initial issue, which is what's required to have this chemistry to happen? Right? The cysteine needs to be deprotonated. So probably the pH of your buffer is going to be elevated some. Does your protein or enzyme like that or not? Maybe, maybe not. Yeah? AUDIENCE: You can also run into selectivity issues-- I mean, having free cysteine residues isn't common, but it could be a potential problem. ELIZABETH NOLAN: Yeah, so you need-- well, it will depend on the protein, right? Is the cysteine free or a disulfide? Is it a native cysteine, or have you done site-directed mutagenesis first to put this cysteine in the position you want? Right? And then what happens if your protein has multiple cysteines building on what Rebecca said, and you want to have this label at a site-specific location? Right? What are you going to do about that? Are you going to have non-specific labeling? Are you going to mutate out the other cysteines? If you do that, what could that mean for your protein fold or function? There's a number of caveats that need to be considered. Nonetheless, it's a possibility to do. In terms of time, this is a pretty extreme example, but I'll just show one example here in thinking about this whole process and what you do, which also builds upon Rebecca's question. So imagine a protein with two subunits. And subunit 1 has a cysteine, and subunit 2 doesn't. So for some reason, you want to do this labeling. This is actually a protein from my group. And we wanted to stick a fluorophore on it. So we have a cysteine on one of the two subunits. You can run this reaction and get this fluorophore modified form here. And then you can see that's the case, looking at SDS-PAGE. So here we're looking at Coomassie stain that shows us total protein, and we see there's two subunits, 1 and 2. So the molecular weights are a little different, and we can separate them on this gel. And then if we look in the fluorescence channel, what do we see? We only see fluorescence associated with subunit 1 and not subunit 2, which tells us our labeling strategy has worked well. Like, what we're showing in this equation. But what's everything that needs to be done? Well, we need to overexpress the protein in some organism. In this case, E. coli. We need to purify the protein. And once we have this purified protein in hand, we need to do the chemical reaction for the labeling. And then we need to purify that product somehow, and that's going to depend on the system you're working at. And then it needs to be analyzed, right? You always want to know what you're working with, right? So was this reaction to 100%? Did we end up with a mixture? If it's a mixture, what to do about that? So what does this mean in terms of time? And this is not for all cases, OK? This is for this exact case involving this protein shown as a cartoon here. So it takes about six days from start to finish to overexpress and purify it. Steps 2 to 4, based on the purification, we do another four days, right? So that's 10 days from start to finish, just to get this protein you'd like to use in your experiment. Right? And you can imagine if somehow a label could be put on in vivo, during this initial step here, that that would save some time at the end of the day. So before moving on to what's done for unnatural amino acid incorporation by what we'll call the Schultz method out of Professor Peter Schultz's group, just to think about biosynthetic methods for a minute. So some common ones are done for structural studies. So for instance, you can imagine feeding an organism something like selenocysteine or selenomethionine. Another example is labeling nitrogens or carbons for NMR, where the organism is fed, say, a labeled amino acid, maybe with N15 or C13 there, right? So that's just a biosynthetic method, where you're changing the growth conditions, rather than doing something to manipulate the genetic code or the ribosome. So what's the conclusion here? What we want is we want a method of site-specific incorporation of unnatural amino acids in vivo. So in a cell and in a desired organism, depending on what you want to do with high efficiency and also fidelity, so getting back to that idea and before. OK, so why do we want to do this in vivo? It allows for studies within cells, and you also can purify protein from cells, so you can do in vitro experiments as well. OK, and you can imagine, if you could have all of the pieces of this machinery in a cell, maybe there's some technical advantage to that. So this is what we're going to consider here. So to this question, can the ribosome incorporate unnatural amino acids into proteins? Effectively, what do we need to think about? One, we need to think about relaxing the substrate specificity of the aminoacyl tRNA synthetase to accommodate some unnatural amino acid, right? Somehow that unnatural amino acid needs to get to the ribosome. So if this can be done, and we can make a tRNA that has an unnatural amino acid attached to it, can this aminoacyl tRNA get to the A-site and do the work? So this is the method we're going to talk about in some detail for the rest of today and into Wednesday, this Schultz method. So the idea is that there's a tRNA that's dedicated for this unnatural amino acid. We see this unnatural amino acid shown here, where the UAA is indicated by probe. We need an aminoacyl tRNA synthetase that will take this unnatural amino acid and attach it to the three-prime end of the tRNA to give us this aminoacylated tRNA. And then what? Imagine this tRNA can make its way to the ribosome. What happens? We need a codon for this aminoacyl tRNA. It needs to carry the anticodon, and we're going to talk about this in some more detail in a minute. So we can have a plasmid that has the DNA with the gene of interest in it, right? This plasmid DNA can be transformed into, say, E. coli that has this machinery here. We can have transcription to give the mRNA that is from this plasmid DNA. And then imagine translations such that this unnatural amino acid is incorporated. So effectively, where we're going is that we need a general method. We want this method to be broadly useful, where we can genetically encode this unnatural amino acid and have it incorporated in response to a unique triplet codon, here. So in thinking about that, what are the pieces that we need? And we'll think about E. coli for the moment, but this could be other. So yeast, mammalian cells, right? Let your imagination run wild with this here. When the incorporation of the UAA in response to a unique triplet codon. So if we're going to do this, what do we need? OK, effectively, we need some new components of the trans-- like protein biosynthetic translation machinery, right? So we need to rewind and think about the whole translation process. OK, so the first order of business is that we need a unique codon. Right? So this only designates or uniquely designates the UAA. And so we need to ask, where does this come from? Because we just went over the fact that the codons are used up for amino acid start and stop. We need a new tRNA. OK, so this tRNA needs to be specific for the unique codon. OK, and we need the corresponding aminoacyl tRNA synthetase, right? And we need this to load the unnatural amino acid onto the unique tRNA here. So what is a key feature of all of this? A key feature is that if we want to do this in some organism, we need this machinery to be orthogonal to the machinery in that organism. We cannot have cross-reactivity, because then there's not going to be any selectivity of this incorporation. So no cross-reaction. So what do we need to consider, in terms of these? We need to think about all of the machinery, right? And I just list some considerations here. So this new tRNA can only allow for translation of the codon for the UAA. It can't be a substrate for any of the endogenous aaRS, because then it will become loaded, potentially, with the wrong amino acid. So think back to lectures 2 and 3. This new aminoacyl tRNA synthetase can only recognize the new tRNA and not endogenous tRNA. So cross-reactivity again. This unnatural amino acid also can't be a substrate for endogenous enzymes. And also keep in mind, there needs to be some way to get this unnatural amino acid into a cell if we want to do this in a cellular context. So there's just transport issue that needs to be kept in mind. Will this unnatural amino acid get into the cell? OK, so what we're going to do is consider these requirements and what was done to build up this methodology during initial work. So the first issue is this unique codon, and what is its identity here? And so if we consider the 64 codons, they're used up with the 20 common amino acids. We have the three stop codons and the one start codon. And so in thinking about this, can ask, do we really need three stop codons? We certainly need our start, and we need codons for the amino acids. But is there some wiggle room here? And so in terms of these stop codons, we have TAA, TAG, and TGA. And these all have names. Ochre, amber, and opal. OK, and so the idea we're going to see is just the question, can we reassign a stop codon? And can we reassign a stop codon such that it's the codon for the unnatural amino acid? And so basically, if we want to reassign a stop codon, how do we choose? Right? So two things to consider. One, how frequently is each stop codon used? So what do we know about that? And then does this stop codon terminate essential genes? So we can imagine that if we were to reassign a stop codon that's used frequently by E. coli or another host, or if we were to reassign a stop codon that's important for terminating the synthesis of essential genes, in either case, the outcome could be pretty bad, right? So what was found in thinking about those issues is this amber stop codon, TAG, one, it's the least frequently used. And just for an example, about 9% in E. coli and about 23% in yeast for terminating genes. And additionally, it rarely terminates essential genes. OK, so based on this, it was decided to reassign TAG as the codon for the unnatural amino acid. OK, so we've gotten through to here. So the question is now, what about requirements 2 and 3? So yeah? AUDIENCE: I have a question. So it seems interesting to choose a stop codon to change because if the stop codon messes up, it seems more catastrophic to us all than one of the other redundant amino acids. Like, why use a stop codon? I think it's interesting. ELIZABETH NOLAN: Yeah, so there is a risk. There's certainly a risk, right? But these considerations were made to try to diminish that risk. Right? So you could make the argument that maybe all of these stop codons aren't essential, right? And what is more deleterious? Will it be to try to use a stop codon that's infrequently used, or to reassign a codon that's for an amino acid that comes up in many, many different proteins in the cellular pool? Right? So there's a judgment call there. But if we consider in E. coli, this TAG stop is for about 9% of proteins. How does that compare to, say, reassigning one of the codons to incorporate a lysine or a valine. I don't know, but that was just want to think about, how frequently is that codon used? Because certainly there's different codon usage in different organisms. Do you have something to say? JOANNE STUBBE: So it depends on what you want to put the unnatural amino acid in for. So if you want it in endogenous levels, it could be a problem. But if you're overproducing your protein-- ELIZABETH NOLAN: Yeah, it may not be a problem. JOANNE STUBBE: Then it's not a problem, because you induce, and then you flood it with that and you get high levels of cooperation. So it depends on what your purpose is. ELIZABETH NOLAN: Yeah, so is JoAnne's point clear to everyone? So you could imagine expressing at an endogenous level, right? Or you could imagine causing the cell to overexpress the proteins, like off a plasmid, like what many of you have done in lab class or maybe in research there for that. Are there many examples of reassigning a different one? JOANNE STUBBE: I think it's really tough. I mean, inside the cell, you really do have problems if you don't re-engineer, because you get truncations also. ELIZABETH NOLAN: Yeah, we're going to talk about that. So there is a big problem about the stop that we're going to talk about, once we get to how we have this done, which is premature termination. AUDIENCE: I'm kind of confused more or less at that point, because stop codon, we're just using because it's not currently-- it doesn't go for anything, it just ends, like, endogenous sequences that are not the UAA so if we replace it, we might not get those. But we'll get the one that we're trying to synthesize. ELIZABETH NOLAN: Right, so we need a codon for the unnatural amino acid. And right now, we're limiting our space to triplet codons, which is what was initially done when this type of methodology was developed. So the question is, we have four options in terms of bases. AUDIENCE: That are not coding, right? ELIZABETH NOLAN: No, no, in terms of our codons, right? So three, right? Four, three, so there's 64 codons, and they're all used up. AUDIENCE: Yeah. ELIZABETH NOLAN: Right, so there's not some extra. AUDIENCE: Yeah. ELIZABETH NOLAN: So then what can be reassigned? AUDIENCE: These stop-- so the stop. ELIZABETH NOLAN: Well, yeah, well, we can't reassign the start. Then there's no proteins, right? There's only one start codon. So the thinking was, is a stop codon dispensable? Right? And then a decision was made, based on basically the frequency of use of the stop codon and whether or not the stop codon terminates essential genes. Is this something foolproof? No, there's major problems in terms of yields that come up as a result. And we'll see that on Wednesday, right? But you need a starting point to get a method underway. So where will we begin tomorrow is talking about where this tRNA and the aaRS come from to do this.
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https://ocw.mit.edu/courses/5-111-principles-of-chemical-science-fall-2008/5.111-fall-2008.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: All right. As you're settling in, why don't you take 10 more seconds to answer the clicker question. This is the last question we'll see in class on the photoelectric effect, so hopefully we can have a very high success rate here to show we are all ready to move on with our lives here. OK, good. So, most of you did get the answer correct. For those of you that didn't, you, of course, can ask your TA's about this in recitation, they'll always have a copy of these slides. But just to point out the confusion can be we've actually switched what the question is here. What the information we gave was the work function, which is what we've been giving before, but now we gave you the kinetic energy of the ejected electron, so you just need to rearrange your equation so now you're solving for the incoming energy, which would mean that you need to add those two energies together. So, hopefully everyone that didn't get this right, can look at it again and think about asking it's just asking the question in a different kind of a way. All right. So, we can switch over to the class notes. So today, we're going to start talking about the hydrogen atom. Now that we have our Schrodinger equation for the hydrogen atom, we can talk about it very specifically in terms of binding energies and also in terms of orbitals. And we talked about on Wednesday the conditions that allowed us to use quantum mechanics, which then enable us to have the Schrodinger equation, which we can apply, and part of that is the wave particle duality of light and matter. So there was a good question in Wednesday's class about the de Broglie wavelength and if it can actually go to infinity. So I just wanted to address that quickly before we move on, and actually address another thing about dealing with wavelengths of particles that sometimes comes up. So, the question that we had in last class, was if we have actually a macroscopic particle, and the velocity let's say starts to approach zero, shouldn't we have the wavelength go to infinity, even if we have a magic board, and even if the mass is really large. So, in most cases, you would think that as the velocity gets very tiny, the mass is still going to be large enough to cancel it out and still make it such that the wavelength is going to be pretty small, right. Because if we think about the h, Planck's constant, here that's measured in 10 to the negative 34 joules per second. So, we would actually need a really, really, really tiny velocity here to actually overcome the size of the mass, if we're talking about macroscopic particles, to have a wavelength that's going to be on the order. So, let's say we're talking about the baseball, have a wavelength of the baseball that's on the order of the baseball. So, if we kind of think about the numbers we would need, we would actually need a velocity that approached something that's about 10 to the negative 30 meters per second. So first of all, that's pretty slow here. It's going to be hard to measure anyway, and in fact, if we're talking about something going 10 to the negative 30, and we're going to observe it using our eyes, so you using visible light to observe something going this slow, we're actually not going to be able to do it because we're limited by the wavelength of light to see how precise we can measure where the actual position is. So let's say we have the wavelength of light somewhere on the order of 10 to the negative 5 meters, being the wavelength of light, we're only going to be able to measure the velocity because of the uncertainty principle to a certain degree. And it turns out it's three orders of magnitude that -- the uncertainty is three orders of magnitude bigger than the velocity that we're actually trying to observe to get to a point where we could see the wavelength, for example, for even a baseball that is moving this slow? So that the more complete answer to the question is that no, we're never going to be able to observe that because of the uncertainty principle it's not possible to observe a velocity that's this slow for a macroscopic object. So, hopefully that kind of clears up that question. And, of course, when the velocity actually is zero, this equation that the de Broglie has put forth is valid for anything that has momentum, so if something does not have any velocity at all, it actually does not have momentum, so you can't apply that equation anyway. And another thing that came up, and it came up in remembering as I was writing your problem set for this week, which will be posted sometime this afternoon, your problem set 2, is when we're talking about wavelengths of particles, and for specifically for electrons sometimes, you're asked to calculate what the energy is. And I just want to remind everyone, so this is a separate thought here, that we often use the energy where energy is equal to h c divided by wavelength. So if we're talking about, for example, we know the wavelength of an electron and we're trying to find the energy or vice versa, is this an equation we can use to do that? What do you think? No. Hopefully you're going to say no. And the reason is, and this will come up on the problems and a lot of students end up using this equation, which is why I want to head it off and mention it ahead of time, we can't use an equation because this equation is very specific for light. We know it's very specific for light because in this equation is c, the speed of light. So any time you go to use this equation, if you're trying to use it for an electron, just ask yourself first, does an electron travel at the speed of light? And if your answer is no, your answer will be no, then you just know you can't use this equation here. So instead you'd have to maybe if you start with wavelength, go over there, and then figure out velocity and do something more like kinetic energy equals 1/2 n b squared to get there. So this is just a heads up for as you start your next problem set. All right. So jumping in to having established that, yes, particles have wave-like behavior, even though no, they're not actually photons, we can't use that equation. But we can use equations that describe waves to describe matter, and that's what we're going to be doing today. We're going to be looking at the solutions to the Schrodinger equation for a hydrogen atom, and specifically we'll be looking at the binding energy of the electron to the nucleus. So we'll be looking at the solution to this part of the Schrodinger equation where we're finding e. Then we'll go on, after we've made all of our predictions for what the energy should be, we can actually confirm whether or not we're correct, and we'll do this by looking at photon emission and photon absorption for hydrogen atoms, and we'll actually do a demo with that, too, so we can confirm it ourselves, as well as matching it with the observation of others. And if we have time, we'll move on also to talking about the other part of the solution to the Schrodinger equation, which is psi or this wave function here. And remember I said that wave function is just a representation of the particle, particularly when we're talking about electrons -- we're familiar with the term orbitals. psi is just a description of the orbital. So, we'll start with an introduction to that if we got to it at the end. So, to remind you, when we look at the Schrodinger equation here, we have two parts to it, so when we solve the Schrodinger equation, we're either finding psi, which as I said, is a wave function or an orbital. And in addition to finding psi, we can also solve to find e or to find the energy for any given psi, and these are the binding energies of the electron to the nucleus. And the most important thing about using the Schrodinger equation and getting out our solutions for potential orbitals and potential energies for an electron with the nucleus, is that what we find is that quantum mechanics, and quantum mechanics allowing us to get to the Schrodinger equation, allows us to correctly predict and confirm our observations for what we can actually measure are indeed the energy levels. Here we're talking about a hydrogen atom and that's what we'll focus on today. And it's incredibly precise and we're able to make the predictions and match them with experiment. Also, when we're looking at the Schrodinger equation, it allows us to explain a stable hydrogen atom, which is something that classical mechanics did not allow us to do. So here's the solution for a hydrogen atom, where we have the e term here is equal to everything written in green. We've got a lot of constants in this solution to the hydrogen atom, and we know what most of these mean. But remember that this whole term in green here is what is going to be equal to that binding energy between the nucleus of a hydrogen atom and the electron. So, let's go ahead and define our variables here, they should be familiar to us. We have the mass, first of all, m is equal to m e, so that's the electron mass. We also have e, which is going to be the charge on the electron. In addition to that, we have that epsilon nought value, remember that's the permittivity constant in a vacuum, and basically that is what we use as a conversion factor to get from units. We don't want namely coulombs to units, we want that will allow us to cancel out in this equation. And finally we have Planck's constant here, which we're all familiar with. So, what actually happens when people work with the solution to the Schrodinger equation for a hydrogen atom is that they don't always want to deal with all these constants here, so we can actually group them together and use them as a single new constant, and this new constant is the Rydberg constant. And the Rydberg constant is actually equal to 2 . 1 8 times 10 to the negative 18 joules. So when we pull out all of those constants and instead use the Rydberg constant, what it allows us to do is really simplify our energy equation. So now we have that energy is equal to the negative of the Rydberg constant divided by n squared. So, what we have left in our equation is only one part that we haven't explained yet, and that is that n value. And it turns out that when you solve the Schrodinger equation, you find that there are only certain allowed values of this integer n. And those allowed values range anywhere from n equals 1, you can have n equal that 2, 3, and it goes all the way up to infinity. But the important part is that there are only certain allowed values, so for example, you can't have 1 . 5 or 2 . 3, there are only these interger numbers. And this n here is what we call the principle quantum number. And what we find is when we apply n and plug it in to our energy equation, is that what we see is now we don't just have one distinct answer, we don't just have one possible binding energy of the electron to the nucleus. We're going to find that we actually have a whole bunch of possible, in fact, an infinite number of possible energy levels, and that's easier to see on this energy diagram here. So, let's start with n equals 1, since that's, of course, the simplest case. So, if we have n equals 1, we can plug it into our energy equation here, and find that the binding energy, the e sub n, for n equals 1, it's just going to be equal to the negative of the Rydberg constant, so we can actually graph that on an energy diagram here, and it's going to be down low at the bottom because that's going to be, in fact, the lowest or most negative energy when n equals 1. But we saw from our equation that there's more than just one possible value for n, so we could, for example, have n equals 2, n equals 3, all the way up to n equaling infinity. So what this tells us here is that this is not necessarily the binding energy of the electron in a hydrogen atom, it's also possible that it could, for example, have this energy, it could have this energy up here, it could have some energy way up here. So we have this infinite number of possible binding energies. But the really important point here is that they're quantized. So it's not a continuum of energy that we can have, it's only these punctuated points of energy that are possible. So as I tried to say on the board, we can have n equals 1, but since we can't have n equals 1/2, we actually can't have a binding energy that's anywhere in between these levels that are indicated here. And that's a really important point for something that comes out of solving the Schrodinger equation is this quantization of energy levels. And thanks to our equation simplified here, it's very easy for us to figure out what actually the allowed energy levels are. So for n equals 2, what would the binding energy be? Someone shout it out. Yup. So, I think the compilation of the voices that I heard was negative r h over 2 squared. We can do the same thing for 3, negative r h over 3 squared is going to be our binding energy. For 4, we can go all the way up to infinity, and actually when we get to the point where it's infinity, what we find is the binding energy at that point is going to be zero. And when we get to infinity, what that means is that we now have a free electron, so now the electron has totally separated from the atom. And that makes sense because we're at the point where there's no binding energy keeping it stable. You'll also know that all of these binding energies here are negative, so the negative sign indicates that it's low. It's a more negative energy, it's a lower energy state. So whenever we're thinking about energy states, it's always more stable to be more low in an energy well, so that's why it makes sense that it's favorable, in fact, to have an electron interacting with the nucleus that stabilizes and lowers the energy of that electron by doing so. So, we actually term this n equals 1 state gets a special name, which we call the ground state, and it's called the ground state because it is, in fact, the lowest to the ground that we can get. It's the most negative and most stable energy level that we have. And when we think about kind of in a more out practical sense what we mean by all of these binding energies, another way that we can put it is to give it some physical significance, and the physical significance of binding energies is that they're equal to the negative of the ionization energies. So, for example, in a hydrogen atom, if you take the binding energy, the negative of that is going to be how much energy you have to put in to ionize the hydrogen atom. So, if, for example, we were looking at a hydrogen atom in the case where we have the n equals 1 state, so the electron is in that ground state, the ionization energy, it makes sense, is going to be the difference between the ground state and the energy it takes to be a free electron. When we graph that on our chart here, it becomes clear that yes, in fact, the ionization energy is just the negative of the binding energy, so we can just look over here and figure out what our ionization energy is. So when we're talking about the ground state of a hydrogen atom, our ionization energy is just the negative of the Rydberg constant, so that easy, it's 2 . 1 8 times 10 to the negative 18 joules. So, that should make a lot of sense intuitively, because it makes sense that if we need to ionize an atom, we need to put energy into the atom in order to eject that electron, and that energy we need to put in better be the difference between where we are now and where we have to be to be a free electron. So in most cases when we talk about ionization energy, if we don't say anything specific to the state we're talking about, you should always assume that we are, in fact, talking about the ground state. So, oftentimes you'll just be asked about ionization energy. If it doesn't say anything else we do mean n equals 1. But, in fact, we can also talk about the ionization energy of different states of the hydrogen atom or of any atom. So, for example, we could talk about the n equals 2 state, so that's this state here, and it's also what we could call the first excited state. So we have the ground state, and if we excite an electron into the next closest state, we're at the first excited state, or the n equals 2 state. So, we can now calculate the ionization energy here. it's an easy calculation -- we're just taking the negative of the binding energy, again that makes sense, because it's this difference in energy here. So what we get is that the binding energy, when it's negative, the ionization energy is 5 . 4 5 times 10 to the negative 19 joules. So we should be able to think about these binding energies and figure out the ionization energy for any state that were asked about. So if we can switch over to a clicker question here and we'll let you do that. And what we're asking you to do is now tell us what the ionization energy is of a hydrogen atom that is in its third excited state. All right. Let's take 10 more seconds on that. OK. Interesting. Usually the majority is correct, but actually what you did was illustrate a point that I really wanted to stress and there's no better way to stress it then to get it incorrect, especially when it doesn't count, it doesn't hurt so bad. So, if you want to switch back over to the notes, we'll explain why, in fact, the correct answer is 4. So the key word here is that we asked you to identify the third excited state. So, what white is n equal to for the third excited state? 4 OK. So that explains probably most of the confusion here and you just want to be careful when you're reading the problems that that's what you read correctly. I think everyone would now get the clicker question correct. So, the third excited state, is n equal to 4, because n equals 2 is first excited, 3 is second excited, 4 is third excited state. So now we can just take the negative of that binding energy here, and I've just rounded up here or 1 . 4 times 10 to the negative 19 joules. So, I noticed that a few, a very, very small proportion of you, did type in selections that were negative ionization energies. And I'll just say it right now you can absolutely never have a negative ionization energy, so that's good to remember as well. And intuitively, it should make sense, right, because ionization energy is the amount of energy you need to put in to eject an electron from an atom. So you don't want to put in a negative energy, that's not going to help you out, you need to put in positive energy to get an electron out of the system. So that's why you'll find binding energies are always negative, and ionization energies are always going to be positive, or you could look at the equation and see it from there as well. All right. So, using the equation we'd initially discussed, the negative r sub h over n squared, we could figure out all of the different ionization energies and binding energies for a hydrogen atom, and it turns out if we change the equation only slightly to add a negative z squared in there, so, negative z squared times the Rydberg constant over n squared, now let's us calculate energy levels for absolutely any atom as long as this one important stipulation, it only has 1 electron in it. So basically we're dealing with hydrogen atoms and then we're going to be dealing with ions. So, for example, a helium plus 1 ion has 1 electron at z equal 2. A lithium 2 plus ion has 1 electron, it has z equals 3, so if we were to plug in, we would just do z squared up here, or 3 squared. Terbium 64 plus, another 1 electron atom. What is z for that? Yup. 65. So again, terbium 64 plus, not an ion we probably will run into, but if we did, we could, in fact, calculate all of the energy levels for it using this equation here. And the difference between the equation, the reason that that z squared comes in there is because if you go back to your notes from Wednesday, and you look at the long written out form of the Schrodinger equation for a hydrogen atom, or any 1 electron atom, you see the last term there is a coulomb potential energy between the electron and the nucleus. So, of course, when we have a charge on the nucleus equal to 1, as we do in a hydrogen atom, the z is equal to 1, so it drops out there, but normally we would have to include the full charge on the nucleus, which is equal to z or the atomic number times the electron. So even if we strip an atom of all of its electrons, we still have that same amount of positive charge in the nucleus. So, this allows us to look at a bunch of different atoms, of course, limited to the fact that it has to be a 1 electron atom. So, now that we can calculate the binding energies, we can think about is this, in fact, what matches up with what's been observed, or, in fact, could we predict what we will observe in different kinds of situations now that we know how to use the binding energy, and hopefully we can and we will. So one thing we could do is we could look at the different wavelengths of light that are emitted by hydrogen atoms that are excited to a higher state. So what we'll do in a few minutes here is try this with hydrogen. So we'll take h 2 and we'll run -- or actually we'll have h 2 filled in an evacuated glass tube. When we increase the potential between the 2 electrodes that we have in the tube -- we actually split the h 2 into the individual hydrogen atoms, and not only do that, but also excite the atoms. So when you just run across an atom in the street, you can assume it's going to be in its most stable ground state, that's where the electron would be, but when we add energy to the system, we can actually excite it up into all different sorts of higher states -- n equals 6, n equals 10, any of those higher states. But that only happens momentarily, because, of course, if you have an energy in a higher energy level, it's going to want to drop back down to that lower or more stable level. And when it does that it's going to give off some energy equal to the difference between those two levels. And that will be associated with a wavelength if it releases the energy in terms of a photon. So that's what we'll look at in a few minutes. There's some important things to point out about what is happening here. Just to visualize exactly what we're saying, what we're saying is when we have an energy in a higher energy level, so let's say energy level, this initial level high up here, and it drops down to a lower final level, what we find is that the photon that is going to be emitted is going to be emitted with the exact energy, and the important term here is the exact. That is the difference between these two energy states. That makes sense because we're losing energy, we're going to a level lower level, so we can give off that extra in the form of light. And we can actually write the equation for what we would expect the energy for the light to be. So this delta energy here is very simply the energy of the initial state minus the energy of the final state. This is a little bit generic, we're not actually specifying the states here, but we could, we know we can calculate the energy from any of the states. So, for example, let's say we excited the hydrogen atom such that the electron was starting in the n equals 6 state, so that's our n initial. And we drop down to the n equals 2 state, or the first excited state. Then we would be able to change our equation to make it a little bit more specific and say that delta energy here is equal to energy of n equals 6, minus the energy of the n equals 2 state. When we talk about that frequency of light that's going to be emitted, it's not too commonly that we'll actually talk about it in terms of energy. A lot of times we talk about light in terms of its frequency or it's wavelength, but that's OK, because we know how to convert from energy to frequency, so we can do that here as well, where our frequency is just our energy divided by Planck's constant, and since here we're talking about a delta energy, we're going to talk about the frequency as being equal to delta energy over h. Or we can say the frequency is going to be equal to the energy initial minus the energy final all over Planck's constant. So this means that we can go directly from the energy between two levels to the frequency of the photon that's emitted when you go between those levels. What we can also do is say something about the wavelength as well because we know the relationship between energy and frequency and wavelength. So, in the first case here, let's say we go from a high level to a low level, let's say we go from five to one. If we have a large energy difference here, are we going to have a high or low frequency? Good. A high frequency. If we have a high frequency, what about the wavelength, long or short? All right. Good. So we should always be able to keep these relationships in mind. So, similarly in a case where instead we have a small energy difference, we're going to have a low frequency, which means that we're going to have a long wavelength here. So now we can go ahead and try observing some of this ourselves. So what we're actually going to do is this experiment that I explained here where we're going to excite hydrogen atoms such that they're electrons go into these higher energy levels, and then we're going to see if we can actually see individual wavelengths that come out of that that correspond with the energy difference. So, our detection devices are a little bit limited here today, we're actually only going to be using our eyes, so that means that we need to stick with the visible range of the electromagnetic spectrum. Actually that simplifies things, because that really cuts down on the number of wavelengths that we're going to be trying to observe here. So it turns out that in the visible range, when you figure out the differences between energy levels, in hydrogen atoms, there's only 4 wavelengths that fall in the visible range of the spectrum. So hopefully, when we turn out the lights, we're going to turn on this lamp here, which has hydrogen in it and we're going to excite that hydrogen. You'll see light coming out, but it's, of course, going to be bulk light -- you're not going to be able to tell the individual wavelengths. So what our TAs, actually if they can come down now, are going to pass out to you is these either defraction goggles, or just a little plate, and you're going to be splitting that light into its individual wavelengths. And the glasses, there aren't enough to go around for all of you, so that's why there's plates. And though glasses do look way cooler, the plates work a little better, so either you or your neighbor should try to have one of the plates in case one of you can't see all the lines. So our TAs will pass these around for us. And I also want to point out, it's guaranteed pretty much you'll be able to see these three here in the visible range -- you may or may not be able to see, sometimes it's hard to see that one that's getting near the UV end of our visible spectrum. So we won't worry if we can't see that. I'll take one actually. Can you raise your hand if you if you still need one? All right, so TA's walk carefully now, I'm going to shut the lights down here. All right, so we do still have some little bits of ambient light, so you might see a slight amount of the full continuum. But if you look through your plate, and actually especially if you kind of look off to the side, hopefully you'll be able to see the individual lines of the spectrum. Is everyone seeing that? Yeah, pretty much. OK. Can anyone not see it? Does anyone need -- actually I can't even tell if you raise your hand. So ask your neighbor if you can't see it and get one of the plates if you're having trouble seeing with the glasses. So this should match up with the spectrum that we saw. And actually keep those glasses with you. We'll turn the light back on for a second here. And hydrogen atom is what we're learning about, so that's the most relevant here. But just to show you that each atom does have its own set of spectral lines, just for fun we'll look at neon also so you can have a comparison point. All right. So, this is probably a familiar color having seen many neon lights around everywhere. So you see with the neon is there's just a lot more lines in that orange part of the spectrum then compared to the hydrogen, and that's really what gives you that neon color in the neon signs. That's the true color of a neon being excited, sometimes neon signs are painted with other compounds. All right, does everyone have their fill of seeing the neon lines? STUDENT: No. PROFESSOR: All right, let's take two more seconds to look at neon then. All right. So our special effects portion of the class is over. And what you see when we see it with our eye, which is all the wavelengths, of course, mixed together, is whichever those wavelengths is most intense. So, when we looked at the individual neon lines, it was the orange colors that was most intense, which is why we were seeing kind of a general orange glow with the neon, which is different from what we see with the hydrogen. All right, so we can, in fact, observe individual lines. There's nothing more exciting to see with your glasses on, while you look nice. You can take those off if you wish to, or you can try to just be splitting the light in the room until the TAs grab your glasses, either is fine. It turns out that we are far from the first people, although it felt exciting, we did not discover this for the first time here today. In fact, J.J. Balmer, who was a school teacher in the 1800s, was the first to describe these lines that could be seen from hydrogen. And he saw the same lines that we saw here today, and although he could not explain, even start to explain why you saw only these specific lines and not a whole continuum of the light. Right, we already have an idea because we just talked about energy levels, we know there's only certain allowed energy levels, but at the time there's no reason J.J. Balmer should have known this, and in fact he didn't, but he still came up with a quantitative way to describe what was going on. He came up with this equation here where what he found was that he could explain the wavelengths of these different lines by multiplying 1 over 4 minus 1 over some integer n, and multiplying it by this number, 3 . 29 times 10 to the 15 Hertz, and he found that this was true where n was some integer value -- 3, 4, 5, or 6. So he could explain it quantitatively in terms of putting an equation with it, but he couldn't explain what was actually going on. But we, having learned about energy levels, having had the Schrodinger equation solved for us to understand what's going on, can, in fact, explain what happened when we saw these different colors. So, what we know is happening is that were having transitions from some excited states to a more relaxed lower, more stable state in the hydrogen atom. And it turns out what we can detect visibly with our eyes is in the visible range, and that means that the final state is n equals 2. Because you see how n equals 1 is so much further away, and actually that's not to scale, it's actually much, much further down in the energy well, such that the energy of the light is so great that it's going to be in ultraviolet very high energy, high frequency range. So we can't actually see any of that, it's too high energy for us to see. So everything we see is going to be where we have the final energy state being n equals 2. So if we think about, for example, this red line here, which energy state or which principle quantum number do you think that our electron started in? STUDENT: Three. PROFESSOR: Good. So, it's going to be in 3, because that's the shortest energy difference we can have, and the red is the longest wavelength we can see -- those 2 are inversely related, so it must be n equals 3. What about the kind of cyan-ish, blue-green one? Yup. so n equals 4 for that one. Similarly we can go on, match up the others. So n equals 5 for the bluish-purple and the violet is n equals 6. And again, that matches up, because the violet, or getting really close to the UV range here has the longest energy, so the highest frequency, and that's going to be the shortest wavelength and we can see here it is, in fact, the shortest wavelength that we can actually see. So, we can see if we can come up with the same equation that J.J. Balmer came up with by actually starting with what we know and working our way that way instead of coming up from the other direction, which he did, which was just to explain what he saw. So, if we start instead with talking about the energy levels, we can relate these to frequency, because we already said that frequency is related to, or it's equal to the initial energy level here minus the final energy level there over Planck's constant to get us to frequency. And we also have the equation that comes out of Schrodinger equation that tell us exactly what that binding energy is, that binding energy is just equal to the negative of the Rydberg constant over n squared. So that means that our frequency is going to equal, if we plug in e n into the initial and final energy here, 1 over Planck's constant times negative r h over n initial squared, minus negative r h over n final squared. So we have an equation that should relate how we can actually calculate the frequency to what J.J. Balmer observed. We can simplify this equation by pulling up the r h and getting rid of some of these negatives here by saying the frequency is going to be equal to the Rydberg constant divided by Planck's constant all times 1 over n final squared -- this is just to switch the signs around and get rid of some negatives -- minus 1 over n initial squared. We can plug this in further when we're talking about the visible part of the light spectrum, because we know that for n final equals 2, then that would mean we plug in 2 squared here, so what we get is 1 over 4. So this is our final equation, and this is actually called the Balmer series, which was named after Balmer, and this tells us the frequency of any of the lights where we start with an electron in some higher energy level and we drop down to an n final that's equal to 2. So it's a more specific version of the equation where we have the n final equal to 2. And it turns out that actually we find that this matches up perfectly with Balmer's equation, because the value of r h, the Rydberg constant divided by the Planck's constant is actually -- it's also another constant, so we can write it as this kind of strange looking cursive r here. Unfortunately, this is also call the Rydberg constant, so it's a little bit confusing. But really it means the Rydberg constant divided by h, and that's equal to 3 . 29 times 10 to the 15 per second. So if you remember what the equation Balmer found was this number multiplied by this here. So, we found the exact same equation, but just now starting from understanding the difference between energy levels. So, sometimes you'll find the Rydberg constant in different forms, but just make sure you pay attention to units because then you won't mess them up, because this is in inverse seconds here, the other Rydberg constant is in joules, so you'll be able to use what you need depending on how you're using that constant. So in talking about the hydrogen atom, they actually have different names for different series, which means in terms of different n values that we end in. So we talked about what we could see with visible light, we said that's actually the Balmer series. So anything that goes from a higher energy level to 2 is going to be falling within the Balmer series, which is in the visible range of the spectrum. We can think about the Lyman series, which is where n equals 1. We know that that's going to be a higher energy difference, so that means that we're going to be in the UV range. We can also go in the opposite direction. So, for example, when n equals 3, that's called the Paschen series, and these are named after basically the people that first discovered these different lines and characterized them, and this is in the near IR range. And again, the n equals 4 is the Brackett series, and that's an IR range. I think there's names for even a few more levels up. You don't need to know those, but just because it's a special case with the hydrogen atom, they do tend to be named -- the most important, of course, tends to be the Balmer series because that's what we can actually see being emitted from the hydrogen atom. So, now we should be able to relate what we know about different frequencies and different wavelengths, so Darcy can you switch us over to a clicker question here? And we can also talk about the difference between what's happening when we have emission, and we're going to switch over to absorption. So, we just talked about emission, so before we head into absorption, if you can answer this clicker question in terms of what do you think absorption means having just discussed hydrogen emission here? So we have four choices in terms of initial and final energy levels, and also what it means in terms of the electron -- whether it's gaining energy or whether it's going to be emitting energy? So, why do you take 10 seconds on that, we'll make it a quick one. All right, great. So already just from knowing the emission part, we can figure out what absorption probably means. Absorption is just the opposite of emission, so instead of starting at a high energy level and dropping down, when we absorb light we start low and we absorb energy to bring ourselves up to an n final that's higher. And instead of having the electron giving off energy as a photon, instead now the electron is going to take in energy from light and move up to that higher level. So now we're going to be talking briefly about photon absorption here. So again, this is just stating the same thing, and it could take in a long wavelength light, which would give it just a little bit of energy, maybe just enough to head up one energy level or two, or we could take in a high energy photon, and that means that the electron is going to get to move up to an even higher energy level. And again, we can talk about the same relationship here, so it's a very similar equation to the Rydberg equation that we saw earlier, except now what you see is the n initial and the n final are swapped places. So instead now we have r h over Planck's constant times 1 over n initial squared minus n final squared. And what you want to keep in mind is that whatever you're dealing with, whether it's absorption or emission, the frequency of the light is always going to be a positive number, so you always want to make sure what's inside these brackets here turns out to be positive. So that's just a little bit of a check for yourself, and it should make sense because what you're doing is you're calculating the difference between energy levels, so you just need to flip around which you put first to end up with a positive number here, and that's a little bit of a check that you can do what yourself. So let's do a sample calculation now using this Rydberg formula, and we'll switch back to emission, and the reason that we'll do that is because it would be nice to actually approve what we just saw here and calculate the frequency of one of our lines in the wavelength of one of the lines we saw. So what we'll do is this problem here, which is let's calculate out what the wavelength of radiation would be emitted from a hydrogen atom if we start at the n equals 3 level and we go down to the n equals 2 level. So what we need to do here is use the Rydberg formula, and actually you'll be given the Rydberg formulas in both forms, both or absorption and emission on the exams. if you don't want to use that, you can also derive it as we did every time, it should intuitively make sense how we got there. But the exams are pretty short, so we don't want you doing that every time, so we'll save the 2 minutes and give you the equations directly, but it's still important to know how to use them. So, we can get from these energy differences to frequency by frequency is equal to r sub h over Planck's constant times 1 over n final squared minus 1 over n initial squared. So let's actually just simplify this to the other version of the Rydberg constant, since we can use that here. So kind of that strange cursive r, and our n final is 2, so 1 over 2 squared minus n initial, so 1 over 3 squared. So, our frequency of light is going to be equal to r times 5 over 36. But when we were actually looking at our different wavelengths, what we associate mostly with color is the wave length of the light and not the frequency of the light, so let's look at wavelength instead. We know that wavelength is equal to c over nu. We can plug in what we have for nu, so we have 36 c divided by 5, and that cursivey Rydberg constant, and that gives us 36 times the speed of light, 2 . 998 times 10 the 8 meters per second, all over 5 times 3 . 29 times 10 to the 15 per second. So, what we end up getting when we do this calculation is the wavelength of light, which is equal to 6 . 57 times 10 to the negative 7 meters, or if we convert that to nanometers, we have 657 nanometers. So does anyone remember what range of light 657 nanometers falls in? What color? STUDENT: Red. PROFESSOR: Yeah, it's in the red range. So that's promising. We did, in fact, see red in our spectrum, and it turns out that that's exactly the wavelength that we see is that we're at 657 nanometers. So it turns out that we can, in fact, use the energy levels to predict, and we could if we wanted to do them for all of the different wavelengths of light that we observed, and also all the different wavelengths of light that can be detected, even if we can't observe them. All right, so that's what we're going to cover in terms of the energy portion of the Schrodinger equation. I mentioned that we can also solve for psi here, which is the wave function, and we're running a little short on time, so we'll start on Monday with solving for the wave function.
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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Now we're going to analyze a more complicated example of drag forces, where we have an object falling in a gravitational field with gravity. We have a resistive force. And this is an object in air. And so our model will be for the resistive force that it's going to be proportional to the velocity squared. Now to get its direction right, opposing the motion, j hat. So this will turn out to be a more complicated analysis. But first again, let's think about the units of our coefficient beta. It has the units of force divided by velocity squared. So we write the units of force, kilogram meters per second squared, and the units of velocity squared, meter squared per second squared. And so we see that we have units of kilogram per meter for our coefficient beta. Now, we'll apply as usual Newton's second law, F equals m a, to get our equation of motion. We're looking at the j direction. And our forces are gravity minus the velocity squared resistive drag force. And that's equal to the derivative of the velocity dv/dt. In this example, it's a one dimensional motion. So I'm dropping any mention of y direction for simplicity. Now we can rewrite this equation as dv/dt. Let's divide through by m is g minus beta over m v square. And this is a linear-- it's a first order differential equation, dv/dt. It's a non-linear equation because the velocity term appears here as a square and there's a constant term. But we can still apply our technique of separation of variables. And so when we write this equation as dv/dt we'll separate out dv's and t's, so we have d d times g minus beta over and m v squared is equal to dt. Now, I'm going to do two things just to clean this up for algebra a little bit later. I'm going to multiply both sides by a minus sign. And I'm going to pull the g out. So I have 1 minus beta over m g v squared. And on the other side, I have minus dt. And now I can get this equation in the form that I'd like to integrate, which is minus dv times 1 minus beta over m g v squared equals minus gdt. Now, the trouble here is this integral is a little bit complicated. So I'd like to make a change of variable. And my change of variable will be u equals the square root of beta m g times v. And that implies that du is beta m g dv. And the limits, if we start our object at rest, so if u0 equals 0, then because v0 equal 0, that's our first limit. Now we have to be a little bit careful because if we drop this object at rest, initially it will be moving very slowly. And so our resistive model doesn't actually apply. However, we're going to neglect that effect even though if we were to do a more complicated analysis, we would have to change our model as the object is following, so it would be a multi-stage motion. First, at the beginning with our only velocity dependent resistant. And then as it gets some initial speed and it's going faster, we change our model. That's why the actual problem can be quite complicated. But we're just trying to keep things simple here. And then u of t is square root of beta mg d of t. And then with this change of variable, my integration, remember I have a dv, so I have to multiply the left side by mg over beta. And I'm integrating with a minus sign du times 1 minus u squared from 0 to this final value u of t And over here I'm just integrating minus g dt. Now again, for simplicity, I'm going to bring this term, the beta over mg over to the other side. So I'll use the magic of our light board by just erasing that and bringing it to the other side, which makes my life a little easier. And now, this interval can be done by the method of integration by parts. It's a nice problem in calculus. And you can verify for yourself that the result is one half natural log of 1 plus u over 1 minus u evaluated at our limits. And over here I have minus g square root of beta over mg. Now, once again, for a little bit of simplicity, I'm going to bring the 2 over to the other side. And now, I evaluate my limits. Now recall that when you have a minus log, we're flipping, because log of b over a equals minus log over ab, so when I put out my limits in, I have natural log of-- now remember, what are our limits? We have 1-- I'm flipping-- minus u is beta mg times v of t. And I have the 1 plus beta mg v of t. And that's equal to minus as 2g square root of beta mg. Now, again, we'll use the fact that e to log of x is x. And so if I exponentiate both sides, I end up with 1 minus the square root of beta mg d of t over 1 plus square root of beta mg v of t is equal to exponential minus 2g beta mg times t. And we'll just move that. OK, now this is a little bit of algebra. I want to solve for v of t. If I bring this side over to there, I'll just do that to make the first step a little simpler to see. So we have 1 minus square root of beta mg v of t equals 1 plus square root of beta over mg v of t times e to the minus this factor 2g square root of beta mg t. Now this is a lot of stuff to carry around. I'd like to introduce a constant here, tao, which I'm going to find to be square root of mg beta 1 over 2g. And so this whole term is going to just be e to the minus tao. It's a nice example for you to work out that the units of tao are the units of seconds. And that's a little exercise to work out. Now, I just have to collect my terms. And what I'll do is I'll collect the T terms on the right and the terms that don't have vt on the left. So I have 1 minus e to the minus t over tao on the left is equal to 1 plus e to the minus t over tao on the right times beta mg v of t. And so I get my solution, v of t equals the square root of mg over beta times 1 minus e to the minus t over tau over 1 plus e to the minus t over tao. Well, it's not a simple solution at all. But let's examine when you have a case like this-- again, it would be a nice exercise to graph this out. But right now we're going to consider the limit as t goes to infinity. And remember that e minus t over t goes to 0 when t goes to infinity. So we just have 1 over 1. And what we get as t goes to infinity is the quantity mg over beta. And this is what we call the terminal velocity. Now what does terminal velocity mean? Well, when object is falling and there's a resistive force, as the object falls faster and faster, the resistive force gets greater and greater until if we go back to Newton's second law and look at it, as v gets faster and faster, eventually these two terms are equal. And when these two terms are equal, that's the statement that the right-hand side has to be zero. So what we mean by terminal velocity is it's the velocity is no longer changing in time. And then we can immediately check our work by going to Newton's second law and see what that case is if we set this quantity equal to zero. In other words, when we set mg minus beta v squared terminal equal to 0 we can solve for v terminal, and we get a square root of mg divided by beta. And that agrees with our lengthy calculation. So we think we're on the right track.
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https://ocw.mit.edu/courses/7-014-introductory-biology-spring-2005/7.014-spring-2005.zip
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So today we are going to continue where we left off last time talking more specifically about variations on the theme of life. And last year I tried to do this lecture using PowerPoint and it was a total disaster so I'm going back to the board. You will have the PowerPoint slides. They'll be on the Web to download to summarize basically what I'm drawing on the board. But it will be slightly different on the board. But I found that for this material it really doesn't work to exclusively use the PowerPoint. So last time we talked about, remember, my life on earth abridged where -- -- we had photosynthesis making glucose or organic carbon plus oxygen? And then the reverse of this was respiration. And then we had elements cycling in the middle. And I said this is very, very abbreviated of how all life on earth works. And so today what I'm going to do is tell you that that's not right. That's grossly oversimplified. And there are some really interesting variations on the theme of how to extract energy and carbon and reducing power and electrons from the earth's system to create life. And it's mostly microbes that have these diverse possibilities. And, again, even what I'm going to talk to you about today is oversimplified. If you go to a microbiology textbook you'll find just about every possible combination of energy sources, carbon sources and electron sources in some microorganisms somewhere to get through life. So I'm giving you, again, the simplified version because otherwise it gets way too complicated. So all of life needs carbon and energy, and a lot of other elements, too, but these are the main axis upon which we're going to order our universe today. So for carbon the choices are inorganic or organic. So this would be CO2 and this might be glucose or sugars, any sugars. And then on the energy axis they can use solar energy, as in photosynthesis, or they can use chemical energy. And within the chemical energy sources they can be inorganic or organic like sugars, etc. And often here you have reduced compounds such as hydrogen sulfide, ammonia, and we'll talk about these. So these are the ways we divide up the possibilities for carbon and energy sources to be alive. All organisms also need to have an energy currency in the cell. And you've talked about this a lot already in the biochemistry lectures so I'm, again, just giving you the impressionist view of this. You know the details. This is just to get you organized. And so all life uses redox reactions. And in your handouts for today there's a primer on redox reactions just in case you want to review that. And one of the key reactions we'll talk about today is the conversion of NADP. If you put energy in you can reduce it to NADPH. So that's a reduction. And the reverse you get energy out when it's oxidized. Now, we're going to be talking about oxidation and reduction today. And then they all use ATP which you've talked a lot about here. And the couple here is ADP. Put energy in. You make ATP which is a high energy intermediate. And in converting it back to ADP that energy can be released. And this is used in the biochemistry of the cell. So all cells have these two energy conversion processes in common. OK, so let's look at just summarizing what we're going to go over today. This is a summary of options for life. See also Freeman, Chapter 25. There is some discussion of this. And we can divide life here between what we call autotrophs. These are organisms that can make their own organic carbon. In other words, they can convert carbon dioxide to organic carbon. Heterotrophs are organisms that can only use organic carbon. They rely on the guts of other organisms in order to get through life. And so now we're going to systematically go through these processes that fall under each one of these. Oxygenic photosynthesis is the one we've been talking about last time and in my abbreviated version of life on earth. And this is carried out by eukaryotic organisms, plants, trees, etc., and also by prokaryotic organisms. Those are the cyanobacteria, microscopic photosynthetic plants. They use CO2 and sunlight. So our first variant on this theme we'll get into is a group of bacteria that do anoxygenic photosynthesis. Oxygenic means they evolve oxygen. These guys use solar energy but they don't evolve oxygen. And we'll get into how that works. And then there's a group of organisms that still use CO2. And in the very similar pathway the Calvin Cycle is photosynthesis. But they use chemical energy in order to make these intermediates to fix CO2. OK, so let's talk about those first. And so we're going to talk about the autotrophs. And all of them share this pathway, CO2 to C6H12. This would be glucose. And it takes ATP to run this reaction and it also takes reduced NADPH -- -- to run this reaction. It also takes this enzyme ribisco which you've talked about I'm sure, ribulose bisphosphate carboxylase. And this is the enzyme that initially takes the CO2 from the atmosphere and binds it to an organic carbon. Now, in a detailed version of this is what's called the Calvin Cycle or the Calvin/Benson Cycle. I don't know which one your book calls it. Calvin got the Nobel Prize but Benson was the graduate student that did all the work, so you should recognize that. Anyway, you studied this in great deal. But an interesting factoid is that ribisco is the most abundant protein on earth. That tells you how important this reaction is for sustaining life on earth. So notice that in order to drive this reaction, which is the Calvin Cycle, it requires energy and reducing power. So where do they get it? Well, there are three ways that autotrophs can get energy and reducing power to drive this reaction. And the first is oxygenic photosynthesis. And the second is anoxygenic. And the third is chemosynthesis. OK, those first three there. So now we're going to go through each of these and look at how they work remembering that all of them are generating ATP and NADPH in order to drive that. So all of the autotrophs have that in common. Well, oxygenic photosynthesis is the one that you know well already. You've studied it in great detail in biochemistry. So we're going to, again, give you the abbreviated version here just so you have a template to map these other ones onto. These are what are known as the light reactions of photosynthesis, the Z scheme taking solar energy, splitting water, evolving oxygen and synthesizing ATP and NADPH. This is all familiar, right? Very familiar. I'm just writing it in a cartoon version. OK, so this is the NADPH and ADP that goes to fuel that process. OK, so now, well, at least I can do it on that board. Let me do it on this board. Anoxygenic -- -- is almost exactly like this process, but instead of splitting water these guys oxidize hydrogen sulfide. So here's our ATP and NADPH. And they use sunlight to do this. So these are called photosynthetic bacteria. And they were around very early on the earth. Long before the earth's atmosphere was oxygenated these were the guys that were able to use solar energy and make organic carbon but without evolving oxygen. Then somewhere along the line some cell evolved, had some mutations and somehow figured out that water, this abundant source of water was a much better electron donor than hydrogen sulfide. And once the biochemistry figured this out, you can see the simple substitution here, the whole earth started going in a different direction. So this is an interesting example of how a small biochemical innovation can dramatically change the whole nature of the planet. Now, these guys are still around on earth. In fact, I'm going to show you some. I'll explain this at the end, but I have some captured in here. See that little purple band? Those are those guys. I've got other little tricks in here but I'll save those. Well, you cannot really see the purple band. But you can come up later and look at it. Those are photosynthetic bacteria. So they're still around on the earth but they're stuck in places where there's no oxygen. So they have a rather restricted niche on the planet now, but they're still extremely important. What did I do? Oh, here it is. So one of the places that they can be found, and if you're interested in them a great place to go find some is out at the Mystic Lakes in Arlington which is a permanently stratified lake so the bottom of the lake is always anaerobic. There's never oxygen there. In a typical lake like that you have a lot of mud on the bottom and you have a lot of hydrogen sulfide coming out of the mud from bacterial processes that we'll talk about. And you have light here. And so you have a gradient here of this is oxygen and this is H2S. And these photosynthetic bacteria have to life somewhere where there's enough light to photosynthesize and enough hydrogen sulfide to use in this part of the reaction. But they're very sensitive to oxygen so they cannot be in the oxygenated part of the lake. So you find them in a layer. It's called the squeeze. They have to have light so they have to be up, but they cannot have oxygen so they have to be down. And they need hydrogen sulfide so they have to be down. So they're layered in lakes. OK. So what about these guys, chemosynthesis? They don't rely on solar energy. Again, they're still driving the Calvin Cycle reducing CO2 from the air into organic carbon, but they're not using sunlight. So what do they do? They get their energy -- -- from redox reactions. And let's just show you an example. Redox reactions couple to the conversion of oxygen to H2O. So oxygen is involved in these reactions. And one organism, for example, can take ammonia and convert it to nitrite. Another type of organism can take nitrite and convert it to nitrate. And there are other organisms that can take hydrogen sulfide and convert it to sulfate. And some can take hydrogen sulfide, oh, no, take iron, ferrous iron, Fe2+ and convert it to Fe3+. So in all of these cases what is happening to these compounds? Are they being oxidized or reduced? I heard an oxidized. Yes, they're being oxidized. So these reduced compounds, relatively reduced compounds can be utilized by oxidizing them. The organism can release the energy that's needed. ATP is generated here. And NADPH is generated by any of these redox couples. So using this energy then the cell takes the reduced NADPH and the ATP and it runs the Calvin Cycle, chemosynthesis. OK. Now, you may think that these are kind of strange, weird bacteria that life in strange pockets of the earth where there's no oxygen. And who cares anyway? They're outdated. They dominated the earth way back in the early stages of the earth but they're not so important now. Well, that's not true. They're incredibly important. In some ecosystems they're the total base of the entire ecosystem. But also on a global scale, as you'll learn, you should have a feeling for this by the end of this lecture, but also when we talk about global biogeochemical cycles you will learn that these microbes are really messengers for electrons in the environment. Without them the redox balance of the earth would not be maintained, OK? You cannot have nothing but oxidizing reactions or nothing but reduction reactions and have a system sustain itself. So it's these microbes that are playing a really important role in maintaining the redox balance of the earth. OK. Now, one system that I'm going to show you in that DVD, that will do much better justice to it than my drawings here, that's a deep-sea volcano in case you didn't recognize it. And this is 2500 meters at the bottom of the ocean, very, very deep. And there is intense heat. I mean just think of a volcano on the surface of the earth. Intense heat and reduced compounds are found in the earth's mantle that are ready to erupt through this deep-sea volcano. And you have sulfate in the sea water that percolates through here. And as it percolates in and gets draw into the volcanic stuff that's coming out of here it's reduced to hydrogen sulfide coming out of the volcano. But you have oxygen in the water in the deep-sea. And we'll be talking about this when we talk about ocean circulation. But the oceans have a global ocean circulation where the surface water that's in equilibrium with the atmosphere actually sinks and travels along the bottom of the ocean. So there is oxygen in the bottom of the ocean, unlike many lakes where you don't have oxygen. And we'll talk about that difference. And in the hot vents the water coming out of here can be very, very hot, but there's a gradient right as it comes out meeting the colder sea water. And so what you have here is a perfect incubator for chemosynthetic bacteria -- -- that use the hydrogen sulfide in chemosynthesis to fix carbon dioxide using the oxygen here. And that forms the base of the entire food web in the deep ocean because there's no light down there. There's no photosynthesis. There's only chemosynthesis. And just a little story that goes back to when I first came to MIT as an assistant professor in 1976. You weren't even born. But when I was young we used to go the Muddy Charles Pub periodically after work and have beers. And there was a professor, in this department actually, John Edmond, who passed away several years ago but who used to be there. It was sort of like our Cheers. And I'll never forget the day he came back from a cruise. He came to the pub. He was a chemist and I'm a biologist. And he said you will not believe what we found on the bottom of the ocean. He had gone down in Alvin, this two-person submersible vehicle. And he started talking about these giant clams and these giant tube worms and all of these things, and I thought he had had one too many beers. I found it hard to believe. Well, it turned out that that was the first discovery of these deep-sea vents and he was on that expedition. And through that collegial relationship I actually ended up with one of the clam shells from the clams there, which is one of the giant clams. Their meat is blood red because they have a special kind of hemoglobin that they use to keep the oxygen tension perfect for these chemosynthetic bacteria. If the oxygen is too high they cannot do this because it will spontaneously oxidize the H2S. So the oxygen tension is very critical. And they have a special kind of hemoglobin that does that. So these claims had symbiotic chemosynthetic bacteria. Well, since then these vents have been discovered everywhere and ecosystems similar have been discovered on the surface. And there are all kinds of different vents. You're going to learn about not only hydrothermal vents, hot vents in this video, but also cold seeps they're called where you have methane bacteria that are really important. OK. So these are the main ways in which organisms can get energy to convert CO2 to organic carbon. Then you have all these heterotrophs, the ones that use the organic carbon, and they have various ways of doing that. You've learned in biochemistry the primary way, which is very powerful, and that is using aerobic respiration to do that. And so we are just going to abbreviate that here. That's our reverse of photosynthesis. So heterotrophs. So we have first aerobic. And let me jump ahead with the slides. OK, there you are. So this is a cartoon version of aerobic respiration. So we'll just put glucose, we'll come down to the Krebs' Cycle. And we are going to let electrons flow here and have oxygen be the final electron acceptor creating water. So we've really just accomplished the absolute reverse of photosynthesis and we've made NADH in doing this and we've made ATP. So these guys are getting the energy out of the glucose that all of the other organisms made. And oxygen is the terminal electron acceptor when there's oxygen around. But there are lots of environments, as we've talked about on earth, where there isn't oxygen. And there are bacteria that can take advantage of those environments. And instead of having oxygen be the terminal electron acceptor there are a number of other elements that they can use, compounds that they can use. For example, there are some that use nitrate and they reduce it to nitrous oxide. N2. Ammonia. All the relatively reduced forms of nitrogen. And so this called anaerobic. And this process is called gentrification. And if it weren't for these bacteria, these anaerobic bacteria that can reduce nitrate, nitrogen would never return to the atmosphere. Remember last time we talked about nitrogen fixation, how specific types of microbes can take N2 from the atmosphere and pull it into the ecosystem? Well, if you didn't have these bacteria doing this process that nitrogen would never get back to the atmosphere. They're central to closing the nitrogen cycle. Then there are some that can use sulfate and reduce it to hydrogen sulfide. As you can imagine, these are critical to creating the hydrogen sulfide that's used in these other processes. There are some that use CO2 and convert to methane. These are methanogenic bacteria, and they're incredibly important in the global carbon cycle and in the methane cycle. Methane is a really powerful greenhouse gas, and we're going to talk about that later. And then there are some that can take Fe3+ and reduce it to Fe2+. And the same for manganese. So you should be starting to sense a sort of symmetry here, right, that these anaerobic bacteria are fulfilling functions on the earth. Let me write these down. These are sulfate reducers, these are methanogens, and these are iron reducers and manganese reducers. So these will all become extremely important when we talk about the global biogeochemical cycles of all of these elements. It's these microbes that make sure that the cycles can continue and don't run into a dead end of oxidation or reduction. OK. Before we go to the movie, I just want to say if you look at Table 25.2 in your textbook, I think it's that one. I'm assuming I'm using the most recent version. You'll see a variation of this theme in which there will be some entries of organisms that don't fall into these categories that I've just shown you. And that is to say that there are organisms that use light energy and organic carbon energy at the same time. For every variation that's possible there's an organism that's evolved to take advantage of it. I've just oversimplified it here, but you should know that. And the bottom line is if it's thermodynamically possible. And, again, this whole lecture could have been done in a thermodynamic mode. We could have looked at which redox couples were energetically possible and then assigned those to particular microbes. But for now I just want you to get the overview. But for anything that's thermodynamically feasible there's a microbe out there that's doing it. And, in fact, microbiologists actually comb through redox tables and put together different redox couples and hypothesize. I ought to be able to find an organism that does this in that environment. And then they go out. And they can almost always actually find it. So they're incredibly versatile. And it gives you a really good strong feeling for the power of thermodynamics in driving the evolution of these biochemical processes. Finally, before we show you the movie I want to show you what this thing is all about. There was a Russian microbiologist back in the previous century named Winogradsky -- -- who wanted to isolate some of these photosynthetic bacteria. And knowing what their characteristics were he went out and got himself some mud and some pond water. And he set up what we've come to call a Winogradsky column. This is a Winogradsky juice bottle, but it works the same. And what you do is you put mud in the bottom and you put pond water here. And the pond water has basically an inoculum. It has representatives of all different types of bacteria. They might be spores. If they don't like the environment they're in they sporulates and then they just don't germinate. But presumably in pond water you have everything that could possibly grow in here. And in the mud you add a source of sulfate. And so you might add calcium sulfate and you might add a little organic matter, you know, plant parts or something just to jumpstart it. And eventually you set up a gradient here of hydrogen sulfide and oxygen. And over time the organisms grow along that gradient. So you'll end up down here with the anaerobic respiration. In fact, the organisms generate this gradient. When you start out the whole thing is oxygenated. And what you should think about in this context is what happens. How do these gradients get generated when you start out with a completely mixed system, everything in there, everything oxygenated? Eventually you have anaerobic -- First you'll just have anaerobic respiration, right? Anything that can use organic carbon and oxygen is going to go like mad, and that's what's going to draw the oxygen down. Then you'll have anaerobic respiration here. You'll have photosynthesis up here, evolving oxygen. You'll have chemosynthetic bacteria here because they need a little bit of oxygen but they also need some of this hydrogen sulfide and photosynthetic bacteria here. Well, they're like down here. Because they need light but cannot have oxygen. And so you can set these up. And this purple band here tells you that you've got your photosynthetic bacteria.
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https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/8.04-spring-2016.zip
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BARTON ZWIEBACH: --that has served, also, our first example of solving the Schrodinger equation. Last time, I showed you a particle in a circle. And we wrote the wave function. And we said, OK, let's see what is the momentum of it. But now, let's solve, completely, this problem. So we have the particle in the circle. Which means particle moving here. And this is the coordinate x. And x goes from 0 to L. And we think of this point and that point, identify. We actually write this as, x is the same as x plus L. This is a strange way of saying things, but it's actually very practical. Here is 2L, 3L. We say that any point is the same as the point at which you add L. So the circle is the whole, infinite line with this identification, because every point here, for example, is the same as this point. And this point is the same as that point. So at the end of everything, it's equivalent to this piece, where L is equivalent to 0. It's almost like if I was walking here in this room, I begin here. I go there. And when I reach those control panels, somehow, it looks like a door. And I walk in. And there's another classroom there with lots of people sitting. And it continues, and goes on forever. And then I would conclude that I live in a circle, because I have just begun here and returned to the same point that is there. And it just continues. So here it is. You are all sitting here. But you are all sitting there. And you are all sitting there, and just live on a circle. So this implies that in order to solve wave functions in a circle, we'll have to put that psi of x plus L is equal to psi of x, which are the same points. And we'll have 0 potential. V of x equals 0. It will make life simple. So the Hamiltonian is just minus h squared over 2m d second dx squared. We want to find the energy eigenstate. So we want to find minus h squared over 2m d second psi dx squared is equal to E psi. We want to find those solutions. Now it's simple, or relatively simple to show that all the energies that you can find are either zero or positive. It's impossible to find solutions of this equation with a negative energies. And we do it as follows. We multiply by dx and psi star and integrate from 0 to L. So we do that on this equation. And what will we get? Minus h squared over 2m integral psi star of x d dx of d dx psi of x is equal to E times the integral psi star psi x dx. And we will assume, of course, that you have things that are well normalized. So if this is well normalized, this is 1. So this is the energy is equal to this quantity. And look at this quantity. This is minus h squared over 2m. I could integrate by parts. If I do this quickly, I would say, just integrate by parts over here. And if we integrate by parts, d dx of psi of x, we will get a minus sign. We'll cancel this minus sign, and will be over. But let's do it a little bit more slowly. You can put dx, this is equal to d dx of psi star d psi dx minus d psi star dx d psi dx. I will do it like this, with a nice big bracket. Look what I wrote. I rewrote the psi star d second of psi as d dx of this quantity, which gives me this term when the derivative acts on the second factor. But then I used an extra term, where the derivative acts on the first factor that is not present in the above line. Therefore, it must be subtracted out. So this bracket has replaced this thing. Now d dx of something, if you integrate over x from 0 to L, the derivative of something, this will be minus h bar squared over 2m psi star d psi dx integrated at L and at 0. And then minus cancels. So you get plus h squared over 2m integral from 0 to L dx d psi dx squared equal E. And therefore, this quantity is 0. The point L is the same point as the point 0. This is not the point at infinity. I cannot say that the wave function goes to 0 at L, or goes to 0, because you're going to infinity. No, they have a better argument in this case. Whatever it is, the wave function, the derivative, everything, is periodic with L. So whatever values it has at L equal 0 it has-- at x equals 0, it has at x equals L. So this is 0. And this equation shows that E is the integral of a positive quantity. So it's showing that E is greater than 0, as claimed. So E is greater than 0. So let's just try a couple of solutions, and solve. We'll comment on them more in time. But let's get the solutions, because, after all, that's what we're supposed to do. The differential equation is d second psi dx squared is equal to minus 2mE over h squared psi. And here comes the thing. We always like to define quantities, numbers. If this is a number, and E is positive, this I can call minus k squared psi. Where k is a real number. Because k real, the square is positive. And we've shown that the energy is positive. And in fact, this is nice notation. Because if you were setting k squared equal to 2mE over h squared, you're saying that E is equal to h squared k squared over 2m. So, in fact, the momentum is equal to hk. Which is very nice notation. So this number, k, actually has the meaning that we usually associate, that hk is the momentum. And now you just have to solve this. d second psi dx squared is equal to minus k squared psi. Well, those are solved by sines or cosines of kx. So you could choose sine of kx, cosine of kx, e to the ikx. And this is, kind of better, or easier, because you don't have to deal with two types of different functions. And when you take k and minus k, you have to use this, too. So let's try this. And these are your solutions, indeed. psi is equal to e to the ikx. So we leave for next time to analyze the [INAUDIBLE] details. What values of k are necessary for periodicity and how we normalize this wave function.
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https://ocw.mit.edu/courses/5-112-principles-of-chemical-science-fall-2005/5.112-fall-2005.zip
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The following content is provided by MIT OpenCourseWare under a Creative Commons license. Additional information about our license and MIT OpenCourseWare in general is available at ocw.mit.edu. Last time, we saw that these electron configurations that you have been writing down are nothing other than a shorthand way of writing down the wave functions for each electron in a multi-electron atom within the one-electron wave approximation, where we let every electron in the system, or in the atom have its own wave function. And, as an approximation, we gave it a hydrogen atom wave function. But we also saw that at most, we gave two electrons a hydrogen atom wave function, the same hydrogen atom wave function. And, of course, you already know that the reason we did that is because of this quantum mechanical concept called spin, quantum mechanical phenomenon called spin. Spin is the intrinsic angular momentum. It is the angular momentum built into the particle. And that angular momentum comes in sort of two polarities, a spin up and a spin down, -- -- corresponding to the two possible values of the spin quantum number, m sub s, that we looked at last time. m sub s can have the value plus one-half, or it can have the value minus one-half. We are about to talk, now, about how spin was actually discovered. And it was discovered by these two gentlemen here, George Uhlenbeck and Sam Goudsmit. They were really very young scientists. They might have been post-docs at the time. What they were looking at was the emission spectra from sodium atoms. And this is 1925, and so they already knew enough about the electronic structure to anticipate at what frequency they ought to see emission from the sodium atoms. And so they got a discharge going, looked at the emission, disbursed it, and they thought that they would see some emission at this particular frequency. But instead, what they saw was some emission a little bit lower in frequency than they expected and a little bit higher in energy than what they expected. In spectroscopy, this kind of emission, here, is called the doublet. A little lower and a little higher. They looked at these results, thought about it and said I think we can understand those two lines, those emissions at the two frequencies there, if the electron, in particular that extra s electron in the sodium, existed in one of two spin states. This was a revolutionary idea. They were quite excited about it. They took the results of their experiment and their interpretation to the resident established scientist at the time that was closest to them, this guy, Wolfgang Pauli. Wolfgang Pauli was not a nice man. They showed him the data and said, this makes sense if the electron is in two different spin states. And Wolfgang Pauli said rubbish. You publish that and you will wreck your young scientific careers. Uhlenbeck and Goudsmit let dejectedly. No sooner than the door slammed shut, Pauli sits down and writes a paper on the presence of the fourth quantum number, m sub s. This is one of the best well-known travesties of science, now well-known. And it actually took, however, another three years and another gentleman, Dirac, who actually wrote down the relativistic Schrödinger equation and solved it. When you do that, out drops this fourth quantum number, m sub s. However, Pauli did contribute to this problem in the sense that he worked on the principles behind these problems with electrons being fermions, etc., which we won't go into. Out of that work came something called this, the Pauli Exclusion Principle. And the essence of that principle is that no two electrons in the same atom can have the same electron wave function and the same spin. Or, another way to say that, no two electrons can have the same set of four quantum numbers. For example, in our electron configuration here of neon, this electron has the quantum numbers 1, 0, 0, plus one-half for m sub s. This electron has the quantum numbers 1, 0, 0, minus one-half for m sub s. These two electrons don't have the same set of four quantum numbers, and that is why we could only put two electrons, here, in this 1s state. Likewise, this electron, here, has the quantum numbers 2, 1, minus 1, plus one-half. This electron has the quantum numbers 2, 1, minus 1, minus one-half. That is the Pauli Exclusion Principle, which prevents us from putting more than two electrons in each one of these states, 2s, 2s, 2px, 2pz, or 2py. Now, what I want to do is try to look at the wave functions and what they look like for the electrons in the multi-electron atom. And, to look at their shapes, what we are going to do is we're going to look at the radial probability distribution function. Remember what the radial probability distribution function tells us? It tells us the probability of finding the electron between r and r plus dr. I plotted those radial probability distribution functions versus r for each one of the electrons in the different states for this multi-electron atom, argon. And what we want to do is we want to compare and contrast these wave functions for the individual electrons in this multi-electron atom to those of hydrogen. Well, first the similarities. If you look at the radial probability distribution for the 1s wave function here, what you see is that the radial probability distribution is zero at r equals zero, as all radial probability distributions. That is not a radial node. That probability distribution increases, goes to a maximum, and then decays exponentially with r. That is exactly what a 1s wave function looks like for a hydrogen atom. If you look at the 2s wave function, it starts at r equals zero, it goes up a bit and then goes to zero. Here is a radial node in the 2s wave function. And then goes back up. Here is the most probable value of r and then decays exponentially. Again, it has the same structure as the 2s wave function in the hydrogen atom. The similarity is that all of these wave functions have the same kind of basic structure as that in a hydrogen atom. They have the same number of nodes, the same number of radial nodes and the same number of angular nodes. The difference between these wave functions and those of the hydrogen atom is that all of these wave functions are closer in to the nucleus. For example, if you looked here at what the most probable value of r is for that 1s electron in argon, it is 0.1 a nought. What is the most probable value for r in the 1s state of hydrogen? a nought. This is ten times closer. This is much closer to the nucleus than it is in the hydrogen atom. And if we went and compared the most probable values for 2s, 2p for those of hydrogen and 3s, 3p for those of hydrogen, we would find that all of these are much closer into the nucleus. Why? Because the nucleus has a larger positive charge on it. The Coulomb interaction here is the charge on the electron times the charge on the nucleus. For argon, that charge is plus 18 times e. That greater attractive interaction holds those electrons in closer to the nucleus. That is the bottom line. That is how these differ. The structure is the same, node structure is the same, they are just all closer into the nucleus because of that greater attractive interaction. Now that I have this radial probability distribution function up here, I also want to use it to just illustrate a concept that I think you already know. That is this concept of a shell. You know about the n equals 1 shell, n equals 2 shell, n equals 3 shell. And the word shell also denotes some kind of spatial information. I want to show you how the spatial information is depicted, here, on this graph. I want you to see that for the n equals 3 states, 3s and 3p, well, the most probable value is not exactly in the same place, but it is in the same place as when you compare it to the 2s and the 2p. You can see how well the n equals 3 shell is separated in space from the n equals 2 shell. Again, the most probable value for 2s and 2p are not exactly in the same place. We saw that 2p is actually a little closer than to 2s, but in terms of comparing it to where the most probable values are for 3s and 3p, that is much closer in. And so this graph, here, gives you an idea of the spatial information that is denoted when we talk about shells. The n equals 3 shell is further out, the n equals 2 shell closer in, and the n equals 1 shell even closer in. I think that is a concept that you mostly know. Here, you see it on the radial probability distribution. Now, we have taken a quick look at those wave functions. Now, it is time to actually look at the energies of the states. We have not done that yet. On the left here, I show an energy level diagram for the hydrogen atom. We saw this before. Here is the n equals 1 state. Here at n equals 2, we have four degenerate states. Here at n equals 3, we have 9 degenerate states. Here at n equals 4, we have 16 degenerate states. But the difference between the hydrogen atom and any other multi-electron atom, starting with helium, are two differences. One is that the energies of these states in the multi-electron atom are all lower than they are in the hydrogen atom. That is, the 1s state here is lower in energy than the 1s state in the hydrogen. The 2s is lower than the 2s state in hydrogen, the 3s is lower, the 2p is lower, the 3p is lower, etc. The energies of those states are all lower. Why? Because of the charge on the nucleus. That potential energy of interaction is greater because the charge on the nucleus is larger. That greater potential energy of interaction lowers the energy of the states. It makes those electrons more strongly bound. Starting with helium, all of these energies are lower than those in the hydrogen atom. That is the first difference. The second difference is, you can see, now, that the 2s state is lower in energy than the 2p state. The degeneracy between 2s and 2p in a hydrogen atom is lifted or is broken, as we say. Likewise, the 3s state is lower in energy than the 3p state, than the 3d state. The degeneracy in those states is lifted, or it is broken. That is now what we have to talk about, why that is the case. Why is 2s, for example, lower in energy than 2p? And the reason for this has to do with the phenomenon called shielding. We have to talk about this phenomenon, shielding, and we have to talk about how that leads to a concept called effective charge. But to do that, I am going to do the following. I am going to realize that each one of these energies here, E sub nl, so now these energies are labeled by both the principle quantum number and the angular momentum quantum number. I am going to realize that these energies here physically are minus the ionization energy because these energies are minus the energy it is going to require to rip the electron off from that particular state. And I am going to set those binding energies equal to a hydrogen atom like energy level here. And now I am going to use the board and explain that just a little bit more. I said these energies, which are now a function of n and l, they are minus the ionization energy from that nl state. And I am going to approximate that as a hydrogen atom kind of an energy scheme. That is, I am going to set this equal to R sub H, the Rydberg constant over n squared. Out here, there is going to be a Z, but this Z is going to be Z effective, Zeff. And it is going to be squared, of course. And that Z is going to depend on that particular nl state that you are in. This Z effective, here, is the effective nuclear charge. It is not the nuclear charge. It is the effective nuclear charge. Why? Well, because of shielding. Let's try to explain that. Let's take helium. Z is equal to plus 2e. We have plus 2e here, and let's do a thought experiment in that we are going to take electron number two, here, and place it kind of close to this plus 2 charged nucleus. And then we are going to have electron number one way out here. In this case, with electron number one way out here, the nuclear charge that electron number one experiences, because it is so far out, kind of looks like a plus one charge. Because this electron, on the average, is canceling one of the positive charges on the nucleus. So, the effective charge here for this electron way out there, we are going to say, is plus one. If that is the case, well, then the binding energy of that electron is one squared times R sub H over n squared. We are going to consider n equal one because we are going to talk about the ground state of the helium atom, here. But you know what this value is going to turn out to be. It is going to turn out to be minus 2.180x10^-18 joules. That is the binding energy of an electron in a hydrogen atom. This is a thought experiment, now. This is helium, this is just one electron way out here, and it cannot discriminate too well between the nucleus and this electron, so the overall effective charge it sees is plus one. But now, we take the other case. The other extreme case is we are going to bring in electron one really close to the nucleus. And electron two is way out here such that it does not do anything as far as this electron is concerned. And so this electron, in this case, is experiencing the total nuclear charge on the nucleus. And so we say in this thought case here, with this electron really close, that the effective charge is equal to plus 2e for this electron. Well, if that is the case, we can calculate the binding energy of this electron. And that is going to be 2 squared R sub H over one squared. We can plug in R sub H. And we are going to get minus 8.72x10^-18 joules. What this is is the binding energy of an electron in helium plus. And so, in this extreme case, with the electron really close, that is the binding energy. With this extreme case, with the electron way far out, this is the binding energy. This is total shielding. This case is no shielding. The electron is much more strongly bound. The reality is that the binding energy of an electron in helium is somewhere in between. That is, the ionization energy for helium, for an electron in now neutral helium here, is 3.94x10^-18 joules. Somewhere in between this extreme case of total shielding and this extreme case of no shielding at all. And that is because, on the average, there is another electron in between that electron and the nucleus. And we can calculate the effective charge, then, from the experimental binding energies by just taking this expression and rearranging it. I am going to solve that expression for the effective charge, Zeff. And, when I do that that, add, of course, this n squared times the ionization energy over the Rydberg constant. n squared is going to be one because we are talking about the ground state. The ionization energy, I said, was 3.94x10^-18 joules. The Rydberg constant is 2.180x10^-18 joules. In the end, I find an effective charge here of plus 1.34. Again, why is that the case? Well, that is the case because here is my electron, here is my helium nucleus and, just on the average, between this electron and the nucleus there is always another electron around. This electron kind of partially shields this nuclear charge. It partially shields it so that we calculated, from using this scheme and using the experimental ionization energies, an effective charge of 1.34. So now, we sort of understand shielding and effective charge. That is a square root, thank you. All right. Now, we have to use this idea of effective charge and shielding to understand why the 2s state is lower in energy than the 2p state. And here comes that. Let's think about the lithium atom. The electron confirmation of lithium is 1s 2 2s 1. The electron configuration of lithium is not 1s 2 2p 1. Why? Because 2s is lower in energy than 2p. But why is that the case? Well, to look at that, we have to look at the radial probability distribution functions for 2s and 2p. Here is a radial probability distribution function for 2s. Here is the radial probability distribution function for 2p. Now, you have to do a thought experiment, here. In this 2s radial probability distribution function, what you see is that there is some finite probability of the electron in the 2s state being really close to the nucleus. For the sake of the argument, here, I am going to say that for this part of the probability distribution function, the effective charge Zeff is going to be plus 3e. I mean, that is an exaggeration because obviously we have some s electrons. But for the sake of this argument, I am going to say the effective charge for this part of the probability distribution function is plus 3e. For this part of the probability distribution function for 2s that is much further out. And I have those 1s electrons closer in. So, for the sake of this argument, I am going to say that these s electrons completely shield the nuclear charge. And so, the effective charge for this part of the distribution is plus 1e. Now, what about 2p? Well, in the case of 2p, you can see that this 2p wave function, although it is a little bit closer in, for the most part it is about in the same place as the second lobe here of the 2s wave function. We are going to say that this effective charge is plus 1e, just like this lobe of the 2s wave function. But now, to get the kind of total effective charge, I am going to have to take the effective charge and average it over this probability distribution function. And so, since in the 2s function I am going to average over a part that has a plus 3 effective charge and a part that has a plus 1 effective charge. Well, if I average over that, that is going to be larger than the effective charge of this 2p wave function because the 2p everywhere is plus one. The effective charge here for the 2s is going to be greater, on the average, than for the 2p. Because of this part of the probability distribution function, this part that is really close to the nucleus, where the electrons can feel more of the nuclear charge or experience more of the nuclear charge than they could if they were a 2p electron. Therefore, since that 2s electron has a greater effective charge, here, what that is going to mean is that the binding energy of the 2s state is going to be lower in energy. It is going to be more negative than the binding energy of the 2p state. The same thing for 3s and 3p. The same reasoning there. And it is all because of this little part of the probability distribution function. Now we can understand why the 2s is, in fact, lower in energy than the 2p. Yeah? Because the 3s, although it does, in fact, have this, it is a little bit further out here, enough to make for this to compensate. But, again, 3s is lower than 3p. We are just going to compare 3s and 3p within the same shell. Yes, it does have to do with the net area underneath. If the probability was much higher, and it is not that much higher, but if it were then you are right, it could cancel it out. Now, therefore, we are ready to write the electron configurations of all the atoms on the Periodic Table. And you already know that we use the Aufbau Principle to do that. Aufbau means building up. What do you do? You take all the allowed states and order them according to their energy. Most strongly bound or most negative energy goes on the bottom, and next most negative energy, on and on. And we will talk about a pneumonic for remembering those energies in a moment. We use the Aufbau principle. We start with the lowest energy state and we put an electron in for hydrogen. There it is, the 1s state. For helium, well, we also put that into the 1s state. Except, as we fill these states, we have to heed the Pauli Exclusion Principle. This electron, here, goes in with the opposite spin. Next lithium, 2s electron. For beryllium, another 2s electron, opposite spin. And then for nitrogen, electron, here, has to go into the 2p state. Now, it does not matter whether you put it in 2px, 2py or 2pz. They are all the same energy. The next electron, carbon, what are we going to do here? Well, here we have to obey something called Hund's Rule. And Hund's Rule says that when electrons are added to states of the same energy, and that is what we are doing here in the 2p, a single electron enters each state before a second electron enters any state. And, those single electrons have to go in so that the resulting spins are parallel. That is, they have the same spin. Do we put in an electron like this? No. Do we put it in like that? No. Do we put it in like this? Yeah, according to Hund's Rule. And then the next electron has to go into that other empty p state before any of these states double up. Then we keep doubling them up. The next electron has to go into the next state 3s. Again, we double them up. Now we are to the 3p state again. One electron goes into 2px. The next electron will go into either 2py or 2pz, according to Hund's Rule, and the spins remain parallel to get the lowest energy configuration. And then the next electron 2pz. And then we start doubling up. And we keep going. You keep going in that way so that you write the electron configuration of all the atoms in the Periodic Table. Let's look at these electron configurations kind of quickly here. Let's start with the third period, the third row here, sodium going across here to argon. Here is the electron configuration for sodium. Notice here that I don't mind if you write that electron configuration as 2p 6 instead of 2px, 2py, 2pz. Because you cannot tell the difference between x, y, and z, anyway, if you are not in a magnetic field. These electrons here in sodium that make up the inert gas configuration of argon, of course, are the core electrons. When we talk about core electrons, we are talking about the electrons that make up the nearest rare gas configuration. And then the valance electron, well, is the electrons that are beyond the nearest but lowest inert gas configurations. And also, when you are writing these electron configurations, say, for sodium, you can write it as the neon configuration and then just show the valence electron, 3s 1. As we go across that third period here everything is very normal. 3s fills up first and then the 3p's fill up. Now we get to the fourth period, from potassium to krypton. Fourth period, what happens here is that those first two electrons go into the s states. They do not go into the 3d states. They don't because those s states are lower in energy than the 3d states. And so these electron configurations are argon 4s 1, argon 4s 2 **[Ar]4s^2**. And then, once we fill those 4s states, we start filling the 3d states. Here is scandium, here is titanium, here is vanadium, everything is normal, 3d 1, 3d 2, 3d 3, and then we get to chromium. We have an exception here, chromium. Chromium is not what you would expect. It is not 4s 2 3d 4. Chromium is 4s 1 3d 5. There is no way for you to know that a priori, unless you do a very sophisticated calculation, but this is experimentally observed that this is the configuration. Why? Because it is lower in total energy than this configuration. It turns out that there is some extra stability in having a half filled 4s shell and a half filled 3d sub-shell compared to having a filled 4s shell and a less than half filled 3d shell. This is an exception that you do have to know. But after chromium, here, manganese behaves well. Iron behaves well. Cobalt okay. Nickel okay. And now we get to copper, and we have a problem. It is not what you would expect. It is not 4s 2 3d 9. Instead, it is 4s 1 3d 10. *Again, this configuration is something you could not have predicted a priori. If you do a sophisticated calculation you can see it, but this is also experimentally observed. We know this to be the configuration. This is not the configuration. This is the lower energy configuration. Here is another exception that you have to know, copper. And then after copper here, zinc, things follow a pattern again. The next electron, then, just fills up the 4s 2 3d Now, at gallium, all of those are filled. We start filling up the 4p states. Everything is fine until we get to krypton. So, you have to know chromium and copper. Now, the fifth row here, starting with rubidium, strontium, ytterbium, zirconium, etc. all the way across here. Starting with rubidium, again, the electrons go into the 5s shell. They do not go into the 4d. That is because with rubidium and strontium, 5s is actually lower than 4d. And it is only once you fill up the 5s states that you start filling up the 4d states right here. Now, there are exceptions along this fifth row. The two exceptions, molybdenum and silver, are the same kind of exceptions as chromium and copper. You have to know the exceptions for silver and copper and molybdenum and chromium. The same identical kind of exception as in the fourth row here. There are other exceptions along this fifth row. You do not have to know those. There is really no way, a priori, for you to know that. Again, it is an experimental observation. A sophisticated hard calculation will also show that. And then, once you are done with cadmium, well, then the 5p's start filling and everything is normal and you get to the xenon inner gas configuration. How do you remember what the ordering is of these states, the energy ordering? Well, here is a pneumonic that maybe some of you have seen before. Start out and write 1s, then write 2s right below it and then 2p to the side of it. And then write 3s, 3p, 3d. And then write 4s, 4p, 4d, 4f. And then write 5s, 5p, 5d, 5f, and 6s, 6p, 6d, and 7s, 7p. And now, to get the energy ordering we are going to draw diagonals. Well, first of all, the 1s is the lowest energy state, and then the next highest energy state is the 2s state, and then the next highest energy state is the 2p. We are going to draw a diagonal. The next state to fill is 2p. The next one is 3s. Now we are going to draw a diagonal again. The next one to fill is 3p. The next one to fill is 4s. Draw another diagonal. The next one to fill is 3d, 4p, 5s. 4d, 5p, 6s. 4f, 5d, 6p, 7s. 5s, 6d, 7p, and that is all that is going to be important. That is one way to remember the relative energy orderings here. And you do have to be able to write this down on an exam. You will have a Periodic Table, but it won't have the electron configurations on it. Now, I am going to tell you something that sometimes people find a little bit confusing. That is, the electron configuration of ions. What I am about to say has no effect on what I just said about how to write the electron configuration for neutrals. This does not affect anything in your writing down the electron configuration for neutrals. This is for ions. The point I want to make is that if you actually look at the energies of the individual 3d states and 4s states across that fourth row, this is what they look like. For example, at potassium, that 4s state is lower in energy than the 3d state. That is why we put the electron in the 4s state when we wrote the neutral. The same thing for calcium. That is why we put that electron into the 4s state and not the 3s state. Lo and behold, right here at scandium, Z equals 21. What happens is that the 3d state actually drops below in energy than the 4s state. These are the energies of the individual state now, and that continues all the way across the Periodic Table. However, that does not affect how you write the electron configuration of the neutrals. For example, if you are writing the electron configuration for titanium, here it is. It is the argon core, 4s 2 3d 2. And, by the way, I don't care whether you write 3d 2 4s 2 or 4s 2 3d 2. You can write them in either order. Now, you might say, well, why is this the electron configuration if at this value for Z, Z equals 22, titanium, the 3d state is lower than the 4s state? Why don't these 4s electrons just hop into the 3d states? Well, they don't do that because this electron configuration actually minimizes the electron repulsions. If these four electrons were in the 3d state, well, then the repulsive interactions would be greater. Because they are in the same state now. And, therefore, the entire energy of the atom will be larger. And it is the entire energy, the total energy of the atom that is important when we look at the electron configurations. In this particular case, if we look at the individual d states and the 4s states, yeah, the d states are lower in energy. But what is important is when we sum up all of the energies of the interactions, what is lower? What I am saying to you is that when we do that, when we sum up all of the interaction energies, this still is the lower energy configuration, even though the 3d electrons at this value of Z, those 3d states are lower in energy than the 4s states. That is why we do not have this hoping over into the 3d states for the neutral. Now, here comes the ion configuration. If you have this configuration for the neutral, and now you ionize titanium to make titanium plus, the electron configuration is argon 3d 2. That is, it is the 4s electrons that come off, that are plucked out. They are plucked out because they are the higher energy electrons now. This is the electron configuration for titanium plus We are going to pull out those higher energy electrons, which are the 4s electrons. Again, this affects only the ion configuration. The same thing happens here on the fifth row. The fifth row starting here with rubidium and strontium, 5s is lower in energy than 4d. That is why that rubidium electron went in the 5s state. The same thing with the strontium electron, it went in the 5s state. But right here at ytterbium, the 4d state goes below in energy the 5s state. Again, that does not affect how you write the electron configuration of a neutral. For example, silver, which is way out here and is one of those exceptions, if you ionize it, if you pluck off an electron, which electron is going to go? 5s. And so the silver plus one configuration is the krypton core, 4d 10. It is that 5s electron that is going to disappear. This is important. Do not let it confuse you with the writing the electron configurations of the neutrals. Okie-dokie. See you Friday.
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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We would like to now apply the momentum principle to examples of recoil. So recall that the momentum principle is that the external force causes the momentum of the system to change. Now, this is a vector equation. So for example, if the external force in the x direction is 0, then the momentum of the system in the x direction, let's say, the final momentum will be equal to the initial momentum. I would like to now apply the momentum principle to an example of recoil. In our recoil example, we have a person jumping off a cart. So let's just look at how this example can work if we draw momentum diagrams. So suppose we choose a ground frame. And in that ground frame, we have a cart and a person. A person is standing on the cart. And this is t initial. And here, they are at rest. Now, the person we're going to assume to jump horizontally off the cart, and the cart will recoil in the opposite direction. So after the jump, we can describe this picture. The person is moving with the velocity, v. The cart is moving with vc. And the person has jumped with the velocity vp. Now, suppose I choose a different reference frame. That instead of choosing a ground frame, as a reference frame moving with the velocity vc. You can imagine that maybe I have a car here and you're inside that car moving with velocity vc, and you're looking at this picture. Then, what would our momentum diagrams look like? Well, if I'm moving in a car this way and in the ground frame the initial picture is the person in the cart is at rest. Then, in my moving frame, it actually looks as if the cart and the person are moving in the opposite direction. So let's write that this way. Here's the initial picture, t initial. And in this frame, the person and the cart are moving with vc minus vc. I put an arrow here to indicate that it's opposite the direction of vc there. But their velocity is minus the velocity of the reference frame. After the jump-- so here's the person now. The cart is at rest, why? Because we're in the reference frame moving with vc. So if you're in a car and you're moving at the same speed that the cart has with the ground frame, then in your frame, this cart looks like it's at rest. What about the person jumping off? Well, let's write it this way. So this is the velocity. I'm going to use a symbol, u. Now, u-- this is what do we mean by that. This is the velocity of the person in the moving frame that's moving with velocity vc-- that's the velocity of the person as seen by a car. Sometimes we call this the velocity of the person relative to the cart. What does that word relative to the cart mean? Well, you can see in this picture. In this moving frame, the cart is at rest, and the person jumps with the speed u velocity u relative to the cart. So these are momentum diagrams for a ground frame in which the person in the cart started at rest, the person jumps off. I can put an arrow here, but it's really information is in that vector. The cart is moving. In a frame moving with the velocity of the cart, then what does my picture looks like? Well, the cart and the person initially are moving opposite directions. Again, you're moving this way. The cart looks like it's moving that way if you're inside the car. And the final state, person, cart is at rest and the person is jumping with the velocity u relative to the cart. Now, our question now is how do we relate these two velocities, u and vp? What are u and vp? vp is the velocity of the person in the ground frame. And u is the velocity of the person in the moving frame. Well, we've already seen our equation for how to get velocities in different frames. We have that vp equals the relative velocity of the two frames plus the velocity in the moving frame. So what we have is, this is the velocity of the person-- let me just clean that up-- of the person in the ground frame. v is the relative velocity of the two frames. So here we have that v is the velocity of the cart, because you're in a frame moving with vc with respect to the ground. And u is the velocity of the person in the moving frame. So this is how we can show the same type of interaction in two different reference frames. Next, we'll figure out what these velocities are.
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https://ocw.mit.edu/courses/7-014-introductory-biology-spring-2005/7.014-spring-2005.zip
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I just got the feedback from the one minute things just a few minutes before class, so I think I may defer commenting on the couple of them until the beginning of next lecture. But there are a couple of things that I think I can say. One is, several people were wondering, how does the cell decide whether to do mitosis or meiosis? But the mitosis, which is ordinary cell division is what happens everywhere in your body, in your intestine, on your skin, in your eye, anywhere except, since that I you make all the different cells you have. The meiosis, which creates the sex sells, or gametes, happens in a very specific place, either in the testes if you are male or the ovary if you are a female. So, there's a dedicated place where meiosis takes place, and just your knowledge of human anatomy and physiology can make a pretty good guess as to where that is. Everywhere else, it's mitosis. There's a very special place where that happens. A couple of you still got confused when I was talking about meiosis and I were showing you a progression through, and I put a double headed arrow, meaning that's was with that period was called. The process is unidirectional. It only goes in one way. It's not a reversible process. There were some questions about the chiasmata or the chiasma: why is there a crossover in chiasma in meiosis but not mitosis? I will tell you about that today very specifically. Why was there a tall and a short pair of chromosomes? That was arbitrary. I made up a simple cell for us to consider these properties in, one with a long, and what with a short, nothing else. I wasn't trying to represent any particular organism. And I'll pick up on a couple of other things. I just want to quickly mention something. Since I've seen you, I flew out to San Diego, gave a talk yesterday morning at a major meeting, hopped on a plane, got back at midnight, and here and again. As part of being a scientist, come teacher, at a place like MIT, you guys tend not to see, but my research life goes on while I'm teaching. And I just want to briefly mentioned one thing I talked about yesterday, because just in my own life it captures a couple of things that I've been trying to tell you, that the textbook is not the ultimate authority. That's just what we think up until today. A new finding to change the way you think about things, and what I'm telling you, and some of your frustrated at, that I'm not just parroting back the textbook is because this is the way it is. If you guys are going to be leaders in whatever field you're in, you're going to be dealing with this process of shifting sands as we gain new knowledge. So, in fact what I was talking about has some relationship to the cell cycle that I talked about. Lee Hartwell particularly helped us understand that there is what's called a G1 phase, which you could think of as a preparation for DNA synthesis for what's known as the S-phase where DNA synthesis actually occurs. And now, you're at 4N. You've doubled the DNA content, and there's G2 where it cleans up from S-phase, gets ready for mitosis, and mitosis is what you, then, separate the daughter chromatids, go back to 2N, and then ultimately the cell divides. And we are back to 2N. And what we were talking about DNA replication, I told you how replicative DNA polymerases test for Watson-Crick shape, and remember that little movie I showed you where they flipped the base pair into a very narrow slot in the protein, and they check that it's there. And they said at that time, that's why polymerases have a problem when they hit a lesion such as a thymine dimer that we get we go out into the sun. And then one of the recent pieces of excitement in the DNA repair field that I work in, was the discovery of a whole class of translesion DNA polymerases that are very flexible, active sites and we are able to copy over a lesion. Right at this point, you will find in the literature all sorts of reviews about polymerase switching, where people are envisioning the replicate of polymerase coming along. It hits a lesion, gets stuck. It recruits one of these translesion polymerases that comes in, copies of the lesion; it switches back to the replica of polymerase, [and then it goes?]. And there are reviews like that coming out in the literature. So, one of the genes that's needed for this sort of error prone kind of translesion synthesis in yeast and in humans is a gene called [rev-1? . You don't have to notice for the exam, this bit. But if you knock out the function of that gene, the yeast aren't mutated by UV or chemicals anymore. So, you know it has an essential function somehow in this process of translesion synthesis. So, one of our big surprises was I was trying it. We were trying to actually fish out partners that might interact with it cause we thought it would be regulated. And we were surprised. The experiments revealed something we hadn't expected. We found out that this protein that was critical for mutagenesis was extremely strongly cell cycle regulated. Well, given the reviews I've told you, or if somebody you're writing a textbook today, this is what they would tell you: polymerase switching during S-phase. So you might have thought it would be high and S-phase, but it isn't. We can barely detect it during S-phase. But instead, the rev-1 levels are at least 50 times in the G2M phase. Actually, it goes right through this part of the cell cycle, 50 times what they are during S-phase. And so, there's a couple of possibilities right now. Either these very tiny amounts that are doing S-phase are what everybody thinks and current models would predict: you're getting polymerase switching. And it makes 50 fold more during this phase for something else. If so, I don't know what that is. The other possibility is that we have to rethink our model, and that it isn't polymerase switching during replication. Actually, the replication for it just keeps moving, and it leaves behind little masses. And when the cell is busy starting to line up as chromosomes, and even while it's pulling them apart, that's when this translesion synthesis stuff comes in, and it cleans up the damage. Now, I don't know what the right answer is. That was one of the things I was talking about yesterday morning. But it's an example of how something that you guys can now hopefully at least understand in principle is being debated right now. And the finding from my lab changed the way I'd been thinking about it at least. I can now see another very real possibility. So, just to move on, so we are going back to Mendel now, who did an awful lot more than I told you. A few of you thought it was pretty frustrating and it had all that. But Mendel was doing other things. You already know what he could do. He knew how to do crosses. He knew how to count. And he could think, which was really important. And he's our ratios instead of just numbers. So it was another class of experiment that he could do, and that was he could do a cross where he looked at more than one trait at once. And he didn't have all that many options for things he could do, but he did what are known as dihybrid crosses, where he followed two traits at once. And I showed you some of the traits that he studied. In fact, that picture I showed you is actually a cross that we can think of right now. He's got smooth, and yellow, wrinkled, and green, and yellow, and you'll see them in all four combinations. So, an example of the kind of crust that Mendel carried out, then, was he took smooth yellow, which actually is the dominant allele, and [both? , which he learned from his other crosses. And he crossed it with wrinkled green, which I'll represent as a little s, little s, little y, little y, where he previously known the little s and little y alleles, the wrinkled and green were the recessive alleles. And then, the F1 generation, wouldn't be surprising to you at this point, I think, that they were all smooth, yellow. And, they all were ss yy. And if you think about the possible gametes you get out of this, you could only get a big S, big Y out of this one, and little s, little y out of that one. So if they came together, it would have to be that. So then, what he did the self crossed the F1, and what he got out of that was the kind of mixture of things that he saw there. He saw smooth yellow, smooth green, wrinkled yellow, and wrinkled green. In doing the same kind of experiments that we talked about before, he counted them, he looked to see if he saw characteristic ratios. He did. The ratios he saw were 9:3:3:1. So, I think we know how his head worked at this point. He was try to figure out if he could explain these results by the kind of model that he was developing where hereditary information came in, in particles. And he could. But he had to make a critical assumption. And if you've just seen the next square going to draw, without realizing the assumption underlies it, then you've missed a great deal of his thinking. And that was that the two genes of sort, or the two traits anyway, because he didn't know they were genes yet independently. And the way we could see that would be to think about what were the kind of gametes that you could make from those F1s. So, we could get a big S and a big Y from here, or a big S and a little y, or a little s and a big Y, or a little s and a little y. And, since this is [selfing? , it's the same thing. So I won't fill this whole thing in, but we get two S's, two big Y's here. Down here we'd have little s, little y like that. I'll fill in a couple up here where we've got big S, big S, big Y, and a little Y. Here we have big S, big S, but two little Y's. Here we'd have big S, little s, big Y little y. And over here we'd have big S, little S, two little y's. So, if you were to make out [then? a table or a chart like this that showed the phenotypes, you'll find that all of these wild type, we can see it here. However, this one would be smooth. But it's got the two alleles. So it would be green. This one would be smooth and yellow. And this, again, would be smooth but green. And down on the corner it is the wrinkled green. And if you follow that out, draw out the rest of that, you'll discover there the 9:3:3:1 ratios. That was what Mendel observed in all of his experiments. But he was particularly bothered by it because he'd shown particular information. The fact [they sorted? randomly wasn't an issue. However, that was before people knew about the chromosomes, which we spent quite a bit of time talking about the other day. As I said, when people saw those chromosomes that gave rise to what's known as the chromosomal theory of inheritance. And what was interesting about chromosomal theory of inheritance was that it predicted a different outcome depending on whether the traits you were studying were encoded by genes on the same chromosome or by genes on different chromosomes. Now, as it happened, what I think Mendel did was he found traits that were well behaved and he could study. And those all happened to be on separate chromosomes. So, his results didn't disclose this issue that was raised by the discovery of chromosomes. The chromosomal theory of inheritance gives, as I said, predicts different outcomes depending whether things are on the same or different chromosomes. So let's consider that by taking the F1. So, say they are on separate chromosomes. And let's take the F1 from the previous cross over here. So, big S., little s, big Y, little y, and then we'll cross it with the homozygous recessive parent. That was a cross that I mentioned the first lecture was very important. It's important enough that it's given a special name. It's called a test across. Well, here we are using it here. Well what are the possible outcomes that could come from this sort of thing? Well, I think the way you could think about it most easily is, what kind of gametes or sex cells could we get out of this? Well, from here we could get a big S and a big Y. Or, we could get a big S. and the little y, a little S and a big Y, or a little S and a little y. So, we could get four different types of sex cells. The gametes we can get out of this sign, there's only one type. So, if we start combining those, I think you can see what the outcome would be. It's so simple. We don't even need to draw out the square. If we had big S, big Y over, they're all going to be over. Well actually, I'll draw up this way because I think it's a little easier to see. So, there would be four possible things that would come out of this test cross. This would be the smooth and yellow. Here would be, what have I done wrong here? No, this is right. This should be smooth and green. This would be wrinkled and yellow. And this would be wrinkled and green. And, do you see what the ratio would be? It would be 1:1:1:1 because we have an equal probability of making any of those. And, I'm going to rearrange these because the way that it'll help us think about it, this is one of the types that we've found in the original cross. That's a parental phenotype, smooth and yellow. It was one of the parents up there. This is the other one that looks like one of the original parents. So, I can divide these into parental or nonparental phenotypes. These are 1:1, and then these others were ones where the progeny differed from the parent. And they're all 1:1:1:1. Now, that's what the chromosomal theory of inheritance would predict if they were on separate chromosomes. That's what Mendel saw when he was doing his crosses. What if they were on the same chromosome? So, we are going to have to go back to the original cross now to think this one through. So, what Mendel started with was a smooth, yellow parent. So that was SYSY. But because these are on the same chromosome, I'm going to depict them this way so we can see they're going to travel as a unit. So that was then crossed with the wrinkled green, which would be a little s, little y. In each case, [UNINTELLIGIBLE] should be F1 in this case will be all smooth and yellow. The recessive traits disappear, but this time they're only, what the progeny will have, they will have gotten one of the possible alleles from this one, which would be the SY on the same chromosome, and one from the other parent, the little s, little y again on the same chromosome and one from the other parent, the little s and y, again, on the same chromosome. So now if you start to ask, what will happen if I do a test cross, you're going to get a different outcome. So here's the test cross in this case. So, we've got this F1. It's smooth and yellow, but it's actually at the chromosomal level now. It's like this, and we're crossing it with the homozygous recessive parent that's now going to look like that. So if you think through, what kind of gametes could we get out of this? Well, there's only two possibilities. We could get this one or this one. And over here, the kind of gametes we could get, there's only one type. So, what are we going to get out of this cross? What we'll end up getting is big SY, over the little sy, or little sy over the little sy like that. That's smooth and yellow, wrinkled and green, the ratio of these 1:1. But you see the difference from what we saw before? These are the parental phenotypes. The chromosomal theory of inheritance doesn't predict that you will get, or it predicts that you will not get the nonparental phenotypes. So, if you wanted to distinguish between these two hypotheses, and figure out where your genes were, you'd do the experiment. And what happened when this was done, again, what often happens in science, you think you've got it straightened out. You got hypothesis A and hypothesis B, and you do experiment planning to do the scientific method and show it. And you get a result that isn't what you expected with either model. And that's in fact what happened in this case. And it led to the discovery of genetic recombination. And to show you actually the experiments where that was discovered, I'm going to switch to another widely used genetic model, which is the Drosophila melanogaster, or the fruit fly, which you see in the summer around rotting fruit. Or if you're in the biology building, one is apt to land on your sandwich because there's just a few stragglers that get out of the Drosophila genetics labs. But the one thing you can see from this is the eyes, pretty cool. It's got red. The wild type is a red eye, and the body's brown. And, you can see the red eyes. If you look at a fruit fly carefully, you'll be able to see, the summer if you take a look, you'll see that they have red eyes. So, they've been a very useful model organism for genetics partly because they grow pretty fast and they are easy to handle in a lab. And so, I'll introduce you to a couple of features. So, the wild type you find in nature is a brown body. We'll call it a plus, and it has normal wings, and which I'll refer to as plus. And some mutant phenotypes that geneticists have been able to find is a black body which I'll refer to as little b. And vestigial wings, which I'll refer to as VG. So, these are fairly easy to score. They look like little tiny wings. It's fairly easy if you're crossing Drosophila, and that looking at progeny to go through and score, what color are their eyes? And do they have ordinary wings or little wings? So, this is the kind of cross that was carried out. We are going to switch now to an organism that has male and female. Up until now, plants are both. They make pollen, and they have the eggs that are going to develop into the seeds once they're fertilized. But, Drosophila are more like us. They come in males and females. So, we're going to have to specify which is which. And I think most of you are probably familiar with this terminology. This is the symbol geneticists use for female. And this is the one that is used for male. So, in this case, let's set up exactly the same kind of thing that we start here. We're going to cross a homozygous parent versus a homozygous recessive parent. That's just exactly the kind of thing we have up here. But I'm going to do at this time using Drosophila genetics speak. And so, we'll be taking female whose wild type for both traits, we're going to cross with a male that's homozygous recessive for both traits. This is the way Drosophila geneticists tend to represent this kind of thing. So it's exactly the same kind of cross we got before. And at this point, what we get out of the F1, it shouldn't be news for you. You can see genetically what it's going to be. They'll have to get one allele from here, one allele from there. So, by definition, the only possibility that you can get out of this, the F1s, will all have one allele dominant, and one recessive allele just like up there. And this will be, since the wild type is the dominant allele, this will be brown and normal, and normal wings. And then, at this point they then will set up a test cross, exactly the same idea. We're going to take the F1. And in this case, we'll take a female whose got this, and we'll cross her with a male who is homozygous recessive. So that will be black over black, and vestigial over vestigial. And out of this, there are four possibilities. We can get parental phenotypes. So, there will be, from that, black over, if you look to the various combinations of how these things could go together, you'll find we can get vestigial plus. Or, we can get black over black, and vestigial over vestigial. For example, this one, coming from those two getting together, and this one coming from this one and this one getting together. So this is just exactly the same kind of thing we've done but done using Drosophila traits. And then, nonparental, so now we get, for example, black over black, and vestigial over plus, black but normal wings, or we could have black over plus, vestigial over vestigial, which would be a brown body, but with vestigial wings. So, if we were trying to do this kind of cross, we are doing this kind of cross, and we are trying to see what the outcome would be, Mendel, every trait [assorts? independently would predict the 1:1:1:1. And the chromosomal theory of inheritance would predict we'd get 1:1 here, and we wouldn't see those at all. Instead, what happened when this experiment was done where numbers that didn't really fit. This part works pretty well. These are numbers from an experiment of this type. So these are pretty close to 1:1. That's OK. That fits with both of the models. The surprise is over here where there is [2/6 to 185? . So, these are in the ballpark of 1:1. But what you can see is the result is not what was predicted by the model. This is the kind of thing I've sort of been trying to tell you through the course. It keeps happening in biology. It's not a QED sort of thing where you can work it out by logic. Very often, you do an experiment, you get a result that doesn't fit within your current framework. And then you have to go back and redo your thinking. So what was going on here, as it turned out, and it all got sorted out, was scientists had just discovered genetic recombination. And here is where, what was going on. So when we talked about Meiosis I, and we had duplicated the DNA, so we now had to chromatids, and I showed you the chiasma when it was drying, it looked sort of like this. And, I said that you always sought these physical attachments that once the homologous pairs were duplicated, they would always be in contact. And I said there was actually a physical interaction going on there. Well what was going on in there was an exchange, a recombination event at the DNA level that's just exactly equivalent to what was happening when I showed you the phage cross. Remember, we had a phage cross, and if we got a recombination in between the two genes, then we could get progeny that were a mixture of the two traits. So, if we were plus plus, for example, these are on the same chromosome. And the other one is like this, black, vestigial, black, and vestigial. If we get a genetic recombination going on here where these two intersect and recombine in between these two genes, what you get out of that, then, is this chromatid is still the same. It's got both wild type alleles. But in this case, now, the black allele has moved from here to the tip of this one, pairing it with that. And the other chromatid over here, the other homologous pair, now has the plus together with the vestigial here and then here. So, if you, then, draw out the possible gametes one will look like one parent. One will look like the other parent. This one can give rise to a nonparental phenotype in a test cross. And that one can give rise to a nonparental phenotype and a test cross. And that's what's going on in this experiment I've described. But the nice thing about this since this sort of thing is happening is one can calculate, then, a recombination frequency in just the same way that we calculated it when we were doing phage cross. And in this case, it was the recombinants, which in this context I'm referring to the nonparentals, the recombinants over the total of the parental types and the nonparental. So in this case, this would be 206 plus 185 over 965 plus 944 plus 206 plus 185. And if I haven't blown the arithmetic, that would be a recombination frequency of 17%. Now, we were back talking about that phage cross. Remember, we did a pair of crosses between gene one and two, and between two and three, and then we were trying to figure out if they were in the linear order, how we could explain it. And we finally worked out the order by crossing allele one and three. And we made a little genetic map that could show the order of gene 1, 2, and 3, based on nothing more than the recombination frequency. So, that's exactly what people were able to do by this kind of measurement. And that, then, led to the generation of chromosome maps. This is a publication from Science in 1994. This is before the human genome was done. And they were mapping a kind of genetic marker that we'll talk about [when?] we do restriction enzymes. But they were able to associate it with [banning? patterns they had seen on the chromosome. This is the sort of thing that cytologists saw, and as scientists were working this out, then they were able to associate these genetic maps of loci, and begin to associate them with the physical maps of banning patterns [in?] chromosomes. Now we have the sort of ultimate genetic map, which is the sequence of the human chromosomes. So now we know exactly, to the base pair, how far different genes are. Part of the way that the human genome was assembled from all these little tiny fragments of DNA that were sequenced were taking advantage of these kind of maps that told the scientists assembling all these little fragments of DNA sequence what order they had to be in, what part of the chromosome they were on, and that kind of thing. There is, this, then, leads us, though, to other issues since I've now started to talk about chromosomes. And one more thing, let me just say before I leave this. So, from this kind of thing, if the recombination frequency is much less than 50%, then they're on the same chromosome. And the word geneticists use to describe this is they say genes are linked. If the recombination frequency is 50%, then they are on different chromosomes. At this point, it's just random assortment. [You go?] one way or the other [as you get?] a number of 50% if you do this kind of calculation. And these are referred to as unlinked. And those of you who are thinking about it can probably imagine that there might be a problem, that if you had a very long chromosome, so the two genes you are studying were very far apart, you might get so many recombination events in between that it would begin to look, the recombination frequency might come close to 50%. And you would perhaps have a difficult time in that genetic cross telling whether the genes are really unlinked. They were on separate chromosomes, or they were linked but very far apart. So this kind of thing could be a little hard to resolve that kind of situation. But there are other things you can do to look at that. So, the chromosomes we've been talking about are what are known as autosomes. These are identical pairs. But there's an exception. And those are the genes that are involved in sex determination. It's the chromosomes that are involved in sex determination. These are known as heterozomes. And, they're on this picture that I showed you where they used this technique of chromosome painting to show it. You can see how all these autosomes are in identical pairs. But this is a male, obviously, because there's the Y-chromosome, and there is an X. If you're a female, you'd have two copies of the X, something I think most of you know. So, if we think about how this works in humans, females have two X's, and males have an X and a Y. In Drosophila, the fruit fly, it's the same thing. Males have two X. Females have two X's, and males have an XY. But there isn't anything magic in nature about females having two of the same, and males having one of each because in birds where it's different enough they use different notation. The females have one of each. And let me make sure I got my notation right. I think it's ZZ, excuse me. And the females have one of each. So, nature tends to use these differences as far as sex determination. And this, then, poses a new kind of problem. And that is, what would happen if you were doing a cross, and the allele that you are studying happened to be a sex chromosome instead of one of these? You might guess that since females have two of one and males have one of each that I would not give the results that are predicted by what we've talked about so far. And so, this led to the discovery of what's known as sex linkage. And that's important. And in fact, as you'll see in a minute, affects stuff that we are familiar with in our lives. I want to just quickly introduce you to this, and show you how it was discovered. It was done by Thomas Morgan in 1910 actually, this discovery. What he was doing was he took a white eyed male, crossed it with a red eyed female wild type, and yet the expected result that the F1s were all red eye. But then when he took the F1 female, which was red, and crossed it with a red eyed male, you got something that was very puzzling at the time. The females were all red. The males, half of them had red eyes, and half of them had white eyes. If you followed the logic up until now, where you try to work this out, you would find that you couldn't generate this pattern by the stuff that we've talked about now. So once again, this led to the need to create a new model, something that expanded our thinking. So, the way the thinking went was, well, there must be something to do with the sex of the fruit fly in this. And so, here was the hypothesis, and that was that the white eyed male had this genotype. They had an allele that caused the white eyeness. But it was located on the X chromosome so that the male would have had a Y chromosome paired with that. And, the red eyed female used in the first cross would have a wild type allele on both X chromosomes. And so, if that were the model, what's going to happen to them when we do this cross that we've described here? Well, let's think it through. So, we've got female whose X plus. We're crossing with the male who's got the X with the white allele over the Y. So, the females can either be, they'll get an X plus XW, and the males, yeah, right. The males are going to get, they will get this allele for Y. So now, if he takes us red eyed female, that was the F1 from the cross up here, which will be this, and crossed with a red eyed male. Now, that means the male has to have the good allele. What are we going to get? Well, for the females in this cross, we've got a couple of possibilities. This one, we could get just the wild type female back, or we could get this one pairing with this one, which will give us this. So, these are all red that fit. But, the males, then, if you see what happens, they have to have, each have to have a Y, and then they can either get this allele or that allele. If they get this allele, they're red. If this allele, they're white. If you stand back and look at that, you will see that is the outcome that was observed. Females were all red. The males are half red, half white. Well, this is a characteristic of, this would be an X-linked trait. And I want to just point out one thing. This female, excuse me, wrong female. This female here is what's referred to as a carrier. She's got this allele that causes a white guy, but she's not expressing it herself. But she's able to transmit it to her sons. And when she has progeny, on average, half of her sons will have the trait. And now, X-linked traits, there are a number of them that we know about. Some of you may know hemophilia. Queen Victoria was a carrier of this gene causing hemophilia, where there is a problem with the clotting mechanism. And if you get a cut, then you can bleed a lot. So, some of her sons had this. A more common one, which has to apply to some people in this room, is red-green color blindness. If you're a male, you have a much higher probability of being colorblind because it's an X-linked trait. And I want to just close by showing you how human geneticists think about the sort of thing. And we have to think about things differently if we are doing human genetics because as most of you know, I think all of us would be very uncooperative subjects in a kind of genetic cross that a fruit fly geneticist [or most? geneticists would like to have us do, in which we'd be put in a cage with a member of the other sex and say, mate. That was your choice in life. Well, that would be a different kind of existence for us all. So, human geneticists don't have that luxury of having pure breeding strains and doing controlled crosses. We all have very strong feelings about the kind of crosses that we want to engage in. And so, what they have to do, they have to make use with what they find. And they use a couple of symbols here that I'll just show you. They look at pedigrees, and then they look for patterns. And, they use a little shorthand for doing this. Males are squares, which I don't know if there's any symbolism to that or not, but if they are affected they show it as a shaded square, and unaffected as an open symbol. So, affected males are solid squares. Affected females are solid circles. So, let's take a look at the sort of pedigree that human geneticists might see. And let's consider something. Let's consider red-green color blindness, which is an X-linked trait. So, let's take a male, which there is probably one in this room at least, maybe more, who have this. So, since the colorblindness trait is on the X chromosome since he's a male, the other pair will be Y. That means if you're a male and you've got it, you're going to display the phenotypes. So this is, George, let's say, who is colorblind. They leave a lot of the romance out of these things, as you'll see, who had progeny with Mary. Let's say, they got married after they graduated from MIT. It was very happy. They had a, let's see, son. He had to get the Y from dad. So, he had to get a good allele from his mother. But they also had a daughter, and she had to get the color blindness allele on an X chromosome because the dad only had one of them, and then a good one from her mom. So at this point, everybody's normal. But you'll notice, this daughter has this trait of being a carrier because even though she doesn't display the trait herself, she's got it in her genome. So, let's say, [UNINTELLIGIBLE PHRASE], a woman who doesn't have any colorblindness allele. So, if we have, let's say, a daughter, and a son, a daughter, and a son, this is not much happening over on the part of the pedigree. Everybody would be normal. Yeah, did I miss something? Pardon? Oh, it's over Y. Excuse me, yep, I'm going too fast here. Over Y, here we go. I don't need to introduce any genetic abnormalities on top of what we're already trying to do here. This is complicated enough. OK, so what happens, let's say then that the daughter that married a guy, and they had four children. I'm going to help us with the genetics of it, from the geneticist, that sort of a perfect family in this case would be four kids representing all four possibilities. And so, first off let's think what would happen with the daughters. Well, the daughters could either have the colorblindness allele paired with this one. Or, they could get a wild type allele with this one. So actually, I'd better put this in. So this is a daughter. So this would be [CB?] over plus, or it could be plus over plus. This one would, again, be a carrier. Now, with the sons, they're both going to get the Y since they're male. But there's a possibility of either getting the good allele, which means he'll be unaffected. But, if you get the other one, he'll be affected. Now, this would be sort of a typical pedigree. And you realize, depending on the number of kids, you might or might not see this. But this, you're going to get to do some more of these and to do some other traits. But let me just sort of point out, if you're a human geneticist, what you would recognize here, the trait's more frequent in males. The frequency of color blindness is about 8% on the X chromosome. So, if you're a female, you've got to have two of them. So that means, you've got 0.64% because you've got to get two together. It's a much smaller probability. The trait skips a generation, [you'd often say?]. You see it here. You see here. But you don't see it in between, and that's because you have this carrier. The affected males don't transmit to their sons because what they give to their sons is the Y. So, they [kept giving?] the colorblindness thing. And then, the heterozygous females who are carriers will transmit the trait to their sons about half the time. And that's a pattern that a human geneticist would look for. And they'd say, ah-ha, it must be an X-linked trait. And you'll see in your problem sets and [recitation? sections are the patterns. OK, I'll see you on Friday.
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https://ocw.mit.edu/courses/8-851-effective-field-theory-spring-2013/8.851-spring-2013.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. IAIN STEWART: So last time, we were talking about regularization and power counting. And in particular, we were-- we argued that there some things that are nice about dimensional regularization when you're doing dimensional power counting, which is what we've been discussing so far, power counting and ratios of mass scales. So I want to continue along that theme today, and in particular move towards really talking about matching calculations in the context of mass of particles. So just continuing with this discussion of dimensional regularization, we also have to pick a scheme. And the scheme that is a nice scheme for dimensional regularization is this MS bar scheme. There are some things that are good about it. Well, it's good because it works under the context of dimensional regularization and preserves-- and it doesn't mess up any of the nice things about that regulator. So it preserves symmetries-- gauge, symmetry, Lorentz symmetry, the things we mentioned last time. In terms of doing calculations, it makes them technically easy, or easier. So a lot of-- if you look at the literature and you look at multi-loop calculations, they're all done on the MS bar scheme. And there's a reason for that, because you've only-- you haven't introduced these extra scales that would make your loop calculations more complicated. And finally, in the context of our discussion of effective field theory, it's nice because it often gives what I would call manifest power counting, where we can power count both the regulator diagrams, the renormalized diagrams. We don't have to worry about whether we've added the counter terms or not. We can just do power counting. So if there's some good, there should be some bad. So what's the bad? Nothing is free in life. So one thing that we kind of lose with dimensional regularization is the physical picture. When we had this cutoff that we introduced, or if we have a Wilsonian picture, it's very clear what you're doing as far as removing degrees of freedom. And in MS bar, that's a little less clear. So if it's just less clear but it's still fine, that would be-- that would just be a technical aspect and we would just pretty much ignore it. So what's one-- I'll give you one example of something in which the picture being less clear could actually mislead you. So if you calculate some renormalized quantities in MS bar, you could have some a-priori knowledge that these renormalized quantities should be positive. Maybe they're supposed to be kinetic energy or some operator, higher-dimension operator, that gives a kind of higher-order kinetic energy term, some kind of physical intuition that it should be positive. Well, that might not be true in MS bar because MS bar loses this physical picture. You're removing just the 1 over epsilons. You're not really having any control over the constants, or the constants are kind of predefined. If those constants are kind of too negative, then you can end up with renormalized matrix elements in MS bar that could be negative, your physical intuition tells you they should be positive. So that's something to be aware of. Where a physical picture could guide you and tell you what to expect, but you might lose that by using MS bar. There's another thing that's kind of technical but is worth knowing. And we will talk about it a little bit more later. You could ask the question, is the Wilsonian picture and the Wilsonian scale separation really equivalent to the scale separation that you do in dimensional regularization? And the answer is, almost but not quite. It certainly is for all the log terms. And I mentioned last time that the log terms that we saw in the two ways of doing the calculation, you could just see a direct correspondence between them. But there's a leftover residual effect called renormalons that does show up in the MS bar scheme and is related to power divergences. And so this effect is carefully hidden. It's actually hidden in the asymptotics of the expansion in alpha s. And it goes under the rubric of something that people call renormalons. So there is a leftover physical effect from kind of having the freedom to drop all the power divergent terms. But it carefully hides itself in a strange place in the gauge theory. And we'll talk more about this later. It's not so important if you're working to order alpha s. But if you start working toward alpha squared or higher, then-- or if you start going to higher orders in perturbation theory, then this could come in and be important. And it does come in in QCD at fairly low orders. At order alpha squared, it can cause numerical effects if you ignore this. And there are ways of taking it into account without losing the nice things here. So we'll talk about that later on. And then there's a final thing-- we had three good points, we should have three bad points-- the final thing we'll deal with right now. And that is the fact that MS bar does not satisfy something that's a theorem called the decoupling theorem. So what is the decoupling theorem? So this goes back to Appelquist and Carrazone. And it says the following. So you're thinking about an effective field theory, you're thinking about deriving a low-energy effect of field theory by integrating out some mass of particle. And the decoupling theorem says that if the remaining low-energy theory is renormalizable, and we use a physical renormalization scheme, then you kind of get what expect, that all the effects of the heavy particles turn up in couplings or effects, they're suppressed. So this seems very physical from what we've described about what effective field theories do. So in that sense, it's-- I'm not going to try to prove it to you or anything, but there is this caveat that you need to use a physical renormalization scheme, of which MS bar is not one. So we decided that we liked MS bar, but then we found some problems. And this problem here, this last problem, we're going to deal with right away. So MS bar is not physical. And hence, it doesn't satisfy this theorem. It's mass-independent. And that is something that we like, but it also is causing exactly this problem. That's how it-- it's why it violates the decoupling theorem. Because it's mass-independent, it doesn't know enough about the mass to know that the effect of massive particles should always decouple. So in MS bar, what you have to do is you have to implement that decoupling by hand. And we've gotten actually so used to this logic that it's often even not mentioned that this is something that we're doing. So go over that a bit. So I'll just go over this in the context of a particular example, which is the most common example, which is QCD with massive particles in MS bar. And I'll show you how it works. So QCD has a beta function of the renormalized coupling mu d by d mu of g of mu. And just to establish some notation, I'm going to remind you of some facts that you've seen before. So this is the beta function at lowest order. I'll call this factor that depends on the group that you're dealing with and how many light fermions you have, I'll call it b0. And this is less than 0. And this is only the first-order term in a series, but it'll be enough for our discussion right now. So this is g to the fifth terms, et cetera. That's the beta function for g. We can also think about using-- since g likes to come squared, we can switch to alpha. QCD is asymptotically free. When we solve this equation for alpha, it looks as follows. And the coupling constant decreases as you go to higher energy. So this is alpha at md, and this is alpha at mW. Alpha at mW is less than alpha at md. This is the lowest-order solution. So just a little bit more in the way of setting the stage, you can also associate to this solution here an intrinsic mass scale dimensional transmutation. If you form the combination mu exponential of minus 2 pi over beta 0-- b0 alpha of mu, and you use this equation over here, then what you'll find is that this combination is also equal to mu 0, same thing alpha of mu 0. So it's independent of what choice of scale you use, mu or mu 0. Therefore, you define it to be a constant, and that constant is called lambda QCD. Once you do that, you can also just take this equation and write it like this. Write that, rewrite that solution like this. So it's another way of specifying a boundary condition, if you like, for the differential equation. One way is to pick a value of the coupling somewhere, another way is to fix this constant. So this guy is independent of mu, as I said. And it's the scale where QCD becomes nonperturbative. And the reason I have to mention this is because we're going to be talking about anomalous dimensions, and I have to tell you when the things that I write down are valid. And it's all going to be valid as long as we're not near this scale. Because if we're near this scale, the coupling gets too large for us to do the perturbation theory that we're doing when we calculate anomalous dimensions. So you can ask the question, what does this thing depend on? And if we look at the right-hand side here, we can already see some things it depends on, because it depends on b0. And if we looked back at what b0 is, b0 depended on the fact that we were at su3 if it's QCD. It also depended on the number of light flavors. So this scale that you fix here, well, it depends on the order of the loop expansion. We wrote down a formula that was valid at first order, but we could extend that formula at higher orders as well. And the formula would change, and the value would change if we went to higher orders. Depends on the number of light flavors. And it actually also starts to depend on the scheme but only at two loops. So it'll depend on MS bar versus MS, for example, but only beyond two loops. Now, the issue with this is that a priori, nothing tells us whether if you have a top quark or an up quark or a bottom quark. In MS bar, nothing a prior tells us what to do with these formulas, what to include in the b0. Just seems like we should include everything that exists, all the known light fermions that couple to the gauge field. But there might be some we don't know about. Should we include those, too? MS bar, with the logic we've presented so far, doesn't tell us what to include there in that nf. So let me phrase it this way. But the top quark and the up quark, which have very different masses, both contribute to b0. And it seems like they do that for any mu. Even though-- even if I'm at a very low-energy scale and the top is very heavy, MS bar is not smart enough to decouple the top quark. If you were to work in a physical scheme, then this decoupling theorem guarantees, actually, that the top quark would decouple, and it would drop out of the beta function in that scheme. But in this MS bar scheme, it doesn't happen. So the solution to this is to build back in the thing that happened in the physical schemes into our MS bar scheme. And we do that by implementing the decoupling by hand. And we just say that when we get to a mass threshold, we're going to integrate out the heavy fermion in this case. So at some scale of order of the mass of that fermion, we're going to integrate out the heavy fermion. And that's an example of during matching, actually. You're moving from one theory to another, because you're changing the field content. You're removing some fields, and that's an example of doing matching. So what this means is that we'll define different b0's depending on what scale we're at. So nf would be 6 if we're at scales above the top quark. But then we drop that nf down to 5 once we're below the top quark mass, et cetera. So it's a discrete jump in what the b0 is. So in some sense, what that means is that this kind of matching is forced upon you in order to preserve physics in MS bar. If you were to do some other scheme like a physical scheme, it's not to say that you couldn't take the same logic. You could do matching to integrate our heavy particles in those schemes as well. And in fact, some sometimes people have. But in MS bar It's really forced upon you. There's no way out, because you otherwise wouldn't get the physics right. So we've got to be careful. If you gain something, sometimes you lose something. And if it's physics that you're losing, you have to build it back in. OK. So how do we actually do this matching? Well, we use what are called matching conditions. So what is a matching condition? At some scale which I'll call mu m-- that could be equal to m, but it could also be equal to twice m or 1/2 m, some scale that's of order m-- we're going to demand something about the theory above and below. And that is that S matrix elements in the two theories are going to agree. Of course, we should be careful to make sure that we work within the realm of things that we can calculate in the low-energy theory. So we should consider only S matrix elements with light external particles. If we're getting rid of the heavy ones, we don't want them on the external lines. So the basic idea is that we set up matching conditions which are conditions that say that S matrix elements should agree between theories 1 and 2, what we called theories 1 and 2 last time. So if we do that for QCD, the example we've been talking about, and we do it at lowest order, then the conditions are actually very easy. So here's the picture. Let's imagine that the top quark is here, bottom quark is here. My picture is not to scale. Charm quark, et cetera. Then what we do when we want to say that the theory above and below, that that threshold is the same, if we just match QCD above and below, it just gives a continuity condition on alpha. So let me write it down at this scale. So here we would have alpha 6. Alpha depends on the number of flavors. Here we have alpha 5. A different definition of the coupling a different number of active flavors, a different value of b0. 3. And what these arrows are representing are just kind of renormalization group evolution, which I'll just call running. And then what the lines are representing is doing some matching. So every time we reach a threshold, we do a matching, we switch the field contact, and we get a new effective field theory with a new coupling constant. So the coupling just depends on what theory content we have. And the matching condition is, say, at this b quark scale that alpha s at 5 at some scale that's of order mb is equal to alpha s at 4. That's the leading order matching condition here. And similarly, the same condition at all scales. And this is mu b, just to be clear, is something that's of order mb, could be equal to mb, could be equal to mb/2, could be twice mb. And sometimes people use these different choices to get uncertainties. OK. So it seems fairly straightforward, just continuity of the coupling. But that has some caveats. But go ahead and ask the question. AUDIENCE: Yeah. So how do you know that that's the physical observable that you care about? What if I was trying to measure the [INAUDIBLE] alpha s? IAIN STEWART: So this is the-- yeah. So if you're going-- so you have to think about it as not the derivative of alpha s, but you have to think about it as an S matrix element-- something two-to-two scattering, right? If you do two-to-two scattering, then it's just going to be proportional to alpha of mu. But the mu dependence in your leading order prediction is not really fixed. You need to go to higher order to do that. So if you're just ensuring continuity at leading order of S matrix elements, then this is all you need. And if you want to construct something that's like a derivative of alpha, you have to think of the derivatives of the S matrix. But you can't take derivatives of mu, because that's a higher-order question. So this is all you need at leading order. AUDIENCE: OK. So maybe my derivative example is [INAUDIBLE] necessarily know the alpha s, you'd get only-- IAIN STEWART: So in general-- right. AUDIENCE: --but I don't necessarily know that-- IAIN STEWART: Yeah. No, in general, you have to-- this is like one particular example. In general, you have to ensure continuity of all S matrix elements. And that will-- in this case, we'll give you conditions also and what's happening with masses and stuff. So it's all the parameters of the theory. There should be conditions for all of them. AUDIENCE: I guess these words seem more complicated than just asking for [INAUDIBLE] some different equation. Does it ever boil down to anything-- do you ever have to do anything more complicated than just like-- IAIN STEWART: I'm just-- yeah, so-- AUDIENCE: --change the beta function [INAUDIBLE] continuity and-- IAIN STEWART: It's really-- it's not complicated. It's just continuity of S matrix elements. But you have to just know that it's S matrix elements and not the Lagrangian. Because those are two different things. So let me give you an example why it's not the Lagrangian. So this kind of thing seems simple, right? Just say, well, the Lagrangian is continuous. The alpha-- the g that appears in the Lagrangian is continuous. But that's not true once you go to higher orders. That's only a leading order statement. If I write down the analog of this condition at higher orders, it looks as follows. So the coupling is actually not continuous at MC bar at higher orders. So demanding continuity of the S matrix elements does not lead to continuity of the coupling. And the matching condition at some scale looks like this. So the words are carefully crafted to be correct. And they sound a little more complicated than they need to be, because I want them to be correct, even if I want-- even if I were to do matching at higher orders in perturbation theory. And it's really at this alpha squared level that you start to see things a little more interesting. OK. So now we've gone two others beyond what I just told you before. So continuity at lowest order-- at the next order, you can retain continuity as long as you pick the particular point where mu b is equal to mb. And you'd have-- then this log would go away, and you'd still have continuity. But even that doesn't work out once you go to one higher order, there's this constant. You could get rid of the logs by picking a scale choice, but there's no scale choice which will make it continuous once you get to alpha squared in this matching condition. So this is the condition that's necessary to ensure continuity of S matrix elements once you get to that level of perturbation theory. OK, so that is why I said S matrix and not just continuity of couplings. AUDIENCE: So this condition is for 2-by-2 scattering, And then you get other conditions-- IAIN STEWART: This is the only condition you need for all the S matrix-- you can use different S matrix elements, and they'll lead to the same condition. AUDIENCE: How did you-- where does it come from? IAIN STEWART: So this, it comes from ensuring that I calculate S matrix elements, say, 2-to-2 scattering, in the theory with five flavors and four flavors. AUDIENCE: At two loops? IAIN STEWART: And I demand that-- yeah, up to two loops. So where would this guy here come from? This guy here would come from the graph with an explicit b quark, which is in the theory with the b quark but not in the theory without. Once you get to this level, then there's more complicated diagrams. Just think of generalizations of this. And they involve constants as well as logs. This guy just involves a log. And really, what you're ensuring is that in the theory with the b quark, which is this five-flavor theory, you get the same S matrix elements as in the theory with four flavors. Since the diagrams differ in those two theories, they're kind of the same at lowest order because you're just doing two graphs. But once you have the b quark and it can go around in the loop, then they differ, and the conditions become more complicated. So any other questions about that? OK. So one other thing that we see from this is related to these logarithms. We also see from these conditions here that there is going to be no large logarithms as long as we pick the scale where we do this matching to be of order mb. So you don't want to pick mu to be the gut scale or something you want to pick it to be some scale so mu is equal to mu b, which is of order mb. Because you don't want, for example, to make this log so large that it starts to overcome the coupling. You want it to be-- you want this to really be a perturbative thing so it makes sense to think about this, and then this is a correction, and that's another correction. OK. So the general procedure is the same idea. So if I just kind of adopt a more general procedure for massive particles [INAUDIBLE] a more general notation. So this is with any operators and couplings. It doesn't have to be gauge theory. And if we have a hierarchy of particles-- let's say we have n of them-- when we want to pass from an L1 to an L2, an L3, to an Ln-- and we're doing the same type of thing we just did over there. So the steps are the following. Consider S matrix elements in theory 1. Do so at a scale which I'll call mu 1 that's of order m1, and match that onto in the same way of assuring the continuity and do-- using these matching conditions, match that onto L2. So this is the technical step by which I said that you could do-- you could in a top-down approach take theory 1 and determine the parameters of theory 2. The matching conditions are determining the parameters. Alpha s 4 is of parameter of theory 2 to alpha s 5. This is theory 1. And we're just determining what this alpha s 4 should be, and this is the condition that relates them. So after you do that step, if you want to go through this picture here, or the analog of that picture for this case over here, then you need to compute the beta functions in anomalous dimensions in this theory. So it's the theory that doesn't have particle 1. And then you evolve/run the couplings down, which just means using the evolution equation for the couplings, whatever they may be. So we had the evolution equation for alpha S a minute ago on the board. And I would just use that to go from a scale of order mb down to a scale of order mc. And then I repeat. And this is the general kind of paradigm of matching and running that you hear about in effective theory all the time. So we just keep going. And at the end of the day, we're going to stop somewhere. And the place we're going to stop is the place we want to do lower-energy physics. So let's say we stop at the n-th level, since that's the last level I wrote. So if you're interested in dynamics at that scale, that's where you stop. And that's where you compute your final matrix elements. So everything up until that stage is just to determine the theory Ln and what are the values of decoupling in that theory-- determined from knowing information at the high scale. Knowing information in theory 1, how do I propagate that knowledge all the way down to low energies consistently without losing information I need? And then once I'm at the lowest scale, I just do my computations of observables. Yeah. AUDIENCE: Professor. IAIN STEWART: Sure. AUDIENCE: [INAUDIBLE] mention at the scale where the particle is there-- actually, I was a bit confused because if we write the full theory, then, say, 2-to-2 scattering in that mass scale, isn't there any like [INAUDIBLE] say resonance effect, and then you want to match that to a theory without the particle? I thought we should match the s-- IAIN STEWART: So there is no resonance effect. And the reason is because the momentum on the external lines that you're taking-- you're thinking of it as small, right? The scale is chosen to be the mass of the particle, the cutoff, mu, the soft cutoff. But not the momentum of the particles. Good question. Any other questions? OK. So we're not really done talking about subtleties here. And we're not really done talking about some of the key ideas that come into these calculations. I've given you a very simple example just the coupling. That's a little bit too simple. So I want to do something a little more sophisticated but still fairly simple and widely used. And that is in the standard model just to take the heaviest particles, which are the top, the Higgs, the W, and the Z, and remove them. So we'll spend a bit of time talking about what happens when you do that. That's not the only thing you could do. And in particular, before people do how heavy the Higgs was, you could think of other possibilities. And before people knew how heavy the top was, people did think of other possibilities. This is actually a reasonable thing to do, though. let me give you one example of another possibility. Or you could say the top is heavier than the W and the Z, so why don't I do the top first, and then I'll just follow your diagram over there. I erased it, but it'd follow the picture. And then I'll do some running in the theory without the top quark, and then I'll get down to the W and the Z scales, then I'll remove the W and the Z. OK. That would be a valid thing to do. But it introduces complications that are actually not worth the effort. And benefit is actually not that great. So I want to emphasize that. When should you think of things as being comparably heavy versus when should you think of things being hierarchically heavy? Well, one complication of removing the top quark is that it breaks su2 cross u1 gauge invariance, because the top quark was in a doublet with the b, and you're trying to keep the b and remove the top, which doesn't sound good. You're trying to keep the gauge particles, the W and the Z, in your theory. So you should still have that gauge symmetry, even if it's spontaneously broken. But you're trying to remove one member of a doublet. And that leads to [INAUDIBLE] terms that you'd have to clue to the effective theory so you can deal with it. And it's just a little bit annoying. But, you know, it's something you can deal with. The real crux of the matter is that if you compare mZ over m top, or mW over m top, that's about a half. So if you think about what you're expanding in when you take external particles that have momentum of order mZ, integrate out particles of order mt, you're expanding in a half, which is not such a great expansion parameter. Usually in effective field theory, you want to expand in something that's at least a third, hopefully a quarter. And a tenth if you really want to have a good expansion. So half is not really that good. So that's the real reason not to do it. And you could ask, well, why do we lose by not doing that? And that, of course, is the real question. Well, you miss some running, right? Because you don't have a theory that has no top quark but still has a W and a Z, and that theory could have anomalous dimensions, and you'd miss any running in that theory. You're kind of collapsing two of the lines in my picture with the arrow between, you're collapsing them down to just a single line. So you're missing the running that would go between mt to mW. Well, of course, that's not a very big hierarchy, and it's logs of two. And that's why we don't care that much. What it boils down to is that you're treating alpha s at mW. And if we want to be generous, we can say it's mW squared over m top squared. So they're logs of four. But you're treating this perturbatively. If you remove both the top and the W at the same time, then this is going to show up in your matching conditions, and you're not thinking of that is something as large. Thinking of it just as order 1. So that's the cost, which is not a big cost. Especially when you give, say, that the cost of going the other way would be expanding in the half. Better to take the cost in the logarithms than expansion in the half. So that's a general kind of rule, that if you're thinking about removing massive particles, you should ask, how close are they to each other? Should I integrate out a slew of them at the same time, or one at a time? That'll depend on exactly the scales in the problem. And there's-- and it means you're organizing the theory differently if you do it, the two different approaches. All right. So we'll do an example of this. We'll take a very simple example, although not completely trivial. So just b quarks changing to charm quarks u bar and d. So on the standard model, which is where we start, let's say we have a W boson. We have an up quark that's left-handed. We have a CKM matrix connecting flavors together. And we have down quarks. So this is-- there's a matrix space here in the flavor, which includes bottom charm up and down. And tree level matching is easy. And some of the complications I want to talk about will come into the loops. So let's first get through the tree level and set up some notation. And then we'll talk about what happens with the loop. So tree level, this is the diagram in the theory 1. Just calculate it at tree level. W propagator, unitary gauge. And then there's some spinners. So we have an antiquark. So we have a v spinner for the antiquark. That's an up. Everything else is a u spinner. I've put in explicitly the P lefts to denote the fact that it's left-handed. So there's some momentum in this diagram, which is momentum transfer, which is the be momentum minus the c momentum and the kind of obvious notation. And of course, by momentum conservation, that's also the d momentum plus the u momentum. And that's the momentum that's going through the propagator. We're going to count momenta as being of order masses. And the heaviest mass in this case is the b quark mass. And we'll take it to be the b quark mass. We can use the equations of motion. So Pb slash on ub, spinner for the b quark is mb ub, et cetera. And we can simplify the diagram by doing that. And of course, the key thing is that we can expand the propagator since the momentum is smaller than the mass of the W for integrating out the W. So the leading term when we do that is this term. And any other terms that we drop are down by that much. So for example, this term here, we just said that the k's are of order mb so it's down. And then we could drop the k squared here, I'll get the mW squared, so we just have that. And then we put that together, we can calculate the Feynman rule in the effective theory. The Feynman rule in the effective theory, we can determine the coefficient of it. So let's start out with just saying it's some four-quark operator that couples together those flavors, conveniently chosen so that they're all different to avoid [INAUDIBLE] factors. And in some conventional normalization for this Wilson coefficient, we call it GF, which is for Fermi. And the Feynman rule would again give spinners, the same spinners as before, with the same Lagrangians for these light quarks. Haven't changed anything about that. We're just removing the heavy particle. And it's conventional also to pull out the CKM factor [INAUDIBLE]. OK. So if you like, you can say that there's a coefficient here, and that coefficient has been fixed to be 1. And what this G Fermi is, then, would just be the leftover factors of the gauge coupling and the mass of the W. And that's the usual convention. So G Fermi is just fixed in this case, not as the thing that is going to get corrected at higher orders in QCD but would get corrected at higher orders in electroweak, if you like. And then you sort of put in with the QCD corrections into some other coefficient that we can call C. And we'll just say that it's 1 at the level of this. And this line agrees with the expansion of that line, and that's the matching of something that's measurable, which is this four-point function. So we can call it an S matrix element. So this theory is very popular. It's called the electroweak Hamiltonian. It's used all over the place. And it involves more than just doing what I did, because I chose a particular set of flavors for a particular channel, and you have to do it for the whole-- for all sorts of other channels as well. So we'll study some aspects of this theory. We won't study every possible aspect. But I've given you a handout that studies a lot more. It's 250 pages long. I'm not asking you to even read all of that. I've pointed you at some pages of that in the reading, which are relevant to the discussion we're having here. If you really want to dig deeper, you can. So people often call this an electroweak Hamiltonian, which is just minus the Lagrangian. So then that's 4 GF over 2. And as a Hamiltonian, we write fields instead of spinners. So we have this four-quark operator. And we would have determined the Hamiltonian from tree level matching, from what we've done. And this would be the result. Standard stuff. How do we want to go further than that? Well, if we want to think about theory 2, which is this theory, and we want to think about it in more detail, we should worry about whether we're-- with just this operator, whether we've got a complete set of structures that could possibly occur. So we should think about the symmetries. And it's useful to, therefore, construct the most general basis of operators in theory 2. And we'll just do that. So if you read this 250-page review, then it's done for the full electroweak Hamiltonian, and we'll just stick with the flavors we're talking about here. So what are the most general bases of operators? How should we think about that? Well, we're constructing some theory that's going to-- that we're matching onto it mu equals mW. And at that scale, when we're determining the theory and determining the coefficients of the theory, we can treat the bottom the charm the down and the up respectively as if they're massless. And really, what I mean by that is that the masses of these particles are only going to show up in the operators and not in the coefficients, which I've-- you see that I talked about over here. The thing that shows up in the coefficients are the mass scales we're removing. And we're not removing the mass scales of these things. So that means, if we can think of them as massless, that we can think about the matching in terms of using something like chirality. And we can use the fact that QCD for massless quarks does not change chirality. And we can use that in constructing our basis. So that means even though the b quark and the charm quark really have masses, for the purpose of constructing the operator basis, we can think of them as massless. And we only have left-handed guys to worry about here. Chirality means more than that. It also-- well, it means effectively that. But it means also that we will only get one gamma matrix here. And we can think about why that is. Chirality means that you get an odd number. You need an odd number to have left on both sides. But you can reduce 3 to 1. So there's an identity which I won't write out for you, but you can reduce 3 back down to 1. So any higher odd number can be reduced back down to 1. So once we use chirality and that fact, then we have a fairly restrictive basis in terms of Dirac structures. You could ask, what are the most general possible ways of contracting spinner indices? Who said I had to put this charm quirk with the bottom quark? I could have put the up quark over here. And maybe in higher orders that happens. Well, at higher orders, you could think about that happening. But you can always get back to the form that I wrote over there using what's called the spin Fierz. So if I write it for fields, it means that there's an identity that I can rearrange this guy the other way. And so these are equivalent operators. And sometimes, you have to use these relations when you're doing matching calculations, because you construct a complete basis, which is the minimal one. And then maybe when you do your calculation, you get this operator, so you have to turn it back into this one. There's two minus signs in this relation. One is from the Fierz identity, and one is because when you do the manipulations you end up anticommuting to fields. So from the statistics. So if you were to use the same relation for spinners, then there would be one less minus sign. Something to be careful, though. So what this tells you, once you put that information together, is that you know that you can write the operators this way, and you also know that this gamma, capital gamma, has that gamma u p left form in terms of the Dirac structure. So the only really thing that can happen is color. And you can contract the color of these 3's-- so you have four 3's, you can contract the color of them in different ways. Well, I have two 3 bars and two 3's. Think about having different color contractions. And that will expand our operator basis by one more operator. There is also a color Fierz. And I can use that color Fierz to get rid of having explicit TA's. So color is a lot like spin in the sense that if I had more TA's, then I can always reduce a higher number of TA's back down to a lower number. So I've had two, I can reduce it down to one or zero. OK. So I only really have to think about having a TA here and a TA there. If I have a TA there and a TA there, there's a spinner-- or there's a color Fierz relation for this guy that you could use. And there's a handout that I posted on the web that has a summary of all these relations. I'm not going to bother writing them down in lecture here. So once I take that into account, then it ends up in two operators. And if I were to write them in a kind of renormalized notation where I introduce the scale mu, then I would write them as follows. Let's call them O1 and O2. And the difference between O1 and O2 is just simply color. So in O1, the color index alpha is contracted between charm bottom and up down and up. And then in O2 it's the the colors contracted the other way. So I don't write explicit TA's, but I do have to consider the other possible way of contracting the indices. Up, down, doesn't matter. OK. So these operators, these are the two operators they have once I satisfy all the symmetries and I have some coefficients. And if you ask really what the coefficients can depend on here, well, they can depend on mass scales like top and the other particles that I'm integrating out, Z. I'm not going to make that explicit in my notation. If I tried to make every possible mass scale like the Higgs or the top, it would just be too much. So let me just denote two things that it can depend on. Mu over mW can show up, and alpha of mu can show up, as well as some ratios of other particles which we will suppress. So then what tree level matching is saying is that you've got one of these operators, and you didn't get the other one. So it says that at tree level, what we did before-- and so say we do matching at mu equals mW and C1 at 1, which is mu over mW equals 1 and alpha mW is 1. And then corrections to that would be suppressed, no large logs, just by alpha. And C2 of 1 alpha mW is 0 plus something of this order. So that's what we determined by doing our tree level matching. Now, when we did our tree level matching we did something. We used external states which were quarks. We matched up the S matrix elements in that way. If you were really interested in this process b goes to c u bar d, you'd not really be interested in measuring quarks. We don't see them. You'd be interested in thinking about the process for mesons-- B meson changes to a D meson, and a pion, for example. And who's to say that the matching that we did for quarks is also the matching that's valid for hadrons? Well, it is. So there's a key fact about matching which is important. One of the things I wanted to emphasize to you is that it doesn't depend on what states you pick. It's independent of the choice of states. And it's independent of the IR regulators that you pick. So when you do matching, you often pick some IR regulators to regulate IR divergences. And the only thing you have to do is pick the same states and same IR regulators in the two theories. And once you do that, your results for your coefficients will be independent of the choice you made. So this is intuitively something I think that's, once you think about it, fairly obvious. What this sentence is saying is that the matching, which is supposed to be a high-energy property, is independent of the low-energy physics. The choice of the states and the choice of IR regulators, that's parameterizing something about low-energy physics-- lower energy than the scale we're trying to determine. These things should only depend on higher-energy physics. So the outcome for them will be independent of this choice. So just to emphasize, even when we use hadronic states, the result that we obtain from quark states is valid. That's just different choice of state. And what we're doing when we do the matching is we're picking a convenient choice of state, something that makes it easy to calculate. If we tried to calculate the matrix elements of B and D and pi mesons, then I'm saying if you're a strong enough, you'd get the same matching coefficient. Of course, you should pick the result that makes the calculation easy, and that would be to use quarks. Any questions about that? OK. So this is something that you should keep in mind. And you should keep in mind that it is-- you do have to make sure that you're using the same states and IR regulator in the two theories. In this case, that's pretty easy. The states are really the same states, because you're defining the states with the Lagrangian machine for the b quarks, say, and the b quark Lagrangian hasn't really changed. So there's no change in the state. And you can set things up so that you have the same IR regulator. And that'll become important in the example that we're just about to do. So let's do an example of carrying out this matching for C1 and C2 in a little more detail. And before we carry out matching, we actually have to renormalize the two theories. Well, we've been talking as if the theory above is the standard model. So imagine that we've already renormalized that. And so we only have to renormalize the effective theory in MS bar. So we'll talk about doing that, remind you of doing that-- maybe something you've seen before. So there's going to be wave function renormalization. So you have this 4/3. I'm going to leave out the prefactor, this prefactor. It's always going to be there. I'm going to stop writing it and start just focusing on the thing in square brackets here. We'll do calculations with Feynman gauge. I'm not going to really do the calculations. I'm just going to quote results to you. And in order to simplify what we have to write down, let me define the set of spinners that you get from taking the tree level matrix element of O1 to be S1 and of O2 to be S2. So this is just the spinners that we wrote down before for the case of S1, exactly what we had before. And then for S2, just a slightly different contraction of color indices. OK. So that's just trying to make the lowest-error result simple to write down so that when I write down the higher-error result, we can focus on the things that are changing and mattering, and not on the complications that come in from what we're talking about. So if we think about diagrams here, we have four-quark operator. And we just have to draw all the loop graphs Right. We have to regulate them in some way, because they are IR divergent. So let's regulate them with-- if we really want to talk about the matching, which is what I really want to do eventually, we should regulate them in some way. And so let's regulate them with off-shell momenta on the external lines. And I'll take it to be a common off-shell momenta p. And I'll just set the masses of the external lines to 0, since they're not going to matter. OK. So then what does it look like? So the matrix element of O1-- the 0 means Bayer-- is going to have divergences. And it has the following structure, times spinner 1. And then there's another piece that involves spinner 1. And then there's a piece that involves spinner 2 that shows up in from these loop graphs. So even though we're calculating the Bayer matrix [INAUDIBLE] of operator 1, the spinner combination 2 shows up, because this is supposedly a gluon, and the gluon moves the color around. And I'm only writing the divergent terms, although later on we'll be interested in the constant terms as well. So the dots are the constant terms underneath logarithms, and all these little round brackets have constant terms in them. Part of the reason for introducing the basis in the way I did is that when I want to quote you the result for O2, it's the same, just switching 1 and 2. And so the statement that you would have from this-- these are ultraviolet divergences, the infrared divergences are regulated. The statement that you would have from this is that you can look at these various terms, and you can ask, what's going on with those divergences? Well, this one here is actually cancelled by wave function renormalization, which I haven't put in yet. And this one here is O1 mixing into O1. So this one here is a counterterm, if you like, for O1. And this one here is a counterterm for O2. And the language you use as you say that O1, which we're calculating, has mixed into O2. So there's two different methods we could use to carry out the renormalization here that are actually equivalent. So method one is called composite operator renormalization. It's useful to know what things are equivalent so that you know what you can get away with ignoring. So what is composite operator renormalization? Well, you think about having a Bayer operator. And you need to introduce some constants to renormalize. These are not related to wave function renormalization. It's actually operator renormalization. You have multiple fields at the same spacetime point, you need additional renormalization constants, that's what these Z's are. When you go to take the matrix element, you have to include the wave function renormalization as well. So the Bayer matrix element has a wave function renormalization as well as this Z. And this is the relation between the renormalized matrix element and the Bayer one. So this guy here is renormalized and amputated. OK. So that's the relation you can calculate O0. That's what we were doing. We just calculated O0. We could remove Z psi. And I just told you if you remove Z psi, it's going to get rid of this. And then you use the remainder of this stuff to get Zij. I'm going to need it the other way around. So let me write the-- so I can write this relation the other way around. And then it looks like this. OK, so that's one method. The other one is related more to how we usually think about things in terms of gauge theory. And that is to have counterterm coefficients for the various operators. So start with the Hamiltonian written in terms of Bayer coefficients and operators that have Bayer fields-- Bayer operators, if you like. And now switch over to renormalized quantities, so you get Z psi squared. We have to switch over from the Bayer coefficient to a renormalized one. So let me do that first. So we get some Z for the coefficient, some renormalized coefficients. And then we get a Z psi squared from the four fermions that are in this operator. Then we get an O. And then we can write this in terms of Ci Oi plus Z psi squared Zij C minus delta ij Cj Oi. And then this would be our counterterms. So we would stick in this guy, that operator we haven't renormalized-- we haven't done our operator renormalization. We're not doing method 1. So this, the matrix element here will still diverge, but we cancel them off with the counterterms. So that's the logic of method 2. And once you do that, the divergences cancel between these two terms, and you're left, if you like-- or even if you don't-- with the coefficients times the operators, which are renormalized, both renormalized, even renormalized separately. And these two are equivalent. These are equivalent ways of thinking about the same thing. And we can even go further and derive the equivalence of them. So if we take this theory and we take the matrix element of the Hamiltonian in this second way of doing things, what do we get? We get this. So even though as Oi, I said, we haven't done operator renormalization, so we still have the divergences there, that's Cj Oj. And if we write the OJ using the relation at the top of the board, then we get Cj Zji inverse Z psi squared Oi 0. And then we can just look at the two sides. They both have the Z psi squared. They both have a Cj, they both have Oi 0. The only thing that's different is this. So we find the Zij for the coefficient is the transpose of the inverse of the Z's for the operators. So that's a more definite way of saying the two would lead to the same results. And this is how you would actually get the relation between the two ways of doing it. So it's not completely trivial. You have to know you have to take the inverse and transpose. But it's the same information. OK. So in our example, the Z is a matrix. It starts out as the unit matrix. And in MS bar, we collect the divergences that are not cancelled by wave function renormalization. And that gives us this little 2-by-2 matrix. And from that little 2-by-2 matrix, we can construct anomalous dimension for the operators. So let's do that. So how do we do that? Well, if we take the renormalization of the Bayer operators, remember those dependent on the regulator, but they didn't depend on the scale mu. So if we take mu d by d mu of Oi 0, which is the Bayer operator at 0. That's what I've written. And then we can write O in terms of the Z, in the equation I raised. And both Z and the renormalized operator depend on the scale mu. And so that gives us this equation. So I'm doing method 1 here, if you like, using the notations from method 1. If I take that equation and rearrange it, I can write it as anomalous dimension equation for the operator. So it's mu d by d mu is one of the terms. I isolate this, and I move that over by taking an inverse of it. And I just call everything that I get there the anomalous dimension of the operator. So gamma ji is all the other stuff, Zjk inverse. And I put the explicit minus sign here so there's no minus sign in that equation. Anomalous dimension is determined by the Z factors. And we determine the Z factors, and which we can stick this equation into that equation and get the anomalous dimension. When we do that, we have to be careful about the fact that alpha s in d dimensions has a little extra piece that's actually important for this discussion, which is this piece. So when we take mu d by d mu, you ask, what depends on mu? And it's the alpha here that depends on mu. And it's still in-- depends on epsilon as well in that equation. And so this term is the term that matters at one loop. And this matters at two loops, and so does that. But at one loop, we could just replace this by the identity, and we can drop this. OK. Put those things together. Anomalous dimension, of course, is something that doesn't depend on epsilon. And it's, in this case, a 2-by-2 matrix. And we can solve this problem. If we want to solve this matrix, then what do you do? Well, you could diagonalize it. That's how you would solve it. So if you want to run the operators in this theory just for completeness, you would diagonalize by forming O1 plus or minus O2, call the coefficients of those new operators, C plus or minus, and an obvious notation. And when you write down an anomalous dimension equations for these guys, there's no mixing anymore at one loop. In general, this is the procedure if you do this diagonalization that you have to carry out again when you get to two loops. It's not like the basis that you pick at one loop will be fine for two loops. We're not so lucky. But at one loop, this is a perfectly valid basis for the problem. And in an hopefully self-evident notation, I either take all pluses or all minuses. OK. And if I write my Hamiltonian, I could write it in the original basis. Then if I switch to the other basis, I'll set up my convention so that it's simple. And that means I picked C plus or minus to be C1 plus or minus C2/2. AUDIENCE: [INAUDIBLE] minus? IAIN STEWART: Whoops, thank you. Yeah, they have the opposite sign. First one's positive, second one's negative. And in this new basis tree level matching, which is your boundary condition for your revolution, is that both coefficients are 1/2. OK. So then we could solve the differential equation the same way we were doing for QCD-- write down a result that would sum logs below the scale of the W mass. OK. So we'll continue along these lines next time, say a few more words about renormalization, which is really just a lead-up of doing the matching, which is what I want to-- kind of the thing I want to spend a little more time on. So we'll finish that up also next time and move on to some other things. So there's a few things to emphasize when we do the matching. And I want to write down the matrix elements and the full theory, which are box diagrams, I'll write down the results for those. We'll compare those to the results of the effective theory, and we'll draw some lessons from that. And we'll carry out the matching, figure out what the next order coefficients would be here. What would those alpha s terms be? What's the procedure I would go through to get those coefficients? OK. So we'll stop there for today.
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https://ocw.mit.edu/courses/7-014-introductory-biology-spring-2005/7.014-spring-2005.zip
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So just trying to remind you that the replication fork looks something like this where 5 prime to 3 prime and 5 prime to 3 prime. This is what's known as the leading strand because DNA, the synthesis of the new strand can go -- Which is going 5 prime to 3 prime is going in the same direction as the movement of the replication fork. The other strand, which is known as the lagging strand, the DNA synthesis is actually going backwards to the movement of the replication fork, which means it has to go and then start up here and go again. And it's continually jumping. And I told you that the little RNA primer is used to start each strand. And then the DNA polymerase is able to elongate that. And then at the end these little nicks in here, the RNA has to be removed, fill in the gap and then it's sealed up by the enzyme DNA ligase, which we'll talk about when we talk about recombinant DNA. Someone asked, I had mentioned why this strategy of using RNA was beneficial, and that has to do with the fact that the fidelity, which is going to be the next thing I'm going to focus on of DNA replication as not, you can get a much higher accuracy if you have the end of a primer already there and then carry out the chemistry in there. No enzyme has ever achieved the accuracy that you see in DNA replication if it's starting a strand. So RNA polymerase, which constantly starts strands to make RNA copies, as we'll talk about, is not as accurate as DNA replication. And by putting a little bit of RNA, because the cell has to start a new strand. Before it gets here there's no strand at all on this lagging strand so it needs to make this little RNA primer. It needs to make a little primer. And by making it out of RNA then it can tell what doesn't belong there. It doesn't matter if it's not quite as accurate as the rest of DNA replications because it's going to take it out anyway and fill it in using the DNA polymerase. And if you think about that maybe you can see one of the reasons that the cell has chosen or nature has chosen through evolution to use little RNAs to begin the strands. OK. Well, in any case, the fidelity of DNA replication is really pretty amazing. Incidentally, just speaking of DNA, many of you wrote some very thoughtful things about Vernon Ingram's visit. I didn't give him a whole lot of warning and he had to go and change his schedule and move meetings around in order to come to talk to you. And it was very nice of you. Many of you wrote some very thoughtful things, which I'm going to pass onto him. I want him to know that many of you appreciated his visit. I also saw a lot of you reacted to his advice about crowded labs. That has been my experience, too. And one thing about the scientific process is it's not just one person. You're in with a group of people, just as Vernon described, and that group of people becomes the creative engine that drives all the science within that lab. And so you're not only picking your project, you're looking for a group of people to work with. And, as Vernon said, if the lab is really doing hot stuff they tend to attract a lot of people. So a crowded lab can sometimes be a really good indicator. No absolutes, and there's an exception to everything, but that was a good piece of advice he gave you if you're looking for UROPs sometimes. OK. So, anyway, DNA fidelity. Remember I said we've gone from, our bodies have somewhere from like 10 to 20 billion miles of DNA in them if we could take all the human DNA and stretch it out? But that fidelity is done at an error rate of about one mistake to every ten to the minus tenth nucleotides replicated. Which I said if you were typing all the time it would be like sort of making one mistake every 38 years. So it's an astonishing degree of fidelity. Something that's beyond anything within our experience. And there are three principles that go. One is polymerase is really good at the base pair recognition telling that an A is paired with a T or a G is paired with a C. And discriminating against everything else there's a phenomenon known as proofreading, and I'll tell you how that works. And then there's a third system called mismatch repair. And all three of these contribute to this very, very low-frequency of errors, one mistake for approximately every ten to the tenth nucleotides replicated. So the first thing is I've pointed out to you several times that if you draw the hydrogen bonds between an A and a T base pair, the two hydrogen bonds or the three hydrogen bonds between a G and C base pair, that the shapes of this pair and that pair are virtually identical. You can pick them up and lay it right down on top. Now, if you actually look at it you'll see you could draw some base pairs between, for example, a G and a T. In fact, you can draw two hydrogen bonds, which is the same as between an A and a T. But the one thing I hope you can see, just from the shapes even without being able to see the individual atoms, is that a GT base pair doesn't have the same shape as the correct base pairs. So when I showed you that little movie the other day where this is the template nucleotide, this is the incoming nucleotide and there's this alpha helix that's swinging up. What's happening in there is that the enzyme is checking the way that the incoming nucleotide is the correct shape to go with the base pair. And you can sort of see it's flipping it right into a very narrow little slot in the enzyme. So it's not only asking for sort of hydrogen bonds, it's asking for the exact shape. If you just did it by thermodynamic grounds you'd make about one mistake in a hundred because that's about the discrimination between the correct base pairs and some of these other ones. This works so well. You get more like one mistake in ten to the fourth or ten to the fifth. We're still quite a distance away from the ten to the tenth, but this is one of the things. It's looking for the correct shape of the base pair. Now, the second thing that helps with fidelity is a phenomena known as proofreading -- -- exonuclease. Things called a nuclease. That means it can degrade DNA. And the exo works at an end. And, furthermore, the directionality of this proofreading was something that puzzled people initially because it's going 3 prime to 5 prime. And when people started to purify DNA polymerases or complexes of DNA polymerases involved in replication there seemed to be a puzzle because the polymerase, as I've told you, goes 5 prime to 3 prime, but the same enzyme complex had an exonuclease that went in the opposite direction. So this seemed very peculiar at first in the sense if you were trying to polymerase DNA in this way why in that same enzyme would you have something that wanted to degrade DNA in the other way? And the answer turned out that this was known as a proofreading exonuclease, as I've put up here. And here's the principle of how it works. Suppose you were replicating the DNA and there was a G. And if you put a C in there it very quickly goes on and continues the replication. If it puts in a T, let's say, this is not a very good base pair. It wouldn't have the right shape. So when the enzyme came up looking for that 3 prime hydroxyl, which would be right at the end of that T, things are not in the right place. And so the polymerase activity slows down. And as that primer terminus, if it sits there for a little bit, it's able to just peel off the DNA, flip up, and there's this function that does just what you'd do if you were typing and you made a mistake. You'd just hit the delete key and take off the last nucleotide that you did. And I have a little movie showing you that. This is a crystal structure. This is the DNA template. And the polymerase catalytic activity site is right here. And in this little movie it's just added an incorrect base pair and the polymerase is sort of stalled. And the actual nuclease function is physically separate on the protein structure. But what you'll see in the movie is that if the polymerase cannot go very well eventually this thing will come up and it will chop off one nucleotide, come back and try it again. Let's see. I think if we do this, oopsy-daisey. Let me see if I can get this to work here. Nope, it's not working. OK. Well, anyway, I'm going to skip it for right now. I don't want to waste time. But, in any case, the end would go up here and it would take off one nucleotide. So there at least are two of the ways that the polymerase is able to work with such fidelity. It selects for the correct base pair shape. And then after it's done in addition it sort of looks back, just as if you were a very slow typist, and every time you typed a letter you looked back and said did I make a mistake? And if you made a mistake then you'd delete and then just try again. And that gets the cell another maybe two orders of magnitude of accuracy. So we're up to about one mistake in ten to the seventh base pairs replicated. The third system, which is called mismatched repair, turns out to be very important for a whole variety of reasons. And before I tell you about it, I want to first introduce the idea of DNA repair in general. One of the things that's wonderful about DNA -- -- as you've learned, is it's got the information in two copies. It's in a complimentary form but it's like having the photograph and the negative. And if your kid sister pokes a hole with a pair of scissors through the picture of your boyfriend or your girlfriend, you're not really in trouble as long as you've got the negative because you can get the information back again. And that same principle applies in DNA repair. So if you have some kind of lesion in DNA, and this might have come from going outside in the sunlight, your DNA absorbs in the UV and it undergoes photoreactions, they tend, for the most part, to just effect one of the two strands of DNA. Or if you smoke, which I hope none of you do, there are many chemicals in smoke that will react with DNA, and they'll modify one strand. And so what the cell has is a system that has many kinds of repair systems, but it has a special type of repair system known as nucleotide excision repair. And you could think of this as a protein machine that constantly scans the DNA looking for little distortions. And if it finds it then what it needs to do is it needs to make cuts, remove the DNA and make a little gap. And now you can see what it can do now. Once it's got a little gap the information over here is a complimentary form. So if a DNA polymerase were to come along it could fill in that gap and seal it up and then you'd be back to ordinary DNA, the lesion would be gone. And I made a silly little PowerPoint thing here to show it. So if you were to, say, damage the guanine with something, say one of the carcinogens you find in cigarette smoke, you could think of this protein machine as being a sort of pair of scissors that have a conditionality in them. As this protein machine scans along the DNA the scissors aren't activated until it recognizes there's a distortion here, at which point then it senses that there's some bump in the DNA. And it's very cleaver the way it does it because the nuclease activities, the things that are going to cut the DNA are actually some distance away, a few nucleotides away from the lesion. So even if this is distorting the DNA, the scissors are able to work out here and out here. It makes two cuts. That was a huge surprise. Nobody expected that when they started to do the biochemistry. And then in principle once you cut it now you can remove this little nucleotide and then a DNA polymerase can just come in, and following those A pairs with T, G pairs with C, copy it along and then would seal it up to get to the end. And I've actually shown you a picture of what happens if a human is missing that system. When I was showing you how profound an effect you could get from just losing one single gene or a mutation affecting one single gene, this disease called xeroderma pigmentosum. They're a variety of different groups. And the one on the left is an example. That's someone who is missing one of the genes that encodes one of the proteins involved in nucleotide excision repair. And this is really, really important for fixing up the damage we get all the time in sunlight. So if you miss that repair system and you got out in the sun then you get all kinds of lesions and people are very susceptible to skin cancer. And I told you fortunately now you don't find people with this disease looking like that because at least in developed countries we recognize it. They're kept out of the sun. And these were the kids who I said are called ìchildren of the moonî because they, for example, go to summer camps where they do everything at night so they won't get exposed to sunlight. But that's what happens to us if we miss that excision repair. And, again, what makes that possible is that the information is there twice in a double-stranded DNA. I also showed you a little movie early on when I was showing you, I'm going to actually run this in QuickTime because it works a little more smoothly, I think. So I showed you this when we were talking about DNA because I wanted you to sort of get that sense of what it was like to kind of fly down the groove of a DNA. But what I didn't emphasize was this protein that was bound to the DNA. That's a protein that's a DNA repair protein. And it's one of these things that looks for lesions in the DNA. And as we fly along the major groove this little green thing is actually the lesion that that protein is looking for. And it sort of puts fingers down into the groove and it's able to sense that. And you can sort of see how this protein is bound to DNA. This is a lesion that we get all the time from oxidative damage. And remember I said oxygen is bad for DNA? So our bodies have to have systems that are able to do that. So DNA repair is very important for life. We'll just finish flying down the major groove one more time here. OK. I'm going to go back to PowerPoint. OK. So mismatched repair is a form of repair that's got that same idea. Let's think about it if we had a replication fork here, and let's say there was a G here and the T got misincorporated, but in this case it wasn't removed by the proofreading which happens about one in ten to the seventh times. Now if that strand is fixed up, excuse me, is continued then you'd end up with a GT base pair. And the next time you copied it this strand would give rise to a GC but this one would give rise to an AT. And then you'd have a mutation that now would have changed. And if it affected an important gene that could be bad for you. So the cell has what's known as a mismatch repair -- -- that works in exactly the same logic as here. That it basically comes along. It scans the DNA. It finds the bump because this is not a proper base pair. And then it fills it in and you're back to ordinary DNA with a GC base pair. There's one little wrinkle. For this system to work it has to do one other thing that's different from that kind of DNA repair. Can anybody see what it is? Why don't you talk to the person next to you and see if you can figure it out. This system must be doing something else in order for this to work. OK, you can ask somebody. What do you think? What if I removed the gene? Would that work? What would happen if I took the gene instead? Say I made the little gap over on this strand instead, cut it here? Yeah. So which one is the one that's right, the old strand or the new strand? The old strand, yeah. See, this is the old and this is the new. And the term that's usually used, it's known as the daughter strand, the new strand. So the other thing this system has to do is it not only has to be able to detect that there's an incorrect little base pair in there, but it also has to know which is the parental strand, the template strand, and which is the daughter strand, the newly synthesized strand. And this system makes the assumption that the strand that's old is the one that's correct and the mistake is on the new one. You guys see that? OK. So that gets another two or three orders of magnitude in accuracy and that's what brings it up. Now, the people who made this, who formulated this model for mismatch repair, complete with the feature that it needed to recognize the old and new strand, that's a bit of a trick, if you think about it because it's DNA on both sides. And there are several different ways used in nature, so I'm not going to go into it, but there's at least a couple of different ways of doing that trick. You could sort of see if you were the replication fork and you talked to that you could certainly, just from the geometry of that, if you wanted, you could probably keep track of who's old and new. E. coli has a very cute trick, but it's not universal so I won't go into it, but the people who did the seminal stuff, I had to just quickly show you a couple of pictures. When I showed you that picture of the DNA 50th, the guy sitting in the front row was Miroslav Radman who was one of the two people. He's a European scientist originally from Croatia. And he collaborated with someone you've heard about before, Matt Meselson, who was up at Harvard. And it was with the Meselson-Stahl experiment that showed the semi-conservative mechanism of DNA repair. This was a little reception. And Matt was talking to Alex Rich who's in the MIT Biology Department. And I was amused because remember how Vernon told you how Francis Crick would run up and down the stairs in the Cambridge lab and he was talking all the time? And I've heard Vernon say you could never really tell whether an idea came from Watson or Crick because they'd just talk, talk, talk all the time. So this was at sort of nice reception at the DNA 50th. And within a couple of minutes, I looked over and there were Miroslav Radman and Matt Meselson talk, talk, talk. They were in the corner drawing pictures on a board. I also showed you actually a picture of one of the genes that's involved in recognizing this mismatch, because there's a protein that recognizes that mismatch and it's given the name of mute S. And when I was showing you some proteins it had one that had a lot of alpha helices. This is actually a picture of mute S. It's a dimer. That's why some of it's green and some of it's blue. And this is DNA viewed end on and it's recognizing a GT mismatch in DNA in that picture. Now, this may sound very esoteric, you know, and obviously important for life and an important part of sort of understanding how life works if you're interesting in studying molecular biology. It may not seem to have very much connection to your real life. But, in fact, in this case mismatch repair does because it affects the frequency with which, if you lose it, then when you replicate your DNA you're going to make more mistakes. And I need to just give you a very quick introduction to cancer so you can see why this is important. Cancer comes from the fact that remember a human cell or a multi-cell like us that has many kinds of different cells starts out from one cell. And I talked about first you get the embryonic stem cells that can become anything. And the cells become successively more and more and more specialized as they go along. So ultimately a cell that's in your retina or in, say, the lining of your colon needs to know that's where it belongs. And it also needs to know that it cannot just keep replicating. So if this is actually showing a little picture of the lining of your intestine. And there's a single layer of cells right along the inside edge of your intestines. This is the cells through which all the nutrient exchange happens and everything else when your body extracts nutrients as food stuff passes through your intestine. And so what happens with cancer is a cell that's normally a part of your body has to obey a whole set of rules. And what you can think of when someone starts to develop cancer is that what started out as an ordinary cell undergoes some kind of successive changes in its DNA that gradually causes it to forget the rules that make it be part of an organized body system. So if we take a look here at all these different cells. But let's imagine just one of the gets a change that makes it forget to stop, or it should know to stop replicating when it touches its neighbors, but if a cell were to lose that control what would happen? Well, it would then begin to proliferate. And then what happens in cancer is the cell will, now there are more of them, and one cell with acquire an additional mutation that will lead to a further loss of growth control. You can see now the cells are starting to become sort of funny shapes. And then one of the cells in here will undergo yet another change. And right at this point, up until now, the cancer has, even though the cells are dividing and have lost some of their growth control they're still staying in the same place. So that would be sort of, you know, like a wart or something like that, or what you would hear as a benign tumor. You can go in surgically and take it away. But then the other thing that can happen is cells can forget where they're supposed to be in the body. And when that happens they say the cells metastasize and become metastatic or a malignant tumor. And what that means is the cell is beginning to, it's acquired yet another change that's made it forget which part of the body it's supposed to be in. And they've signified it here as being a change in this cell that then leads to, you can see here right now it's starting to invade into the whole intestine. Or if one of those cells comes off lose in your bloodstream it can land somewhere else in your body and then start to grow there. And that's what happens when somebody has metastatic cancer. You cannot really cure it because now there are cancer cells all over the body. And that usually is a very difficult situation to get any kind of cure on. So to put this in perspective, you needed to have a number of changes to go from an ordinary cell to a metastatic cancer cell. So each one of these changes there was some kind of change in the DNA. Either there was a mutation or maybe a chromosome was lost or something like this so that you need a series of successive genetic alterations. So there was a very key insight that a number of people had after we understood the mechanism of mismatch repair. Because some people realized that if a human cell had lost mismatch repair then the frequency of each one of these changes would go up. It wouldn't affect what the change was. It wouldn't actually have anything to do, if you lost mismatch repair it wouldn't affect directly the ability of this cell to stop dividing when it touches its neighbors. But it would increase the chances that a mutation somewhere would have that effect. And if every one of these steps goes now a hundred or a thousand times faster, you can see that if somebody loses mismatch repair in a cell then the chances of that cell coming into a cancer are very high. So there was a kind of human cancer, it's a susceptibility to colon cancer called hereditary nonpolyposis colon cancer. You don't need to remember the name. It's often abbreviated HNPCC for people who cannot remember the name. But it was a kind of susceptibility to cancer that ran in families. So it was thought to be genetically determined in some way. And one of the interesting things was a number of the people who had this disease would show a kind of instability of the genome if they looked in the tumors. They just looked at the DNA. It seemed to be undergoing changes at a much faster rate. And the insight that came out was that the people who had this disease had, for example, a mutation affecting what we can think of as a human homolog of mute S. And we'll talk about genetics of humans in a small number of weeks, but I think most of you know that for most genes, except for the genes associated with the sex chromosomes, you get one copy of a gene from mom and another copy of a gene from dad. So under most circumstances we would have two good copies of this gene encoding a human homolog of mute S. What does that human homolog of mute S do? The same thing as the bacteria. It recognizes a mismatch in DNA and fixes it up. So it turned out that what the people with this disease have is they have one of the genes. The gene they got from mom or the gene they got from dad is broken. So they're still OK. They have one copy of mismatch repair in every cell. But if a cell ever had lost that copy of the good version now that cell and all of its descendents would mutate at something like a hundred or a thousand times the normal probability. And so they would progress down this pathway. And so the polyposis means that if they look in the colons of people who have this disease they find lots and lots of little growths or polyps that are on their way to progressing down this disease. Even in these people it takes quite a while. And so once they knew that they were able to go in and through colonoscopies find these cancers and remove them. And most of you will not have that disease, but this is now a kind of cancer that's pretty much preventable as long as it gets detected. It can take in a normal person as long as 20 years or something for an initial cell that underwent this initial change to go all the way down to becoming metastatic. So when you get older, and this certainly applies to most of your parents or in this age group, you should have ask them if they've had a colonoscopy. It's not the world's most fun procedure because, you know, they stick a probe and look inside your intestine, but it isn't that bad. And what they do is if they see one of these little polyps they can catch it before it's progressed far enough to be metastatic. And then there's no problem. I had my first one done about, I don't know, three or four years ago and they found one. And they took it out and I'm fine. But if it had been left there and allowed to progress then some years down the line I would have gotten colon cancer. And I'm going to have to go back and get checked again in another year or two. But it is something that you should check with your parents because everybody should have a colonoscopy. My hope is by the time you guys reach an age when this comes they'll probably have some kind of little blood test or something where you won't have to go through this indignity. But right at the moment it's something everyone should do, I think. I just wanted to make one other comment about basic research because there's another thing here. Actually, my lab was the first lab to clone the mute S gene. We cloned it, we sequenced it, and we looked in the databases. And at that time in the late eighties there was nothing else that looked like it. I thought it would be like, there were some sort of similar mutants, and here's what it looked like. This is a culture of E. coli. And there are about ten to the ninth cells per mil. And we plated about ten to the ninth or ten to the eighth on a plate with a drug on it. And you can see they almost all died, but there were maybe three or four that survived. And then their descendents were able to grow up and form a colony. This is how we recognized something was defective and what we now know as mismatch repair. If you took this mutant of E. coli and plated it out, you'd see you got a lot more drug-resistant colonies. That's the difference that I was describing, the importance of mismatch repair. If you don't have mismatch repair you can see, you get a lot more mistakes that show up as mutants. So I was studying that. And we cloned the mute S and mute L genes which are another gene that's involved in this. Didn't see anything in the database, but there were very similar mutants in streptococcus pneumonia that people had isolated. Remember streptococcus pneumonia in the transformation experiments? So I thought, well, maybe these are the same genes on an evolutionary basis. So I phoned some labs, and I found one that was sequencing what turned out to be a homolog of mute S. We tried to publish our papers in a medium fancy journal because I thought this was a pretty cool result that two bacteria that were evolutionarily very diverged had this conserve mechanism for mismatch repair, but the reviewer said, you know, this is a pretty specialized topic, it's not of general interest, it should go in, the phrase they use is ìa more specialized journalî. So it was published in the Journal of Bacteriology which is a really wonderful journal, but it basically deals with bacteria. And about a week after that paper came out my phone rang and it was a guy from Emory. And he said, ìI work on mouse. We were sequencing a gene,î it doesn't matter what, ìand we sequenced in the wrong direction. And we seem to have something called mute S. Do you know anything about mute S? And a couple of days after that I got a phone call from somebody at NIH. And they said the same thing, ìWe were trying to sequence this gene in humans. We kind of sequenced in the wrong direction and found mute S. So within a week of the paper coming out I knew there were mouse and human homologs. And that led from these sorts of studies, which my first graduate student worked on, to the identification of the human homologs. And then not me but others made the connection between mismatch repair and cancer. But this is the way a lot of things happen with basic research. This doesn't look like anything that's very important. And it sure doesn't look like it's going to lead to an insight into cancer, but this is very much the way it goes. I've had this happen twice with another set of genes in my life that turned out to be important for cancer as well. And, as I said, what happens, if you lose mismatch repair, then all these alterations happen much more quickly and the cells can become cancerous. I've included a couple of outtakes because I actually made this slide with my son's pillowcase on our dining room counter. And our cats, who you saw at some point earlier in the year, thought this was the weirdest thing they had ever seen, when I brought these plates home. So, OK, anyway. All right. So one other thing to tell you about DNA replication before I move on, and that is -- -- the initiation of DNA replication. In E. coli there's one great big piece of DNA. And it's all one giant circular chromosome. And if you realize what I've told you about DNA replication, I've talked to you only about once you have a replication fork established how you keep it going. But, as you might guess, a really important point of biological control is the initiation of DNA replication. And so the way cells do that is they have a special sequence in their DNA. It's written just with Gs and Cs and As and Ts, but it's a word sort of written in a different language than the kind of genetic code we're going to be talking about in the next couple of lectures. And what it means is ìstart replication hereî. And so in E. coli these terms are called origin DNA replication. And, for example, in E. coli it's a stretch of DNA that's about 250 base pairs long. And it's got a sequence that lets proteins bind and they kind of are able to make a little bubble like this. And it's at the edges of this little bubble where it's able to start a replication fork. And one of the secrets to control of cell division is that cells are able then to control whether the protein that sees the origin is there or not. And it won't start a new round of replication unless everything is right. Then it can make the things that initiate a new round. And after that it finishes. Our eukaryotic cells with a lot more DNA use the same thing. The same idea, but there tend to be multiple origins. And you get a little bubble and another little one down here. And once you get the replication forks established then these kind of merge. And then eventually we end up with the two strands of DNA. But I just mention that in passing because it's an example of how even though the DNA is nothing but Gs and Cs and As and Ts, you can kind of write words in there that mean different things. Some of them on the genetic code tell you what the order of amino acids in the cell are, but everything else has to be encoded in the DNA, too. And here's a really nice example of how that works. Now, we're going to switch at this point from worrying about how DNA is replicated to how information is stored and interpreted. And there's a figure that most of you have probably seen, DNA goes to RNA goes to protein. This is the usual direction of information flow. The information for making proteins is encoded in the DNA, as we'll talk about in more detail, and an RNA copy of some piece of that, one gene's worth usually, gets made in RNA. And then that information in the RNA is used to direct the sequences of amino acids that appear in a protein. And this is a four letter alphabet, if you want, A, G, T and C. This is a four letter alphabet, A, G, U and C, where the uracil and the thiamine have the same base pairing capacity. And this is a 20 letter alphabet. All those 20 amino acids that you were looking at, at the chart over at the back of the exam. So from the point of view of information storage and information flow there are some interesting things that had to come up in order for the information to flow in that way. But before I do that I want to just get you to think about DNA as an information storage device. This is MIT. I'm almost sure in this room there are some people that are experts in high density information storage. And even if you're not most of us have now a lot of experience with it. Your computer can do gigabytes of information. Your iPod probably has a 40 megabyte hard drive in it or something like that. So you have some experience with high density information storage. So here's the question. How much DNA would it take to encode everybody who's alive on earth today, 6 billion and a bit people? And let's argue that all we need is a single cell's worth of DNA because everybody started out a single fertilized egg and went on. Yeah? OK. Enough DNA to fill one human being. Anybody else got any sense? All right. This is, I think, the most amazing demo. I did this when I was teaching for the first time. The amount of DNA it would take to encode everybody who's alive on earth, one cell of everybody who's alive on earth today is this little thing in here, which you probably cannot see even, but I took a picture of it. There are about six times ten to the minus twelfth grams of DNA in a human cell. And if you multiple that out by 6 billion people it comes out to 36 milligrams of DNA. And I weighed out 40 something milligrams of DNA. So there's actually more DNA there than you need to encode everybody who's alive on earth today. And I don't know how this hits you, but I've been working on DNA my entire life. And every time I do this, you know, I think I understand this molecule, but I don't really think I do at some more fundamental level. It's absolutely amazing how much information is stored in that molecule. So the one point I will, actually, I think it's close enough. Why don't we just call it a day, and I'll pick this stuff --
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https://ocw.mit.edu/courses/8-701-introduction-to-nuclear-and-particle-physics-fall-2020/8.701-fall-2020.zip
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PROFESSOR: Welcome back to 8.701. So in this lecture, we talk about the Higgs mechanism. As you might know, the Higgs boson was discovered in 2012 by the LHC Experiment, but the theoretical discovery of the Higgs boson happened much, much earlier than that did. In the mid 1960s, Peter Higgs and a few others proposed a mechanism which gives rise to masses of the gauge boson, the w and the z boson. And the Higgs boson, or the Higgs field then can also be used to give masses to the fermions. So let's have a look at this and start with a simple observation. When we have written down our Lagrangian for a simple spin one field, gauge field, like a photon, we find that we want to have local gauge invariant for this equation, which means that we can do a local gauge transformation of our fields, and the physics should be unchanged of this. So the physics, meaning that the description by the Lagrangian, should be invariant under this transformation. All right? The problem, however, is that, if you want to have a spin one gauge field which is massive, you have to have terms in your Lagrangian, like this one here, where you have a mass term for your fields. So in general, this is not possible without breaking gauge invariants, and this is a guiding principle of our Lagrangian theory. So this is a real bummer. So if you want a stopping point, so you have a beautiful theory which describes all the interactions, but one important characteristic, the masses of the particle is missing. But you are able to actually do this, not by adding specific mass term but by breaking the symmetry, by breaking the local gauge symmetry. There's various ways to do this, and one of the ways is to use spontaneous symmetry breaking. So what is spontaneous symmetry breaking? Imagine you have a symmetry of rotation, a symmetry like this pen here, and by applying some force on top, this pen would bend. And by bending this pen, it bends in one specific direction, that breaks the rotational symmetry. Another way to look at this is to just let this pen drop. Let it go to its ground state, lowest possible energy state, and it will land somewhere on the table and by doing this breaking the symmetry, and it does this spontaneously. Let's look at spontaneous symmetry breaking in a toy model first. So what we're going to do here just add a complex scalar field and the corresponding potential for this field. Potential is shown here, and this general potential can have multiple forms. The first form would just be this parabola here. This is a solution where mu square, this term, mu square, is greater than 0. In this term, there's this unique minimum. The minimum is here at 0, and because of that, the mass of this field would be equal to 0, and the mas of our gauge field would also be equal to 0. But what happens now if we have through this potential a breaking of the symmetry? So in this case here, the vacuum itself, the lowest energy state, breaks the symmetry. You go away from the 0 point, and you're breaking the symmetry. So this minimum is at v over square root 2. v is the vacuum expectation value of this field, and you can simply rewrite then this field itself by evolving it around its minimum. And so you find two fields here, this chi and this h. The h is already kind of pointing towards the Higgs boson, and the vacuum expectation value. Now, if you add this back into your Lagrangian-- and I'll do this again. This is shown here, but also on the next slide-- you can start to identify terms which look like mass terms for your particle. And the first one is here which can be identified as a mass term for our gauge field. The mass is e time v. e times v is the strength of the coupling of this gauge field e times the value of the vacuum expectation value. All right? So this is interesting. So we used this new scalar field to break spontaneously the symmetry, and then the mass term appears which is proportional to the strength of the coupling and the vacuum expectation value. So the mass is generated through the spontaneous symmetry breaking and the coupling to the field. You also find a mass term for the six field here, for this h field. This is not the Higgs boson. It's just a field which looks like it, and so this mass term is here. But remember that mu squared is less than 0, and then this chi, or the so-called Goldstone boson, its mass is 0. But then we have those terms left over here, which we cannot really interpret it very well. And it's possible to remove them by choosing a specific field. So we do a gauge transformation by just relabelling things, and then the new Lagrangian is independent of this field. Just as a reminder, Goldstone, the Goldstone boson, you find those Goldstone bosons in many places in physics. And Jeffrey Goldstone is a retired faculty at MIT, so you might, in the spring or next summer, you'll him walking across the corridors. So we find our new Lagrangian which has our mass terms here, which has a term for the Higgs field, and has our potential for the Higgs field. So this specific gauge we just decided to use, this so-called unitarity gauge, and it's important to note that the Lagrangian itself contains all physical particles. But this chi, the Goldstone boson, is gone, and the lingo we sometimes use here is that the Goldstone boson has been eaten by the physical bosons. And the way it has been eaten is through the longitudinal polarization of this boson. It's the equivalent of saying that it has acquired mass. So the pocket guys here for spontaneous symmetry breaking is such that spontaneous symmetry breaking of a u1 gauge symmetry by a non-zero vacuum expectation value of a complex scalar results in a massive gauge boson and one real massive scalar field. So we created mass, but as a side product, we also have an additional field. And that field itself has a mass term, so it's massive. The second scalar we had just disappeared. The Goldstone boson has been eaten by the longitudinal component of the gauge field itself. All right. That was a simplified toy model. Let's look at the standard model. So now, here, we have to generalize from u1 to su2 or su-n gauge groups. The scalar field is now an n-dimensional fundamental representation of that group for the standard model. That will be su2. The gauge fields are n squared minus one-dimensional joint representations, like our photon, for example, or our unmixed w boson field. And the Lagrangian looks very similar to the one we just before with our potential. Again, we have this mu squared term. We also have a lambda term here, and then we require local gauge invariants again. OK. So now, for the standard model, again, su2 cross u1 gauge groups. We introduce a complex field, complex six field in su2. It's a duplex, meaning that it has four components. It's complex and has two components, which there's a total of four components to it. So we already know that we want to use m square less than 0 for our potential to allow for spontaneous symmetry breaking to occur. The minimum is then at 1 over square root 2. 0 for the upper component, and v, the vacuum expectation, for the lower component. That's a choice already. Again, why mu square less than 0? Because if we would have chosen to use a positive value for mu square, we wouldn't have spontaneously broken the symmetry. OK? So we need to have potential which looks like this Mexican hat here. All right. So now, what happens now to our w and z bosons. We have discussed electroweak mixing already. Good. So that was the first step. Now, we will understand where the mass terms actually come from. So now, we just did the spontaneous symmetry breaking, and now we are looking at what happens now if we also couple the Higgs field to the bosons. So again, we write this this way, and then we just try to find terms. It's really like a mechanical writing of the individual terms, and you find again terms which have a vacuum expectation value here and the coupling here. And you find the coupling to the u1 term and the coupling to the su2 and the coupling to the u1 term representative to the coupling to the original photon field and our gauge field for su2. All right. The rest is rewriting and identifying terms. If you do this-- and this is like a couple of pages of writing, fine-- but if you do this, you find, again like before, that you find the first and second component of our su2 gauge field. It gives us the charge at the boson, the w plus and the w minus. And then the z boson and the photon mixtures of the third component and the field b. All right? So these are all physical fields, and then we try to identify the mass terms you find for the w. That the w mass is proportional or equal to the coupling strands of the su2 group times the vacuum expectation value over 2. And the mass of the z boson is given by both couplings times the square root of the sums of the square times v over 2. OK. If you're trying to look for a mass term for the photon, you find none, meaning that the photon is massless. And then we can look again at weak mixing angles, and they are now defined directly through the couplings in those two gauge groups. The masses of the w and the z bosons are related via this weak mixing angle, cosine theta w. All right. Those elements we already saw before. Now, we find that the masses of the gauge bosons are given by spontaneous symmetry breaking via the vacuum expectation value and the strength of the coupling of the gauge field to the Higgs field. All right. So in summary, we started with a complex scalar field. A representation of su2 is four degrees of freedom. The Higgs vacuum expectation value breaks the symmetry spontaneously. The w plus and w minus and the z boson require mass, and the three Goldstone bosons are each absorbed into the w's and the z bosons. We also find an additional scalar Higgs boson that remains. And so that was the understanding in the '60s and '70s, and the standard model was further developed. And then it took us all the way to 2012 to actually find this new scalar particle, the Higgs boson itself.
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https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/8.05-fall-2013.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: All right. So today, we'll continue our kind of review that included, of course, the last lecture, the variational principle that's supposed to be new stuff you didn't see in 804. And today, as we continue, we'll talk about position and momentum for about 30 minutes or 40 minutes, and then begin the study of spin. That will be spin-1/2 with a Stern-Gerlach experiment and the mathematics that comes out of it. Now, we will talk about the Stern-Gerlach experiment in quite some detail so that you can appreciate what was going on there. And then we will extract a few of the mathematical lessons that this experiment tells us about quantum mechanics. Immediately after that, which will be probably middle of next lecture, we will pivot. And as we learn this mathematics that the Stern-Gerlach experiment is telling us or asking us for, we will go in some detail on the necessary mathematics for quantum mechanics. We'll talk about vector spaces, linear operators, Hermitian operators, unitary operators, [INAUDIBLE], matrix representations, all kinds of things. That probably will be about two weeks, three lectures at least. So it will be a nice study. And in that way, people that don't have a background in linear algebra will feel more comfortable with what we're going to be doing. And I think even for the people that have a background in linear algebra, you will gain a new appreciation about the concepts that we meet here. So today, we begin, therefore, with position and momentum, and these are operators in quantum mechanics. And they have letters to denote them. x, we put a hat with it, that's a position operator. p, we put a hat on it. And the position and momentum operators don't commute. And the commutator is given by ih bar. Now, we have been dealing so far with wave functions. Our wave functions, where these functions of x and t, they represent the dynamics of your system, the dynamics of your particle as it moves in time. But time, as you are seeing in quantum mechanics, is a little bit of a spectator. It's an arena where things happen. But the operators, and most of the interesting things, are going on without reference to time. Time evolution, you have an expansion of a wave function in terms of energy, eigenstates, at a given time. And then you can evolve it easily with the way we've learned, adding e to the minus i et over h bar for each energy eigenstate. So time will play no role here. So when I talk about the wave function, at this moment you could put the time, but we will talk about the wave functions that have no time dependence. So, say, a psi of x wave function. So this psi of x may be the true wave function at time equals 0, or you could just simply think of it as the psi of x. Now, this wave function means that we're treating x in a particular way, and we say that we're working in the x representation, the position representation. Now, this means that we have an easy way to figure out what this operator does when it acts on this function. So what it acts on this function, it will give you another function, and the definition of this is that the position operator acting on the function psi of x is defined to be another function, which is the function x times psi of x. Well, we're talking about these wave functions and operators on wave functions. And a recurrent theme in quantum mechanics is that we will think of wave functions, sometimes we call them states. Sometimes we call them vectors. And we basically think of wave functions as vectors. And things that act on wave functions are the things that act on vectors. And the things that act on vectors, as you know in mathematics, is matrices. So we're compelled, even at this early stage, to get a picture of how that language would go if we're talking about these things. So how do we think of a wave function as a vector? And how do we think of x as a matrix? So there's a way to do that. It will not be totally precise, but it's clear enough. So suppose you have a wave function, and we're interested in its values from 0 up to a. This wave function is a function of x between 0 and a. So it's the psi of x for x between a and 0. That's all the information. What we're going to do is we're going to divide this thing, this line, this segment, into a lot of pieces. And we're going to say, look, instead of writing a function like sine of x or cosine of x, let's just give the values and organize them as if this will be a vector of many components. So let's divide this in sizes epsilon, such that N times epsilon is equal to a. So there are N of these intervals. So we think of psi as a vector whose first component is psi at 0. The second is psi at epsilon. The third is psi at 2 epsilon. And the last one is psi at N epsilon. And depending on how much accuracy you want to work with, you take epsilon smaller and larger, keeping a constant. And this would be like summarizing all the information of a function in a vector. Now, that's intuitively a nice way to think of it. May look, with your background in classical physics, a little strange that we sort of put the value at 0 along the x-axis, first component, the value at epsilon along the y, the value of 2 epsilon along the z. But we need more axes. So you need many axes here. In this case, this is a N plus 1 column vector. It has N plus 1 entries, because 0 up to N, that's N plus 1 entries. But that's a fine way of thinking of it. Not exact because we have an epsilon. In this way of thinking about the wave function, we can then ask, what does the matrix x hat look like? So x hat is an operator, and it acts this way. So here is how it looks like. We would think of x hat as an N plus 1 times N plus 1 matrix. And its entries are 0 everywhere, except in the diagonal, where they are 0 epsilon, 2 epsilon, up to N epsilon. And here is a big 0 and a big 0. This, I claim, is the way you should think of the x operator if you thought of the wave function the way we wrote it. And how do we check that? Well, x operator acting on psi should be this acting on that. And then, indeed, we see that if x hat is acting on psi of x, what do we get? Well, it's easy to multiply a diagonal matrix times a vector. Here you get 0 times psi of 0. You get a vector, so let me make this thinner. Then I get epsilon times psi of epsilon, 2 epsilon times psi of 2 epsilon, up to N epsilon times psi of N epsilon. And indeed, that matrix looks like the matrix associated with this wave function because here is the value at 0 of this wave function. Here is the value at epsilon of this wave function, and so on. So this has worked out all right. We can think of the wave function as a column vector, and then the position operator as this vector as well. Now, given that we know how the x operator is defined, we can also think easily about what is the expectation value of x on a wave function. Something that you really know, but now maybe becomes a little clearer. Here you're supposed to do psi star of x times the x operator acting on psi of x. But we have the definition of this, so this is, as you imagine, dx-- and I should put primes maybe, well, I don't have to put primes-- dx psi star of x x psi of x, which is what you would have done anyway. Well, given that we've started with this, we can ask also, is there eigenstates of the x operator? Yes, there are. but then fortunately, are a bit singular. So what should be an eigenstate of x? It's some sort of state. Intuitively, it has a definite value of the position. So it just exists for some value of x. So it's naturally thought as a delta function. So let me define a function, psi sub x0 of x. So it's a function of x labeled by x0, and define it to be delta of x minus x0. So I claim that is an eigenstate of x hat. x hat on psi x0 of x is equal, by definition, to x times psi x0 of x, which is x times delta of x minus x0. And when you multiply a function of x times a delta function in x, it is possible to evaluate the function that is being multiplied by the delta function at the place where the delta function fires. It has the same effect on integrals or anything that you would do. So here, this is equal to x0 times delta x minus x0. You evaluate the x at x0. And finally, this is x0 times that function psi x0 of x. And therefore, you've shown that this operator acting on this function reproduces the function-- that's the definition of eigenstate as an operator-- and the eigenvalue is the number that appears here, and it's x0. So this function is an eigenstate of x hat with eigenvalue, e.v., x0. The only complication with this eigenfunction is that it's not normalizable. So it doesn't represent the particle. It can be used to represent the particle, but it's a useful function. You can think of it as something that can help you do physics, and don't insist that it represents a particle. So this is the story for position. And the position gets actually more interesting as soon as you introduce the dual quantity, momentum. So what is momentum here? So momentum is an operator, and this operator must be defined. Now, you had a shorthand for it in 804, which is p hat equal h bar over i d dx. And this shorthand means actually that, in what we call the position representation where we're using wave functions that depend on x, well, the momentum is given by this operator. And the story of why this was the case was sort of something that was elaborated on in 804, the work of de Broglie, that saw that the wavelength of the wave has to do with the momentum of a wave. And finally, people understood that this would measure the momentum of the wave. So this is the operator. And therefore, in the representation that we're working-- representation is a word that has a lot of precise meaning, but now I'm just using it in the sense that, well, we're working either with x's or with p's. And we're working with x's. That's why p looks like something to do with x. So what is p hat on a wave function? Well, that's what this means. It's another wave function obtained by taking the x derivative. So that's the definition of it acting on a wave function. The one thing that must be verified, of course, is that this definition is consistent or implies this commutation relation. So you've defined it as an operator. x, we've defined it as an operator. But most of us think that it doesn't look like an operator is multiplying. But it is an operator. So this one does look like an operator. It's a differential operator. And you can try to see if this equation is true. And the way to test these commutators is something that, again, I don't think is unfamiliar to you, but let's go through it, is that you try to evaluate this product of operators acting on a wave function. And if things work out well, we'll see you should get ih bar times that wave function. If that is the case, you say, OK, I've proven that equation, because it's an operator equation. The left-hand side of that equation is the product in different orders of two operators, therefore it's an operator. The right-hand side is another operator. It's the operator multiplied by ih, anything that you'll get. Well, if this is an operator identity, the operator on the left must be equal to the operator on the right, which just means that, acting on anything, they must give the same answer. So if I managed to prove that this is equal to this, I've proven that for anything that is the answer. And therefore, I can write the top one. And let me just do it, even though this may be kind of familiar to many of you. It's good to do this slowly once in your life. So let's go through this. So this says x operator p operator on psi minus p operator of x operator on psi. When you have several operators, like ABC acting on psi, this really means let C act on psi, and then let B act on C psi, and then let A act on that. The operators act one by one. The closest one acts first. So here I'm supposed to let B act on psi, but that means that thing. So now x is acting on h over i d psi dx. On this one, I have p acting on x psi, because that's what x hat psi is. Here, this is multiplication by x of a function of x. So this is just h over i x d psi dx. And here, I have h over i d dx of this whole thing x psi. And you can see that when you act here, you act first on the x, and you get something. And then you act on the psi, and you get this same term. So the only contribution here is equal to minus h over i, the d dx on x times psi, which is ih bar psi, which is what I wanted to show. So this is true. And therefore, you could say that this definition is consistent with your definition of x, and they represent this operator. One more thing you could try to do, and it's fun to do it, is we had a matrix representation for x. Can I think of p as a matrix? How would you do it? What kind of matrix would p look like? Well, yes? AUDIENCE: You just generate a finite difference equation. PROFESSOR: You could do it, exactly, with taking finite differences. So for example, if you think that you want to produce the wave function psi prime at 0, psi prime at epsilon, psi prime, that's what the derivative gives you, you'll write this as 1 over epsilon, say, psi at epsilon minus psi at 0. That's the derivative at 0 roughly. It would be psi at 2 epsilon minus psi at 0 over 2. And you could build it. You could build it. I'm not going to do it. You may want to do it and try and see how the derivative operator looks as a matrix. And then if you really want to spend some time thinking about it, you could try to see if this matrix and this matrix commute to give the right answer. And as you try it, you will figure out all kinds of funny things that we will talk about later in the course. So you can represent the momentum operator as a matrix indeed, and there are interesting things to say about it, and it's a good subject. So let's continue with the momentum and ask for eigenstates of the momentum. So eigenstates of p, you know them. They're e to the ipx things. So let's write them with some convenient normalization. This is an [INAUDIBLE] wave function that depends on x with momentum p. And we'll write it, as a definition, as e to the ipx over h bar, and I'll put it a 2 pi h bar here. It's kind of a useful normalization. Then p hat on psi p of x, well, p hat is supposed to take h over i d dx, and take h over i d dx of psi p. And h over i cancels the i over h. When you take the d dx, you get p out, and you get the same wave function. So indeed, you get p times psi p of x. So indeed, this is the eigenstate of the momentum operator, and it has momentum p. Well, what is the use of this? Well, say you have a representation, what we call the position representation, of the wave function and operators. Let us think now of the momentum representation. So what does all that mean? Well, there is the Fourier transform operation in which we have psi of p. Well, let me write it this way, actually. I'll write any psi of x physically can be represented as a sum of momentum eigenstates. Therefore, that's Fourier's theorem, minus infinity to infinity dp e to the ipx over h bar square root of 2 pi h psi tilde of p. That's Fourier transformation, defines psi tilde of p. And Fourier's theorem is the fact that not only you can do that, but you can invert it so that psi tilde of p can also be written as an integral, this time over x from minus infinity to infinity e to the minus ipx over h bar, also 2 pi h bar psi of x. So let's ponder this equation for a couple of minutes. Well, as a physicist, you think of this, well, this is telling me that any wave function could be written as a superposition of momentum eigenstates. Here are the momentum eigenstates. And for each value of momentum, you have some coefficient here that tells me how much of that momentum eigenstate I have. Now, here is the opposite one. Psi tilde of p and psi of x are related in this way. So these coefficients, if you want to calculate them, you calculate them this way. But now let's think of it as a change of representation. The physics is contained in psi of x. All what you wish to know about this physical system in quantum mechanics is there in psi of x. But it's also there in psi of p, because they contain the same information. So there are different ways of encoding the same information. What is the relation between them? This, we thought of it as a vector, vector in position space, an infinite dimensional space that is talking about positions. This is another vector in momentum space. Think of it now the infinite line. So this is an infinite vector with all those points little by little, from minus infinity to plus infinity, all of them there, gigantic vector. And here is another gigantic vector with p from minus infinity to infinity. And in between, there's an integral. But now, with your picture of quantum mechanics, you see an integral, but you also see a matrix. And what is this matrix? Think of this as some sort of psi sub p. And this as some sort of matrix, px psi x, in which if you have a product-- you'll remember when you multiply matrices, a matrix on a vector, you sum over the second index. That's the product for matrix. And then the first index is the index here. So here is what it, more or less, is like. Psi tilde of p [? subtend ?] by this, and this matrix depends on two labels, p and x, and it's that. So it's a matrix full of phases. So how do you pass from the coordinate representation of the information, a vector of all values of the wave function in all positions? By multiplying with this matrix of phases that is here, and it gives you this representation. So different representations means using different vectors to represent the physics. And this vector is a very nice one. And because of these properties of the momentum operator and all these things, this vector is also a very nice one. And there's an integral transform or some sort of infinite matrix product that relates them. And we shouldn't be uncomfortable about it. That's all fine. So we say that we have, for example, psi of x as one representation of the state and psi tilde of p as another representation of the same physics. We can do one more thing here, If I continue. We can take that boxed equation on the blackboard up there and act with h bar over i d dx on psi of x. So that is equal to h i d dx, and I'll write what psi of x is, is minus infinity to infinity dp e to the ipx over h bar square root of 2 pi h bar psi tilde of p. Now, when we act on this, as you know, h bar over i d dx just acts on this and produces the factor of p. So this is equal to minus infinity to infinity dp e to the ipx over h bar over square root of 2 pi h bar p times psi tilde of p. So look at this equation again. This double arrow is to mean that there are equivalent physics in them. They have the same information. It's the same data encoded in a different way. And that different way, this arrow is Fourier transformation. And this Fourier transformation is explained here. So now you have Fourier transformation the same way. So here we have-- what we've learned is that h over i d dx of psi is represented in momentum space by p psi tilde of p. And this was p hat acting on psi of x. So the corresponding thing in momentum space of p hat acting on psi of x is p multiplying psi tilde of p, which is to say that we can think of the abstract operator p hat acting on psi tilde of p as just p psi tilde of p. So in momentum space, the operator p hat acts in a very easy way. In coordinate space, it takes derivatives. In momentum space, it's multiplicative. So in position space, x is multiplicative. But in momentum space, x would not be multiplicative. x would also be a derivative. So I leave it for you as an exercise to show that or convince yourself in several ways, that x hat is really i h bar d dp in p space, in i h bar d dp. All right. So that's really all I wanted to say about position and momentum operators at this moment. They will come back when we'll introduce bra-ket notation in detail. We'll revisit this a little. But the main concepts have been illustrated. Are there questions? We're about to leave this, so if you have any questions at this moment. Yes? AUDIENCE: Could you explain again how you used this [INAUDIBLE] h bar over i d dx assign to [INAUDIBLE]? PROFESSOR: Right. So the question was, why did I associate these things? So it really goes back here to what the meaning of this arrow is. The meaning of this arrow is Fourier transformation. So this psi tilde and psi of x are related in this way. That's Fourier transformation, and that's what we mean by this arrow. It also means that whatever physics you have here, you have it there. So really, when you have something acting on a state, for example, if you have some operator acting in here, well, you get a new wave function. And there should be one on the right that corresponds to it, that has the same information as the one in which you've acted with something. So what we claim here is that, also in the sense of Fourier transformation or having the same information, h bar over i, the derivative of psi, is encoded by this. So we say, thinking abstractly, what is this? This is the momentum operator. Therefore, I'm going to say that the momentum operator really is the same momentum operator, whether it acts on wave functions that you show them to mean this way or wave functions that, because you're in another mood, you decide to give them to me in momentum space. So as you change your mood, the operator takes different forms but is doing the same thing. It's totally reversible. It's acting on that-- you see, the operator is always the same, but you give me the data in two different ways, then the operator has to do the thing in a different way. So that's what it means that the operator has different representations. In the [INAUDIBLE] representation, it looks like a derivative. In the momentum representation, it looks like multiplying. Other questions? Yes? AUDIENCE: So by saying that they sort of represent [INAUDIBLE] to the same positions, does that mean that h bar over i p e to the xi and p psi p are like the same [INAUDIBLE]? PROFESSOR: That h bar over d dx psi and p-- yeah. They are the same data, the same state represented in different ways. Yeah. All right. So time for a change. We're going to talk about Stern-Gerlach and spin. Now, spin will keep us busy the biggest chunk of this semester. So it will be spin-1/2, and we're really going to go into enormous detail on it. So this is just the beginning of the story that will be elaborated at various stages. So at this moment, I will talk about this experiment that led to the discovery of spin, and if you try to invent the theory that describes this experiment, what you would possibly begin doing. And then we go through the mathematics, as I mentioned to you, for maybe a week and a half or two weeks, and then return to the spin with more tools to understand it well. So the subject is the Stern-Gerlach experiment, Stern-Gerlach experiment. So the Stern-Gerlach experiment was done in Frankfurt, 1922. It was an experiment that, in fact, people were extraordinarily confused. It was not clear why they were doing it. And for quite a while, people didn't understand what they were getting, what was happening with it. In fact, Pauli had thought that the electron has like two degrees of freedom and didn't know what it was, those two degrees of freedom. Kronig suggested that it had to do somehow with the rotation of the electron. Now, Pauli said that's nonsense. How can an electron rotate and have angular momentum because it has a rotation? It would have to rotate so fast, even faster than the speed of light to have the angular momentum, and then this little ball that would be the electron would disintegrate. And it made no sense to him that there would be such a thing. So Kronig didn't publish this. Then there were another two people, Uhlenbeck and Goudsmit, at the same time, around 1925, had the same idea, angular momentum of this particle. And their advisor was Ehrenfest, and said it doesn't make too much sense, but you should publish it. [LAUGHTER] And thanks to their publishing, they are given credit for discovering the spin of the electron. And Pauli, a couple of years later, decided, after all, I was wrong. Yes, it is spin, and it's all working out. And 1927, five years after the experiment basically, people understood what was going on. So what were these people trying to do? First, Stern and Gerlach were atomic physicists, and they were just interested in measuring speeds of thermal motion of ions. So they would send beams of these ions and put magnetic fields and deflect them and measure their velocities. And eventually, they were experts doing this kind of thing. And they heard of Bohr, that said that the electron has angular momentum and is going around the proton in circles, so it might have angular momentum. They said, oh, if it has angular momentum because it's going around the proton, maybe we can detect it. And when they did the experiment, they got something. And they said, well, we're seeing it. But it was not that. They were not seeing the orbital angular momentum of the electron because that electron in these silver atoms actually has no angular momentum, as we will see, no orbital angular momentum. It only has spin. So they were actually seeing the spin. So it was a big confusion. It took some time. Basically, they took the beam, and they split it with a magnetic field, and the clean split was something nobody understood. So they called it space quantization, as of it's separated in space. Space is quantized. A pretty awful name, of course. There's nothing quantized about space here. But it reflects that when you don't know what's really happening, your names don't come out too well. So what we have to understand here, our goal today is to just see what's happening in that experiment, quantify a bit the results, and then extract the quantum mechanical lessons from it. So let us begin with the important thing. You don't see the spin directly. What you see is magnetic moments. So what's that? So what are magnetic moments? Magnetic moments, mu, is the analog, the magnetic analog of an electric dipole. A mu is called a magnetic dipole. You say it has a magnetic moment. And the magnetic moment is given by I times the area. What does that mean? Well, a precise discussion would take some time. But roughly, you can simplify when you think of a loop that is in a plane, in which case there's an area associated to it. And if the loop is this one, the area is defined as the normal vector to the oriented loop. So an oriented loop has an area vector. And the orientation could be focused the direction of the current. There is some area. And the magnetic moment is given by this thing. It points up in the circumstances when this current goes like that. So that's a magnetic moment. A little bit of units. The way units work out is that mu B-- magnetic moments and magnetic fields have units of energy. So magnetic moments you could define as energy, which is joules, divided by tesla, or ergs divided by gauss, because mu B has units of energy. So how do magnetic moments originate in a charge configuration? Well, you can simply have a little current like that. But let's consider a different situation in which you have a ring of charge, a ring of charge of some radius R. It has a total charge Q, and it has a linear charge density lambda. It's uniform, and it's rotating with some velocity v. If you wish, it also has a mass M. There are all kinds of [? parameters. ?] How many? Mass, charge, radius, and velocity. Here we go. We have our solid ring of charge rotating, and we want to figure out something quite fundamental, which is the origin of this principle. We said, you really never see spins directly. You never see this intrinsic angular momentum directly. You see magnetic moments. But then actually what happens is that there's a universal relation between magnetic moments and angular momentum. This is a key concept in physics. Maybe you've seen it before. Maybe you haven't. Probably you might have seen that in 802. So how does that go? Let's calculate the magnetic moment. So the current is the linear charge density times the velocity. The linear charge density is the total charge divided by 2 pi R times the velocity. Now the area, to give the magnetic moment, we'll have mu is equal to I times the area. So it would be this Q times 2 pi R v times the area, which would be pi R squared. So the pi's cancel, and we get 1/2 QvR. OK. 1/2 QvR, and that's fine and interesting. But OK, depends on the radius, depends on the velocity. So here is the magnetic moment is supposed to be going up. But what else is going up? The angular momentum of this thing is also going up. So what is the magnitude of the angular momentum L? L is angular momentum. Well, it's the mass times the momentum-- it's the mass momentum cross R, so MvR. The momentum of R cross p for each piece, contributes the same, so you just take the total momentum. This really is 0, but add them up little by little, and you've got your MvR. So here you have vR, so here you put 1/2 Q over M MvR. And you discover that mu is equal to 1/2 Q over M L. So maybe write it better-- Q over 2M L. I'm sorry, this is the normal. The M shouldn't change, M. And I box this relation because an interesting thing has happened. All kinds of incidentals have dropped out. Like the velocity has dropped out. The radius has dropped out as well. So if I have one ring with this radius and another ring with a bigger radius, the relation between mu and L is the same, as long as it's rotating with the same speed. So this is actually a universal relation. It is not just true for a little ring. It's true for a solid sphere or any solid object axially symmetric. It would be true. You could consider any object that is axially symmetric, and then you start considering all the little rings that can be built. And for every ring, mu over L is the same, and they all point in the same direction. Therefore, it's true under very general grounds. And that is a very famous relation. So now you could speculate that, indeed, the reason that a particle may have a magnetic moment if it's made by a little ball of charge that is rotating. But that was exactly what Pauli didn't like, of course. And you would like to see what's really happening with particles. So when you think of a true quantum mechanical particle, let's think of a particle in general, a solid particle rotating. We'll change the name to S for spin angular momentum. Because that little part, this is just one particle. We're not thinking of that little particle going around a nucleus. We're thinking of that little particle rotating. So this is a little piece of that little particle that is rotating. So you could ask, if, for the electron, for example, is it true that mu is equal to e over 2 mass of the electron times its spin? So this would be a vindication of this classical analysis. It might be that it's related in this way. So actually, it's not quite true, but let's still improve this a little bit. In terms of units, we like to put an h bar here and a 2Me. And put spin here, angular momentum, divided by h. Because this has no units, h bar has the units of angular momentum, x times p. It's the same units, so units of angular momentum. So h bar would be convenient. So that over here, you would have units of a dipole moment, or magnetic moment, magnetic moment units. So what does happen for the electron? Well, it's almost true, but not quite. In fact, what you get is that you need a fudge factor. The fudge factor is that, actually, for elementary particles, you have a g, which is a constant, which is the fudge factor, e h bar 2 over M of the particle S over h bar. Sometimes called the Lande factor. You must put a number there. Now, the good thing is that the number sometimes can be calculated and predicted. So when people did this, they figured out that for the electron the number is actually a 2. So for the electron, g of the electron is equal to 2. Now that, you would say, cannot be an accident. It's twice what you would predict sort of classically. And the Dirac equation, the relativistic equation of the electron that you have not studied yet but you will study soon, predicts this g equal to 2. It was considered a great success that that equation gave the right answer, that people understood that this number was going to be 2. So for the electron, this is 2. So this quantity is called-- it's a magnetic dipole moment-- is called mu B for Bohr magneton. So how big is a mu B? It's about 9.3 times 10 to the minus 24 joules per tesla. AUDIENCE: Professor. PROFESSOR: Yes? AUDIENCE: [INAUDIBLE]. So where exactly does the fudge factor come in? Is it just merely because [INAUDIBLE]? PROFESSOR: Right. So the classical analysis is not valid. So it's pretty invalid, in fact. You see, the picture of an electron, as of today, is that it's a point particle. And a point particle literally means no size. The electron is not a little ball of charge. Otherwise, it would have parts. So an electron is a point particle. Therefore, a point particle cannot be rotating and have a spin. So how does the electron manage to have spin? That you can't answer in physics. It just has it. Just like a point particle that has no size can have mass. How do you have mass if you have no size? You get accustomed to the idea. The mathematics says it's possible. You don't run into trouble. So this particle has no size, but it has an angular spin, angular momentum, as if it would be rotating. But it's definitely not the case that it's rotating. And therefore, this 2 confirms that it was a pointless idea to believe that it would be true. Nevertheless, kind of unit analyses or maybe some truth to the fact that quantum mechanics changes classical mechanics. Turns out that it's closely related. For the proton, for example, the magnetic moment of the proton is quite complicated as well because the proton is made out of quarks that are rotating inside. And how do you get the spin of the proton and the magnetic moment of the proton? It's complicated. The neutron, that has no charge, has a magnetic moment, because somehow the quarks inside arrange in a way that their angular momentum doesn't quite cancel. So for example, the value for a neutron, I believe, is minus 2.78 or something like that. It's a strange number. Another thing that is sort of interesting that is also true is that this mass is the mass of a particle. So if you're talking about the magnetic moment of the proton or the neutron, it's suppressed with respect to the one of the electron. The electron one is much bigger because, actually, the mass shows up here. So for a neutron or a proton, the magnetic moment is much, much smaller. So, in fact, for an electron then, you would have the following. Mu is equal to minus g, which is 2, mu B S bar over h. And actually, we put the minus sign because the electron has negative charge. So the magnetic moment actually points opposite. If you rotate this way, the angular momentum is always up. But if you rotate this way and you're negative, it's as if the current goes in the other direction. So this is due to the fact that the electron is negatively charged. And that's the final expression. So OK, so that's the general story with magnetic moments. So the next thing is, how do magnetic moments react when you have magnetic fields? So that is something that you can calculate, or you can decide if you have a picture. For example, if you have a loop of charge like this, and you have magnetic field lines that go like this, they diverge a bit. Let me see you use your right-hand rule and tell me whether that loop of current will feel a force up or down. I'll give you 30 seconds, and I take a vote. Let's see how we're doing with that. And I'll prepare these blackboards in the meantime. All right. Who votes up? Nobody. Who votes down? Yeah, [INAUDIBLE]. Down, exactly. How do you see down? Well, one way to see this, look at the cross-section. You would have this wire here like that. The current is coming in on this side and going out this way. Here you have the field lines that go through those two edges, and the magnetic field is like that. And the force goes like I cross B. So I goes in, B goes out. The force must be like that, a little bit of force. In this one, I cross B would be like that, a little bit of force. Yep. Has a component down because the field lines are diverging. So what is the force really given by? The force is given by the gradient of mu dot B. This is derived in E&M. I will not derive it here. This is not really the point of this course. But you can see that it's consistent. This is saying that the force goes in the direction that makes mu dot B grow the fastest. Now mu, in this case, is up. So mu dot B is positive, because mu and the magnetic field go in the same direction. So mu dot b is positive. So the force will be towards the direction-- that's what the gradient is-- that this becomes bigger. So it becomes bigger here, because as the field lines come together, that means stronger magnetic field. And therefore, mu dot B would be larger, so it's pointing down. If you have a magnetic field that is roughly in the z direction, there will be a simplification, as we will see very soon. So what did Stern and Gerlach do? Well, they were working with silver atoms. And silver atoms have 47 electrons, out of which 46 fill up the levels and equal 1, 2, 3, and 4. Just one lone electron, a 5s electron, the 47th electron, it's a lonely electron that is out in a spherical shell, we know now with zero orbital angular momentum. It's an S state. And therefore, throwing silver atoms through your apparatus was pretty much the same thing as throwing electrons, because all these other electrons are tied up with each other. We know now one has spin up, one spin down. Nothing contributes, no angular momentum as a whole. And then you have this last electron unpaired. It has a spin. So it's like throwing spins. So moreover, throwing spins, as far as we're concerned, Stern and Gerlach wouldn't care. Because of these relations, it's throwing in dipole moments. And they would care about that because magnetic fields push dipole moments up or down. Therefore, what is the apparatus these people had? It was sort of like this, with an oven, and you produce some silver atoms that come out as a gas, a collimating slit. Then you put axes here-- we put axes just to know the components we're talking about. And then there's magnets, some sort of magnet like this, and the screen over there. So basically, this form of this magnet that I've tried to draw there, although it's not so easy, if I would take a cross-section it would look like this. So the magnetic field has a gradient. The lines bend a bit, so there's a gradient of the magnetic field. And it's mostly in the z direction, so z direction being pointed out here. So there's the magnetic field. The beam then comes here. And the question is, what do you get on this screen? Now, I have it a little too low. The beam comes there and goes through there. So the analysis that we would have to do is basically an analysis of the forces. And relatively, we don't care too much. The fact is that there's basically, because the magnetic field is mostly in the z direction and varies in z direction, there will be a force basically in the z direction. Why is that? Because you take this, and you say, well, that's roughly mu z Bz, because it's mostly a magnetic field in the z direction. And mu is a constant, so it's basically gradient of Bz. Now, that's a vector. But we're saying also most of the gradient of Bz is in the z direction, so it's basically dBz dz. Now, there is some bending of the lines, so there's a little bit of gradient in other directions. But people have gone through the analysis, and they don't matter for any calculation that you do. They actually average out. So roughly, this gradient is in the z direction. I'm sorry, the gradient is supposed to be a vector. So you get a force in the z direction. And therefore, the thing that people expected was the following. You know, here comes one atom, and it has its magnetic moment. Well, they've all been boiling in this oven for a while. They're very disordered. Some have a z component of magnetic-- the magnetic moment is pointing like that, so they have some component, some down. Some are here. They have no component. It's all Boltzmann distributed all over the directions. Therefore, you're going to get a smudge like this. Some ones are going to be deflected a lot because they have lots of z component of angular momentum or z magnetic moment. Others are going to be deflected little. So this was the classical expectation. And the shock was that you got, actually, one peak here, an empty space, and another peak there. That was called space quantization. Stern and Gerlach worked with a magnetic field that was of about 0.1 tesla, a tenth of a tesla. And in their experiment, the space quantization, this difference, was 1/5 of a millimeter. So not that big, but it was a clear thing. It was there. So everybody was confused. They thought it was the orbital spin, angular momentum that somehow had been measured. At the end of the day, that was wrong. It couldn't have been that. People understood the Bohr atom, realized, no, there's no angular momentum there. The idea of the spin came back, and you would have to do a calculation to determine what is the value of the spin. So the exact factor took a while to get it right. But with the idea that mu z is equal to minus 2 Bohr magenton Sz over h bar, which we wrote before. Well, mu z, if you know the strength of your magnetic field, you can calculate the deflections. You know what mu B is. So therefore, you get the value for Sz over h. And experiments suggested that Sz over h was either plus or minus 1/2. And this kind of particle, it has Sz over h bar equal plus or minus 1/2, is called the spin-1/2 particle. So again, from this equation, this can be measured. And you then use this, and you get this value. So the experiment is a little confusing. Why did this happen? And how do we think of it quantum mechanically? Now 804 sort of began with these kind of things. And you know by now that what's happening is the following, that somehow, mathematically, every state is a superposition of a spin up and a spin down. So every particle that goes there has half of its brain in the spin up and half of its brain in the spin down. And then as it goes through the magnetic field, this thing splits, but each particle is in both beams still. And they just have this dual existence until there's a screen and there's detectors. So they have to decide what happens, and then either collapses in the top beam or lower beam. Nothing happens until you put the screen. That's what we think now is the interpretation of this experiment. But let's use the last few minutes to just write this in terms of boxes and get the right ideas. So instead of drawing all that stuff, we'll draw a little box called a z hat box, a Stern-Gerlach apparatus. In comes a beam, out would come two beams, Sz equal h bar over 2 and Sz equal minus h bar over 2. And the convention is that the plus goes up and the minus goes down, which I think is probably consistent with that drawing. And that's the Stern-Gerlach apparatus. It measures Sz, and it splits the beam. Each particle goes into both beams until there's a device that measures and decides where you go. So you can do the following arrangements. So here's arrangement number 1, a Stern-Gerlach device with z. Then you block the lower one and let the top one go as Sz equal h bar over 2. And then you put another Stern-Gerlach machine, z hat, that has two outputs. And then you ask, what's going to happen? And the experiment can be done and, actually, there's nothing here coming out, and all the particles come out here with Sz equal h bar over 2. What are we going to learn from this? In our picture of quantum mechanics, we're going to think of this as there are states of the electron that have-- and I will write them with respect to z-- they have plus h bar over 2 and states that have minus h bar over 2. And what we will think is that these are really old basis states, that any other state, even one that points along x, is a superposition of those two. This is a very incredible physical assumption. It's saying this system is a 2-dimensional complex vector space, two vectors, two unit, two basis vectors. And from those two, all linear combinations that are infinite represent all possible spin configurations. And what is this saying? Well, as we will translate it into algebra, we will say that, look, here is a state plus. And when you try to measure, if it had any minus component, it had nothing. So we will state that as saying that these states are orthogonal. The minus state and the plus state have zero overlap. They are orthogonal basis states. And, for example, well, you could also do it this way. That would also be 0. And you could also say that z plus and z plus is 1, because every state that came in as a plus came out as a plus. They had perfect overlap. So these are two orthonormal basis vectors. That's what this seems to suggest. And it's a little strange, if you think, because there's a clash between arrows and the notion of orthonormality. In 3-dimensional vectors, you think of this vector being orthogonal to this. But you wouldn't think of this vector as being orthogonal to that one. And here is the spin is up, and this is the spin down. And those two are orthogonal. You say, no, they're anti-parallel. They're not orthogonal. No, they are orthogonal. And that's the endlessly confusing thing about spin-1/2. So these states, their pictures of the spins are arrows. But don't think that those arrows and the dot product give you the orthogonality, because this is up and down. If you would be doing the dot product of an up and down vector, you would not get 0. But this is 0. Then you do the following experiment. So let's do the next one. And the next one is, again, the z filter. Take this one, block it. Then you put an x filter. And what actually happens is that you would get states with Sx, now, h bar over 2 and Sx equal minus h bar over 2, because it's an x filter. The magnetic field is a line in the x direction. Now, all these things have Sz equal h bar over 2. And what happens in the experiment is that 50% of the particles come out here and 50% come out there. So a spin state along the x direction has some overlap with a spin state along the z direction. Normal vectors, a z vector and an x vector, are orthogonal. Not here for spins. The spin pointing in the z and the spin pointing in the x are not orthogonal states. They have overlaps. So this means that, for example, the x plus state and the z plus state have an overlap. This is notations that-- we're going to be precise later. But the same thing with the x minus state, it has an overlap, and somehow they're about the same. Finally, the last experiment is this, z hat, block again, x hat, but this time block one. So here is a state with Sx equals minus h bar over 2. Here is a state with Sz equal h bar over 2. And now you put the z machine again. And what happens? Well, there's two options. People who were inventing quantum mechanics no wonder thought about them. Here they could say, look, I filtered this thing, and now all these electrons have Sz equal h bar over 2. And now all these electrons have Sx equal minus h bar over 2. Maybe, actually, all these electrons have both Sz equal h over 2 and that because I filtered it twice. So it maybe satisfies both. So if all these electrons would have Sz equals h over 2 and this, then you would only get something from the top one. But no, that's not what happens. You get in both. So somehow, the memory of these states coming from Sz equals h over 2 has been destroyed by the time it turned into a state with Sx equal minus h over 2. And a state cannot have simultaneously this and that. That's two properties, because you get 50% here and 50% there. So we'll discuss next time a little more about these relations and how can the states be related, the ones that we use as the basis vectors and all the others along x and others that we could build some other way. All right. See you next week. There's office hours today, 5:00 to 6:00, Monday, 4:30 to 5:30.
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https://ocw.mit.edu/courses/8-422-atomic-and-optical-physics-ii-spring-2013/8.422-spring-2013.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation, or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Good afternoon. We continue our discussion of quantum states of light. We talked at length about coherent state, and when you talk about quantum states of light, each mode of the electromagnetic field is an harmonic oscillator. We also encountered, naturally, the number states. And we realized-- yesterday, actually, in the last class-- that those number states have non-classical properties. For instance, they have a g2 function, the second order correlation function, which is smaller than 1, which is impossible for classic light, as you're proving in one of your homework assignment. So at that point, we have encountered coherent states, which are as close as possible to classical states. And we have found the number states as non-classical states. Well, are there other interesting states? I wouldn't ask you this question if the answer would not be yes, and this is what we want to discuss today. We want to talk about non-classical states of light, which we can engineer, actually, in the laboratory, by sending laser light through nonlinear crystals. Those go by the name, squeezed states. Just to give you the cartoon picture, in our two-dimensional diagram, with the quasi-probabilities, we have coherent states, where the area of this disk, delta x delta p, is h-bar over two. It's uncertainty limited. What we can do we now, is-- we cannot go beyond this. This is the fundamental limit of quantum physics. However, we can take this circle and we can squeeze it. We can squeeze it horizontally, we can squeeze it into an elongated vertical shape, or we can squeeze it at any angle. That's what we call, squeezed states. And those states have non-classical properties. They are important for metrology they are important for teleportation. There are lots and lots of reasons why you want to know about them. But again, as so often, I feel I cannot convey to you the excitement of doing squeezing in the quantum domain. And many, many physicists now, they hear about squeezing just in the quantum domain. But I want to start with classical squeezing. I will actually show you video of an experiment on classical squeezing. You can see squeezing with your own eyes. But this is just sort of to set the stage, to also get a feel of what squeezing is. And then we'll do quantum mechanical squeezing. But maybe-- tongue in cheek-- I would say, since classical harmonic oscillators and quantum harmonic oscillators have a lot in common, the step from classical squeezing to quantum of mechanical squeezing is actually rather small. It's nice to squeeze light. It's nice to have those non-classical states. But the question is, how can you detect it? If you can't detect it, you can't take advantage of it. And the detection has to be face-coherent. I will tell you what that is. And it goes by the name, homodyne detection. And finally, we can take everything we have learned together, and discuss how, in the laboratory, teleportation of a quantum state is done. There is a nice teleportation scheme, and I want to use that as an example that the language and the concepts I've introduced are useful. Concepts like, squeezing operator, displacement operator-- those methods allowing us to, in a very clear way, discuss schemes which lead to teleportation. That's the menu for today. Let's start with classical squeezing. For squeezing, we need an harmonic oscillator, means for parabolic potential, we have potential v of x. And then we study the motion of-- that should be x squared-- the motion of a particle in there. Before I even get started any equation, let me explain what the effect of squeezing will be about. If you have an harmonic oscillator, you have, actually, the motion of a pendulum. It has two quadrature components, the cosine motion and the sine motion. And they are 90 degrees out of phase. What happens now is, if you parametrically drive the harmonic oscillator-- you modulate the harmonic oscillator potential-- it's to omega. I will show you mathematically, it's very, very easy to show, that depending on the phase of the drive, you will actually exponentially amplify the sine motion, and exponentially damp the cosine motion. Or if you change, vice versa. So by driving the system, you can amplify one quadrature component, and exponentially die out the other quadrature component. And that is called, classical squeezing. Let's do the math. It's very simple. Our equation of motion has the two solutions I've just mentioned. It has a solution with cosine omega 0, and one with sine omega 0 t. And we have two coefficients. The cosine is called, c. The sine coefficient is called, s. I have to call it c 0, because I want to call that c and s. So what we have here is, we have the two quadrature components of the motion in an harmonic oscillator. And graphically, we need that for the electromagnetic field, as well. When we have our two axes, like, you know, the complex plane for the cosine of probabilities, I call one the s-axis. One is the c-axis. That's just something which confuses me. If you have only one-- just give me one second. Cosine-- Yeah. If you have only cosine motion, the s component is 0, and the harmonic oscillator would just oscillate here. If you have only a sine component, you stay on the x-axis. And now, if you have an equal amount of cosine and sine, then you can describe the trajectory to go in a circle. OK. This is just the undriven harmonic oscillator. I don't want to dwell on it any longer. But what we are doing now is, we are adding a small parametric drive. Parametric drive means we modulate the spring constant, or we replace the original harmonic potential, which was this, by an extra modulation term. So we have a small parameter, epsilon. And as I pointed out, the modulation is at twice the resonance frequency. Now we want to solve the equation of motion for the harmonic oscillator, using this added potential. The way how we want to solve it is, we assume epsilon is very small. So if the pendulum is swinging with cosine omega t, it will take a while for the epsilon term-- for the small term-- to change the motion. So therefore, we assume that we can actually go back and use our original solution. And assume that over a short term, the epsilon term is not doing anything. So for a short time, it looks like an harmonic oscillator with a sine omega 0, and cosine omega 0 t oscillation. But over any longer period of time, the small term will have an effect. And therefore, the coefficients c of t, c, and s are no longer constant, but change as a function of time. We want to solve, now, the equation of motion. That means we use this, here, as our ansatz. And we calculate the second derivative. We assume that the coefficients c and s are changing slowly. Therefore, the second derivative of c and s can be neglected. By taking the derivative of the second derivative of the cosine term and the sine term, of course we simply get, minus omega 0 squared, x of t. And now we have the second-order derivatives. Since we neglect the second-order derivative of c and s, the other terms we get when we take the second derivative is, first derivative of c times first derivative of cosine. First derivative of s times first derivative of sine. So we get two more terms, which are, minus omega 0 c dot, times sine omega 0 t. Plus omega 0 s dot, times cosine omega 0 t. This is the second derivative of our ansatz for x. This has to be equal to the force provided by the potential. So taking the potential-- We need, now, the derivative of the potential, for the potential of use across this line. The first part is the unperturbed harmonic oscillator, which gives us simply, omega 0 squared times x. And the second term, due to the parametric drive, is 2 sine omega 0 t. And now, for x, we use our ansatz for x, which is the slowly-changing amplitude c times cosine omega 0 t, plus s times sine omega 0 t. Those two terms cancel out. So now we have products of trig function. Sine 2 omega times cosine omega. Well, you know if you take the product of two trig functions, it becomes a trig function of the sum or the difference of the argument. So if you take sine 2 omega 0 times cosine omega 0, and we use trigonometric identities, we get an oscillation at 3 omega 0, which is 2 plus 1. And one at the difference, which is omega 0. Let me write down the terms which are of interest to us. Namely, the ones at omega 0. So let me factor out epsilon omega 0 squared over 2. Then we have the term c times sine omega 0 t, plus s times cosine. And then we have terms at 3 omega 0, which we are going to neglect. Now we compare the two sides of the equations. We have sine omega 0 term. We have cosine omega 0 term. And the two sides of the equations are only consistent if the two coefficients of the sine term, and the sine term, are the same. So therefore, we obtain two equations. One for c dot, one for s dot. And these are first-order differential equations. The solution is clearly an exponential. But one has a plus sign, one has a minus sign. So the c component, the c quadrature component, is exponentially amplified with this time constant. Whereas the sine component is exponentially de-amplified. This finishes the mathematical discussion of classical squeezing. We find that s of t, and c of t, are exponential functions. In one case, it's exponentially increasing. In the other case, it is exponentially decreasing. And that means that, well, if we go to our diagram, here-- and let's assume we had an arbitrary superposition of cosine and sine amplitude. This is cosine. This is sine. We had sort of a cosine oscillation, and a sine oscillation. Which means that, as a phasor, the system was moving on an ellipse. If the sine component is exponentially de-amplified, and the cosine component is exponentially amplified, that means whatever we start with is squashed horizontally, is squashed vertically. And is amplified horizontally. In the end, it will become a narrow strip. So this is classical squeezing. You may want to ask, why did I neglect the 3 omega 0 term. Well, I have to, otherwise I don't have a solution. Because I have to be consistent with my approximations. So what I did here is, I had an equation where I have the clear vision that the solution has a slowly varying c and s coefficient. And then I simply use that. I take the second-order derivative, and I have only Fourier components with omega 0, the sine, and cosine. Now I've made an approximation, here. For the derivative of the potential, the first line is exact. But in order to match the approximation I've done on the other side, I can only focus on two Fourier components resonant with omega 0, which I have here. So in other words, the 3 omega 0 term would lead to additional accelerations. Which I have not included in the treatment. So it's consistent with the ansatz. It's consistent with the assumption that we have resonant oscillations with a slowly changing amplitude. There will be a small [INAUDIBLE] for your omega 0, but it will be small. Any questions about that? Let me then show you an animation of that. Classroom files. [VIDEO PLAYBACK] -We have Dave Pritchard, professor of physics at MIT, demonstrating what squeezing is. Right now, we see a wave that's going around in a circle. What's next? What's going to happen now, Professor Pritchard? -Well, if we drive it in twice the basic period, then we will parametrically amplify one quadrature component, and we will un-amplify the other one. So now I'm going to start doing that. And then you notice that its motion turns into an ellipse. We've amplified this quadrature component, but we've un-amplifed that one. And that's squeezing. [END VIDEO PLAYBACK] PROFESSOR: Feel free to try it at home. [LAUGHTER] PROFESSOR: Actually, you may start to think about this demonstration. What he has shown was, when you have a circular pendulum which goes in a circle or an ellipse, and you start pulling on the rope with a certain phase, that one quadrature component will be de-amplified. The other one will be amplified. And as a result, no matter what the circular or the elliptical motion was, after driving it for a while, it will only swing in one direction. And this is the collection you have amplified. There is one thing which should give you pause. I have discussed, actually, a single harmonic oscillator. What Dave Pritchard demonstrated was actually two harmonic oscillators. The harmonic oscillator has an x motion and a y motion. However, you can say, this was just sort of a trick for the demonstration, because when you have a circular motion, initially, you have the sine omega and the cosine omega 0 component present simultaneously. And you can see what happens to the sine and the cosine component in one experiment. So in that sense, he did two experiments at once. He showed what happens when you have, initially, a sine component, and what happens when you initially have a cosine component. OK. So we know what classical squeezing is. And what we have learned, also-- and this helps me now a lot to motivate how we squeeze in quantum mechanics-- you have realized that what is really essential here is, to drive it to omega 0. What we need now to do squeezing in the quantum domain, if we want to squeeze light, we need something at 2 omega 0. So let's now squeeze quantum mechanically. Go back here. The second sub-section is now, squeezed quantum states. What we want to discuss is, we want to discuss a quantum harmonic oscillator. We want to have some form of parametric drive at 2 omega 0. And this will result in squeezed states. Now, what does it require, if you want to bring in 2 omega 0? Well, let's not forget our harmonic oscillators are modes of the electromagnetic field. If you now want to couple a mode of the electromagnetic field, at 2 omega 0, with our harmonic oscillator at omega 0, we need a coupling between two electromagnetic fields. So therefore, we need nonlinear interactions between photons. So this was a tautology. We need nonlinear physics, which leads to interactions between photons. Linear physics means, each harmonic oscillator is independent. So we need some nonlinear process six which will be equivalent to have interactions between photons. The device which we will provide that is an optical parametric oscillator. I could spend a long time explaining to you how those nonlinear crystals work. What is the polarization, what is the polarizability, how do you drive it, what is the nonlinearity. But for the discussion in this class, which focuses on fundamental concepts, I can actually bypass it by just saying, assume you have a system-- and this is actually what the optical parametric oscillator does, is you pump it with photons at 2 omega 0. And then the crystal generates two photons at omega 0. Which of course, is consistent with energy conservation. And if you fulfill some phase-matching condition, it's also consistent with momentum conservation. But I don't want to go into phase-matching at this point. Technically, this is done as simple as that. You have to pick the right crystal. Actually, a crystal which does mixing between three photon fields cannot have inversion symmetry, otherwise this nonlinear term is 0. What you need is a special crystal. KDP is a common choice. And this crystal will now do for us the following. You shine in laser light. Let's say, at 532 nanometer, a green light. And then this photon breaks up into two photons of omega 0. This is how it's done in the laboratory. The piece of art is, you have to pick the right crystal. It has to be cut at the right angle. You may have to heat it, and make sure that you select the temperature for which some form of resonant condition is fulfilled, to do that. But in essence, that's what you do. One laser beam, put in a crystal, and then the photon is broken into two equal parts. And these are our two photons at omega 0. OK. I hope you enjoy the elegance-- we can completely bypass all the material physics by putting operators on it. We call this mode, b. And we call this mode, a. So the whole parametric process, the down conversion process of one photon into two, is now described by the following Hamiltonian. We destroy a photon in mode b, a 2 omega 0. And now we create two photons at omega. We destroy a photon at 2 omega 0, create two photons at omega 0. And since the Hamiltonian has to be Hamiltonian, the opposite, the time-reverse process, has to be possible, too. And that means we destroy two photons at omega 0, and create one photon at 2 omega 0. So now we forget about nonlinear crystals, about non-inversion symmetry in materials. We just take this Hamiltonian and play with it. By simply looking at the Hamiltonian, what is the time evolution of a photon field under this Hamiltonian. We figure out what happens when you send light through a crystal, and what is the output. And I want to show you now that the output of that is squeezed light, which is exactly what I promised you with these quasi-probabilities. We have a coherent state, which is a nice circle. We time-evolve the coherent state, our nice round circle, with this Hamiltonian. And what we get is an ellipse. And if you want intuition, look at the classical example we did before, which really tells you in a more intuitive way what is happening. OK. We want to make one simplifying assumption, here. And this is that we pump the crystal at 2 omega 0 with a strong laser beam. So we assume that the mode, b, is a powerful laser beam. Or in other words, a strong coherent state. We assume that the mode, b, is in a coherent state. Coherent states are always labeled with a complex parameter, which I call beta, now. Well, it's mode b, therefore I call it, beta. For mode a, I've called it, alpha. The coherent state has an amplitude, which I call, r over 2. And it has a phase. We know, of course, that the operator, b, acting on beta, gives us beta times beta, because a coherent state is an eigenstate of the annihilation operator. But when we look at the action of the operator b plus, the photon creation operator, the coherent state is not an eigenstate of the creation operator. It's only an eigenstate of the annihilation operator. But what sort of happens is, the coherent state is the sum over many, many number states with n. And the creation operator goes from n, to n plus 1, and has matrix elements which are square root n plus 1. So in other words, if n is large, and if we don't care about the subtle difference between n, and n plus 1, in this limit the coherent state is also an eigenstate of the creation operator, with an eigenvalue, which is beta star. This means that we have a coherent state which is strong. Strong means, it has a large amplitude of the electric field. The photon states which are involved, n, are large. And we don't have whether it be n, or n plus 1. This is actually, also, I should mention it here, explicitly-- this is sort of the step when we have a quantum description of light. And we replace the operators, p and p dega, by a c number, then we really go back to classical physics. Then we pretend that we have a classical electric field, which is described by the imaginary part of beta. So when you have an Hamiltonian, where you write down an electric field, and the electric field is not changing-- you have an external electric field. This is really the limit of a quantum field, where you've eliminated the operator by a c number. This is essentially your electric field. And we do this approximation, here. Because we are interested in the quantum features of mode a-- a is our quantum mode, with single photons, or with a vacuum state, and we want to squeeze it. b is just, they have parametric drive. With this approximation, we have only the a operators. This is our operator. Any question? AUDIENCE: [INAUDIBLE] would give us a [INAUDIBLE], right? PROFESSOR: Yes, thank you. That means, here should be a minus sign, yes. OK. I've motivated our discussion with this nonlinear crystal, which generates pair of photons. This is the Hamiltonian which describes it. And if you want to have a time evolution by this Hamiltonian, you put this Hamiltonian into a time evolution operator. In other words, you-- e to the minus iHt is the time evolution. If you now evolve a quantum state of light for a fixed time, t, we apply the operator, e to the minus iHt, to the quantum state of light. What I've just said is now the motivation for the definition of the squeezing operator. The squeezing operator, S of r, is defined to be the exponent of minus r over 2, a squared minus a dega squared. This is related to the discussion above. You would say, hey, you want to do that time evolution, where is the i? Well, I've just made a choice of phi. If phi is chosen to be pi over 2, then the time evolution with the Hamiltonian, above, gives me the squeezing operator, below. So with that motivation we are now studying, what is the squeezing operator doing to quantum states of light? Any questions about that? I know I spent a lot of time on it. I could have taught this class by just saying, here is an operator, the squeezing operator. Trust me, it does wonderful things. And then we can work out everything. But I find his unsatisfying, so I wanted to show you what is really behind this operator. And I want you to have a feeling, where does this operator come from, and what is it doing? In essence, what I've introduced into our description is now an operator, which is creating and destroying pairs of photons. And this will actually do wonderful things to our quantum states. What are the properties of the squeezing operator? What is important is, it is unitary. It does a unitary time evolution. You may not see that immediately, so let me explain that. You know from your basic quantum mechanics course, that e to the i operator A is unitary, when A is Hermitian. So the squeezing operator-- with the definition above-- can be written as, I factor out 2 i's over 2 a squared minus a dega squared. And you can immediately verify that this part, here, is Hermitian. If you do the Hermitian conjugate, a squared turns into a dega squared. a dega squared turns into a squared. So we have a problem with a minus sign. But if you do the complex conjugate of i, this takes care of the minus sign. So this part is Hermitian. We multiply it with i, therefore this whole operator. Thus a unitary transformation in [INAUDIBLE]. Any questions? OK. So after being familiar with this operator, we want to know, what is this operator doing? I can describe, now, what this operator does, in a Schrodinger picture, or in a Heisenberg picture. I pick whatever is more convenient. And for now, this is the Heisenberg picture. In the Heisenberg picture, what is changing are the operators. Therefore, in the Heisenberg picture, this unitary transformation transforms the operators. And we can study what happens when we transform the operator, x. The unitary transformation is done by-- the operator, x, is transformed by multiplying from the left side with S, from the righthand side with S dega. You are familiar with expressions like, this, and how to disentangle them. If you have an e to the i alpha, e to the minus alpha, if you could move the alpha past x. So if A and x commute, i A, minus i A would just give unity. So therefore, this expression is just x, unless you have non-Hermitian commutators between A and x. I think you have solved, in your basic mechanics course, many such problems which involve identities of that form. Then there are higher order commutator, the commutator of A with the commutator of a x. Unless one of those commutator vanishes, you can get an infinite series. Our operator, A, is nothing else than the annihilation operator, a squared minus the creation operator, a dega squared. So we can express everything in terms of a, and a dega. The position operator in our harmonic oscillator can also be expressed by a, and a dega. By doing elementary manipulations on the righthand side, and recouping terms, you find immediately that the unitary transformation of the Heisenberg operator, x, gives you an x operator back. But multiplied with an exponential, e to the r. And if we would do the same to the momentum operator, which is a minus a dega over square root 2, we will find that the unitary transformation of the momentum operator is de-amplifying the momentum operator by an exponential factor. If we would assume that we have a vacuum state in the harmonic oscillator, and while classically, it would be at x equals 0, p equals 0, quantum mechanically, we have single-point noise in x, and single-point noise in p. Then you would find that the squeezing operator is amplifying the quantum noise in x. But it squeezes, or reduces, the noise in p. If we apply this squeezing operator to the vacuum state, we obtain what is usually called, squeezed vacuum. And it means that, in this quasi-probability diagram, the action of the squeezing operator is turning the vacuum state into an ellipse. What happens to energy, here? The vacuum state is the lowest-energy state. If you now act with a squeezing operator to it, we obtain a state which has-- the same energy? Is it energy-conserving, or very high energy? AUDIENCE: Higher [INAUDIBLE]. PROFESSOR: Yes. Why? AUDIENCE: It's no longer the [INAUDIBLE]. PROFESSOR: Sure, yeah. It's a vacuum state. We act on the vacuum state, but we get a state which is no longer the vacuum state. The reason why we have extra energy-- the squeezed vacuum is very, very energetic. Because the squeezing operator had a dega squared, a squared. Well a squared, the annihilation operator acting on the vacuum, gives 0. But what we are creating now, we are acting on the vacuum, and we are creating pairs of photons. So we are adding, literally, energy to the system. And the energy, of course, comes from the drive laser, from the laser 2 omega 0, which delivers the energy in forms of photons which are split into half, and they go into our quantum field. In the limit of infinite squeezing-- I will show it to you, mathematically, but it's nice to discuss it already here. In the limit of infinite squeezing, what is the state we are getting? AUDIENCE: Eigenstate of momentum. PROFESSOR: Eigenstate momentum. We get the p equals 0 eigenstate. What is the energy of the p equals 0 eigenstate? AUDIENCE: Infinite. It has to contain all number states. PROFESSOR: It contains all number states? OK, you think immediately into number states, which is great. But in a more pedestrian way, the p equals 0 state has no kinetic energy. But if a state is localized in momentum, p equals 0, it has to be infinitely smeared out on the x-axis. And don't forget, we have an harmonic oscillator potential. If you have a particle which is completely delocalized in x, it has infinite potential energy at the wings. So therefore in the limit of extreme squeezing, we involve an extreme number of number states. Actually, I want to be more specific-- of photon pairs. We have states with 2n, and n can be infinitely large. But we'll see in the classical picture, what we get here when we squeeze it is, we get the p equals 0 eigenstate, which has infinite energy, due to the harmonic oscillator potential. If we would allow with the system now, after we have squeezed it, to evolve for a quarter period in the harmonic oscillator, then the ellipse would turn into an vertical ellipse. So this is now an eigenstate of x. It's the x equals 0 eigenstate. But the x equals 0 eigenstate has also infinite energy, because due to Heisenberg's uncertainty relation, it involves momentum states of infinite momentum. Questions? AUDIENCE: [INAUDIBLE] a is the photon field, right? So p is roughly the electrical field, right? PROFESSOR: Yes. AUDIENCE: So it's kind of that the electric field counts 0, and x is kind of the a, the-- and it-- because of [INAUDIBLE]. The electrical field is squeezed? PROFESSOR: Yes. AUDIENCE: It means we have no electrical field? PROFESSOR: We'll come to that in a moment. I want to do a little bit more math, to show you. I wanted to derive for you an expression of the squeeze state, in number basis, and such. Your question mentioned something which is absolutely correct. By squeezing that, we have now the p-axis is the electric field axis. So now we have, actually, in the limit of infinite squeezing, we have an electric field which has no uncertainty anymore. By squeezing the coherent state into a momentum eigenstate, we have created a sharp value for the electric field. We have created an electric field eigenstate. Well you would say, it's pretty boring, because the only electric field state we have created is electric field e equals 0. But in the next half-hour, we want to discuss the displacement operator, and I will tell you what it is. That we can now move the ellipses, and move the circles, anywhere where we want. So once we have an electric field state which has a sharp value of the electric field at e equals 0, we can just translate it. But before you get too excited about having an eigenstate of the electric field, I want you to think about what happened after one quarter-period it of the harmonic oscillator frequency. It turns upside down, and your electric field has an infinite variance. That's what quantum mechanics tells us. We can create electric fields which are very precise, but only for a short moment. So in other words, this electric field state which we have created would have a sharp value. A moment later, it would be very smeared out, then it has a sharp value again, and then it's smeared out again. I mean, that's what squeezed states are. Other questions? AUDIENCE: That's why [INAUDIBLE]. PROFESSOR: That's why we need homodyne detection. Yes, exactly. If we have squeezed something, which is sort of narrow, that's great for measurement. Now we can do a measurement of, maybe, a LIGO measurement for gravitational waves with higher precision, because we have a more precise value in our quantum state. But we have to look at it at the right time. We have to look at it synchronized with the harmonic motion. Homodyne detection means we look only at the sine component, or at the cosine component. Or if I want to simplify it, what you want to do is, if you have a state like this, you want to measure the electric field, so to speak, stroboscopically. You want to look at your system always when the ellipse is like this. The stroboscopic measurement is, as I will show you, in essence, a lock-in measurement, which is phase-sensitive. And this will be homodyne detection. So we can only take advantage of the squeezing, of having less uncertainty in one quadrature component, if you do phase-sensitive detection, which is homodyne detection. But now I'm already an hour ahead of the course. OK. Back to basics. We want to explicitly calculate, now, how does a squeezed vacuum look like. We actually want to do it twice, because it's useful. We have to see it in two different basis. One is, I want to write down the squeezed vacuum for you in a number representation. And then in a coherent state representation. The squeezing operator is an exponential function involving a squared, and a dega squared. And of course, we're now using the Taylor expansion of that. We are acting on the vacuum state. I will not do the calculation. It's again, elementary. You have n factorial, you have terms with a dega acts on c, well, you pay 2 photons. If it acts again, it adds 2 more photons, and the matrix element of a dega acting on n is square root n plus 1. You just sort of rearrange the terms. And what you find is, what I will write you down in the next line. The important thing you should immediately realize is, the squeeze state is something very special. It is the superposition of number states, but all number states are even because our squeezing operator creates pairs of photons. This is what the parametric down-conversion does. We inject photons into the vacuum, but always exactly in pairs. And therefore, it's not a random state. It's a highly correlated state with very special properties. OK. If you do the calculation and recoup the terms, you get factorials, you get 2 to the n, you get another factorial. You get hyperbolic tangent-- sorry, to the power n. And the normalization is done by the square root of the cos function. And the more we squeeze, the larger are the amplitudes at higher and higher n. But this is also obvious from the graphic representation I've shown you. Let me add the coherent state representation. The coherence states are related to the number states in that way. If we transform now from number states to coherent states, the straightforward calculation gives, now, superposition over coherent states. Coherent states require an integral. e to the minus e to the r over 2, divided by-- Anyway, all this expressions, they may not be in its general form, too illuminating. But those things can be done analytically. I just want to mention the interesting limiting case of infinite squeezing. AUDIENCE: When you do the integral over alpha, is this over like, a magnitude of alpha, or a real part, or [INAUDIBLE]? PROFESSOR: I remember, but I'm not 100% sure that alpha is real, here. I mean, it sort of makes sense, because we start with the vacuum state. And if we squeeze it, we are not really going into the imaginary direction. So I think what is involved here are only real alpha. AUDIENCE: For negative r, we should get [INAUDIBLE]. PROFESSOR: For negative r, we need imaginary state. AUDIENCE: So we should [INAUDIBLE]. PROFESSOR: Let me double-check. I don't remember that. You know that, sometimes, I admit it, the issue-- if you research material, prepare a course some years ago, you forget certain things. If I prepared the lecture, and everything worked out yesterday, I would know that. But certain things you don't remember. As far as I know, it's the real axis. But I have to double-check. The limiting case is interesting. If r goes to infinity, you can show that this is simply the integral, d alpha over coherent states. We have discussed, graphically, the situation where we had-- so these are quasi-probabilities. In that case of infinite squeezing, we have the momentum eigenstate, p equals 0. This is the limit of the infinitely squeezed vacuum, and in a coherent state representation, it is the integral over coherent state alpha. I'm pretty sure alpha is real here, seeing that now. There is a second limit, which happens simply-- you can say, by rotation, or by time evolution-- which is the x equals 0 eigenstate. And this is proportional to the integral over alpha when we take the coherent state i alpha, and we integrate from minus to plus infinity. OK. So we have connected our squeezed states, the squeezed vacuum, with number states, with coherent states. Now we need one more thing. So far we've only squeezed the vacuum, and we have defined the squeezing operator that it takes a vacuum state and elongates it. In order to generate more general states, we want to get away from the origin. And this is done by the displacement operator. The definition of the displacement operator is given here. The displacement by a complex number, alpha, is done by putting alpha, and alpha star, into an exponential function. In many quantum mechanic courses, you show very easily the elementary properties. If the displacement operator is used to transform the annihilation operator, it just does that. If you take the complex conjugate of it-- so in other words, what that means is, it's called the displacement operator, I just take that as the definition. But you immediately see why it's called the displacement operator when we do the unitary transformation of the annihilation operator, we get the annihilation operator displaced by a complex number. So the action, the transformation of the annihilation operator is the annihilator operator itself, minus a c number. So therefore, we say, the annihilation operator data has been displaced. So this is the action of the displacement operator on an operator-- on the annihilation operator. The question is now, how does the displacement operator act on quantum states? And the simplest quantum state we want to test out is the vacuum state. And well, not surprisingly, the displacement operator, displacing the vector state by alpha, is creating the coherent state, alpha. This can be proven in one line. We take our displaced vacuum, and we act on it with the annihilation operator. If we act with the annihilation operator on something, and we get the same thing back times an eigenvalue, we know it's a coherent state. Because this was the definition of coherent states. So therefore, in order to show that this is a coherent state, we want to show that it's an eigenstate of the annihilation operator. So this is what we want to do. The proof is very easy. By multiplying this expression with unity, which is DD dega, we have this. And now we can use the result for the transformation of operators. Namely, that this is simply the annihilation operator, plus alpha. If the annihilation operator acts on the vacuum state, we get 0. If alpha acts on the vacuum state, we get alpha times 0. So therefore, what we obtain is that. When the annihilation operator acts on this state, we get alpha times the state, and therefore the state is a coherent state with eigenvalue alpha. In a graphical way, if you have a vacuum state the displacement operator, D alpha, takes a vacuum state and creates a coherent state alpha. If you want to have squeezed states with a finite value, well, we just discussed the electric field. Related to the harmonic oscillator, we want squeeze states, which are not centered at the origin, which have a finite value of x or p. We can now create them by first squeezing the vacuum, and then displacing the state. AUDIENCE: What's the physical realization of the displacement operator? PROFESSOR: What is the physical realization of the displacement operator? Just one second. The physical representation of the displacement operator-- we'll do that on Monday-- is the following. If you pass an arbitrary state through a beam splitter-- but it's a beam splitter which has very, very high transmission-- and then, from the-- I'll just show that. If you have a state-- this is a beam splitter-- which has a very high transmission, T is approximately 1, then the state passes through. But then from the other side of the beam splitter, you come with a very strong coherent state. You have a coherent state which is characterized by a large complex number, beta. And then there is a reflection coefficient, r, which is very small. It sort of reflects the coherent state with an amplitude r beta. If you mix together the transmitted state and r beta-- I will show that to you explicitly, by doing a quantum treatment of the beam splitter-- what you get is, the initial state is pretty much transmitted without attenuation. But the reflected part of the strong coherent state-- you compensate for the small r by a large beta-- does actually an exact displacement of r beta. It's actually great. The beam splitter is a wonderful device. You think you have a displacement operator formulated with a's and a dega's, it looks like something extract. But you can go to the lab, simply get one beam splitter, take a strong laser beam, and whatever you send through the beam splitter gets displaced, gets acted upon by the beam splitter. Yes. AUDIENCE: You showed the displacement operator, when you acted on the vacuum state, will displace the vacuum state to a state alpha. Does it still hold if you acted on, like another coherent state. Or in this case, a squeeze state like that? PROFESSOR: Yes. I haven't shown it, but it's really-- it displaces-- When we use this representation with quasi-probabilities, it simply does a displacement in the plane. But no. To be honest, when I say it does a displacement on the plane, it reminds me that we have three different ways of defining quasi-probabilities. The w, the p, and the q representation. I know we use it all the time, that we displace things in the plane. But I'm wondering if the displacement operator does an exact displacement of all representations, or only of the q representation. That's something I don't know for sure. AUDIENCE: I was thinking it could also, like-- I mean, are you going to be able to displace all types of light, like thermal light, or any representation of light that you could put in, is the same displacement operator going to work? Or is its domain just the vacuum and coherent states? PROFESSOR: The fact is, the coherent states-- I've shown you that it's a vacuum state. I know that's the next thing to show, the displacement operator if you have a displacement by alpha followed by displacement by beta, it is equal to displacement by alpha plus beta. So displacement operator forms a group, and if you do two displacements, they equal into one area of displacement, which is the sum of two complex numbers. What I'm just saying, if you do the first displacement you can get an arbitrarily coherent state. So therefore, the displacement operator is exactly displacing a coherent state by the argument of the displacement operator. And now if you take an arbitrary quantum state and expand it into coherent states-- coherent states are not only complete, they are over-complete. All you have done is, you've done a displacement. Now the over-completeness, of course, means you have to think about it, because you can represent states in more than one way by coherent states. But if you have your representation, you just displace it, and this is the result of the displacement operator. So since the q representation is simply, you take the statistical operator and look for the elements in alpha, and if you displace alpha, the q representation has been moved around. So I'm sure that for the q representation, for the q quasi-probabilities, the displacement operator shifted around in this place. For the w and p representation, I'm not sure. Maybe there's an expert in the audience who knows more about it than I do. OK. We have just five minutes left. I want to discuss now the electric field of squeezed states. And for that, let me load a picture. Insert picture. Classroom files. Let us discuss, now, the electric field of squeezed states. Just as a reminder, we can discuss the electric field by using the quasi-probability representation. And the electric field is the projection of the quasi-probabilities on the vertical axis. And then the time evolution is, that everything rotates with omega in this complex plane. We discussed it already. For coherent state, we have a circle which rotates. Therefore, the projected fuzziness of the electric field is always the same. And as time goes by, we have a sinusoidal-bearing electric field. Let me just make one comment. If you look into the literature, some people actually say, the electric field is the projection on the horizontal axis. So there are people who say, the electric field is given by the x-coordinate of the harmonic oscillator, whereas I'm telling you, it's the p-coordinate. Well, if you think one person is wrong, I would suggest you just wait a quarter-period of the harmonic oscillator, and then the other person is right. It's really just a phase convention. What do you assume to be t equals 0-- it's really arbitrary. But here in this course, I will use the projections on the vertical axis. OK. If you project the number state, we get always, 0 electric field, with a large uncertainty. So that's just a reminder. But now we have a squeezed state. It's a displaced squeezed state. If you project it onto the y-axis, we have first some large uncertainty. I think this plot assumes that we rotate with negative time, so I apologize for that. You can just invert time, if you want. So after a quarter-period, the ellipse is now horizontal, and that means the electric field is very sharp. As time goes by, you see that the uncertainty of the electric field is large, small, large, small-- it modulates. It can become very extreme, when you do extreme squeezing, so you have an extremely precise value of the electric field, here, but you've a large uncertainty, there. Sometimes you want to accurately measure the 0 crossing of the electric field. This may be something which interests you, for an experiment. In that case, you actually want to have an ellipse which is horizontally squeezed. Now, whenever the electric field is 0, there is very little noise. But after a quarter-period, when the electric field reaches its maximum, you have a lot of noise. So it's sort of your choice which way you squeeze. Whether you want the electric field to be precise, have little fluctuations when it goes through 0, or when it goes through the maximum. So what we have done here is, we have first created the squeezed vacuum, and then we have acted on it with a displacement operator. OK. I think that's a good moment to stop. Let me just say what I wanted to take from this picture. The fact that the electric field is precise only at certain moments means that we can only take advantage of it when we do a phase-sensitive detection. We only want to sort of, measure, the electric field when it's sharp. Or-- this is equivalent-- we should regard light is always composed of two quadrature components. You can say, the cosine, the sine oscillation, the x, and the p. And the squeezing is squeezed in one quadrature, by it is elongated in the other quadrature. Therefore, we want to be phase-sensitive. We want to pick out either the cosine omega T, or the sine omega T oscillation. This is sort of, homodyne detection. We'll discuss it on Monday. Any question? OK. Good.
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https://ocw.mit.edu/courses/8-851-effective-field-theory-spring-2013/8.851-spring-2013.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. IAIN STEWART: So last time we were talking about the subleading one of our MQ Lagrangian HQET. Hopefully more people remember that we have a makeup lecture today, or will not be stopped by the snow. So today we're going to continue this discussion of power suppressed Lagrangians. And I'll explain to you why we think of these as giving power corrections to observables. And we'll talk about a couple different observables where these specific operators are playing an important role. And then we're going to turn to the topic of renormalons, which is fun stuff. How many people here know what a renormalon is? AUDIENCE: [INAUDIBLE] IAIN STEWART: Yeah. By the end of the lecture, you all know what a renormalon is. OK, so this is where we left off. We were talking about this symmetry reparameterization invariance. And what we showed last time is that the Wilson coefficient of this operator is 1 to all orders in perturbation theory because of that symmetry. The Wilson coefficient of this operator is not. And I told you that at leading log order, it would be given by some expression like this. But in general, this is just the lowest order expression. And it gets perturbatively corrected. It's actually known at three loops. So let's continue. We've talked about reparameterization invariance for the Lagrangian. Reparameterization invariance also has important consequences for operators. So just like we wrote down sub-leading Lagrangians, we should also write sub-leading currents. So there's 1 over mq supressed currents. And if you look at those currents and you construct them, then you find, again, that reparameterization invariance plays an important role. And again, there's relations between Wilson coefficients. And what happens in this case is actually that the leading order operator had a Wilson coefficient, the lowest order operator. So the Wilson coefficients of the subleading operators get related to that of the leading operator. And I may give you a problem on that on your next problem set. But I haven't quite made up my mind yet. So rather than go through that, I want to talk about some observables where these operators played a role. So we've seen one example of reparameterization invariance. And I think from that one example, you would be able to do other examples. Let's talk about masses. These operators here actually have an impact on the masses of hadrons. If we think about the mass of a heavy meson, h, there's a heavy quark in that heavy meson. So we can pull out that heavy quark. And then there's some remainder. And we can characterize higher order terms in this formula. So let me first explain what this lambda bar is. Our Lagrangian here, if we're working at lowest order in the theory, our Lagrangian would be L0 of HQET plus, for the light quarks, they're basically just a QCD Lagrangian, so for the light quarks and the gluons. So for the heavy quark, we have HQET, and then there would be higher order terms that are 1 over m suppressed, which are the ones up there. So the question is, if we have this Lagrangian, what is the mass of a heavy state like the b meson or the d meson or the d star? And we can characterize that by knowing the structure of the Lagrangian. So first of all, what about this term, this lambda bar? That term actually comes from these parts of the Lagrangian. So if we take these parts of the Lagrangian, we can calculate the Hamiltonian. Let me call that H0. And the definition of this lambda bar is the thing that you get by taking that Hamiltonian and letting it act on the state. So you can think that this is-- if you like, you can think that this state is just an eigenstate of the Hamiltonian. You get the energy. But since the Hamiltonian has no heavy quark mass in it, what you're getting is some parameter, which is traditionally called lambda bar, that doesn't have a heavy quark mass in it. And so there's no heavy quark mass in here. So it's independent of mq. It's also independent of spin. So B versus B star and flavor B versus D. So whether it's a B meson, D meson, as long as we're in this limit where both the bottom mass and the char mass are heavy, if we're thinking of this expansion, then this is a universal thing just determined by these Lagrangians. It's traditionally called lambda bar. And it's telling you that if you think about this thing as having an order mq piece, this is the order mq to the 0 piece. The symmetry of the theory is telling us that this is a universal constant. It does depend on the state here, although that state could be connected to other states by spin symmetry and flavor symmetry. There was this quantum number we talked about last time, which we called SL for the spin of the light degrees of freedom, and then pi for the parity of the light degrees of freedom. So it's the same within every multiplet that we get from a fixed SL pi. But there's a different value for each multiplet. So you'd have a different value of lambda bar for the B, D, B star B than for the lambda B and the lambda C, for example. So that's this term. We can also characterize what's going on with this term by using this Lagrangian here, the L1. So at order 1 over MQ, we can figure out what kind of contributions to the mass we're getting. h1 is just minus L1. You can think of it now perturbatively. So that just flips the sign of what we had before. And we get two parameters from taking matrix elements here. They get the following names. And let me just do everything here in terms of-- it's not totally crucial, but it's a little easier to interpret certain things if we work in the rest frame. So I'll just work in the rest frame. So if you work in the rest frame, you should just think of this as like giving some kind of measure of the kinetic energy that the heavy quark gets wiggled by. And that's called lambda 1. This, again, is a matrix element that doesn't know about the heavy quark mass. And the heavy quark mass in an explicit MQ. So this guy has dimension 2 if you work out the dimensions. And then you can do the same for the other operator. And here you have to think a little bit harder. And if you think a little bit harder, you can characterize what the coefficients are for different spins. So there's some Wilson coefficient. And then we have this operator. And this here is like sigma dot B. I told you that before. And so you should think of what's going on here as you have sigma, which is giving the SQ, and then you have a B field. And then it's a question of what vector could be left over that the B field could know about. And it's basically SL. So this guy here is leading to the SQ. And this guy here is leading to the SL. And so that's where this SQ.SL comes from. So you can think about other possibilities, but you have to have the right symmetry structure under time reversal and parity and things like that. And that forces you to use a spin here for the B and not something like the V. So this SQ.SL is something that you can write out as, remembering the definition of it and its relation to J squared. You can write out that factor like that. And then we can derive here how this lambda 2 contributes to the differing states, because we have a different contribution for the B and the B star from the lambda 2 guy, because it was violating the spin symmetry. So the lambda 1 guy is going to violate the flavor symmetry. That's this 1 over MQ. And so charm quarks-- charm mesons and bottom mesons will get different contributions from that. They have the same lambda 1, but they have a different connection because it gets suppressed by a different factor. And then again for the lamba 2, Lambda 2 is a little bit different because it actually has mu dependents. That's not quite right. So it actually has MQ dependents because the coefficient here has mu dependents. So this operator doesn't have any MQ dependence, but the coefficient does, remember. The mu dependence cancels and there's a left over MQ dependence to the lambda 2, but it's logarithmic MQ dependence. If I define it this way-- so that's why I wrote MQ there. So you could, as we talked about last time, you could sum up the logs of large logs and have this guy-- you could re-sum logs, if you want. So you could think of these as you can re-sum logs inside this lambda 2. So if we ignore those logs or we imagine that we re-sum them and we just look at the remainder, then we have the expectations based on power counting for how big these things are. And we just [? expect ?] that they're both given by the dimension. And the only [? dimensionful ?] parameter around is lambda QCD because the MQs can only occur in logarithms, and the Wilson coefficient in nowhere else. And these are non-perturbative parameters. Lambda bar is also a non-perturbative parameter. These ones have a little more dynamics in them. So then we could go out and write for our states the results by putting these things together. So we'd have M capital B plub M little B plus lambda bar lambda 1 over 2 and little B. And if we go through the spin structure there, we get a 3/2 and then lamba 2 logarithmic MB dependence over MB. So when we start with the B star, the first three terms are the same. And then this guy comes in with a plus. And it's lambda 2 over MB over 2MB. And then the D states are similar in turn. Lambda bar lambda 1 over 2M term, and then these two terms again. Same spins, so it's lambda 2. And the only difference is, now we're evaluating the lambda 2 as logarithm and charm dependence, OK. So there's some formulas that are correct, actually-- including all the way up to 1 over M corrections. And you can see that there's still sort of a structure to these things. This is a universal correction for all of them. These corrections are universal between B and B star. And so the splitting between B and B star is just given by this lambda 2. That's what causes the states to split. So up to that level, they're the same. And that's why the B and B star are very degenerate and the masses are very close. So you can form combinations to cancel things out. So you can take a kind of spin average mass where you take 3 times the vector plus the scalar divided by 4. And if you do that, then you're canceling out the lambda 2. And then you can also form differences. So if you looked at MV star squared minus MB squared, then the only thing that's causing a difference is lambda 2. Numerically, this is 0.49 GV squared. And from our formula up there, it's 4 lambda 2 of MB, so something like four times 0.12 GV squared. So that's the value of this lambda 2 is 0.12 GV squared, which is about lambda QCD squared. So our dimensional analysis is working. And then you can do the same thing for the D meson. Everything here works beautifully, actually. Extract the value of lambda 2 and you see there's a slight difference because that difference can be attributed to the logarithms. So if you take lambda 2 MB over lambda 2 MC, then the only thing that differs is the Wilson coefficient. So that's something you can predict. So experimentally, if you take the ratio of those two things there, you get a prediction like this one. And if you plug in theory like leading log RGE, then you're getting alpha S of MB. Oh, sorry. [INAUDIBLE] And this is 1.17. So that's for three light flavors, which is the right thing to do for the beta function. So it's not so bad for a leading log prediction. That's kind of the accuracy you would expect. All right. So we're really understanding quite a bit from our [INAUDIBLE] [? instruction ?] of this effective theory about the states, and even make predictions for the ratios of the mass divided by that as something preservative, which is kind of non-trivial. Perturbative up to the order we were working. There would be corrections if we continued in our series here. There could be some non-perturbative corrections. But up to 1 over M, the ratio of this to that, which is what I'd wrote over there, is perturbative. All right. So that's some phenomenology. It's kind of baseline phenomenology that we can do with the Hamiltonian or with Hamiltonian derived from our subleading Lagrangian. There's other phenomenology that we can do. And I'll mention some of the most important phenomenology. There's another class of predictions that we can make where we have a lot of predictive power by using the effective theory. And I want to talk a little bit about that. So you can look at semi-leptonic decays. And there's two different types of semi-leptonic decays that you can look at-- so-called exclusive decays and inclusive decays. So exclusive is making transitions between the meson states and inclusive is making a transition where you allow any charm state, not just the lowest order ground state charm, but you could have in this state here, XE, you could have a D pi or a D star pi pi or other things like that. So it doesn't even have to be a single hadron state. So exclusive refers to a particular channel. Inclusive refers to some overall channels. So the theory in these two is quite different. In this one, you would have form factors for the current between the states. And I already mentioned to you that heavy quark symmetry reduces the number of form factors. You get this single Isgur-Wise function. And so heavy quark symmetry is very powerful here. And you can actually also work out 1 over MQ corrections. And people have done that. And it turns out there aren't any. So there's something called Luke's theorem, which actually was a result derived, I think, when Luke was a graduate student that proved there was no 1 over m Q corrections here. And then you can work out alpha s corrections. And so this kind of-- you can just keep going. People have worked out 1 over m Q squared corrections. And this kind of formalism is used to measure V c b because these decay rates would depend on V c b, and that's kind of what you're after if you're doing this. So you'd like to parameterize and figure out the hadronic physics as much as possible, and aid the experiments in getting V c b, and exactly this framework is used. People nowadays-- what they do is, the 1 over m Q squared corrections need to be computed on the lattice. So lattice QCD computes those corrections. The perturbative corrections need to be computed by continuum people like me, so those are computed. And they put these things together to get V c b from these decays. You could also do it with this inclusive guy. And here it's more interesting because you can basically do everything with pen and paper. So here, you can use an Operator Product Expansion, OPE. And actually, HQET constrains the form of the operator product expansion, as well. And indeed, when you work out the leading power corrections, they turn out to not enter until two orders down. So again, you're protected from first-order corrections, so you just get corrections that occur at 1 over m Q squared. And furthermore, they depend only on the two parameters we already defined. So it's not like you even get any new nonperturbative matrix comments. You just get the two we were talking over here, were the masses. OK? So there's perturbation corrections, and there's power corrections, but you're actually not even getting any new information relative to the masses, any new matrix elements. OK? So we'll talk a little bit more about this second one with the OPE. There's a lot of work that's gone into this one, both from the theory side, as well as from experiment. So when you think about the decay rate for this B to X sub c l nu bar, it's a doubly differential decay rate in Q squared over the lepton pair, so that's Q mu, E l-- because you can pick it as a second variable-- and then you don't know the mass of the state X sub c because it could be different things. It could be a D. It could be a D star. It could be a D pi. So that's another variable. So you have a triply differential spectrum, and you can compute it with an operator product expansion. I'm not going to go through the details of carrying out the operator product expansion. If you want to do reading about that, there's some supplemental reading that I haven't assigned to you where you can read about that in the book Heavy Quark Physics by Manohar and Wise. So for our purposes here, let's think of what operator product expansion means. It is simply that we want to carry out an expansion in lambda QCD over m Q. And if you go into the details, there's an important role played by the fact that you're summing over all states, and that's allowing you to connect the partonic calculations to the hadronic calculations, basically by probability conservation, but I'm not going to go so much into that. So when you do this operator product expansion, you can think of it in terms of diagrams. And usually, you draw these diagrams as forward diagrams. So here's kind of the matrix elements squared. Here is the final state. There's the b quark going in and another b quark going out, so that's like a matrix element squared. The amplitude squared. And you think about doing an operator product expansion for that. And you match onto operators in the effective theory. Just to give you a schematic structure for those operators. So there's a leading-order operator. There's a subleading-order operator that has an extra covariant derivative in it. That's kind of what you would do. And then you have Wilson coefficients. The leading-order operator here is just b bar b. That's what it turns out to be. This operator here would look familiar because it's D transverse squared, so that's like our lambda 1 operator, and that's where the lambda 1's are going to come from. And then there would be a lambda 2 term, and I just didn't write it. So there's a magnetic guy here, too. So when you look at the leading-order operator, b bar b, that counts the number of b quarks, OK? That's just a number operator. So to all orders in perturbation theory, b bar b is 1. Yeah? AUDIENCE: How do you cut these-- like, you're actually calculating [? B to A, right? ?] [INAUDIBLE] IAIN STEWART: Right. So you should think of actually looking-- so this is a charm quark. And so you would write down that propagator, and then you have a large injection of momentum from the b quark, right? Use momentum conservation and expand the propagator. Thinking about the propagator, if you like, thinking about this guy here having-- so you could write this guy's momentum as m b v plus k, all right? And you would expand in k. So at first order, you just drop all the k's, and that's basically giving you this term. At second order, you'd keep the k's, so the k squared would be identified with this D t squared. AUDIENCE: Right, but how do you cut the effective [INAUDIBLE]?? IAIN STEWART: The effective theory diagram, it's already been cut, if you like. So this is just a real thing, and I've already taken the imaginary part. So this is all real. There's no cut to make in the effective theory. The thing I'm integrating out here-- so this is a good comment. The thing I'm integrating out here is hard and off-shell. So when I go over to the effective theory and I get rid of the off-shell stuff, I have a real part here. There's nothing to cut in the effective theory. Sometimes, as you know, there are things to cut in the effective theory, but here there's nothing to cut on this side. OK, so this is one tall order. So that actually means that we don't even have a nontrivial matrix element here. We just have a Wilson coefficient. So everything in the first-order term is completely calculable. No nonperturbative parameter because of symmetry. And this guy here is the power correction. And if you look at what the Wilson coefficient is-- so it's a function of alpha s. You could think of it like this, if you want. And all the kinematic variables, like kinematic variables like these guys-- and it is equal to the free b quark decay, including the loop corrections, of course. So not only the tree-level result, but adding loop corrections to the right-hand side here. If you add the loop corrections and you drop all the k squared's, then you get just the free b quark decay, no nonperturbative matrix elements, so you can just calculate that first term, order by order. Just doing a [? partonic ?] calculation gives you the right thing to describe this decay. And the OPE is telling you that that formally is exactly the right thing. Even if you just started to do it, it's actually technically the right thing to do here. It wouldn't be the right thing to do for the exclusive decays, but it is the right thing to do here for the inclusive. So what is the range of validity of this sort of-- is there any restrictions on these kinematic variables? And there is. We're treating the kinematic variables as if they're hard. And that means basically that we're thinking that they scale like powers of the heavy quark mass. So there can be regions of phase space where that wouldn't be true, but as long as we stay away from those corners, edges, then what I said is right. And it's a little more powerful than that. So stay away from edges. Or there's another way of thinking about it, and that is that you can integrate over regions of the Dalitz plane for the phase-space variables. That includes all the way up to the edges. And as long as the size of your integration region is order m Q, then you're also fine. So think of, like, something with, like, m B squared up here, d m x squared, for example. And you can-- it doesn't have to be exactly m B squared. It could be m B squared over 2, m B squared over 4, something that you're counting as order m B squared. And again, you would be fine with doing this type of operator product expansion. So what happens is, if you restrict yourself to be close to the edges, then restricting yourself to be close to the edges introduces new scales in the problem. And if there's new scales in the problem, just having a power counting that separates out lambda, QCD, and m B would not be enough. You would have to do more detail. And there is actually some interesting things that happen there, and we probably will talk about at least one example later on when we talk about [? SCT. ?] OK, so I already said it, that there's a nontrivial fact happening at NLO. Well, I already implied that there's no 1 over m b corrections, and so that's an important outcome from this result. If you want to derive this, you need the effective theory. You basically use the equation of motion to get that to be true. So this is NLO in the power corrections, and then NNLO as I said, is just lambda 1 and lambda 2 at order lambda QCD squared over m b squared. So phenomenologically, this is actually wildly successful. This is-- as far as I know, this is the case in QCD or in any theory where people have actually carried out the operator product expansion in the most detail. So people have gone up to 1 over m Q to the 4th. So they calculated at least two loops for everything and maybe even three loops for the [? NF ?] pieces. And experimentally it's been explored to death. You have these three variables, and they've constructed of order 80 moments from these three different kinematic variables, sliced and diced the decay rate in all sorts of imaginative and crazy ways. And they've really tested that this OPE [? always ?] works beautifully, OK? So you can think of that you're getting a consistent picture, 80 different observables, just from a few simple predictions and your perturbative results. You also get v c b. So everything fits together with the framework agreement I've discussed with you, and you get a result for v c b, which I'm not going to write on the board. OK? So that gives you-- I didn't go through the details. I will give you some reading to learn more about this OPE if you're interested, and sort of [? seize ?] how some of the things I mentioned, like using the equation of motion, how that actually works out. It's all done very nicely in this book, Heavy Quark Physics. But I just wanted to give you a flavor with this effective theory that there's something very useful you can do with it once you have it. And in this case, kind of the main phenomenological thing that you're after is v c b and what I've described to you, but you could also think about b decays, where you're looking for new physics. And again, if you can construct how the decay rate is looking, then you can have a hope of finding new physics in the coefficient of those decay rates, effectively in the Wilson coefficients of decay rates. OK? So we're going to turn to something else, but let me pause and see if there's any questions. AUDIENCE: Is this the best way to [? get ?] [? v c b? ?] IAIN STEWART: The two are actually pretty competitive, the exclusive and inclusive. Basically because the lattice is doing a good job of the matrix elements that you need in the exclusive. And again, you have a lot of kinematic information you can use from the experiment on the shape. So they're pretty competitive, actually. AUDIENCE: Do they agree? IAIN STEWART: They agree-- I think the disagreement is at, like, kind of the 1.8 sigma level, so they agree. There's an interesting story in v u b. So you could do the same kind of thing, not for a charm quark, as we did here, but for a light quark. And then actually, if you do b to pi l nu and you look at the inclusive version of that, and you can do the same type of OPE-- a little bit different, actually, but you can do it in OPE there, as well. Then you go through and you get a v u b in the two different ways, and the disagreement's at the 2 and 1/2 sigma level, maybe even-- yeah. It fluctuates with time, but on average it's 2 and 1/2 sigma. [CHUCKLES] So that's interesting. There's not an understanding of what's going on there. That's more interesting than-- this one basically agrees. OK, so rather than go further into this, I want to turn now to my promised topic to you, to tell you about renormalons. Actually, renormalons have something to do with power correction, so it's not disconnected from what we've been talking about. So what are the ideas here? So we've already seen in our discussion of renormalization that there is a freedom in defining the perturbative series. And we kind of focused on the m s bar scheme as being the simplest thing. But I told you about some subtleties with m s bar, and now we're going to address one of them that's related to this thing called renormalons. So we have some freedom, if you like, in adjusting the cutoffs. And we had this cutoff, mu, and it was dividing up what was perturbative and nonperturbative in m s bar. But we could have done something else. We could have used a different type of cutoff. We could have used a Wilsonian cutoff, and that would have divided up things a little bit differently. And you should ask the question, are any possible way of dividing things up equivalent? We've already saw when we were doing calculations that, for the logarithms, they were equivalent. But it turns out that, if you don't-- that the powers, sort of the power separation of power divergences can actually have an impact. And that's related to what renormalons are. And m s bar and dim reg basically avoids thinking about power corrections. You just set them to zero. And so what can happen is, in m s bar, your matrix elements are having a little bit too much UV physics, and your Wilson coefficients are actually sensitive to the IR. They're not sensitive-- you don't really see it when you look at the coefficient. You just see a number, and it looks pretty good, but there's actually an asymptotic structure to those Wilson coefficients that comes from higher orders of perturbation theory. There's things hiding there. That's what this discussion is going to be about. So if you do actually not make a good choice, then it can be the case that when you go to higher orders in the perturbative expansion, you get lousy convergence. And that goes hand-in-hand, actually, with kind of another thing, which is harder to visualize, but I'll explain it. And that is that you have trouble extracting the nonperturbative parameters. And the reason that you're having trouble-- so you could think about doing some calculation like that OPE over there. And you extract a value for lambda 1 from all your fits to all these moments I was describing to you. And then you go to one higher order in alpha s, and you extract another value, and all of a sudden, this guy changes by a factor of 2 and you wonder what's going on. And you wonder, well, what's nature telling you about the kinetic energy if I change the order of my perturbation theory and all of a sudden I'm extracting a factor of 2 different value for that matrix element. And that's related actually to this poor convergence. These things go hand in hand. If there's pure convergence in the series, then when you extract matrix elements you can get different values. Phenomenologically, you can imagine how that would be related, but it's physically related, too. It has to do with the fact that there's a poor convergence here because you haven't divided up the IR physics and the UV physics fully correctly, and that's reflected in this inability to extract these guys in a convergent fashion. And we can actually quantify that using something called renormalon techniques. So one way of saying it is that, if you make a poor choice, you're plagued by something called renormalons, so they're actually something bad and you don't want them. So what goes wrong is that the short distance of coefficients have hidden power law sensitivity to the IR. You wanted these to be UV things, but they are sensitive to the IR. And it comes in dimensional regularization in the powers. We've been very careful about separating out logarithms, but we weren't as careful about powers, and it comes back to haunt us if we look carefully at the theory. And there's a corresponding sensitivity to the UV in the matrix elements. So let me give you an example that's not working through formalism, but just numerics. So let's look at b to u e bar nu at lowest order. The up quark is massless. And we'll think about this-- like we were talking about for the OPE for charm, we'll think about it inclusively so that we can actually just look at this and it makes physical sense. The same thing that I told you about the charm case applies for the up quark case. The lowest-order prediction is just calculate this guy and include loop corrections. The leading power prediction is the same story as for the charm. So that means that physically it's relevant to think about this decay rate. So what does it look like? There's some G Fermi. Integrating out the W boson. There's some factors of pi. The mass dimensions of this guy are 1. G Fermi squared is minus 4, so you need 5 powers, and the thing that's the setting the mass dimensions is m b, so you get 5 powers of m b. And if you look at the perturbative series and you said set mu to equal m b, this is what it looks like. So epsilon-- I'm just introducing something which is a counting parameter, and it's 1. I could also make alpha s the counting parameter, but I actually want to stick a number in for alpha s, so alpha s is going to disappear in the next line, and I'll just keep the epsilon, which is just telling me the contribution is coming from this order. So you have some choice here for how you define the b quark mass, OK? You could use the pole mass, pole in the b quark propagator. Or you could use the m s bar mass. Or you could use some other definition. And it's raised to the 5th power. So whatever definition you pick, it can be pretty sensitive to that. So let me tell you what the results look like in three different schemes for that mass. So let's first do the pole scheme. This is the same. We have m b pole to the 5th power, and then we'd write out what the series looks like in that scheme. You look like you're doing pretty good when you're at one loop. You went down by a factor of 5. But then you go into two loops, and you find that your correction at two loops is pretty much the same size as your correction at one loop. So you say, well, let's use m s bar. That changes the perturbative series. And maybe you think you're doing a little bit better because at least, well-- (CHUCKLING) but this guy got bigger, and this guy is bigger than that guy, so it's also not really working. Well, if the phenomenology is as accurate as I told you, you must imagine that there is something that does work better than that. Indeed there is. So there's other mass games, and I'll talk about one of them, continuing this table, where we switch to something called the 1S mass. Now [INAUDIBLE] looks like that. We're very happy. Or at least we're much happier than we were over here. So what is the 1S mass? The 1S mass is basically that you take half the perturbative mass for the epsilon system. So the 1S mass is m b b bar system calculated perturbatively, divided by 2. OK, we'll talk more about why this is working in a minute. Let me also write down-- these are conversion formulas because you calculate the decay rate once and for all, so you calculate the first line. Then you know how to convert between schemes, and that's what's causing the perturbative serious to differ. So m b pole is equal to m b m s bar, and there's a series that relates them that looks like that. And I could put in numbers here. And so that's what's changing the numbers when I go from this line to this line. Basically, kind of the 5 hours of 0.09 that you get takes you from here to here. We could also think about switching from the pole scheme to this m b 1S scheme. We have a different series. It still doesn't look very convergent. The numbers look small, but-- this is supposed to be a 6-- but this number here is bigger than that number. So that when I stick this series and combine it together with this series for the pole mass, then I get that series at the top of the board. And the problem is not in the m b 1S. The problem is in the m b pole. And the problem in the m b pole is being reflected in this series that relates them and its poor convergence. AUDIENCE: Can you explain [? what ?] [? m ?] [? b ?] [? bar ?] [? being ?] [? perturbative ?] [? means? ?] IAIN STEWART: Yeah, so you calculate-- so you can think of calculating the Coulomb potential between b and b bar, and that will give you alpha s corrections to just the mass. So you have 2 m b plus Coulomb potential plus, you know, radiative corrections to that. And you keep dressing it up and just perturbatively calculate as if it was a Coulomb problem, a QED problem, the mass of the b b bar state, and then you divide by 2. AUDIENCE: [INAUDIBLE] IAIN STEWART: Yeah AUDIENCE: But you just pretend? IAIN STEWART: There's nonperturbative corrections to it, but what we're extracting from it is the series. OK? I'll explain more why it's going to work, why this is, like, a reasonable choice, and we'll come to that. OK, so the lesson here is simply that the choice of mass scheme has a big impact on the perturbative series. And we haven't yet figured out why some things work and why some things don't, but we will. It's absolutely crucial to use the right one. Otherwise, the predictions will not be good. Well, physically, we can actually argue right away why m b pole is not so good. And that's because, physically, there is no pole. There's no pole in the quark propagator. And that is because of confinement. So we use this notion of pole and a quark propagator when we're doing perturbation theory. But nonperturbatively, it's not a well-defined notion, or at least there doesn't really exist poles in quark propagators. So it's only perturbatively a good notion. So I'll say it's only perturbatively meaningful. And in reality, it's ambiguous, it's an ambiguous notion. And you can think that the amount by which it's ambiguous is related to hadronization, which is set by the scale lambda QCD. So there's an ambiguity in what you mean by a pole mass physically due to nonperturbative effects. Now, when we set up HQET, we actually used m pole. So we're using a questionable physical parameter. So when we did these phase redefinitions, we were actually using the pole mass scheme because we were expanding about mass shell. So if you were to do other things, you can think about getting to other choices and implementing them in the HQET. No problem with doing that. But there does exist another operator that we sort of dropped without thinking too hard about it. So if we switch to using a different mass scheme, there's some delta m, which you can think of as a series in alpha s, which we wrote on the board a moment ago. And where that delta m would show up is that you would have kind of an L delta m operator, which would just be exactly delta m, and then Q v bar Q v. That would be left over when we cancel the masses in the Lagrangian, if you like. That's one way of thinking about it. So we have an extra term in the Lagrangian in HQET. And now you can ask, well, if we have an extra term, we'd better worry about power counting. And it turns out that the way we can understand why m s bar wasn't working is related to power counting. So in m s bar, if you look at what delta m is, delta m is m bar itself times alpha. But by power counting, you don't want something that's growing with m b in the Lagrangian. m b is suppressed by alpha, but it's alpha at m b that is providing [? some ?] [? suppression, ?] but that's only 0.2, and that's actually just not enough suppression to get rid of the big m b here. You don't get-- you go from 5 GV-- you know, there's some number in front. Maybe you get down to 1 and 1/2, 2 GV, but you're not getting down to a small enough value. So parametrically, this is just not good, both parametrically, because it grows with m b, and numerically it's too big. It's not order lambda QCD, which would be OK. So parametrically and numerically, delta m is just too big for HQET power counting in this m s bar scheme. You'd include an operator that effectively is on average larger and more important than your kinetic term because the v dot D operator-- v dot D is counting like lambda QCD, and it had v dot D with two Q's. Now you have something delta m with two Q's, and if it's on average larger, then you're messed up. So that's why actually physically the m b bar is not a good choice. You can think about it from the effective theory from that way. m b bar is actually a mass that's-- you're supposed to think of it as a mass that you use for high-energy physics, for physics really high energy, above the b quark mass scale. Then it's a good parameter. We're now doing physics below the b quark mass scale, and this is just one way of seeing why it's not such a good parameter there, because the perturbative corrections are just too big. So what about this 1S mass? So here, it turns out that if you calculate delta m, you get-- in the same way that you think about corrections to hydrogen being order alpha squared, you get m b 1s alpha squared. And numerically, that's small enough when you put the coefficient in. So it still doesn't make us feel very good because it grows with m b, but if I just care about numerics and I'm doing b quark physics, it works because the alpha squared is enough suppression that this is a good mass scheme, and that's why the perturbation theory treating as order lambda QCD numerically, if not parametically. Is OK. Now, if that doesn't sit pretty with you, you can do something even more fancy, and we'll talk about one example of more fancy a little later on, where you basically, instead of having m show up there, you have some parameter R, and you have something that you can just pick. And so you could pick a scheme here where R is of order lambda QCD-- by however, maybe you pick it be 1 GV or 500 MeV or whatever you like. So you can make up a scheme where this is true, and then both parametrically and numerically you're OK. And these schemes also work just as well as the 1S scheme. So we can get more fancy and also satisfy our formal requirement. All right, so that's kind of phenomenological numbers and a bit of physics. Let's come back to some mathematics and talk about what renormalons are mathematically. Is there a mathematical way of characterizing what's going wrong? What if I didn't think about this physics? Could I do a calculation and see that something's going wrong? And the answer is yes. So first I have to teach you a few things, if you don't know them already, about the asymptotics of perturbative series in quantum field theory. And these are actually not convergent series. So they are what are called asymptotic series. So what's the definition of an asymptotic series? So we say a function has an asymmetric series, which I'll write as follows. Some coefficients. And we'll just call the expansion parameter alpha. You can think of it as alpha s, if you like, but this is just math, so you don't have to think about it as anything but the expansion parameter. And we say that a function has an asymptotic series if and only if the following is true. f of alpha minus the partial sum up to some level n is less than kind of the scaling that you would get from the next term in the series, which is n plus 2, for some numbers K N plus 2. So that's actually a quite different definition than what you would have for a convergent series. For a convergence areas, you would say that you could pick any epsilon you like here. You could make the N big enough that this would get close to that. Here, I'm just saying that it's less than this with some power of alpha. So imagine that I pick alpha to be 0.1. The thing is that this K here could still grow with N, and that actually will happen. So the truncation area in some sense is not being bounded in these asymptotic series. And perturbation theory and quantum field theory is generically of this structure, and I'll show you some examples of that. It's asymptotic rather than convergent. So when you do a QFT calculation, a kind of typical result if you work out some asymptotics is that you would have these fn coefficients that are some power of a, and then times an n factorial. And that's kind of how they're scaling as n is getting large. You can think about the number of diagrams as just growing, but even when diagrams are not growing you can get these n factorials. So that means if you're fixing alpha as some value of the parameter, and no matter how small you take it, at some order in perturbation theory, you're basically running out of gas, and your truncation-- you start to-- your predictions start to grow and diverge. So the corresponding values of these K's, if you had these f's that were of that form, would be that the K's are basically growing in the same way. So you'd need larger and larger numbers in order to satisfy the definition of asymptotic over there, and having these large numbers you're never able to satisfy convergent. So the series has zero radius of convergence. And then you wonder, why have we been teaching you quantum field theory? [CHUCKLES] That's true. So what do we know? Even though these series are asymptotic, we can still make use of them. So typically what happens is that the series will decrease. For a while, it'll look like it's convergent, and then it'll start diverging. And you can characterize where it starts to diverge. And I'm not going to go through all the algebra, but I'll just tell you some facts. And so you could imagine that you're doing some perturbation series, and it looks like it's going well. Things are converging. But then, [? some ?] [? things ?] don't start to go so well. So think of what I'm plotting here as the partial sum of your perturbation theory at N-th order. I sum up all the connections up to N, so it's like the series that I was writing over here. It's this thing. Now, in perturbative QED, perturbative QCD, we're never getting more than a few terms. So you could say, well, OK, maybe I don't care so much about this. And in QED, that's actually the case. You don't really care so much about that. But in QCD, actually, the turnover happens already around three loops, and so as soon as you start including two loop corrections this is something you have to care about. Now, you can say, well, the best thing I can do is stop doing perturbation theory at some N star because that's kind of where it looks the best. Like, I've gone toward something in it. And that actually is not such a bad thing. And it turns out you can characterize actually the mistake you're making by stopping there, and the mistake you're making by stopping there is of the following form. So it's an exponential in 1 over alpha. So you would never be able to see kind of the correction that you'd need to get to the correct value, which I'm imagining is on the axis here, in perturbation theory because this doesn't have a perturbative expansion. And I'll actually show you that kind of the bad behavior here and this gap are related to power corrections. And they're exactly related to this connection that I was telling you between perturbative corrections and large-order asymptotics and power corrections. This type of exponential is something that we can relate to power corrections, and we'll see how that pans out, although we might-- probably won't get there until next class. OK, so this is kind of a prelude, and now we're going to go into more detail. And it turns out, in order to go into more detail here, that, much as you make Fourier transforms to explore another space, we're going to do a transform to something called Borel space. So it turns out that, when you have a divergent series or an asymptotic series, there's still degrees of divergence. And we can classify how divergent it is by using something called the Borel transform. Yeah? AUDIENCE: [INAUDIBLE] will the series still be asymptotic? IAIN STEWART: So no, it's not. Yeah, so 1S mass-- let's see. Yeah, it is still asymptotic, but it's-- in a way that I'll describe once we understand what this Borel is, it's much less divergent than the other series, than the pole series. If you think about the pole series as being degree-0 divergent, then this is like a few orders down less divergent. So once we define less divergent of a divergent series, then I'll be able to make that more precise. It's a question of when power corrections come in. And if you look at, like-- so physically, the 1S mass is a physical thing for an epsilon state. If you look at how power corrections come in, then they're suppressed because there's an ambiguity between power corrections in perturbation theory. If the power corrections are suppressed, you have less ambiguity in the perturbation theory. That's another way of thinking about it. But the answer to your question. AUDIENCE: [INAUDIBLE] IAIN STEWART: It basically affects this a. All right, lots of good questions. Let's see how we get there. So when I transform, I'm going to call the transform function capital F, and I'm going to call its argument B for Borel. And so what's the definition of the Borel transform? We're going to define it as follows. There's a first-order term, and I'm just going to put a delta function there. The real thing that matters is the series of terms that come next. And instead of having F to the n alpha to the n, I'm going to divide by an extra n factorial, and that's the definition of, given a set of F's, how I construct what's called the Borel transform. And because of the n factorial, I get improved convergence, so the n factorial is making it converge better. So if there's a transform, there should be an inverse transform. So here's the inverse transform. It's an integral from 0 to infinity, b over alpha F of b. And that would get me back to the F. So if you have a convergent series and you think of this transform, then you just can go back and forth, and you get back to the original-- the function that you started with. So if you have some series that's like that, you get back-- you could calculate that series, and you get some F of alpha just before Borel transforming. If you do this Borel transform, then you calculate this integral, you get back the same F of alpha from the inverse transform. Let me give you an example that shows you that, if you have a divergent series, it's not too crazy to define the sum of the divergent series by using these transforms. So I claim that, for a divergent series, where F of b and the inverse transform exist-- so if we start with a divergent series, but these things exist, then we could just use this transform as a way of defining F of alpha. So let me give you a simple example of that to convince you that it's not such a bad thing to do. So let's consider the following series. Just an alternating series, but alpha is a parameter greater than 1. That series doesn't converge. Of course, for alpha less than 1, it would converge. And if I asked you what it should be, you would say, well, I calculate for alpha less than 1, I analytically continue, and it's 1 plus alpha. Well, that's one way of getting there. Another way of getting there is using this Borel transform. So if you calculate it here, F of b, you sum up 0 to infinity minus b to the n-- once you put the minus 1 in there and there's an extra n factorial. So that's e to the minus b, perfectly well-behaved function. And then you have to do an interval, 0 to infinity, db. You could have e to the minus b over alpha times the F of b, which is e to the minus b, and that just gives you alpha over 1 plus alpha. And this is a perfectly well-defined interval for large alpha. Large alpha is even making it better. So the integral is perfectly well-defined, and we get an answer that we like. OK, so this Borel transform is a useful way of dealing with divergent series. The real question is, what happens if the inverse transform doesn't exist? And that's what's going to be the thing that we're most interested in. So if the F of the integral, integral over the F of b doesn't exist, then the integrand can tell us about the severity of the singularities, I.e., the severity of the divergence. And that's the real power of this Borel method. So let's do another example, where actually things will diverge. So let me take F n to be a to the minus n, n plus k factorial. And if we do that Borel transform of that-- if you have a series, you can do it. And there's only one piece of it that we care about, so I'm just going to write that piece. So there's this piece that has a kind of pole-like structure, and it has a pole at b equals a. And what you say is you call this pole a renormalon. This is the renormalon. You call it a b equals a renromalon. So you're characterizing the renormalon by where that pole is. A renormalon is a flavor of a particle. This has nothing to do with a particle. Usually, you think of poles as having something to do with particles, and that's where the name comes from. It's due to 't Hooft. Blame him. [CHUCKLES] So if a is less than 0, your integration contour is positive, and the pole is on the other side, so you don't have a problem. The inverse transform exists. These are called UV renormalons. They still can guide the perturbation theory, and they can be important for thinking about why the perturbation theory is behaving the way it is, but you can do the inverse transform. The real ones that are kind of problematic are a greater than 0 because then the pole is on the integration contour, and we can't do the transform. So we're integrating d b from 0 to infinity, and we just have a pole sitting on the axis, so what do we do? So you could think about characterizing how poorly behaved a perturbation theory is by looking at these poles, and that's kind of what I meant by having some kind of notion of how divergent a series is. So you look in this Borel space, and you look at the real axis, and those poles actually can be on both sides. The most severe pole is the one that's closest to the origin, and that has the smallest value of b. And you can see if you have a small value of b, then you're multiplying here with some large numbers. So if a was 0.1, you'd be multiplying by 10 to the n. So these guys are more severe. As you go to larger values, you get additional suppression from this a to the n, so if a was-- if you went all the way out to 100, then you'd have 1 over 100 to the n. That's good, but factorial eventually wins. So it's still divergent, but we look like it was a lot better. OK? So you can characterize how severe the series is diverging by the location of these poles. And you can characterize the ambiguity in doing an integration by-- if you just think about one pole, you can characterize the ambiguity, so let's just imagine there's one pole. You can characterize the ambiguity by going above or below. So you have two possible ways of defining the interval. You don't know which to pick. One is to go above. The other is to go below. And then you can think about what the ambiguity is. And the ambiguity is the contour 1. Call this c1. This is c2. The ambiguity is c1 minus c2, and that is circling the pole. So you can look at the residue of the pole, and that gives you the ambiguity. So what we'll do next time is we'll look at an example in perturbative QCD. We'll see that there's a series that has a renormalon, and we'll find it. And then we'll look at the residue, and we'll find actually that out will pop lambda QCD. So we'll see, we'll calculate, for some perturbative series, that the pole mass really has an ambiguity of order lambda QCD by using this technique of going around a residue of a pole. OK? AUDIENCE: [INAUDIBLE] when we say that the b quark has a 4 GV mass, what's the mass? [INAUDIBLE] IAIN STEWART: Yeah, so if you say the b quark has, like, 4.2 GV mass, that's the m s bar mass. If you say it has a 4.73 GV mass, that's like the m b 1S mass. And if you look at how the PDG measures the b quark mass, they measure it in one of these schemes that I'm telling you about, like the 1S scheme. They [INAUDIBLE] result for that. You can convert between the 1S and the m s bar mass. The m s bar mass is a perfectly well-defined quantity it doesn't actually have a renormalon, OK? That's not its problem. The pole mass has a renormalon. The m s bar mass, its problem is related to power counting in the low-energy theory. It's a good mass for high-energy physics. It's a bad mass for low-energy physics. That's the problem with m s bar. It's not it's not a technical problem, that it has a-- [INAUDIBLE] severity of renormalon. It can have a higher-order renormalon, but it doesn't have the same severity of renormalon as the pole mass. The pole mass basically has the most severe possible renormalon, which is, in some units, 1/2, so there's an extra 2 to the n. We'll talk about that next time. So basically, it's the u equals 1/2 renormalons that really are causing a problem in perturbative [? QCD ?] because they're the most important-- there's higher-order ones. Even alpha s actually has a normalon, alpha s in the m s bar scheme, I suspect. [CHUCKLES] But even up to five loops you won't see it, or it's very hard to see. Let's stop there.
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https://ocw.mit.edu/courses/7-91j-foundations-of-computational-and-systems-biology-spring-2014/7.91j-spring-2014.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation, or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Any questions from last time about Gibbs Sampling? No? So at the end, we introduced this concept of relative entropy. So I just wanted to briefly review this, and make sure it's clear to everyone. So the relative entropy is a measure of distance between probability distributions-- can be written different ways, often with this D p q notation. And as you'll see, it's the mean bit-score, if you're scoring a motif with a foreground model, pk, and a background model, qk it's the average log odd score under the motif model. And I asked you to show that under the special case where qk is 1 over 4 to the w, that is uniform background, that the relative entropy of the motif ends up being simply 2w minus 8. Did anyone have a chance to do this? It's pretty simple-- has anyone done this? Can anyone show this? Want me to do it briefly? How many would like to actually see this derivation? It's very, very quick. Few people, OK. So I'll just do that really quick. So summation Pk log Pk over qk equals-- so you rewrite it as a difference, the log of a quotient is the difference of the log. Of summation log Pk plus summation Pk log qk. OK, and then the special case that we're dealing with here is that qk is equal to a quarter, if we're dealing with the simplest case of a one-base motif. And so you recognize that that's minus H of Pp, right? H of p is defined as minus that, so it's minus H p. And this here, that's just a quarter. Log 2 of a quarter is minus 2. You can take the minus 2 outside of the sum, so you're ending up with minus 2-- I'm sorry. How come Sally didn't correct me? Usually she catches these things. So that's a minus there, right, because we're taking the difference. And so then we have a minus 2 that we're pulling out from this, and you're left with summation Pk. And summation Pk, it sums to 1. So that's just 1. And so this equals minus minus 2, or 2 minus H of p. And there are many other results of this type that can be shown in information theory. Often there are some simple results you can get simply by using this by splitting it into different terms, and summing. So another result that I mentioned earlier, without showing, is that if you have a motif, say, of length 2, that the information content of that motif model can be broken into the information content of each position if your model is such that the positions are independent. So you would have, in that case-- let's just take the entropy of a model on [? dinucleotides. ?] It that would be minus summation Pi Pj log Pi Pj, if you have a model where the two are independent, and this sum would be taken over both i and j. And so if you want to show that this is equal to-- I claim that this is equal to the i-- Anyway, if you have different positions, in general-- this would be the more general term-- where you have two different compositions at the two positions for the motif. And then you can show that it's equal to basically the sum the entropies at the two positions. OK, you do the same thing. You separate out the log of the sum, in terms of the sum of the logs, and then you do properties of summations until you get the answer. OK, so this is your homework, and obviously it won't be graded. But we'll check in next Thursday and see if anyone has questions with that. So what is the use of relative entropy? the main use in bio-informatics is that it's a measure that takes into account non-uniform backgrounds. The standard definition of information basically works when the background is uniform, but falls apart when it's non-uniform. So if you have a very biased genome, like this one shown here which is 75% A T, then the information content using the standard method would be two bits of this motif, which is P C equals 1. But then, that would predict, using the formula, that a motif occurs 2 to the information content-- once every 2 to the information content bases-- that would be 2 to the 2, which would be 4 bases, and that's clearly incorrect in this case. But the relative entropy, if you do it, there will be four terms, but three of them just have a 0. And then one of them has a 1, so it's 1 times log 1 over 1/8, in this case, and that's will be equal to 3. And so the relative entropy clearly gives you a more sensible version. It's a good measure for non-uniform backgrounds. Questions about relative entropy? All right, so then we said you can use a weight matrix, or a position-specific probability matrix, for a motif like this five-prime splice site motif, assuming independence between positions. But if that's not true, then a natural generalization would be an inhomogeneous Markov model. So now, we're going to say that the base at position k depends on the base at position k minus 1, but not on anything before that. And so, the probability of generating a particular sequence, S1 to S9, is now given by this expression here, where you have for every base after the first, you have a conditional probability. This is the conditional probability of seeing the base, S2, at position minus 2, given that you saw S1 at position minus 3, and so forth. And again, you can take the log for convenience, if you like. So I actually implemented both of these models. So just for thinking about it, if you want to implement this, you have parameters-- these conditional probability parameters-- and you estimate them as shown here. So remember, conditional probability of A given B is the joint probability divided by the probability of B. And so in this case, that would be the joint probability of seeing C A at minus 3, minus 2, divided by the probability of seeing C at minus 3. You could have the ratio of the frequencies, or, in this case, the counts, because the normalization constant will cancel. Is that clear? So I actually implemented both the weight matrix model and a first-order Markov model of five-prime splice sites, and scored some genomic sequence. And what you can see here, the units are in 1/10th-bit units, is that they both are partially successful in separating real five-prime splice sites-- shown in black from the background, shown in light bars-- but in both cases, it's not a perfect separation. There's some overlap here. And if you zoom there, you can see that the Markov model is a little bit better. It has a tighter tail on the left. So it's generally separating the true from the decoys a little bit better. Not dramatically better, but slightly better. Yes, question? AUDIENCE: From the previous slide, could you clarify what the letter R and the letter S are? PROFESSOR: Yes, sorry about that. R would be the odd ratio-- so it's the ratio of the probability of generating that sequence under the foreground model-- the plus model, we're calling it-- divided by the probability under the background, or minus, model. And then, I think I pointed out last time that when you get products of probabilities, they tend to get very small. This can cause computational problems. And so if you just take the log, you convert it into a sum. And so we'll often use score, or S, for the log of the odds ratio. Sorry, should have marked that more clearly. So Markov models can improve performance, when there is dependence, and when you have enough data to estimate the increased number of parameters. And it doesn't just have to be dependence on the previous base-- you can have a model where the probability of the next base depends on the two previous bases. That would be called a second-order Markov model, or in general, a K-order Markov model. Sometimes, these dependencies actually occur in practice. With five-prime splice sites, it's a nice example, because there's probably a couple thousand of them in the human genome, and we know them very well, so you can make quite complex models and have enough data to train them. But in general, if you're thinking about modeling a transcription factor binding site, or something, often you might have dozens or, at best, hundreds of examples, typically. And so you might not have enough to train some of the larger model. So how many parameters do you need to fit a K-order Markov model? So question first, yeah? AUDIENCE: [INAUDIBLE] If you're comparing the first-order Markov models with W M M, what is W M M? PROFESSOR: Weight matrix, or position-specific probability matrix. Just a model of independence between the two. Coming back to this case. So let's suppose you are thinking about making a K-order Markov model, because you do some statistical tasks and you find there's some dependence between sets of positions in your motif. How many parameters would there be? So if you have an independence model, or weight matrix or position-specific probability matrix, there are four parameters at each position, the probabilities of the four bases. This will be only three free parameters, because the fourth one-- but let's just think about it as four, four parameters times the width of the motif. So if I now go to a first-order Markov model, now there's more parameters, because I have these conditional probabilities at each position. So how many parameters are there? For a first-order Markov? How many do I need to estimate? Yeah, Kevin? AUDIENCE: I think it would be 16 at each position. PROFESSOR: Yeah, 16 at each position, except the first position, which has four. OK, and what about a second-order Markov model, where you condition on the two previous positions? 64, right? Because you have two possible bases you're conditioning on, that's 16 possibilities times 4. And so in general, the formula is 4 to the k plus 1. This is really the issue-- if you have only 100 sequences, and you need to estimate 64 parameters at each position, you don't have enough data to estimate those. So you shouldn't use such a high order model. All right, so let's think about this-- what could happen if you don't have enough data to estimate parameters, and how can you get around that? So let's just take a very simple example. So suppose you were setting a new transcription factor. You had done some sort of pull-down assay, followed by, say, conventional sequencing, and identified 10 sequences that bind to that transcription factor. And these are the 10 sequences, and you align them. You see there is sort of a pattern there-- there's usually an A at the first position, and usually a C at the second, and so forth. And so you consider making a weight matrix model. Then you tally up-- there's eight A's, one C, one G, and no T's at the first position. So how confident can you be that T is not compatible with binding of this transcription factor? Who thinks you can be very confident? Most of you are shaking your head. So if you're not confident, why are you not confident? I think-- wait, were you shaking your head? What's the problem here? It's just too small a sample, right? Maybe T occurs rarely. So suppose that T occurs at a frequency of 10%, what's the probability of that in natural sequences? And we just have a random sample of those. What's the probability we wouldn't see any T's in a sample of size 10? Anyone have an idea? Anyone have a ballpark number on this? Yeah, Simona? AUDIENCE: 0.9 to the 10th. PROFESSOR: 0.9 to the 10th, OK. And what is that? AUDIENCE: 0.9 is the probability that you grab one, and don't see a T, and then you do that 10 times. PROFESSOR: Yeah, exactly. In genera it's a binomial thing, but it works out to be 0.9 to the 10th. And that's roughly-- this is like a Poisson. There's a mean of 1, so it's roughly E to the minus 1, so about 35% chance that you don't see any T's. So we really shouldn't be confident. T probably doesn't have a frequency of 0.5, but it could easily have a frequency of 10% or even 5%, or even 15%. And you might have just not seen it. So you don't want to assign a probability 0 to T. But what value should you assign for something you haven't seen? Sally? So, it turns out there is a principled way to do this called pseudocounts. So basically, if you use maximum likelihood estimation, you get-- maximum likelihood, it turns out, is equal to the observed frequency. But if you assume that the true frequency is unknown, but was sampled from all possible, reasonable frequencies-- so that's a Dirichlet distribution-- then you can calculate what the posterior distribution is in a Bayesian framework, given that you observed, for example, zero T's, what's the distribution of that parameter, frequency of T? And it turns out, it's equivalent to adding a single count to each of your bins. I'm not going to go through the derivation, because it takes time, but it is well described in the appendix of a book called, Biological Sequence Analysis, published about 10, 15 years ago by a number of leaders in the field-- Durbin, Eddy, Krogh, and Mitchison. And there's also a derivation of this in the probability and statistics primer. So basically you just do this poster calculation, and it turns out to be equivalent to adding 1 count. So when you add 1 count-- and then, of course, you re-normalize, and then you get a frequency-- what effectively it does is it will reduce the frequency of things that you observe most commonly, and boost up the things that you don't see, so that you actually end up assigning a probability of 0.07 to T. Now, if you had a larger sample-- so let's imagine instead of 8 1 1 0, it was 80 10 10 0, you still at a single count. So you can see in that case, you're only going to be adding a very small, close to 1%, for T. So as you get more data, it converges to the maximum likelihood estimate. But it does something more reasonable, more open-minded, in a case where you're really limited in terms of data. So the limitation-- you always want to be aware when you're considering going to a more complex model to get better predictability-- you want to be aware of how much data you have, and whether you have enough to accurately estimate parameters. And if you don't, you either simplify the model, or if you can't simplify it anymore, consider using pseudocounts. Sometimes you'll see smaller pseudocounts added-- like instead of 1 1 1 1, you might see a quarter, one pseudocount distributed across the four bins. There's arguments pro and con, which I won't go into. So for the remainder of today, I want to introduce hidden Markov models. We'll talk about some the terminology, some applications, and the Viterbi algorithm-- which is a core algorithm when using HMMs to predict things-- and then we'll give a couple examples. So we'll talk about the CpG Island HMM, which is about the simplest HMM I could think of, which is good for illustrating the mechanics of HMM. And then a couple later, probably coming into the next lecture, some examples of real-world HMMs, like one that predicts transmembrane helices. So some background reading for today's lecture that's posted on the course website, there's a nature biotechnology primer on HMMs, there's a little bit in the textbook. But really, if you want to understand the guts of HMMs, you should read the Rabiner tutorial, which is really pretty well done. For Thursday's lecture, I will post another of these nature biotechnology primers on RNA folding. This one is actually has a little bit more content, takes a little bit longer to absorb probably than some of the others, but still a good introduction to the topic. And then it turns out the text has a pretty good section on RNA folding, so take a look at chapter 11. So hidden Markov models can be thought of as a general approach for modeling sequence labeling problems-- you have sequences, they might be genomic sequences, protein sequences, RNA sequences. And these sequences have features-- promoters, they may have domains, et cetera, linear motifs. And you want to label those features in an unknown sequence. So a classical example would be gene finding. You have a genomic sequence, some parts are, say, exon, some are introns. You want to be able to label them, it's not known. But you might have a training set of known exons and introns, and you might learn what the sequence composition of each of those labels looks like, and then make a model that builds things together. And what they allow you to do, though, with HMMs is to have transition probabilities between the different states. You could model states, you can model the length of different types of states to some extent-- as we'll see-- and you can model which states need to follow other states. They're relatively easy to design, you can just simply draw a graph. It can even have cycles in it, that's OK. And they've been described as the LEGOs of computational sequence analysis. They were developed originally in electrical engineering four or five decades ago for applications in voice recognition, and they're still used in voice recognition. So when you are calling up some large corporation, and instead of a person answering the phone, some computer answering the phone and attempting to recognize your voice, it could well be an HMM on the other end, which is either correctly recognizing what you're saying, or not. So you can thank them or blame them, as you wish. All right, so Markov Model example-- we did this before, imagine the genotype at a particular locus, and successive generations is thought of as a Markov chain. Bart's genotype depends on Homer's, but is conditionally independent of Grandpa Simpson's, given Homer's. So now what's a hidden Markov model? So imagine that our DNA sequencer is not working that week, we can't actually go in and measure the genotype. But instead, we're going to observe some phenotype that's dependent on genotype. But it's not dependent in a deterministic way, it's dependent in a more complex way, because there's an impact of environment, as well, let's say. So we're imagining that your genotype at the apolipoprotein locus is correlated with cholesterol, but doesn't completely predict it. So you're homozygous, you tend to have higher LDL cholesterol than you are heterozygous. But there's a distribution depending on how many doughnuts you eat, or something like that. Imagine that we observe that grandpa had low cholesterol, 150, Homer had high cholesterol, and Bart's cholesterol is intermediate. Now if we had just observed Bart's cholesterol, we would say, well, it could go either way. It could be homozygous or heterozygous. You would just look at the population frequency of those two, and would use that to guess. But remember, we know his father's cholesterol, which was 250, makes it much more likely that his father was homozygous, and then that, in turn, biases the distribution [? of it. ?] So that'll make it a little bit more likely that Bart, himself, is homozygous, if you didn't know. So this is the basic idea-- you have some observable phenotype, if you will, that depends, in a probabilistic way, on something hidden. And that hidden thing has some dependent structure to it. And you want to, then, predict those hidden states from the observable data. So we'll give some more examples coming up. And the way to think about these models, or at least a handy way to think about them, is as generative models. And so this is from the Rabiner tutorial-- you imagine an HMM used in order to generate observable sequences. So there's these hidden states-- think of them as genotypes-- observable-- think of them as the cholesterol levels. So the way that it works is you choose an initial state from one of your possible hidden states, according to some initial distribution, you set the time variable equal to 1. In this case, it's T, which will, in our case, often be the position in the sequence. And then you choose an observed value, according to some probability distribution, but it depends on what that hidden state was. And then you transition to a new state, and then you emit another one. So we'll do an example. Let's say bacterial gene finding is our application, and we're going to model a bacterial gene-- these are protein coding genes, only it's got to have a start coat on, it's got to have an open reading frame, and then it's got to have a stop code on it. So how many different states do we need in our HMM? What should our states be? Anyone? Do you want to make that-- Tim? AUDIENCE: Maybe you need four states, because the start state, the orf state, the stop state, and the non-genic state. PROFESSOR: OK, start, orf, stop and then intergenic, or non-genic. OK now, remember these are the hidden states, so what are they going to emit? They emit observable data, what's that observable data going to be? Sequence, OK. And how many bases of sequence should each of them emit? AUDIENCE: Well I guess we don't know. PROFESSOR: You have a choice. You're the model builder, you can do anything you want. 1, 5, 10-- any number of bases you want. And they can emit different things, if you want. This is generative, you can do anything you want-- there will be consequences later, but for now-- I'm going to call this-- go ahead. AUDIENCE: You could start with the start and the stop states maybe being three. PROFESSOR: Three, OK. So this is going to emit three nucleotides. How about this state? What should this emit? AUDIENCE: Any number. PROFESSOR: Any number? Yeah, OK, Sally? AUDIENCE: If you let it emit one number, and then add a self-cycle, then that would work. PROFESSOR: So Sally wants to have this state emit one nucleotide, but she wants it to have a chance of returning to itself. So that then we can have strings of N's to represent intergenic. Does that make sense? And these, I agree. Three is a good choice, here. If you had this one emit three, as well, then your genes would have to be a multiple of three apart from each other, which isn't realistic. You would miss out on some genes for that. So this has to be able to emit arbitrary numbers. So you could either have it emit an arbitrary number, but it's going to turn out to make the Viterbi algorithm easier if it just emits one and recurs, as Sally suggested. And then we have our orf state. So how about here? What should we do here? AUDIENCE: It can be three, and then you put the circle [INAUDIBLE]. PROFESSOR: So I'm going to change the name to Codon, because it's going to emit one codon-- three nucleotides. And then recur to itself. And now what transitions should we allow between states? AUDIENCE: So start to four, orf to stop, then stop to N, and then to start. PROFESSOR: Any others? Yeah? AUDIENCE: N could go to stop, as well. PROFESSOR: I'm sorry, N could go to stop? AUDIENCE: Yeah, so that the gene [INAUDIBLE]. PROFESSOR: OK, so that's a question. We're thinking of a gene on the plus strand, a gene could well be on the opposite strand. And so we should probably make a model of where you would hit stop on the other strand, which would emit a triplet of the inverse complement of the stop code, [INAUDIBLE] et cetera. That's true, excellent point. And then you would traverse this whole circle in the opposite direction. But it wouldn't be the same state. It would be stop-- because it would emit different things. So you'd have minus stop, stop minus strand. And then you'd have some other states there. And I'm not going to draw those, but that's a point. And you could have a teeny one-codon gene if you want, but probably not worth it. All right, everyone have an idea about this HMM? So this is a model you have to specify in order for this to actually generate sequence. This model will actually generate annotations and sequence. You have to specify where to start, so you have to have some probability of starting, but the first base that you're going to generate is going to begin intergenic, or start, or codon, et cetera. And you might give it a high probability of this, and then it'll generate a label. So for example, let's say N. And then it'll generate a base, let's say G. And then you look at these probabilities, so the transition probability here, versus this-- you either generate another N, or you generate a start. And let's say you go to start, and then you'll generate three bases, so A T G. And then you would go to the codon state, you would emit some other triplet, and so forth. So this is a model that will generate strings of annotations with associated bases. Still doesn't predict gene structure yet, but at least it generates gene structures. All right, so we are going to, for the sake of illustrating the Viterbi algorithm, we're going to use a simpler HMM in that. So this one only has two states, and its purpose is to predict CPG islands in a vertebrate genome. So what are CPG islands? Anyone remember? What is a CPG island? Anyone heard of this? I'm sure some of you. Well, the definition here is going to be regions of high CNG content, and relatively high abundance of CPG dinucleotides, which are unmethylated. So what is the P here? The p means that the CG we're talking about is C followed by G along the particular DNA strand, just to distinguish it from C base paired with G. We're not talking about a base pair here, we're talking about C and G following each other along the strand. So this dinucleotide is rare in vertebrate genomes, because CPG is the site of a methylase, and methylation of the C is mutogenic-- it lends to a much higher rate of mutations. so CPGs often mutate away, except for the ones that are necessary. But there are certain regions, often near promoters, that are unmethylated, and therefore, CPGs can accumulate to higher frequencies. And so you can actually look for these regions and use them to predict where promoters are. That's one application. So they have higher CPG dinucleotide content, and also higher C and G content. The background of the human genome is about 40% C G only, so it's a bit AT rich, and so you see these patches of, say, 50% to 60% C G that are often associated with promoters, with promoters of roughly half of human genes. So we're going to-- I always drop that little clicker thing. Here it is. We're going to make a model of these, and then run it to predict promoters in the gene. So here's our model. We have two states, we have a genome state-- this sort of generic position in the genome-- and then we have an island state. We have the simplest possible transitions, you can go genome to genome, genome to island, island to genome, or island to island. So now you can generate islands of arbitrary size, interspersed with genomic regions of arbitrary size. And then each of those hidden states is going to emit a single base. So a CPG island in this model is a stretch of I states in a row flanked by G states, if you will. Everyone clear on this set up? Good. So here, in order to fully specify the model, you need to say what all the parameters are. And there are really three class of parameters. There are initiation probabilities-- so the green here is the notation used in the Rabiner tutorial, except they call them [? pi ?] [? j's. ?] So here, I'm going to say it's a 99% chance you start in the generic genome state, and a 1% chance you start in an island state, because islands are not that common. And then you need to specify transition probabilities. So there's four possible transitions you could make, and you need to assign probabilities to them. So if the average length of an island were 1,000 bases, then a reasonable value for the I to I transition would be 0.999. You have 99.9% chance of making another island, and 0.1% chance of leaving that island state. If you just run that in this generative mode, it would generate a variety of lengths of islands, but on average, they'd be about one kb long, because the probability of terminating is one in 1,000. And then if we imagine that those one kb islands are interspersed with genomic regions that are about, say, 100 kilo bases long on average, then you would get this [? five ?] [? nines ?] probability for P G G, and 10 to the minus fifth as a probability of going from genome to islands. That would generate widely spaced islands, that are on average 100 kb apart, that are about one kb in length. Is that making sense? And now the third type of probability we need to specify are called emission probabilities, which are the B J K in Rabiner notation. And this is where the predictive power is going to come in. There has to be a difference in the emissions if you're going to have any ability to predict these features, and so we're going to imagine that the genome is 40% C G, and islands are 60% C G. So it's a base composition that we're modeling. We're not doing the dinucleotides here, that would make it more complicated. We're just looking for patches of high G C content. So now we've fully specified our model. The problem here is that the model is written from the hidden generating the observable, and the problem we're faced with, in practice, is that we have the observable sequence, and we want to go back to the hidden. So we need to reverse the conditioning that's in the model. So when you see this type of problem, how do you reverse conditioning? In general, what's a good way to do it? You see P A, given B, but the model you have is written in P B, given A. What do you do? What's that? AUDIENCE: Bayes theorem. PROFESSOR: Yeah, Bayes theorem. Right, let's do base theorem here. Remember, definition of conditional probability, if we have P A given B-- so this might be the hidden states, given the observables-- we want to write that in terms of P B given A. So what we do first? How do we derive Bayes rule? You first write the definition of conditional probability-- right, that's just the definition. And now what do I do? Split the top part into what? AUDIENCE: A times P of B given [INAUDIBLE]. PROFESSOR: P B given A. That's just another way of writing joint probability of P A B, using the definition of conditional probability again. So now it's written the other way. That's basically the idea. So this is the simple form. And like I said, I don't usually call it a theorem, because it's so simple-- it's something you can derive in 30 seconds. It should be called maybe a rule, or something. There is a more general form. This is where you have two states, basically B or not B that you're dealing with. And there's this more general form that's shown on the slide, which is when you have many states. And it basically is the same idea, it's just that we've rewritten this term, this P B, and we split it up into all the possible states. The slide starts from PB given A and goes the other-- anyway, so you rewrite the bottom term as a sum of all the possible cases. All right, so how does that apply to HMMs? So with HMMs, we're interested in the joint probability of a set of hidden states, and a set of observable states. So H, capital H, is going to be a vector that specifies the particular hidden state-- for instance, island or genome-- at position 1, that's H1 all the way to H N. So little h's are specific values for those hidden state. And then, big O is a vector that describes the different bases in the genome. So O1 is the first base in the genome, up to O N. One can imagine comparing two H vectors, one of which, H versus the H primes, what's the what's the probability of this hidden state versus that? You could compare them in terms of their joint probabilities with this model, and perhaps favor those that have higher probabilities. Yeah? AUDIENCE: [INAUDIBLE] of the two capital H's have any different notation? Like the second one being H prime, or something like that? Or are they supposed to be-- PROFESSOR: With the H, in this case, these are probability statements about random variables. So H is a random variable, which could assume any possible sequence of hidden states. The little h's are specific values. So for instance, imagine comparing what's the probability of H equals genome, genome, genome, versus the probability that H equals genome, genome, island. So the little h's, or the little h primes, are specific instances. The H's is a random variable, unknown. Does that help? OK, so how do we apply Bayes' rule? So what we're interested in here is the probability that H, this unknown variable that represents hidden states, that it equals a particular set of hidden states, little h1 to h N, given the observables, little o1 to little oN, which is the actual sequence that we see. And we can write that using definition of conditional probability as the joint probability patient of H and O, over the probability of O. And then Bayes' rule, we just apply conditional probability again. It's P H times P O, given H, over P O. So it turns out that this P O-- so what is P O equals O1 to O N in this model? Well, the model specifies how to generate the hidden states, and how the observables are generated from those hidden states. So P O is actually defined as the sum of P O comma H equals the first possible hidden state, plus the same term for the second. You have to sum over all the possible outcomes of the hidden states, every possible thing. So if we have a sequence of length three, you have to sum over the possibility that H might be G G G or G G I, or G I G, or G I I, or I G G, et cetera. You have to sum over eight possibilities here. And if the sequence is a million long, you have to sum over 2 to the one millionth possibilities. That sounds complicated to calculate. So it turns out that there's actually a trick, and you can calculate it. But you don't have to. That's one of the good things, is that we can just treat it as a constant. So notice that the denominator here is independent of the H's. So we'll just treat that as a constant, an unknown constant. And what we're interested in, which possible value of H has a higher probability? So we're just going to try to maximize P H equals H1 to H N-- find the optimal sequence of hidden states that optimizes that joint probability, the joint probability with the observable values, O1 to O N. Is that making sense? Basically, we want to find the sequence of hidden states, we'll call it H opt. So now H opt here is a particular vector. Capital H, by itself, is a random vector. This is now a particular vector of hidden states, H1 opt through H N opt. And it's defined as the vector of hidden states that maximizes the joint probability with O equals O1 to O N, where that's the observed sequence that we're dealing with. So now what I'm telling you is if we can find the vector of hidden states that maximizes the joint probability, then that will also maximize the conditional probability of H given O. And that's often the language of linguistics is used, and it's called the optimal parse of the sequence. You'll see that sometimes, I might say that. So the solution is to define these variables, R I of H, which are defined as the probability of the optimal parse of the subsequence from one to I-- not the whole long sequence, but a little piece of it from the beginning to a particular place in the middle, that ends in state, H. And so first. We calculate R, R 1 1, the probability of generating the first base, given that it ended in hidden state 1. And then we would do the hidden state 2, and then we basically have to figure out a way, a recursion, for getting the probabilities of the optimal parses ending at each of the states at position 2, given the values at position 1. And then we go, work our way all the way down to the end of the sequence. And then we'll figure out which is better, and then we'll backtrack to figure out what that optimal parse was. We'll do an example on the board, this is unlikely to be completely clear at this point. But don't worry. So why is this called the Viterbi algorithm? Well, this is the guy who figured it out. He was actually an MIT alum. He did his bachelor's and master's in double E, I don't know, quite awhile ago, '50s or '60s. And later went on to found Qualcomm, and is now big philanthropist, who apparently supports USC. I don't know why he lost his loyalty to MIT, but maybe he'll come back and give us a seminar. I actually met him once. Let's talk about his algorithm a little more. So what I want to do is I want to take a particular HMM, so we'll take our CPG island HMM, and then we'll go through the actual Viterbi algorithm on the board for a particular sequence. And you'll see that it's actually pretty simple. But then you'll also see that it's not totally obvious why it works. The mechanics of it are not that bad, but the understanding really how it is able to come up with the optimal parses, that's the more subtle part. So let's suppose we have a sequence, A C G. Can anyone tell me what the optimal parse of this sequence is, without doing Viterbi? With this particular model, these initiation probabilities transitions and emissions? Do you know what it's going to be in advance? Any guesses? AUDIENCE: How about genome, island, island. PROFESSOR: Genome, island, island. Because, you're saying, that way the emissions will be optimized, right? Because you'll omit the C's and G's. OK, that's a reasonable guess. Sally's shaking her head, though. AUDIENCE: The transitional probability from being in the genome is very, very small, and so it's more likely that it'll either only be in the genome or only be in the island. PROFESSOR: So the transition from going from a genome to island or island to genome is very small, and so she's saying you're going to pay a bigger penalty for making that transition in there, that may not be offset by the emissions. Right, is that your point? Yeah, question? AUDIENCE: Check here-- when we're talking about the optimal parse, we're saying let's maximize the probability of that letter-- PROFESSOR: The joint probability. AUDIENCE: Sorry, the joint probability of that letter-- PROFESSOR: Of that [INAUDIBLE] state and that letter. AUDIENCE: OK, so that means-- PROFESSOR: Or that set of [INAUDIBLE] states and that set of bases. AUDIENCE: So when we're computing across this three-letter thing, we have to say the probability of the letter, then let's multiply if by the probability of the transition to the next letter, and then multiply it again [INAUDIBLE] and that letter. PROFESSOR: So let's do this. If A C G is our sequence-- I'm just going to space it out a little bit. Here's our A at position one, here's our C at position two, and our G at position three. And then we have our hidden states. And so we'll write genome first, and then we have island here. And so what is the optimal parse of the sequence from base one to base one, that ends in genome? It's just the one that starts in genome, because it doesn't go-- right, so it's just genome. And what is its probability? That's how this thing is defined here. This is this R I H thing I was talking about. This is H, and this is I here. So the probability of the optimal parse of the sequence, up to position one, that ends in genome, is just the one that starts in genome, and then emits that base. So what's the probability of starting in genome? It's five nines, right? So that's the initial probability-- 9 9 9 9 9. And then what's the probability of emitting an A, given that we're in the genome state? 0.3. So I claim that this is the value of R1 of genome, of the genome state. OK, that's the optimal parse. There's only one parse, so there's nothing-- it is what it is. You start here, there's no transitions-- we started here-- and then you emit an A. What's the probability of the optimal parse ending an island at position one of the sequence? Someone else? Yeah, question? AUDIENCE: Why are we using the transition probability? PROFESSOR: This is the initial-- Oh, I'm sorry. Correct. Thank you, thank you-- what was your name? AUDIENCE: Deborah. PROFESSOR: Deborah, OK, thanks Deborah. It should be the initial probability, which is 0.99, good. Initial probability. What about island? Deborah, you want to take this one? AUDIENCE: 0.01. PROFESSOR: 0.01 to be an island. And what about the emission probability? We have to start an island, and then emit an A, which is probably what? AUDIENCE: 0.2. PROFESSOR: 0.2, yeah, it's up on the screen. Should be, hopefully. Yeah, 0.02. So who's winning so far? If the sequences ended at position one? Genome, genome's winning. This is a lot bigger than that, it's about 150 times bigger, right? Now what do when we go-- we said we have to do recursion, right? We have to figure out the probability of the optimal parse ending at position two in each of these states, given the optimal parse ending at position one. How do we figure that out? What are we going to write here, or what do we have to compare to figure out what to put here? There's two possible parses ending in genome at position two-- there's the one that started in genome at position one, and there's the one that started at island. So you have to compare this to this. So you compare what the probability of this parse was times the transition probability, and then the emission in that state. So what would that be? What would this be, if we stay in genome? What's the transition? Now we've got our five nines, yeah, good. So five nines. And the emission is what? AUDIENCE: Genome at 0.2. PROFESSOR: 0.2, right. And times this. And what about this one? What are we going to multiply when we consider this island to genome transition? AUDIENCE: [INAUDIBLE] 0.01? Because the genome is still 0.2. PROFESSOR: It's still 0.2. So we take the maximum of these, right? We're doing optimal parse, highest probability. So which one of these two turns this bigger? Clearly, the top one is bigger. This one is already bigger than this, and we're multiplying by the same data, so clearly the answer here is 0.99 times 0.3 times-- my nines are going to have to get really skinny here-- 0.99999 and 0.2. That's the winner. And the other thing we do, besides recording this number, is we circle this arrow. Does this ring a bell? This is sort of like Needleman-Wench or Smith-Waterman, where you don't just record what's the best score, but you remember how you got there. We're going to need that later. And what about here? What's the optimal parse-- what's the probability of the optimal parse ending in island at position two? Or what do I have to do to calculate it? Sorry? AUDIENCE: You have to calculate the [INAUDIBLE]. PROFESSOR: Right, you consider going genome to island here, and island to island. And who's going to win that race? Do you have an idea? Genome to island had a head start, but it pays a penalty for the transition. The transition is pretty small, that's 10 to the minus fifth. And what about this one? This has a much higher transition probability of 0.999. And so even though you were starting from something, this is about 150 times smaller than this. But this is being multiplied by 10 to the minus fifth, and this is being multiplied by something that's around 1. So this one will win, island to island will win. Everyone agree on that? And what will that value be? So it's whatever it was before times the transition, which is what? From island to island? 999. Times the emission which is? 0.3. Island is more likely [? to omit ?] a C. Everyone clear on that? And then, we're not that until we circle this arrow here. That was the winner, the winner was coming from island, remaining on island. And then we keep going like this. Do you want me to do one more base? How many people want me to do one more base, and how many people want me to stop this? I'll do one more base, but you guys will have to help me a little bit. Who is going to win-- now we want the probability of the optimal parse ending in G, ending at position three, which is a G, and ending in genome, or ending in island. So for ending in genome, where is that one going to come from? Which is going to win? This one, or this one? AUDIENCE: Stay in genome. PROFESSOR: Yeah, stay in genome is going to win. This one is already bigger than this, and the transition probability here-- this is a 10 to minus 3 transition probability. And this is a probability that's near one, so the transitions are going to dominate here. And so you're going to have this term-- I'm going to call that R 2 of G. That's this notation here. And times the transition probability, genome to genome, which is all these nines here. And then the emission probability of a G in the genome state is 0.2. And who's going to win here for the optimal parse, ending at position three, in the island state? Is it going to be this guy, to island, or this one, changing from genome to island? Island to island, because, again, the transmission probability is prohibitive-- that's a 10 to the minus fifth penalty there. So you're going to stay in island. So this one won here, and this one won here. And so this term here is R 2 of island times the transition probability, island to island, which is 0.999 times the emission probability, which is 0.3 of a G in the island state. Everyone clear on that? Now, if we went out another 20 bases, what's going to happen? Probably not a lot. Probably the same kind of stuff that's happening. That seems kind of boring, but when would we actually get a cross? What would it take? To push you over and cause you to transition from one to the other? AUDIENCE: Slowly stacked against you or long enough? PROFESSOR: Yeah, that's a good way to put it. So let me give you an example. This is the Viterbi algorithm, written out mathematically. We can go over this in a moment, but I just want to try to stay with the intuition here. We did that, now I want to do this. Suppose your sequence is A C G T, repeating 10,000 times. Can anyone figure out what the optimal parse of that sequence would be, without doing Viterbi in their head? Start and stay in genome. Can you explain why? AUDIENCE: Because it's equal to the-- what are the [? widths? ?] Because it's homogeneous to the distributor, as opposed to enriched for C N G, and it just repeats without pattern. Or it repeats throughout without [? concentrating the C N Gs ?] anywhere. PROFESSOR: Right, so the unit of the repeat, this A C G T unit, is not biased for either one. So there will be 2.3 emissions, and 2.2 emissions, whether you go through those in G G G G or in I I I I. Does that make sense? So the emissions will be the same, if you're all in genome, or if you're all in island. And the initial probabilities favor genome, and the transitions also favor staying in genome. Right, so all genome. Can everyone see that? So do you want to take a stab at this next one? This one's harder. Let me ask you, in the optimal parse, what state is it going to end in? AUDIENCE: Genome. PROFESSOR: Genome, you've got to run [? with a 1,000 ?] T's. And genome, the emissions favor emitting T's. So clearly, it's going to end in genome. And then, what about those runs of C's and G's in the middle there? Are any of those long enough to trigger a transition to island? What was your name again? AUDIENCE: Daniel. PROFESSOR: Daniel, so you're shaking your head. You think they're not long enough. So you think the winner's going to be genome all the way? Who thinks they're long enough? Or maybe some of them are? Go ahead, what was your name? AUDIENCE: Michael. PROFESSOR: Michael, yeah. AUDIENCE: The ones at length 80 and 60 are long enough, but the one at length 20 is not. PROFESSOR: OK, and why do you say that? AUDIENCE: Just looking at power of 3 over 2, 3 times 10 to the 3 isn't enough to overcome the difference in transition probabilities between the island and genome. But 3 times 10 to the 10 and 1 times 10 to the 14 is, over the length of those sequences. The difference in probability of making that switch once at the beginning [INAUDIBLE]. PROFESSOR: OK, did everyone get what Michael was saying? So, Michael, can you explain why powers of 1.5 are relevant here? AUDIENCE: Oh, that's the ratio of emission probability between the C states and the G states, between island and genome. So in island it's 0.3, and in genome it's 0.2 over [INAUDIBLE]. PROFESSOR: Right, so when you're going through a run of C's, if you're in the island state, you get a power of 1.5, in terms of emissions at each position. What about the transitions? You're sort of glossing over those. Why is that? AUDIENCE: Because that only has to happen once at the beginning. So the ratio between the transition probabilities is really high, but as long as the compounded ratio of the emission probability is high enough over a [INAUDIBLE] of sequences, that as long as that compound emission is greater than that one-off ratio at the beginning, then the island is more [INAUDIBLE]. PROFESSOR: Yeah, so if you think about the transitions, I to I, or G to G, as being close to 1-- so if you think of them as 1, then you can ignore them, and only focus on the cases where it transitions from G to I, and I to G. So you say that 60 and 80 are long enough. So your prediction is at the optimal parse is G 1,000 I 80 G another 2,020-- you said that one wasn't going to-- and then I 60 G 1,000. Michael, is that what you're saying? Can you read this? AUDIENCE: Yeah. PROFESSOR: OK, so why do you say that 10 to the 10th is enough to flip the switch, and 10 to the 3rd is not? AUDIENCE: If I remember the numbers from the previous slide correctly-- PROFESSOR: A couple of slides back? AUDIENCE: So if you look at the ratio of the probability of staying in the genome, and the probability of going from the genome to the island, it's-- PROFESSOR: 10 to the 5th. AUDIENCE: Yeah, 10 to the 5th. So whatever happens going over the next [? run ?] of sequences has to overcome the difference in ratio for the switch to become more likely. PROFESSOR: Right. So if everyone agrees that we're going to start in genome, we've got a run of 1,000 A's, and genome is favored anyway-- so that's clear, we're going to be in genome at the beginning for the first 1,000, and be in genome at the end, then if you're going to go to island, you have to pay two penalties, basically. You pay the penalty of starting an island, which is 10 to the minus 5th-- this is maybe a slightly different way than you were thinking about it, but I think it's equivalent-- 10 the minus 5th to switch island, and then you pay a penalty coming back, 10 to the minus 3. And all the other transitions are near 1. So it's like a 10 to the minus 8 penalty for going from genome to island and back. And so if the emissions are greater than 10 to the 8th, favor island by a gradient of 10 to the 8th, it'll be worth doing that. Does that make sense? AUDIENCE: I forgot about the penalty of [INAUDIBLE], but it's still the [INAUDIBLE]. PROFESSOR: Yeah, it's still true. Everyone see that? You have to pay a penalty of 10 to the 8gh to go from genome to island and back. But the emissions can make up for that. Even though it seems small, it seems like 60 bases is not enough-- it's multiplicative, and it adds up. Sally? AUDIENCE: So it seems like the [INAUDIBLE] to me is going to return a lagging answer, because we're not going to actually switch to genome in our HMM until we hit the point where we should [? tip, ?] which would be about 60 G's into the run of 80. PROFESSOR: So you're saying it's not actually going to predict the right thing? AUDIENCE: Do you have to rerun the [INAUDIBLE] processing to get it actually in line to the correct thing? PROFESSOR: What do people think about this? Yeah, comment? AUDIENCE: That's not quite the case, because you [? pack it ?] or you [? stack it ?] both in the genome and island possibilities, and your transition is the penalty. So it's the highest impact penalty. So when you island to island to island in that string of 80, the transition will only be valid starting at the first one. [INAUDIBLE] PROFESSOR: OK, we're at position 1,000. I think you're on the right track here. So I'm going to claim that the Viterbi will transition at the right place, because it's actually proven to generate the optimal parse. So I'm right, but I totally get your intuition. This is the key thing-- most people's intuition, my intuition, everyone's intuition when they first hear about this is that it seems like you don't transition soon enough. It seems like you have to look into the future to know to transition at that place, right? And obviously you can't look into the future, it's a recursion. How does it work? Clearly, this is going to be the winner. So let's go to position 1,001, that's the first C. And this guy is going to come from here, this guy is the winner overall-- G 1,000 is clearly the winner. But what about this guy? Where's it coming from? G 1,000, it's coming from there. And in fact, the previous guy came from G 1,000. I 1,000 came from G 999, and so forth. Now, here's the interesting question. What happens at 1,002? Sally, I want you to tell me what happens at 1,002. Who wins here? AUDIENCE: Genome. PROFESSOR: Genome. Who wins here? AUDIENCE: Island. PROFESSOR: Island. It had been transitioning, genome has got a head start, so the best way to beat an island is to have been genome as long as possible, up until position 1,000. And that was still true at 1,001. It's no longer true after that. It was actually better to have transitioned back here to get that one extra emission, that one power of 1.5 from emitting that C in the island state. If you're going to be in island anyway-- this is much lower than this, at this point. It's about 10 to the 5th lower. But that's OK, we still keep it. It's the best that ends in island. Do you see what I'm saying? OK, there were all these-- island always had to come from genome at the latest possible time, up until this point, and now it's actually better to have made that transition there, and then stay in island. So you can see island is going to win for awhile, and then it'll flip back. And the question is going to be down here at 1,060, going to 1,061. Who's bigger here? This guy was perhaps-- well, we don't even know exactly how we got here. But you can see that this parse here that stays in island is going to be optimal. And the question is, would it be just staying in genome? And the answer is yes, because it gained 10 to the 10th in emissions, overcomes the 10 to the 8th penalty that it paid. Now what do you do at the end? How do you actually find the optimal parse overall? I go out to position whatever it is, 4,160. I've got a probability here, probability here, what do I do with those? Right, but what do I do first? You pick the bigger one, whichever one's bigger. We decided that this one is going to be bigger, right? And then remember all the arrows that I circled? You just backtrack through and figure out what it was. Does that make sense? That's the Viterbi algorithm. We'll do it a little bit more on this next time, or definitely field questions. It's a little bit tricky to get your head around. It's a dynamic programming algorithm, like Needleman-Wench or Smith-Waterman, but a little bit different. The runtime-- what is the runtime, for those who were sleeping and didn't notice that little thing I flashed up there? Or, if you read it, can you explain where it comes from? How does the runtime depend on the number of hidden states and the length of the sequence? I've got K states, sequence of length L, what is the runtime? So I'm going to put this up here. This might help. So when you look at the recursion like this, when you want to think about the runtime-- forget about initialization and termination, that's not [INAUDIBLE]. It's what you do on the typical intermediate state that determines the runtime. That's what grows with sequence length. So what do you have to do at each-- to go from position I to position I plus 1? How many calculations? AUDIENCE: You have to do N calculations for 33. Is that right? PROFESSOR: Yeah, so 33. Yeah, the notation is a little bit different, but how many-- let me ask you this, how many transitions do you have to consider? If I have an HMM with K hidden states? AUDIENCE: K squared. PROFESSOR: K squared, right? So you're going to have to do K squared calculations, basically, to go from position I to I plus 1. So what is the overall dependence on K and L, the length of the sequence? OK, it's K squared L. It's linear in the sequence. So is this good or bad? Yes, Sally? AUDIENCE: Doesn't this assume that the graph is complete? And if you don't actually have [INAUDIBLE] you can get a little faster? PROFESSOR: Yeah, it's a good point. So this is the worst case, or this is in the case where you can transition from any state to any other state. If you remember, the gene finding HMM-- I might have erased it, I think I erased it-- if you can see this subtle-- No, remember Tim designed an HMM for gene-finding here, which only had some of the arrows were allowed. So if that's true, if there's a bunch of zero probabilities for transitions, then you can ignore those, and it actually speeds it up. That's true. It's a good point. Everyone got this? So this the worst case. K squared L-- is this good or bad? Fast or slow? Slow? I mean, it depends on the structure of your HMM. For a simple HMM, like the CPG island HMM, this is like blindingly fast. K squared is 4, right? So it takes the same order of magnitude as just reading the sequence. So it'll be super, super fast. If you make a really complicated HMM, it can be slower. But the point is that for genomic sequence analysis, L is big. So as long as you keep K small, it'll run fast. It's much better than sequence comparison, where you end up with these L squared types of things. So it's faster than sequence comparison. So that's really one of the reasons why Viterbi is so popular, is it's super fast. So in the last couple minutes, I just want to say a few things about the midterm. You guys did remember there is a mid-term, right? So the midterm is a week from today, Tuesday, March 18. For everybody, it's going to be here, except for those who are in 6874-- and those people should go to 68 180. And because they're going to be given extra time, you should go there early. Go there at 12:40. Everyone else, who's not in 6784, should come here to the regular class by 1:00 PM, just so you have a chance to get set up and everything. And then the exam will start promptly at 1:05, and will end at 2:25, an hour and 20 minutes. It is closed book, open notes. So don't bring your textbook, but you can bring up to two pages-- they can be double sided if you want-- of notes. So why do we do this? Well, we think that the act of going through the lectures and textbook, or whatever other notes you have, and deciding what's most important, maybe helpful. And so this, hopefully, will be a useful studying exercise, so figure out what's most important and write it down on a piece of paper if you are likely to forget it-- maybe complicated equations, things like that, you might want to write down. No calculators or other electronic aids. But you won't need them. If you get an answer that's e squared over 17 factorial-- you're asked to convert that into a decimal. Just leave it like that. So what should you study? So you should study your lecture notes, readings and tutorials, and past exams. Past exams have been posted to the course website. p-sets as well. The midterm exams from past years are posted. And there's some variation in topics from year to year, so if you're reading through a midterm from a past year, and you run across an unfamiliar phrase or concept, you have to ask yourself, was I just dozing off when that was discussed? Or was that not discussed this year? And act appropriately. The content of the midterm will be all the lectures, all the topics up through today-- hidden Markov models. And I'll do a little bit more on hidden Markov models on Thursday. That part could be on the exam, but the next major topic-- RNA secondary structure-- will not be on the exam. It'll be on a p-set in the future. Any questions about the midterm? And the TAs will be doing some review stuff in sections this week. OK, thank you. See you on Thursday.
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https://ocw.mit.edu/courses/5-112-principles-of-chemical-science-fall-2005/5.112-fall-2005.zip
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The following content is provided by MIT OpenCourseWare under a Creative Commons license. Additional information about our license and MIT OpenCourseWare in general is available at ocw.mit.edu. Let's get going, here. Remember where we were? We were trying to figure out the structure of the atom. At the beginning of the course, we saw classical physics, classical mechanics fail to describe how that electron in the nucleus hung together. Then we started talking about this wave-particle duality of light and matter. We saw that radiation and matter both can exhibit both wave-like properties and particle-like properties. And it was really important, this observation of Davisson and Germer, and George Thompson, this observation that electrons exhibited inference phenomena. That is when you took electrons and scattered them from a nickel single crystal. The electrons scattered back as if they were behaving as waves. There were diffraction phenomena or interference phenomena, bright, dark, bright, dark patterns of electrons. Actually, that Davisson and Germer paper is on our website. You are welcome to take a look at that. It was just that observation, coupled with de Broglie's insight into Schrödinger's relativistic equations of motion that led Schrödinger to say, well, maybe what I need to do is I need to treat the wave-like properties of that electron in a hydrogen atom. Maybe that is the key. In particular, maybe that is the key because the electron has a de Broglie wavelength that is on the order of the size of its environment. Maybe, in those cases, I need to treat the particle as a wave and not as a particle with classical mechanics. He wrote down this wave equation, an equation of motion for waves, this H Psi equals E Psi, where we said last time we are going to represent the electron, our particle, by this Psi, the wave. We are going to call it a wavef unction because we are going to put a functional form to it very soon. And there was some kind of operator, here, called the Hamiltonian operator, that operated on this wave function. And, when it did, you got back the same wave function times a constant E. And this constant, as we are going to see, is going to be the binding energy of the electron to the nucleus. But then, we took a little detour and I said, well, let's see if we can derive, in a sense, the Schrödinger equation. And that is what we started to do. And I am really doing this for fun, you are not responsible for it, but I am doing it because I want you to see just how easy this is. To illustrate this, I am just going to take a one-dimensional problem. I am going to let my electron be represented by this wave, one-dimension, Psi of x. 2 a cosine 2 pi x over lambda. And then I said, suppose I want an equation of motion, I want to know how that Psi changes with x. Well, you already know that if I take the derivative of Psi with x, that is going to tell me how Psi changes with x. And we did that last time. And then I said, well, I want to know the rate of change of Psi with x. I am going to take the derivative again. I have the second derivative of Psi of x. And that is what we got last time. And then, I noticed that in the second derivative, and you noticed, too, somebody said this was recursive, that we have our original wave function back in this expression. I can rewrite that whole second derivative here just as minus 2 pi squared, quantity squared, over lambda, psi of x. So far, this is just any old wave equation. Nothing special about this. This anybody could, and had, written down before. What is special is that Schrödinger realized, here, that if this is going to be a wave equation for a particle, then maybe this lambda here, maybe I ought to put in for lambda what de Broglie told me. And that is h over p. Maybe this lambda here is the wavelength of a matter wave, so let me write this expression in terms of the momentum of the particle, where the momentum has this mass m in it. And so when he did this, this became minus p squared over h bar squared, we said hbar is h over 2pi, times psi of x. Hey, this is getting good because now we have a Psi of x over here. But then what he said was, well, I want to write this momentum in terms of the total energy. Total energy is always kinetic plus potential. The kinetic energy, we said the other day, can be written in terms of the momentum. The kinetic energy is p squared over 2m-- plus the potential energy. And I am going to make this as a function of x, the potential energy. Now, I am just going to solve this for p squared. p squared is equal to 2m times the total energy minus this potential energy. Now, I am going to plug this into here right in there. And, when I do that, I am going to get the second derivative of Psi of x with respect to x equals minus 2m over h bar squared times E minus U of x times Psi of x. Just simple substitution for p squared there. Nothing else. Now, I am going to do some rearranging. And the rearranging, on the right-hand side, is I am going to have only E times Psi of x. E times Psi of x looks like the right-hand side of the Schrödinger as I wrote it down. That is good. When I rearrange this, I get minus hbar squared over 2m times the second derivative of Psi of x with respect to x plus U of x times Psi of x equals E times Psi of x. And now, I am going to pull out a Psi of x here, so that is minus hbar squared over 2m second derivative with respect to x plus U of x, the quantity times Psi of x equals E times Psi of x. And guess what? We've got it. We got it because all of this is what we define as the Hamiltonian. All of this is h hat. H hat, operating on Psi of x, gives us E times Psi of x. This Hamiltonian, as you will learn later on, is a kinetic energy operator. This is the potential energy operator operating on psi. That is the Schrödinger equation. It is hardly a derivation. It is taking derivatives. It is a wave equation. The insight came right here, this substitution of the de Broglie wavelength in an ordinary wave equation. This is the insight, getting that momentum in there with the mass, making this, then, an equation for a matter wave. That is it. You just "derived the Schrödinger equation." Easy. Bottom line here is that the Schrödinger equation is to quantum mechanics like Newton's equations are to classical mechanics. When the wavelength of a particle is on the order of the size of its environment, the equation of motion that you have to use to describe that particle moving within some potential field U of x or U, you have to us this equation of motion and not Newton's equations. Newton's equations don't work to describe the motion of any particle whose wavelength is on the order of the size of the environment. It just does not work. Now, just as an aside, classical mechanics really is embedded in quantum mechanics. That is, if you took a problem and solved it quantum mechanically, and you solved the problem, a problem for which the wavelength of a particle was much, much greater than the size of the environment, which is the classical limit, quantum mechanics would give you the right answer. In other words, say you took some problem where the wavelength of the particle is larger than the size of the environment. That is, a problem where you would normally use classical mechanics. But if you use quantum mechanics, you would get the right answer, if you could solve the problem because the equations are very difficult. But, in principle, you would get the right answer. However, if you took a quantum mechanical problem, that is, a problem where the wavelength of the particle is on the order of the size of the environment and you used classical mechanics, well, you won't get the right answer. Because classical mechanics is, in a sense, a subset. It is contained within quantum mechanics. It is a limit of quantum mechanics. We have to learn a new kind of mechanics, here, this mechanics for the motion of waves. Now, for a hydrogen atom, we have to think of the wave function in three dimensions instead of just one dimension, here. And so we are going to have to describe the particle in terms of three position coordinates. Usually you use Cartesian coordinates, x, y, and z. But this problem is solvable exactly if we use spherical coordinates. How many of you had spherical coordinates before and know what we are talking about? Not everybody. If I gave you an x, y and z for this electron in this atom, where the nucleus was pinned at the origin. If I gave you an x, y and z coordinate, you would know where that electron was. But, alternatively, I could tell you the position of this electron using spherical coordinates. That is, I could tell you what the distance is of the electron from the nucleus. I am going to call that r. That will be one of the variables. I could then also tell you this angle theta. Theta is the angle that r makes from the z-axis. That is the second coordinate. And then, finally, the third coordinate is phi. Phi is the angle made by the following. If I take the electron and drop it perpendicular to the x,y-plane and then I draw a line here, well, the angle between that line and the x-axis is the angle phi. Instead of giving you x, y, and z, I am just going to give you r, theta, and phi in spherical coordinates. And now, this wave function is also a function of time, and I will talk about that a little bit probably next time. So, that is the wave function that is in some way going to represent our electron. I haven't told you exactly yet how Psi represents the electron, and I won't tell you that for another few days. Question? No. I actually thought this was the way the book set it up. I may have it backwards. I think this is the way our book sets it up. It won't make a difference in any of the problems we solve here. But, in general, if you are given a problem to solve, you are going to have to look and see how they define their coordinate system. Actually, somebody else asked me that the other day. Now, what we have to do is actually set up the Hamiltonian for the hydrogen atom. That is, we have to set up the Hamiltonian specifically for the hydrogen atom. And we have to set it up in terms of spherical coordinates, r, theta, and phi. And when we do that, and we are not actually going to do that, that is what the Hamiltonian looks like. Actually, this is in three dimensions, and so if I were doing it in the x, y, z, what I would have here is a second derivative with respect to y, plus a second derivative with respect to z. That is what it would look like in three dimensions in Cartesian coordinates. But when I transform from Cartesian coordinates to spherical coordinates, an exercise that takes five pages, and everybody should have that experience once in their life, but maybe now is not the right time for that experience, you do get this. And essentially, the Hamiltonian is a sum of second derivatives with respect to each one of the coordinates, because essentially this term is a second derivative with respect to r. This term is a second derivative with respect to theta. This term is a second derivative with respect to phi. That is the specific Hamiltonian. These are all kinetic energy operators, as you will learn later on. And then there is this term here, U of r. This U of r, what is that? What is U(r)? Potential energy. It is the Coulomb potential energy of interaction. It is isotropic, meaning it is the same at all angles. The only thing it depends on is the distance of the electron from the nucleus. It only depends on r. It doesn't depend on theta and phi. So, that is the Schrödinger equation for the hydrogen atom. It is a differential equation, second-order ordinary differential equation. In 18.03, you are going to learn how to solve that. In 5.61, or 8.04 I think it is, in physics, in quantum mechanics, you are going to solve that. You can do that. It is not hard. But what do I mean when I say solve? What I mean is that we are going to calculate these Es, these binding energies, this constant in front of the psi. That is called an eigenvalue, for those of you who might know something already about these kinds of differential equations. This E is the binding energy of the electron to the nucleus. We are going to look at those results in just a moment. And the other quantity we are going to solve for, here, is psi. What we are going to want to find is the actual functional form for the wave functions. That, we can get out of solving this differential equation. The actual functional form for psi. The functional form is going to be more complicated than what I wrote here, but we can get that. We will look at that in a few days from now. And do you know what those wave functions are? They are what you studied in high school as orbitals. You talked about an s-orbital, p-orbital, d-orbital. Orbitals are wave functions. That where they come from, orbitals, from solving the Schrödinger equation. Now, specifically, orbitals are the spatial part of a wave function. There is also a spin part to the wave function, but for all intents and purposes, we are going to use the term orbital and wave function interchangeably because they are the same thing. Question? We are going to get to that in just a moment, and if you think I don't answer it then we will go back to it. We are going to solve this. And this equation is going to predict binding energies, and it is going to predict these wave functions in agreement with our observations, and it is going to predict that the hydrogen atom is stable. Remember when we tried to predict the hydrogen atom using classical ideas? We found it lived a whole whopping 10^-10 seconds. Not so when we treat the hydrogen atom with the quantum mechanical equations of motion. Let's write down the results for solving the Schrödinger equation. And the part I am going to concentrate on today is these binding energies. And, on Friday, we will talk about the wave function, solving the Schrödinger equation for those wave functions. What does the binding energy look like? Well, the binding energies, here, coming out of the Schrödinger equation look like this, minus 1 over n squared times m e to the 4 over 8 epsilon nought squared time h squared. m is the mass of the electron. e is the charge on the electron. Epsilon nought is this permittivity of vacuum we talked about before. This is a conversion between ESU units and SI units. h is Planck's constant. Planck's constant is ubiquitous. It is everywhere. What we typically do is we lump these constants together. And we call those constants a new constant, the Rydberg constant, R sub H. And so our expression is equal to minus R sub H over n squared. The value of R sub H is equal to 2.17987x10^-18 joules. That is a number that you are going to use a lot in the next few days. But what you also see, here, is an n. What is n? Well, n is what we call the principle quantum number. And its allowed values are integers, where the integers start with 1, 2, 3, all the way up to n is equal to infinity. Well, let's try to understand this a little bit more by looking at an energy level diagram again. I am just plotting here energy. Here is the zero of energy. Here is the expression for the energy levels that come out of the Schrödinger equation. When n is equal to one, that is the lowest allowed value for n, binding energy is essentially equal to minus the Rydberg constant. **E = -(R)H** But our equation says the binding energy of the electron can also be this value because when n is equal to 2, well, the binding energy now is only a quarter of the Rydberg constant. It is higher in energy. When n is equal to 3, well, the binding energy of that electron to the nucleus is a ninth to the Rydberg constant. When it is equal to 4, it is a sixteenth. When it is equal to 5, it is a twenty-fifth. When it is equal to six, it is a thirty-sixth. So on, so on, and so on until it gets n equal to infinity. When n is equal to infinity, then the binding energy is zero. When n is equal to infinity, the electron and the nucleus are no longer bound. They are separated from each other. They don't hang together when n is equal to infinity. Now, this is rather peculiar. This is saying that the binding energy of the electron to the nucleus can have essentially an infinite number of values, except it is a discrete number of infinite numbers of values. In the sense that the binding energy can be this or this or this, but it cannot be something in between, here or here or here. This is the quantum nature of the hydrogen atom. There are allowed energy levels. Where did this quantization come from? It came from solving the Schrödinger equation. When you solve a differential equation, as you will learn, and that differential equation applies to some physical problem, in order to make that differential equation specific to your physical problem, you often have to apply something called boundary conditions to the problem. When you do that, that is when this quantization comes out. It drops out of solving the differential equation. What are boundary conditions? Well, remember I told you in the spherical coordinate system, r, theta, phi? Well, phi sweeps from zero to 360 degrees. Well, if you just went 90 degrees further, so if you went to 450 degrees, you really have the same situation as you had when phi was equal to 90 degrees. What you have to do in a differential equation, which usually has a series as a solution, is that you have to cut it off. You have to apply boundary conditions. You have to cut it off at degrees. Because otherwise you just have the same problem that you had before. This is a cyclical boundary condition. And it is that cutting it off to make the equation be really pertinent or apply to your physical problem, that is what leads to these boundary conditions, mathematically. That is where it comes from. That is where all the quantization comes from. Now, let's talk a little bit about the significance here of these binding energies because somebody asked me about it already. When the electron is bound to the nucleus with this much energy, we say that the electron is in the n equals 1 state, or equivalently, we say that the hydrogen atom is in the n equals 1 state. We use both kinds of expressions equivalently. When the hydrogen atom or the electron is in the n equals 1 state, we call that the ground state. The ground state is the lowest energy state. The electron is most strongly bound there. The binding energy is most negative. And the physical significance of that binding energy is that it is minus the ionization energy. Or, alternatively, the ionization energy is minus the binding energy. It is going to require this much energy, from here to here, to rip the electron off of the nucleus. The binding energy here is minus the ionization energy. That is the physical significance of it. Now, did you ask me about the work function? Okay. The work function is the ionization energy when we talk about a solid. That is just a terminology, that is historical. When we talk about ripping and electron off of a solid, we call it the work function. When we talk about ripping it off of an atom or a molecule, we call it the ionization energy. And the other important thing to know here is that when we talk about ionization energy, we are usually talking about the energy required to pull the electron off when the molecule or the atom is in the lowest energy state, the ground state. That is also important. But our equations tell us we also can have a hydrogen atom in the n equals 2 state. If the hydrogen atom is in the n equals 2 state, it is in an excited state. Actually, it is the first excited state. The electron is bound less strongly. It is bound less strongly, and consequently, the ionization energy from an excited state hydrogen atom is less because the electron is not bound so strongly. And, of course, we could have a hydrogen at n equals 3, n equals 4 and n equals 5, any of these allowed energy levels. Not all at the same time, but, at any given time, if you had a lot of hydrogen atoms, you could have hydrogen atoms in all of these different states. Now, the other point that I want to make is that this Schrödinger result, here, for the energy levels predicts the energy levels of all one electron atoms. What is a one electron atom? Well, helium plus, that is a one electron atom or a one electron ion. Helium usually has two electrons, but if we pull one off it only has one left, so it is a one electron atom. Lithium double plus is a one electron atom because we pulled two of its three electrons off. One electron is left, and that is a one electron atom or an ion. Uranium plus 91 is a one electron atom. We pulled 91 electrons off, we have one left, and that is a one electron atom. The Schrödinger equation will predict what those energy levels are, as long as you remember the Z squared up here. For hydrogen, Z is 1, and so it doesn't appear. But Z is not one for all of these other one electron atoms. Where does the Z come from? It comes from the Coulomb interaction. That potential energy of interaction is the charge on the electron times the charge on the nucleus, which is Z times e. That is where the Z squared comes from. We have to remember that. Now, how do we know that the Schrödinger's predictions for these energy levels are correct, are accurate? Well, we have to do an experiment. The experiment we are going to do is we are going to take a bulb here, pump it out, and then we are going to fill it up with molecular hydrogen. And then in this bulb, there is a negative and a positive electrode. What we are going to do is crank up the potential energy difference between those two so high until finally that gas is going to ignite, just like that. And we are going to look at the light coming out. And what we are going to do is we are going to disperse that light. We are going to send it through a diffraction grating, essentially. And that is like an array of slits. What you are going to see are points of constructive and destructive interference. But in the constructive interference, you are going to see the color separated. There is going to be purple, blue, green, etc. The reason is that the different radiation, the different colors have different wavelengths. They have different wavelengths, then they have slightly different points in space at which constructive interference occurs, and so the light is dispersed in space. We are going to analyze what the wavelengths of the light that are being emitted from these hydrogen atoms are. See, the discharge pulls apart the H two and makes hydrogen atoms. We have to do this by taking a diffraction grating. The TAs are going to give you a pair of diffraction grating glasses. You are supposed to come and look at the discharge here. You can see it. I will turn it a little bit for those of you on the sides here. And we will see what we see. If you cannot see the lamp, you are welcome to get up and move around so that you can see it. Can you do this light, too? This one over here, sir. Can you move this one away? If you look up at the lights in the room, you can see the whole spectrum, because that is white light. I am going to turn the lamp over here so that you can see this a little better. The spectrum that you should see is what is shown on the board. If you look to the right of the lamp, here, you should see a purple line. The purple line is actually very light, so only if you are up close are you going to see the purple line. You can see the blue line very well. That is very intense. The green line is also very diffuse. Again, only if you are close are you going to be able to see the green line. And then, the red line is very bright. And now I am going to move this over here so you have the opportunity to see it. Again, you are welcome to get out of your seats and move around so that you can, in fact, see it. What you should be seeing, hey, interference phenomena works. Useful pointer. What you should see, depending on which one of the constructive interference patterns you are looking at, there should be a purple line, there should be a blue line, very intense, purple is weak, green is rather week, and the red is very intense. What is happening, here? Well, what is happening is that in this discharge, there is enough energy to put some of the hydrogen atoms in a high energy state. And we will call that state E sub i, the energy of the initial state. Actually, that is an unstable situation. That hydrogen atom wants to relax. It wants to be in the lower energy state. And what happens is that it does relax. It relaxes. The electron falls into the lower energy state, but, because it is lower energy, it has to give up a photon. And so the photon that is emitted, the energy of that photon has to be exactly the difference in energy between the energy of the initial state and the final state. That is the quantum nature of the hydrogen atom. The photon that comes out has to have that energy difference exactly. And, therefore, the frequency, here, of the radiation coming out is going to correspond to that exact energy difference. And, for the different energies, for the different transitions, you are going to have very specific values of the frequency of the radiation or the wavelength of the radiation. For example, some of those hydrogen atoms in this discharge have been excited to this excited state, which we will call B. The energy difference between B and this ground state here is small. Therefore, we are going to have some radiation that is low frequency because delta E, the difference in the energy between the two states is small. And then there are going to be some hydrogen atoms that are going to be excited to this state, the A state. And, relatively speaking, that energy difference is large. There are going to be some hydrogen atoms that are going to be relaxing to this ground state. And when they do, since that energy difference is large, the frequency of the radiation coming off is going to be high. Or, correspondingly, the wavelength of the radiation coming off is going to be low because the wavelength is inversely proportional to the frequency. And, likewise, for this transition we are going to have some long wavelength radiation emitted. Well, let's see if we can understand specifically the spectrum, here, in the visible range for the hydrogen atom. What I have done is to draw an energy level diagram again. Here is the energy of n equals 1, n equals 2, n equals 3, etc. Here is the n equals infinity, up here. It turns out that this purple line is a transition from a hydrogen atom in the n equals 6 state to the n equals 2 state. That blue line is a transition from n equals 5 to n equals 2, the green line is a transition from n equals 4 to n equals 2 and the red line from n equals 3 to n equals 2. Notice that since this energy difference here is small, this line has a long wavelength. Since this energy difference is larger, this line has a shorter wavelength. All of these transitions that you are seeing in the visible range have the final state of n equals 2. Now, how do we know that the Schrödinger equation is making predictions that are consistent with the frequencies or the wavelengths of the radiation that we observe here? Well, we have got to do a plug here. This is the frequency that we would expect. It is the energy difference between two states over H. Here is the initial energy. Here is the final energy. We said that the Schrödinger equation tells us that the energies of the state are minus R sub H over n. For the initial energy level, it is minus R sub H over n sub i squared. We plug that in. For the final energy level, well, it is minus R sub H over n sub f, the final quantum number squared. We plug that in. We rearrange some things. And then this is the prediction for the frequency. I said that this level right here is the n equals 2 level, so we will plug in the two for the final quantum state. And so this is then the prediction for the frequencies of the radiation to the n equals 2 level. You just plug in n equals 6, n equals 5, n equals 4, n equals 3. And you know what? The frequencies that are predicted match what we observe to one part in 10^8. There really is precise agreement between the results of the Schrödinger equation and what we actually observe in nature, therefore, we think it is correct. And there are no experiments that cast any doubt, so far, on the Schrödinger equation. Now, the set of transitions that we just looked at are these transitions. Plotted here is another energy-level diagram for the hydrogen atom. Here is the n equals 1 state, n equals 2, n equals 3, n equals 4. And we looked at all the transitions that end up in the n equals 2 state. This set of transitions. It is called the Balmer series. Now, the n equals 2 state is not the ground state. There is a transition from n equals 6 to n equals 1, the ground state. It is this one right here. But, you see, that is a very high energy transition. Actually, this transition, from n equals 6 to n equals 1, is in the ultraviolet range of the electromagnetic spectrum. And so you cannot see it with the experiment that we did, but it is there. And then, of course, the hydrogen atoms that relaxed to n equals 2, well, they actually eventually relax to n equals 1. And there is a transition there, it is over here, n equals 2 to n equals 1. But, again, that is a high energy transition. It is also in the ultraviolet range of the electromagnetic spectrum. And so all of these transitions that end up in n equals 1 are in the UV range. They are called the Lyman series. All the transitions that end up in n equals 2 are in the visible. They are called the Balmer. n equals 3, the Paschen, is in the infrared. Brackett is in the infrared. Pfund is in the infrared. And so we cannot see these easily. Now, these are named for the different discoverers. The reason there are so many discovers is because these are all different frequency ranges. Different frequencies of radiation require different kinds of detectors. And so, depending on exactly what kind of detector the experimentalist had, that will dictate then which one of these sets of transitions he was able to discover. But not only does this work for emission, but this also works for absorption. That is, we can have a hydrogen atom in the ground state, a low energy state. And if there is a photon around that is exactly the energy difference between these two states, well, that photon can be absorbed. If that photon is a little bit higher in energy, it won't be absorbed. If it is a little lower, it won't be absorbed. It has to be exactly the difference in the energies between these two states. Again, that is the quantum nature of the hydrogen atom. And now, if you are calculating the frequency for absorption, what I have done here is I have reversed n sub i and n sub f. This is 1 over n sub i squared instead of 1 over n sub f squared, which was the case in emission. And I have reversed this so that you get a positive value for the frequency of the radiation absorbed. We will have two different equations for absorption and for emission. So, those are the Schrödinger equation results. I forgot to announce that there is a forum this evening from 5:00 to 6:00. You should have signed up in recitation. If you didn't, and still want to come, send us an email and come. And we need the diffraction glasses back as you are exiting. See you Friday.
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https://ocw.mit.edu/courses/5-95j-teaching-college-level-science-and-engineering-spring-2009/5.95j-spring-2009.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK. So misconceptions. So misconceptions, alternative conceptions, understanding the student view of the world, why is that important? Well one of the best ways to introduce the importance of the idea is a story. Now I'm not 100% sure the story is true, but it's so good that it ought to be true even if it isn't. I think it is actually true. So this was two, let's say physics, TAs, physics grad students, who were in a philosophy class. And the teacher in the philosophy class said well-- they were talking about things being relative, and morals being relatives, and relativity in philosophy-- as an example of relativity, and things being relative depending on the situation, if you drop a pencil on the Earth, it will fall to the ground. But if you drop the pencil on the moon, it won't fall to the ground. It will just float. And so most of the people in the class just nodded, basically thinking, oh that's a good example. But the two physicists in the class were a bit shocked and said well wait a minute, wait a minute. That can't be true. Surely, something is wrong here. So they raised their hand. They said, well wait a minute. If you drop a pencil on the moon, surely it's going to fall to the ground, to the ground of the moon. And the other people in the class said, no, no, that's not right. That's not right. So they thought, OK, well we have some way of convincing them that the pencil's not going to float. And they said, OK well you all saw the pictures of the astronauts walking on the moon, right? And they said, oh yeah. Well, when the astronauts walk, did they just float away, or did they stay on the moon? Oh, they stayed on the moon. Well, how'd that happen? They thought, oh we have them now. And what was the response? The response was, oh, well that was because the astronauts were wearing heavy boots. So here was a chance to do a bit of educational research. So they thought, oh my god. Maybe is this just this class? Let's actually do a survey. So they took out the university phone directory and just randomly dialed people. And they called them up and said, OK if you drop a pencil on the moon, will it float away or will fall to the ground? And most of the people said, it's going to float away. So then they said, oh yeah, but what about the astronauts? And they gave them the whole thing about the astronauts again. And about half of the people said, oh yeah, that's because the astronauts were wearing heavy boots. So again, they were shocked and flabbergasted. And I think that reaction is a very healthy reaction. But what it shows is that the students-- so in this case they weren't officially their students-- but students-- they could easily be your students-- can have a very different view of the world, of the physical world, than you have. And if that's true, that means that as you're teaching them, what you're teaching is actually being interpreted in this completely different way. I think Goethe said that-- he said that mathematicians, or maybe it was Frenchmen. It was either Frenchmen or mathematicians. Maybe it was mathematicians. They have their own language, and everything you tell them, they first translate into their own language, and only then do they understand it. So that's the same thing. The students are translating the things you're saying into their own language. But to understand what they're translating into, you have to know what their language is. So in other words, you have to understand not just the misconceptions, but in general the conceptions that students have, so that you can actually plan your teaching. And here is a diagram to illustrate that. So this is a model of teaching to see where all the pieces fit in. So here, this is the system, which is how students think. This is the output, which is what students can do after they go through our teaching. And this is the input, which is what we do, what kind of classroom structure we set up, what assignments we do. And by this, I mean how students think individually and collectively, how students work together, as well. So generally speaking the way-- so this is the system. And call this the result. And this is the teaching. Our problem is trying to figure out the teaching. And how do you figure out the teaching? Well you start with figuring out what result you want. What do you want students to be able to do? What is your goal for the class? And then you have to run it back through the system. Right so you figure out the result, and you run it through the inverse of s. And then there's your teaching. That's the teaching equation let's call it. Now those of you from engineering fields know that basically the only thing you can do anything with is linear systems. And unfortunately anything that's even slightly non linear basically is a bag of tricks, and there's nothing you can do. And you're just hosed. Well unfortunately this system is incredibly non-linear. So it's probably kind of hard to work out s inverse, even if you can work out r. But it's hard, but it doesn't mean you can't start. And a necessary condition for even doing anything in this direction is you have to know what is in s, how do students think. Another necessary condition is you need to work out r. And that's actually something-- what is the result you want? So that we're going to talk about in the session on course design. That's something that's often forgotten in teaching, and it's fundamentally important. That if you don't plan your goals, if you don't plan where you want to go, your chance of getting there is pretty low. So assume you've done that. The next thing you want to do is you want to understand how students think, so that you can plan your teaching to reach them, in other words, so you have some chance of working out s inverse. Now you can't actually write down some equation for s inverse, but you can invert it mentally in your head. And a necessary condition for doing that is to understand s. So you want to understand where students are coming from, so that you can work out t. So that's the signals and systems model of teaching. Now, an example again to show the importance of doing that and what happens if you don't actually understand where the student's are is yet another example from freshman physics. I'll leave this here. So this is one of the examples from the readings. So not everyone may have read that reading. It's the reading by Reif. And it's this problem. So this is a pendulum at five points in its arc. So these are the endpoints. And the question people were asked was to draw the acceleration vector at these five locations. So what's the acceleration right here, here, here, here, and here? So I'm not so interested in what the actual answer is as so much as how people did on this. So this is actually really shocking. So of students in intro physics who've had acceleration-- so they'd done acceleration-- zero out of 124 students could do that problem. So that's 0%. So of grad TAs, it was 15%. So these are grad TAs in physics who TA that course. And PhD students-- so was PhD students on their qualifying exam. It went up to 22%. OK so now this is a fundamentally important idea, acceleration. But you'd expect PhD students on their qualifying exam to be at 100%. And you'd like the TAs in the course to be at 100%. Because there's no chance of them teaching the students an idea that they themselves don't understand. As Polya said, there's no method of teaching yet invented that will allow the students to understand what their teacher does not. And that's true. There's no amount of sort of trickery with style and things you can do to convey an understanding that you don't have. So this is a very shocking number to me. And-- AUDIENCE: Who said that? PROFESSOR: Pardon? AUDIENCE: Who said that? PROFESSOR: George Polya. So Polya was a great mathematics teacher from Hungary and then eventually at Stanford. And he's written-- I'll put references to his books on the website. But they are some of the best books on math teaching ever written. And so he said that in one of his books. And it's I think very true. So what this example shows is that, a, the teachers didn't understand it. But that really fundamental ideas can be completely skipped over for years and years and years. Yes, [INAUDIBLE]. AUDIENCE: It's a special example of some vague, conceptually difficult problem. I'm just wondering-- so I read the article. Did they focus on how to figure the answer out? Because a lot of physics majors don't. PROFESSOR: OK, so the question is, how difficult is this problem? It's an interesting problem, because it's difficult if you try to do it by just blasting out the equations. So they actually asked a bunch of faculty members the same question. And one of them just got completely tangled in knots, physics faculty, trying to write out the differential equation and figure out what the acceleration is by brute force. So yeah, it is hard if you do it that way. But if you actually understand-- I'll show you-- what the acceleration means, then it's no problem. So for example, here, the thing isn't moving. So it has v equals zero. But, what velocity is it going to have just a bit later. Well it's going to be going that way. So that means acceleration is that way. AUDIENCE: But were they given enough time to do this, to do this thinking problem? PROFESSOR: Yeah, so they were given quite a while. I think it was-- when the study on the faculty members, I think it was a big long think aloud protocol. So they were given like half an hour. In fact, the time is what hurt them. Because they all thought, oh I have so much time. I'm going to just blast it out with equations. And if you were only given one minute, you might have actually had a chance. Because you realized, oh I have to think about this intuitively. So here, same thing by symmetry. And then this one is hard. And this one is hard. But this one right at the center is much easier. So this one catches a lot of people. But here it's moving in a circle, and it's at its maximum speed. So here it's moving at its maximum speed. So it's not going to be going faster that way or that way. So its speed is actually at a max, which means there's no change in speed. So you're not going to get any acceleration in that direction. But because it's moving in a circle, somebody's yanking it upwards. So it has to be accelerating upwards to move in a circle, because it was going this way. And then later, it's going that way. So it has to be accelerating upwards. So there's an acceleration vector upwards, like that. So this is a bit exaggerated. It's generally not that big. And then the question is, with these guys, well it's just interpolation. It's somewhere between that and that. So it's going to be something like that and something like that, depending on where exactly they are. But this is the one that tripped up a lot of people. And in fact, a lot of people got tripped up because they memorize the following thing, which is an object in simple harmonic motion has no acceleration at the equilibrium point. And that's true, except this isn't a simple harmonic motion. It's almost. But it's a pendulum. It's not a pure spring. So it has circular motion. So if you reason about it from the motion, it's actually very quick. But if you try to blast it out with equations, you're basically hosed. So I wanted to check also, how deep are these misconceptions? So I actually did a survey of students in Cambridge in the physics major. And so one of the questions I asked them was the following-- so I made a survey of nine questions. And I wanted to see how they reasoned intuitively on these nine questions. And to force intuitive reasoning, I only gave a few minutes to get around the problem. So they knew that basically writing out a bunch of equations wasn't going to help, because you had no time to actually write them all out. So the question was, which is going to go faster down the plane? So these are two identical planes, inclined planes with angle alpha. And there is either going to be a disk like a coin rolling down the plane, or a ring, like somebody's wedding ring rolling down the plane. And the question is, which goes faster, or which has a bigger acceleration. So this we'll call the hoop, and this is the disk. And they're going down the same angle. AUDIENCE: The mass is the same. PROFESSOR: I didn't say that. The question is, is the mass the same? So I'm going to give you that choice. So the question is, which goes faster in acceleration or velocity? So the choices were the hoop-- So to answer the question about mass, one of the choices is that it depends on mass. Depends upon radius or depends on alpha. So it could depend on the mass, the radius, or on alpha. It could be the same. The disk could be faster. Or the hoop could be faster. OK, so just for fun-- and I realize this is not a physics class-- but it's fun to actually try to guess the answer to this, even if you're not a physicist. So choose A, B, C, D, E, or F in cooperation with one or two neighbors. And then I'll talk about my analysis of this problem. OK, take another 30 seconds and collect a vote. Again, it doesn't really matter whether you get it right or not. I just want you to think about it, so you realize it is a subtle question. OK, so who votes for the hoop being faster? Who votes for the disk being faster? Who votes that they're the same? That it depends on the mass, you'd like to know the mass? Fair enough. That you'd like to know the radius? Maybe 12. That you'd like to know the inclined plane angle? Nobody like that one, OK. And now who's sure of their answer. OK, so that's a much lower number. So now I can show you what numbers came up when I asked the students this. So this was 11%. 35%. 13%. This one is 11% as well. Radius was 26%. And 3%. So kind of similar, kind of similar to your percentages. And again, only about a third of the people were sure of their answer. OK, so why is this interesting? Well first of all, all the students had the technical knowledge to do the problem. These were all physics majors in various years, one through four, so freshman through senior. But the freshman had seen the material to actually do this and calculate it if they need to. So now this problem illustrates well first of all that even an intuitive question like this can actually trip people up, even if they've done all the calculations. So the question is, how are they reasoning? Now in Cambridge I had the advantage of also doing tutorials, because that's how half the teaching goes. So in tutorials I actually I set this problem for my students, and I got a chance to talk to them about it. So I got to figure out, how are they reasoning about it? So I asked them, well what do you think is going to happen? And then please explain why, more than you can get just from multiple choice. And the answers were very revealing. What they said was, many of them said, well, just as heavier objects fall faster, so should heavier objects roll faster. So that was basically to justify they wanted to know either the radius or the mass, because that would affect the mass, and that would also affect the mass. So, let me put that up again, because that's a really interesting statement. So actually what I call this, this is like the American theory of the British accent. Some of you know my explanation of that. So the American theory of the British accent is that if you go to an English person and you step on their feet in the middle of the nigh, they'll actually cry out in an American accent. Which is to say that the American, sort of folk theory of the British accent is that the British accent is a fake that people just put on, and in moments of stress they forget to do the put on act. OK, so now why do I bring that up? Well because it's very similar. So if you ask the students-- suppose you didn't tell them this question about inclined planes. You just say, do heavier objects fall faster? You don't know about air resistance or anything. They say, oh no definitely not. Galileo showed that. But what this shows is that in a new situation, in other words where they don't have their automatic reactions, all of a sudden the folk theory comes out again. They really do think heavier objects fall faster. Right, so how can you reconcile that with if you ask them, they'll say no they don't? Well that's because they had so much experience with every time they said that, someone said, whack. Heavier objects fall at the same speed. Galileo showed that, Leaning Tower of Pisa. So it becomes this linguistic string which hasn't actually been incorporated into their way of thinking about how the world works. They think force is related to velocity, heavier objects fall faster. And to get that to come out, you have to, for example, equivalent to stepping on them in the middle of a night, surprising them. So you surprise them with a completely new context, which is the rolling. Rolling, they haven't thought much about rolling. So all of a sudden, they're now thinking about rolling. And they're not thinking about all the policing rules that they were told about what happens to heavier objects. And so now their true way of looking at the world comes out, which is that heavier objects fall faster. So again, if you don't realize that that's what they're doing, your teaching is going to be completely pointless. And what's going to happen is the following, which is that here are two models. The students will develop two models of the world in their mind. One let's call it the school model. And another is their intuitive model. And let's say they start out a bit different from each other. And now you do a whole bunch more teaching, and you don't take account of the fact that this belongs in their intuitive model. And you teach a bunch of stuff about rolling down the plane. You do it with a bunch of equations. By the way, I should tell you what the right answer is. The right answer is this one. It actually doesn't matter how heavy they are, because it's a gravitation problem. But suppose you just calculated it out. Well what have you done? You've overlaid on top of this, you've put in-- so this is their intuitive model. You've done a whole bunch of calculations which contradict it. And the two models diverge farther and farther from each other. So now next time they have to solve a problem, they have to decide whether I'm going to use the school model or the intuitive model. And they learn, OK, well the intuitive model doesn't work very well. Let me use the school model. So the school model basically keeps diverging. And the intuitive model and the school model never meet again, which means that they can't use their intuition for solving any of the problems you give them, or that they're going to solve later. So that is an educational tragedy. So the divergence, and in my view, that is the fundamental reason why most teaching produces no results a year later. Because what you've done is you've developed the school model, the symbolic crunching, whatever it may be. But you haven't actually yanked the intuitive model and the school model together. You need to actually work on bringing them together. So now the natural question is, how do you do that? So let me give you another example of a similar misconception and a way of yanking the two together. Yes, question. AUDIENCE: Would you mind explaining why the disk rolls faster? PROFESSOR: Oh, yeah, why is it the disk? Yeah, sure. So first of all, why doesn't it depend on let's say the mass? So that's one thing to clear up right away. And one way to reason about that, maybe the best way in a physics class is to do it by dimensions. But another way is the following thought experiment. Which is, suppose I have two disks-- let's see, I might be poor today. I only have one disk in my pocket. Well here's two disks. The chalk is also a disk. So now, suppose I have one piece of chalk, and I want to roll it down the incline plane. And now I ask about a second piece of chalk. Let's say they're identical pieces of chalk. They're both going to go at the same speed. OK, so now suppose I stick a bit of glue on this chalk and that chalk, and I stick them together. They're going to go, again, at the same speed, as if the glue weren't there. Because the glue isn't being stressed. They're just moving identically. So what I've shown by that experiment is that a heavy piece of chalk moves just the same as a light piece of chalk. So it's quite plausible from that reasoning that mass doesn't matter. Why doesn't radius matter? Well you could do a couple of scaling arguments to show that the extra torque you get from being bigger is exactly canceled out by the extra mass because you're bigger. So the radius doesn't matter. Mass and radius don't matter. So then it's just a question of shape, which is a dimensionalist thing. And what it is intuitively is how concentrated is the mass near the center versus near the edge. So suppose I actually made a extreme version of the hoop. And actually I did this for the students in Cambridge. We manufactured-- and you could actually imagine what it is. So imagine this were hollow inside. That would be your ring or your hoop. Well can you make anything that would move even slower? Because I'm planning the hoop moves slower than the disk. Yes, what you do is you put giant ears on the thing. So it rolls like those dumbbell things you do for weightlifting. So it rolls on this radius, but it has these giant ears. Well it's just going to move so slowly, because it's just being yanked by this really weak force. And it has to accelerate this giant mass. No chance. So it moves really slow. In fact, that thing it can take 20 seconds to go down a meter long inclined plane. So that's an extreme version. So it rolls on this, and this is these big ears. So that's really slow. That's medium slow. And that's faster. And if you put all the mass right at the center, it would be even faster, because the rolling wouldn't suck up any of the energy. So that's the intuitive explanation. OK so now an example of a question showing a kind of similar misconception, a related misconception. But I'm going to show it to you to illustrate what you can do about this. So what can you do about it? So the question is the following. So that's a steel ball, which I have in my hand, and I drop on a steel table, say from a meter up high. Forget about air resistance. It bounces off the steel table. And the question is I'm interested in the forces on it. So what are the forces. So here's my table. So while it's falling, I'm interested in the forces on it at three different points in its flight. First while it's falling. I've let it go, and it hasn't yet hit. At the instant that it's stationary during its bounce. And while it's rising. So when you ask the students this-- and I've done this several times-- mostly they know it's just gravity while it's falling, mg. So if you actually use this question, don't ask-- this is chronologically next, but don't ask that next. Ask this one, when it's rising. And what you'll find-- what do you think many students say, not all but many? What are the forces on it while it's rising? Yeah, [? Adrian. ?] AUDIENCE: So of them will say the force goes up. PROFESSOR: Right, some will say the force goes up. They'll say mg up. I see people laughing, and that's a good reaction, because that shows you're learning something about how students think. That's kind of amazing. Why would they possibly say that? Well let's talk about this one, and you'll see that it's actually a similar mode of reasoning to this. So what you'll find is even students who get this correct-- so this isn't right. It's mg down. But there's a deep seated misconception that if something's moving upwards, it has to have a force pushing it upwards. And then the force somehow gets used up, and then it stops. OK, well what about here? So now they'll agree that there's weight pushing down. So that's the mg. And then the question is, what's the other force? And they call it by different names. In America, it's called the normal force, which is a terrible name. And in England it's called the reaction force, which is a terrible name for another reason. Which is that it implies that it's the result, somehow a reaction. So it's a terrible name. But let's ask the students, how big is this force? OK, so it's upwards. Everyone agrees that it's upwards. But how big, r or n, depending on which country you're in? So take a minute or two and for yourself figure out how big the reaction force or the normal force is, in comparison with the weight. So this is a small steel ball dropped from say a meter and bouncing off a steel table. So two questions, first figure out what you think. And second, figure out what you think students are going to say. OK so you're doing both, trying to get into the student mind. Let's do a quick vote. So what's the ratio between that normal force and the weight? So who votes for equal? First we'll do what you think, and then we'll do what students think. So who thinks they're equal? Two to one? Greater than 10, but not quite 1,000? Greater than 1,000 to 1? OK now what do you think students say? How many say one to one? How many say two to one? How many say greater than 10? Greater than 1,000? And how many are not sure what the students are going to say? So actually when you ask the students it depends on the population. But generally, what I find is it's 90% for one to one, And about 10% for two to one. And no one says anything else. OK so now why would students say one to one? What's the reasoning going on in their mind? Has no velocity. So that is exactly the same reasoning as this, with mg upwards. But what's interesting is even the students who know it's mg downwards still say, well the velocity is zero. So v equals zero. Therefore, it's in equilibrium. Therefore, there's no forces acting on it. Therefore, the reaction force, or the normal force, and the weight have to cancel. N equals mg because it's stopped. So that what I call that is I call that the f equals mv theory of physics. Now maybe like in the linguistic books, I'll put a star by that saying don't say that. That's not how you translate that sentence into real life. But why f equals mv? Why is it so deep seeded? Well we have a gazillion hours of experience with it. Suppose I move a object, and I push it. Stop pushing it, and it stops moving. If I have a light object, I just push it a little less and it moves. And I stop pushing, and it stops moving. So f is related to m, and f is related to v. This seems to explain most of our world. It's hard to get around that. So the question is, how do you actually try to force the students to see that that's actually completely junk what they said, and that it can't be equal? And the way I want to do it is I want to do it in a way that recombines their intuitive model and their school model. I want to meld them together. So the way I do that is the following, which is I get a rock. So let's call this a rock. And I ask one of the students, could I please borrow your hand? And so somebody gives me their hand. And I say, OK, I'm going to put the rock in your hand. Now, what's the weight of the rock? mg. OK, great. So what's the force of the rock on your hand? mg. Great. Now I hold the rock about the same place I'm going to drop the steel ball, and I say, OK on the count of three, I'm going to drop the rock on your hand. One, two, and then I'm ready. What are they going to do? Like that. And soon as they try to do that, I grab their hand and say, wait a minute. Let's reason about this with physics. Why are you moving your hand? You just told me that when the thing bounces, it's still mg. The reaction force and the normal force is still mg. And you just showed me that mg on your hand didn't hurt your hand. You could hold the rock on your hand no problem. So why are you moving your hand? So then we'll have a discussion, and many of them will then change to two to one. They'll say, oh, well it's actually two to one. And I say OK, two to one, well I have an answer for that. Two rocks. So I have two rocks, and I put them on their hand. So now compared to the first rock, mg, 2mg. So now I take the first rock, and I'm going to drop it on your hand. And one, two, three. And they try to move. And I say wait, wait. You just told me 2mg was fine. This felt fine to you, right? And they say, I know it's going to hurt like hell. So now what they're doing is they now have some kind of contradiction that's evident between the school model and the intuitive model. And now is your opportunity to introduce and explain, yeah you're right. Your intuitive model is actually correct in this case. And your school model reasoning is incorrect. So actually what we're going to do is try to move the school model towards a piece of your intuitive model that is correct. We're trying to get away from the f equals mv part of the intuitive model and emphasize the part that does know. And the part that says, I don't want to have that rock in my hand is correct. So then they're very ready to do a school model calculation of how big the force is. And it turns out to be something like 100,000 or 10,000 times the weight. So it's huge because the impact time is so short. And then to make that plausible to them, we can actually calculate the impact time or estimate it. But then you can actually say, look you already know this. You intuitively understand that. Because if you-- again I'll stand on the table. Suppose you're going to jump off an object like this, what do you do when you land? What does everyone do? Bend your knees. Right, so you do that. And what does that do? That increases the contact time. And why are you increasing the contact time? Because that decreases the acceleration. The acceleration is a change in velocity divided by the contact time. So you can't affect the change in velocity very much. That's either v or 2v or something like that. But the content time you can change. So with the steel ball, contact time is actually very short. Because the speed of sound is so fast. So the impact force is really high. So yes, your intuitive model was correct. And now we're yanking the school model towards it. So you actually bring the two together. And you fight the divergence. So you want to look for examples like that whenever you can. And to summarize the lesson-- because it was pointed out that I should summarize more often-- the summary of the entire theme about misconceptions is that you need to understand that so that you can teach properly. Because there's no way to deduce this without doing that, which we're going to talk about later. But even knowing that, if you don't know this, you can't run the system backwards and figure out what goes here. And so then, how do you actually learn about this? Well you've done one way. You've done one way already. Let me write down two or three ways for you. So the question is, how to learn about s, the students' system. Well one way is to read. So in many fields, there's lots of research about what particular misconceptions students have. And I've given you some of those readings on the website. But there's a whole vast body of literature, and many fields have that. So you can look at that. Another way, and this is-- so this is shut your mouth. Now why is that? Well there's no way to really understand what students are thinking unless you ask them. And if you don't let them talk, you're not going to actually hear what they're thinking. You'll actually put words into their mouth. Because you won't imagine that they think f equals mv. You never would imagine that. So you have to shut your mouth and listen. So office hours is a chance for that. And related to that is the feedback sheet. If students get in the habit of asking you questions about what was confusing, you start to build a model of the students' system, what worked and what didn't work and what was confusing for them. So all of these ways are ways to build models of s, which is absolutely necessary for deducing t. Any questions? OK so if you could do two things. One is to spend one minute filling out the feedback sheet. And the other is everyone who has a homework equation treatment, please raise your hand. OK that's great. So now just find one person near you to swap with. Question? Yeah, OK. So that's great. So just find one person who has one near you to swap with and give each other comments. And then just write your name over the next week and bring it in next time with you. So basically this is a way-- and if there's no one near you, or they've already swapped, just make a group of three. It's no big deal. Does anyone not have someone to swap with? OK great. And so introduce yourself to the other person. It doesn't matter if you're in different fields. It often can be an advantage, because you'll get a fresh set of eyes on it. And I'll give you some directions by email about what to look for and put it on the website. But generally just give each other suggestions and work together on what you thought was a good and what wasn't good. And just write your name on the person you commented on. And then you'll bring them both in. You'll bring the commented sheet from the other person in the next time. Or you'll just give it back to them and bring it. This is not a big police exercise. I just want you to meet somebody else and have practice discussing these things and learn from each other. So questions about that? So if you could finish filling out the sheets and bring the sheets either here or here, I'll pack up. And meanwhile we'll have any questions that people want to talk to me about individually outside, because I think there's another class that 's going to come streaming in with like 150 people. So remember, learn about the student system. It's probably the most important thing to do to make your teaching stand out and have long lasting, good effects. NARRATOR: Answers from lecture four to questions generated in lecture three. PROFESSOR: So first, questions from before. One of the comments-- well, let's see, two of them were that could we have less time for questions, but also they're very helpful. So there's always a tension there. So what I'm going to try to do is shorten some of the total answers to the questions and then maybe put some answers to the questions towards the end of the lecture as well. I do like to try to answer most of the questions. And the reason being is that it provides also pacing. And you'll find that too. If you find yourself generating four times as many questions as you have time to answer the next time, it probably means you did way too much stuff the previous time. So it gives you feedback signal on how much material to discuss in the class time. And so that's one reason I'm reluctant to just put all the answers on the web. Because that's sort of like teaching with slides where you can just flip through all the equations really fast, and it seems like you got through a lot of equations. But actually no one really understood the equations. So it's sort of a symptomatic treatment for a more fundamental problem which may be that I'm doing too many things and not allowing enough time for questions in the class itself. A general question, which is, how do you evaluate your own teaching? When you feel good or bad about how class just went? And what metrics do I use popularity, covering what I set out to? How do I know? So one major way is I use these. It's not really that I say, oh did people hate it or like it? I mean, I want to know if people hated it or liked it. And I just I don't just add up the likes and the hates and say OK, 10 likes minus 3 hates equals 7. That's great. You're batting 80%. But I try to get a sense from reading the comments of what people were confused about, what was interesting to people, and sort of integrate that up and imagine the whole class as one giant student mind. And I want to know how I connect the material to the collective student mind. So I evaluate it that way, sort of intuitively from the comments that people make on the sheet. Which is why I hand out the sheet every time, and I think the sheet is so valuable. Now another way, covering the materials I set out to. Well I'm probably sort of on one extreme of that, which I don't really care how much material I covered. But let me explain why I'm on that extreme. So some people only care about how much material is covered. And some people are more on my end. So I'll give you the reason. And most people I would say on this edge. So I don't need to really justify that, because you've heard lots of reasons for that. But it's the other side, why do I not care? Well I don't care so much just because I think class time and a lecture course is such a small part of a student's life that if you haven't connected to the students, if you haven't made it click for them and got them interested, made them want to continue studying, then you've lost most of the effect of the teaching. The effect of the teaching isn't really going to be so much in the class. Because what you can do is you can kindle interest, kindle a way of looking at the world. But it's really up to the students to carry that on afterwards. And if you've actually bored them, or they've decided that this whole material is useless, they're not going to do that. And then it's all just going to go away anyway. So actually what I'm much more interested in is how much I've kindled those interests. And that's what I try to judge using the sheets. And I think that's actually what produces the long term change in ways of thinking. Another comment was that the rock experiment rocked, which I thought was a good pun. If you're not willing to write out detailed derivations on the board, is it more important to have more detail published lecture notes? And yeah I think that's right. In fact, that's generally necessary. But it doesn't have to be your own published notes. It could just be a book you refer people to. So we've sort of gone backwards from Gutenberg. In the days before Gutenberg, everyone read out their notes. That was what monasteries did. You read out the bible. And at the end of a year or two of doing that, you had 50 bibles, because everyone took notes on what you read. That was the old photocopier, back in say the year 1300. And then when Gutenberg came along the universities then did the same thing. That's what universities were. They basically grew out of the monasteries. Then Gutenberg came along and said, oh actually books are cheap now, well much cheaper than before. Books used to cost roughly say $5,000 a book in today's dollars. But then with Gutenberg and automatic printing, they could be much cheaper. So you can-- and we've developed printing technology much farther. So actually for a while until word processing came along and typesetting on your computer, people would actually just refer people to books and say, OK you're a student. You are expected to read these books. But now that people can make their own notes, people have sort of forgot how to tell people to use books. And instead people, the lecturers, just tend to write their own notes out. And the students actually seem for some crazy reason to basically except notes specifically for that topic and don't want to read books. So it gets really strange. You get in these bizarre cycles where you make notes. And as long as you have notes that seem like they're key to the lecture, that's great. Students will read them. And then you assemble all the notes for the very same lectures into a book, and now suppose the book gets published. The students will say, well now we want notes again. You can't say, well there's the book which was the notes. Because it's not specifically targeted. So part of the cure for that is reading memos. Or any way of teaching students to do reading is to shift back to taking advantage of Gutenberg. And then once we take advantage of Gutenberg, we can worry about taking advantage of all the web technology and word processing technology that we have. But yeah, it is important, to answer the question directly, to have the details somewhere for the students. And the best place generally is in printed material, whether in a book or in something you write. Because writing it on the board is very noisy. By which I mean that it's very easy to mis-copy. It's easy for you to make mistakes. It's easier for the student to make mistakes when they're taking notes. It's best to have it all in print, and also it's best to put the high level stuff in lecture, and have all the details with the students in their own comfortable space when they have time to go back and forth. OK, so then there were several questions about how do you meld the intuitive model and the school model? Are their general principles? So I talked about how for example with the rock example that the school model is that when the rock isn't moving, the force is zero because it's in equilibrium. So the net force is zero. So the reaction force equals the weight. So that's the school model. The intuitive model, somewhere in there is an intuitive model that you don't want the rock to fall on your hand. So what's a general ideas behind melding those two models? So that demonstration I showed of borrowing someone's hand and dropping a rock on it, well not quite dropping the rock on it, but threatening to, Is one way. But what are the general principles behind that? So the general principles are that you want to connect to where all the mental hardware is. So generally symbolic processing it's a sort of surface phenomenon in the brain. It's linguistic. It's only about 100,000 years old, whereas perceptual reasoning, visual reasoning, visceral reasoning, those are hundreds of millions of years old. Organisms have been seeing things for hundreds of millions of years. So there's a huge difference in the amount of hardware that we have. So symbolic hardware say has evolved for 10 to the five years. And perceptual, rough estimates 10 to the five versus 10 to eight years. So there's about 1,000 times more evolutionary help for these things. So if you want to meld-- and this is basically where the intuitive model, the internal model lives. So if you want to change it, you have to connect to where the model lives. You have to do things that are perceptual. So visual arguments, visceral arguments, auditory demonstrations, things they create conflict. Somehow some kind of conflict is also another general principle, where you expose some kind of obvious contradiction. And that's so beneficial, because that's self teaching. As long as there's a contradiction, the students know that the problem isn't done. So in the rock example, there's a contradiction between, I don't want the rock dropping on my hand on the one hand-- that's the perceptual view-- with the symbolic one, which is net force is zero so the reaction force equals mg. So there's a contradiction. As long as you can't get the same answer by two different ways, you know you still have more to learn. So it's self-teaching. You don't have to wait for the teacher to say, oh wait a minute. That's not the right answer. You know on your own. So know on your own. You internally feel that there's something wrong. So you're tapping in, I would say contradictions and puzzles. They're somewhat symbolic, but they're also kind of visceral. You feel them like oh, my god that can't be right. So if you can tap into that as well that's another way of trying to meld the models. It's interesting. I think it's a fundamentally important point about teaching these two models. And I've never seen it discussed anywhere. So you're the sort of first people to hear it officially as far as I can tell. So any questions actually about that model? Because I think it is so important. Yes. AUDIENCE: I'm just curious, what if you're in a field that doesn't lend itself to perceptual-- like if you're in a field that tends to be more symbolic. And physics is really nice because it has this physical aspect to it. So I'm just curious if you've thought about other-- PROFESSOR: So the question was, what happens if you're in a field that doesn't lend itself to perceptual reasoning so well? AUDIENCE: I guess I wouldn't [INAUDIBLE] PROFESSOR: Yeah there's visual in everything. And I think it's hard to imagine a field that doesn't have it. Can you give an example? AUDIENCE: Well, some computer science. And if you spend a lot of time with computers, then you start to develop it. But you could think, well I know that it's not possible that it could just randomly decide to give a wrong answer. There has to be a reason. How the program develops, you can develop intuition about where the bugs might be. But when you're just starting out, that's not something that we necessarily have built in to us in an intuitive way. PROFESSOR: That's true. AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah so computers and programming and computer science was offered as an example. Yeah I agree with you. Some fields are harder. And what makes computer science harder is that computers are symbol processing devices. So it pushes them over here. But even there I think you can do stuff. Like for example, what you could do? Well there are some things like for example, loop invariants, and you can actually turn those into perceptions. So once you have get a way of looking at programs. You say, oh that can't be right because it has a fencepost error in it or it keeps keeps decrementing the counter, but there's nothing to stop it from having a floor. So you start to see problems like that. And you have tools like different kinds of magnifying glasses to look at programs. So you could actually shift it that way. And really good programmers actually do look at programs perceptually. AUDIENCE: See, that's encouragement to build up [INAUDIBLE] PROFESSOR: Yeah, so you want to encourage that. And it's true-- pardon? AUDIENCE: It's not built in? PROFESSOR: I would say that the field itself pushes things towards symbolic. Just because it is a symbol manipulation device. So that's what makes computers so much better at what they do compared to humans. That's why we use computers. Because we're so bad at symbol manipulation, and computers are so good at it. So our ways of solving problems are going to be very different than computers. But the problem that we have when we're using computers, we have to solve with human hardware. So when the computer isn't doing what we want, we have to try to find some way of understanding what's going on. And it can't be just purely symbolic, because we're bad at that. So you can't-- I mean in desperation sometimes you'll trace through every step of a program and see what went wrong. But generally that just overflows you with tons of data. It's much better to actually try to take a step back, break the program into parts, see big chunks, and make sure the big chunks work. And experts do that, not sort of automatically because they've been trained as experts. So in any field the experts see differently than the novices, because their perception is different. So yeah, when you're teaching computer science to students, you want to teach them ways that are perceptual. You want to teach them visceral feelings. So one way to do that for example is running time of algorithms. Different sorting algorithms take different times. So suppose you have-- for all the non CS people, a huge area of study in CS is sorting. Suppose you have a giant list of numbers or list of words you want to alphabetize. Text indexing uses it. All kinds of stuff using sorting. So how fast can you sort? Well the naive sort of simplest thing to program is called bubble sort. And if you have n numbers, it takes something times n squared, so flip operations. So time is of order n squared. But then there's merge sort, which is a divide and conquer algorithm, which takes n log n. Now you could just say that. But it's also very helpful for students to have a feel, because generally students don't have an understanding of functions is what I found. They don't have a good feel for what functions do and mean. So what's the difference between n squared and n log n? Well one way is to actually program a sort algorithm and then try it. Try both algorithms and make your list bigger and bigger. When do you get fed up waiting for the damn thing to finish? So that's a visceral feeling. And actually that will tell you that there's a huge difference between n log n and n squared. Because this guy, basically when n is around thousand-- well, actually now computers are much faster, let's let's say 10,000-- you're pretty much going to be fed up with that. Because that's 100 million sort operations. It's going to take five or 10 seconds, maybe even bit longer if it's written in a higher level language. So maybe you're fed up around n equals 10,000. Here 10,000 times log n. Well, the log of anything is never bigger than 10, as a rough rule of thumb The reason being that if the log of the thing were bigger than 10, then n itself will be so big your computer couldn't even fit it. So you're not even dealing with numbers whose logs are bigger than 10. It depends on the base, but roughly speaking in base 10 logarithm, it's never going to be bigger than 10. So this is say at most 10. And so if you have a 10,000 only times a 10 here. Whereas here you have 10,000 times 10,000. Here you can go up easily to a million before you get fed up. So the fed up is a connection to a visceral reasoning. And actually it then makes the difference between these really, really apparent. So there's ways you can do it even there. Question? AUDIENCE: I think that in the science fields that, if you really understood what you're teaching, then you should be able to basically put it into a simplification. Because then it's basically summarizing everything, and it's just so much better and closer. I mean, if you write for example a term paper or a proposal or whatever, you can write a one page text basically, or you can make a figure or a nice sketch. It's the same information. PROFESSOR: Right so the comment was that if you really understand something, you can make a really compact pictorial representation. That's generally true. And that's why it takes really long time to make those. So when you're making talks, for example, when we talk about this later in the term, when you're making talks and slides, a really good model for making slides is-- here's a slide. This is the four to three ratio sort of. So here is a sentence. So this is some assertion, some message you want to get across. And here is the evidence. The body of the slide is the evidence. And that's a picture. The best is to have some kind of visual evidence. But what you find when you do this-- and it is, I think, the best way to make slides-- it takes a long time to really find the best picture. And even to find a good picture because it's just so tempting to write a bunch of words out. But actually if you look for that picture and you really try to capture the main idea in a picture, you will be much more successful at conveying the messages that you want to convey. So in general, try to talk to the perceptual system. Now there was a question related to that which is that maybe I'm too harsh or too negative about the symbolic ways of doing things. And the particular example was so if you remember last time, I gave this example. So this is a pendulum, and this is the extreme of the pendulum motion. And the question was what is the acceleration vector at these various points in motion. And I said, well the way to do it is to reason intuitively about it, to figure out what it is here and here, understand what acceleration is, and do this one by interpolation. And the comment was well, even if you can't do that, you can still use the Lagrangian. If I know the Lagrangian for the system, which for the non-physicist is sort of the elephant gun. You can solve pretty much anything with the Lagrangian. So you can solve it with the Lagrangian and work out the forces everywhere. And that's true. But I still maintain that it's better to have-- I'm not saying it's bad to have a Lagrangian analysis. The more ways of doing things is better. But there's a fundamentally important reason why it's important to have the intuitive way of doing it. And that is a search argument. So if you remember from before, we talked about chunking. The experts have big chunks, and the novices have small chunks. Well we looked at the effect of that of perception, which is that one of the proxies for it was the chess masters could remember whole positions, and the novices couldn't. Because their chunks were too small, and if you remember only seven chunks, which is typical, you can't remember a whole position. Whereas the chess master's chunks are like three, four, five pieces. And they can easily fit a whole position in seven chunks. Well there's another, perhaps even more fundamental consequence of the different chunk size, and that is the exponential explosion in search space. So suppose you're here. This is your state now, and you're looking for a solution to the problem. So suppose the solution is here. But you don't know that yet. You're looking for the solution. Now if your chunk size is really big-- well let's say suppose your chunk size is really big-- you'll try a few possible chunks. OK, do any of those get me closer? Well this one looks a bit likely. Let me try expanding that one, maybe that one. Try this one. And then eventually you get your way over here. So here suppose there's four possible approaches you could try, and because the chunks are big, there's only three levels. So this gives you four to the three items to search. OK, now suppose you make your chunks really, really small. Suppose your chunks are only half the size. So you now at every half branch you have to search for possibilities. So this is the large chunk. So here we had one, two, three levels. Here we're going to have six levels, in other words, four to the six possibilities to search. So the difference doesn't show up in simple problems. Because you may have only one level to go through versus two levels to go through, and either method works. But when you're trying to solve interesting new problems, you want your chunks to be as large as possible so your search space collapses to something manageable. So you don't want to have to reason with the Lagrangian as the default mode. It's fine for that particular problem. But if that problem is embedded in a bigger problem, you want to just know what the answer's going to be roughly. So you can continue making progress and decide should I go this route or this route. So that's why I'm not negative on the standard ways. But I think the standard ways are actually genuinely bad for problem solving in new domains. And you want to augment your perception so that you guide your search in the right direction with big chunks. So that's fundamentally important. So I think that was sort of the main themes in the questions, and I'll look over them and see which ones to answer probably at the beginning of next time. I'll try to leave more time for questions today, so we don't generate so many questions next time to overflow it. OK, so before I continue, any questions about these general principles of thinking about your teaching? Yes? AUDIENCE: So the approach is trying to focus more on the perceptual way to understand thing, but on the other hand, the symbolic manipulations are useful. So how do you make sure that the students will themselves engage in the symbolic. How do you encourage them to do the symbolic reasoning themselves without scaring them? PROFESSOR: So how do you encourage the students to do the symbolic reasoning? Because it is important. And I agree it is important. Just pure perception isn't quite enough. I think most courses I would say, and most teaching, is of 99% symbolic and 1% perceptual. And it depends on the field, but a rough estimate is I think it should be say something like 20% symbolic and maybe 80% perceptual initially. And then as you go to more and more advanced levels, if you've done it mostly perceptual earlier on, and you've developed the intuition, then you can increase the amount of symbolic. So I think the way you do it is you build it into the structure of the curriculum. So one particular course might just have one mix. And the more advanced students who got the perceptual material earlier are now ready for the symbolic. So there's actually a graph that I show for this. So that's the fraction of symbolic. And that's the fraction of perceptual material as you sort of go on in the major. So I think this is the ideal structure for a major. Because this is going beyond course design to whole design of a major in a curriculum. So early on, say for the freshmen, it should be mostly perceptual. And there's many reasons for that. First of all their symbolic capacities-- we're all bad at symbolic capacities to start with. As freshmen, they're worse than the seniors. They haven't had time for example get sophisticated understanding of differential equations. Their algebra's rusty. It's not really practiced much. So don't rely on those modes of learning. Rely on the perceptual modes. And the perceptual modes give you the big chunks. And so then once you have the big chunks, well then sure they're ready for the symbolic more and more. Because the symbolic fits into the big chunks and helps guide the search. It helps actually do the search. I have to calculate this. Then I calculate this. Well each calculation you do you need the symbolic. So by the time they're graduate students, hopefully the perceptual stuff they've had it up here. It's really solid. And now they're ready for partial differential equations and things like that. But earlier on, you don't want to avoid partial differential equations. You want to solve them in intuitive ways. And an example of that is early on, when you're doing fluid mechanics, you would solve the cones using dimensional analysis, as we did before. You wouldn't solve the Navier-Stokes equations. And later on, for example, in a graduate class on computational fluid mechanics, you'd actually try to simulate that and figure out what the drag coefficient is numerically. So that's the mix between perception and symbolic that I think is a very good way to construct a whole curriculum. Does that answer your question? OK.
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https://ocw.mit.edu/courses/3-091sc-introduction-to-solid-state-chemistry-fall-2010/3.091sc-fall-2010.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So these are the scores from test number two. The celebration. And as you can see, the average has gone down. So as I've said in the past, I think everybody in this room has the intellectual capacity to be here. Some people choose to be there. I don't know why they choose to be there, why they choose not to go to recitation, why they choose not to try the homework, why they choose not to go to office hours, why they choose not to read. But I guarantee you, when you make that choice you will be here. And I have no requirements whatsoever to pass a certain number of people. I can give everybody A's if I want. And I can take a large number of people-- you know, look at this. The way this is right now, the failure rate will be abnormally high. So I would warmly recommend that if you're down here, that you look in the mirror and ask yourself a few questions. OK. Let's get to the lesson. We're going to start a new unit today. The new unit is going to be the second half of the classification of solids. We looked at solids and we reasoned that there were ordered solids and we've looked at crystals and their arrangements and so on. And today, we're going to start talking about disordered solids. And they're characterized by no long-range order. They have short-range order. You might know who your next nearest neighbor is, or your second next nearest neighbor, but you certainly don't know who your 10th nearest neighbor is or who your 100th nearest neighbor is. And these solids are called amorphous solids. And there's a simple one salable Angle-Saxon word for this. It's called glass. So we're going to talk about glasses. What kind of materials can form glasses? Well, obviously, materials that have trouble crystallizing. And they come from a variety of walks of life. So we have in organic compounds. Some inorganic compounds can form disordered solids. And a good example is silicates. And this is what you know as window glass, glass that's used in bottles, cookware, and so on. There are organic compounds that can form disordered solids and a variety of polymers. Things like food wrap and so on, are also prone to forming disordered solids. Some elements. Some simple elements can form disordered solids. A good example is sulfur. Sulfur can form solid in crystalline form, but more often than not, it can form disordered solids. And what I'm going to show you towards the end of the lecture is metal alloys. Metal alloys can form disordered solids, and you're going to see metallic glasses. And they're typically 80% metal and 20% metalloid. That is somewhere along that red staircase on your Periodic Table that divides the metals from the nonmetals. So a good example here is iron 80-- this is on mole basis-- and boron 20, boron being the metalloid. And some, some do both. Some compounds can form both crystal and amorphous. So one example of that is SiO2, silica. When it forms crystalline solid, we call it quartz. Crystalline SiO2 is quartz, and amorphous SiO2 is the silicate glass that we know for windows and bottles. And so as I've said many times, the term "glass" is related to atomic arrangement. It has nothing to do with the ability to see through something. Transparency has no place here. So what are the conditions? I mean, if you have silica and it can form crystalline or amorphous forms, how does it decide which to do? And so I'm going to look at conditions promoting glass formation. And I'm going to form glasses on the basis of solidification. So I'm going to start from the liquid and go to the solid. So I'm asking the question, why, on some occasions, does the liquid go to a solid that's amorphous and in other occasions it will go to a solid that is crystalline? As in the case of quartz. So there are primarily three factors that it boils down to. And I like to make the analogy to the game of musical chairs. So imagine I've got chairs placed around this central table. You know how the game goes. There are more people than chairs. And the music starts and you walk around, and then at some point the music stops and people race for the chairs. And some people are going to get a seat and they get booted out of the game. And then one of the chairs is removed and then so it goes. You know this game. I know you don't want to admit to playing it, but your little siblings play it. But of course, you don't play it. But imagine if you were to play it. The same situation here. I've got silicate floating around in the liquid state, and the chairs are the crystal lattice sites. So the question is, what promotes the ability to get to the lattice sites or not? So obviously, one of the first things we think about is atom mobility. Atom or compound mobility. This is in the liquid phase. And the way I want to write this, I want to talk about promoting glass formation. So obviously, high mobility is going to enhanced crystal formation, isn't it? If you've got high mobility, you're going to be able to find the right lattice site. So I'm going to talk about if atom mobility promotes crystallization, the reciprocal of atom mobility will promote glass formation. The second thing is the arrangement of the chairs. If the chairs are in a simple line, it's easy to get to the chairs. All other things being equal. But if the chairs are arranged in some complex formation, it's harder to get to the chairs. So that has an analogy and that's the complexity of the crystal structure. The more complex, the more likely you are to form glasses. And then the last thing, an analogy to the musical chairs, is the rate at which the music stops. If the music stops suddenly you could get trapped in the liquid state. If I gradually turn down the volume, you know, OK, wait a minute. The music's going to stop. You start moving towards the seat. So that's the analogy. Here is the cooling rate. So a high cooling rate is going to make it more difficult for the system to find the crystal structure because as the cooling starts, the atoms start thinking, gee, it's time to form the solid. But the thermal energy is removed from the system before the atoms have had a chance to find a crystal structure. So the reciprocal of atom mobility, we write in a positive term and call that the viscosity. Something, a fluid, a liquid, that has low atom mobility has high viscosity. So I'll take this out and instead represent it this way. This is the way to think about the formation of the amorphous state. And why are we studying glasses? Well, in the old days it was bottles and food and cook ware and so on. Today, fiber optics. Fiber optics is based on silicate chemistry. Very important. So understanding silicate chemistry has high-tech implications. So let's look at, for the first study, the silicates. The silicates for their value in fiberoptic technology. So these are based on SiO2. And some nomenclature. This is called silica. So the oxide of an atom is named by adding the term "a." So silicon gives us silica as the oxide. So here's the atom silicon, the basic element. And if we fully oxidize-- see, here I've taken silicon, I put two oxygens, so I made a neutral compound-- that's silica. And then if I fully oxidize-- the maximum number of oxygens I can put around is four-- and this is going to have a net charge of 4 minus. And this fully oxidized anion is called the silicate. And that's where we get the name of the family of glasses. So this is fully oxidized. So now let's look at the structure. Silicate glasses. sp3 hybridized. Just like carbon above it. sp3 hybridized. So that gives us the four struts off of the central silicon at 109 degree angles to one another. And let's put some flesh on the bones here. So I'll put a silicon here in the center. One, two, three, four. But instead of putting silicons everywhere, I'll put oxygens at the end of the struts. And then I'll put the second silicon, and you can see the oxygen is acting as a bridge between the two silicons. One, two, three, four. Let's do one more. Another silicon. One, two, three, four. So again, you can see the oxygen acts as a bridge between the two silicons. So you see four oxygens per silicon, but I've got two silicons per oxygen. Hence, the structure looks like SiO4, but the stoichiometry of the compound is SiO2 because the oxygens bridge. So this doesn't give you any clue as to what the structure should be, does it? You really have to write this thing out. Now, here's where the thing gets interesting. This is a three-dimensional network. All of these oxygens bridge to silicons. And I was talking here about viscosity. You can see that this thing in the liquid state is huge. It's like a giant battleship, and this is just one. So now we can have another silicate. It can form chains. It can form meshes. And these meshes entangle. So viscosity very high. So when it goes to solidify-- David, if we can go to the document camera, please-- so here you see the silicate. So the yellows represent silicon. And you can see the oxygens, here, are at 109 degree angles. So this is the SiO4. We'll do some nanotechnology here. So here's the SiO4. One, two, three, four. And now I'm going to make the bridge. But look. The bond between the oxygens and the silicon is defined in three dimensions. It's a 109 degree angle. But the bond between the two oxygens is not defined. It's only defined in two dimensions. So you can see that if I put all of these oxygens on the same plane-- so somebody borrowed this and look, they lost an oxygen for me. Got a missing oxygen, here. If you find this thing, please bring it to my office. So now all five oxygens are on the same plane. One, two, three, four, and five. And then up here are the two silicons. They're in the same plane. The oxygens are in the same plane. What I'm showing you here is the beginning of crystalline quartz. Crystalline quartz, this stuff. But now what happens if we cool quickly? This bond is only specified in two dimensions, which means I can hold this bond at the proper value and this is free to rotate. This is free to rotate. Look at the autofocus on this thing. Who designed this? Boy, what a stupid machine. So anyway, so here we are. Here is the point. This bond is free to rotate, and when it rotates, look-- now this oxygen is no longer in the plane with this oxygen. And as a result of that freedom to rotate, we can end up running out of thermal energy. And this is nowhere near its regular crystalline array. So this gives you the indications of the high viscosity, moderate cooling rate will end up giving you the amorphous silica. But in all cases, this bond is the linkage through. So we end up with this three-dimensional network. So what do we do with to show that we have some evidence for this? David, may we cut back to the slides, please? So how do you characterize? I'm going to use x-rays. So here's an x-ray diffraction pattern of the-- the upper one is cristobalite, which is one of the polymorphs of crystalline SiO2. And you can see you have distinct peaks indicative of satisfying Bragg's Law. There's some width to them, but that's because it's a real crystal. Down here, this is the amorphous silica. You don't see all of these features. You see only one very broad peak. Now if I said it's crystal-- pardon me-- if it's glass, you'd say, well, it has no long-range order. So why do we have even this one feature? What's this one feature indicative of? Doesn't matter how much disorder there is, I still know that no matter what, I'm always going to have these four oxygens as my nearest neighbors. So these four oxygens minus any long-range order gives us this one line here. So we have evidence for it. Now let's look at the energetics, because clearly, these are two different states. One of these is lower energy than the other. Which one is it? How to think about the problem? There's a simple way to do it with the tools we have in 3.091. Energetics can be given by bond formation. Energetics via bond density. When things form bonds the energy of the system decreases. So which one is going to have higher bond density? Well, the higher bond density clearly is going to be exhibited by the one that has the tighter packing. And which one has tighter packing? The disordered solid or the ordered solid? The ordered solid has much more dense packing. So the ordered crystal exhibits tighter packing, therefore, this means more bonds per unit volume. So that means, to me, that the energy of the crystalline state must be more negative than the energy of the glassy state. That makes sense so far. But now I want to show you one other thing that we've just inferred from this little exercise. If we get more bonds per unit volume, then can you see that we've made an association between the binding energy of any arbitrary ensemble and it's molar volume? So volume now, is a very easy thing to measure, isn't it? You can see it with the naked eye. So the volume of something with a given composition-- and we're talking about a mole, we're talking about equal numbers of atoms-- so the molar volume is indicative of binding energy. It's a one-to-one correspondence. If you like, the Vmolar is a measure of disorder, meaning the higher the molar volume, the greater the disorder. Or put the other way, the smallest molar volume is that exhibited by the crystal. So we have some traces here that come from the reading. So this comes from the archival lecture notes that were written by my predecessor, Professor Witt. There's a little typo here that obviously heating means increasing temperature, not decreasing temperature. Make a little correction here. So what we're plotting is the volume, the molar volume of silicate glass as a function of temperature. And imagine we start with some blob of glass-- of some known mass-- so we can divide and figure out what the molar volume is, and we started cooling it. We cool down until we get to the crystallization temperature of quartz. And when we get to that temperature crystalline quartz forms and along with it, a tremendous decrease in the volume. Unlike water ice, which is a rare exception, where the ice occupies a larger volume than the liquid, for most systems, the solid is more compact than the liquid. And that's what you see here. There's an abrupt drop here at the melting point. And then when we continue to cool, there's some thermal contraction. And as you can imagine a hot solid occupies a greater volume than a cold solid. And so this is the cooling curve and you can retrace it, heat up, and when we get to the melting point, there's a tremendous expansion and then off we go. So that's the classical form of the crystallization of a material that is undergoing the normal process of liquid to solid transformation. I'm going to use some tea colors today. So we went down the red line. Now let's go down the green line. So we're going to go down the green line. We go down the green line, we're going to use the same stuff, this silicate network, only we're going to cool a little bit faster. And we cool a little bit faster, we can zoom right on past the normal melting point and create a supercooled liquid. And that supercooled liquid gets lower and lower in temperature and all the while the volume is shrinking. Again, a hot liquid what occupies a smaller volume than cold liquid. You know this from a mercury-- the old days, you remember these old thermometers? You probably have never seen one of these things. It's all done digitally. But we used to have these liquid and bulb thermometers. You'd have numbers on here, and what happens is the temperature goes up, the liquid in here rises to a higher temperature. And at a lower temperature this contracts. Isn't the solid expanding, too? Uh-huh. So what's the other thing you learned from this? It must mean that coefficient of thermal expansion, which we affectionately will call CTE, the coefficient of thermal expansion of the liquid, must be much greater than the coefficient of the thermal expansion of a solid. Otherwise, the two would expand and you wouldn't get any sensible measurement out of this, right? Well, we see that on a curve. We see that on a curve, because when we plot volume versus temperature, when we're up here in the liquid regime we have a steep slope. This slope here is dv by dt. And when we get down on the solid regime, it's a gradual slope. It's still a slope, but it's a gradual because the coefficient of thermal expansion down here is very low. And that's what you're seeing here. So at some temperature the system changes from a supercooled liquid to a solid. But it's a disordered solid because we've quenched in all of that remaining liquid disorder. And take a look at this-- you've changed from the slope that's characteristic of the coefficient of thermal expansion of a liquid down here to the gentle slope coefficient of thermal expansion of solid. You know what this proves? This proves that glass is a solid. There are many people out there, even in the popular press, who will say, glass is just a very, very viscous liquid. Nonsense. Look at this. It has this coefficient of thermal expansion. And don't fall for any of that nonsense they tell you when you go to the cathedrals in Europe and the glass is thicker at the bottom because it's been dripping for 400 years, and that proves-- you know why the glass is thicker at the bottom? Because they made it by spinning. And when they spun it, it was graded in thickness from the center out. And now, if you're the glazier and you're putting the glass in a window, which way would you put a pane of glass that had variable thickness? Would you put the thickest part up or the thickest part down? It's thicker on the bottom because that's the way it was made. Lord help the tour guide when there's an MIT student taking 3.091 on that tour. All right, so here we are. Look at what you have here? You see this? At this temperature-- let's say down here where the lines end, it's room temperature. At room temperature the volume of the crystalline solid is low. The volume of the amorphous solid is higher. That's proving this. Vmolar is a measure of the disorder. And sure enough, the glass that was cooled quickly ends up quenching in more of the liquid state disorder, and you see that in terms of what I call the excess volume. The excess volume is a measure disorder because the crystalline solid non-zero volume. So we can define, we can go from this directly over and say that v, that's the excess-- that's a pun. You know, instead of writing this? This is the way they do in high school. This is the excess volume. Bologna. We don't write like that. Excess volume is equal to v of the glass minus v of the crystal. And the greater the excess volume, the greater the degree of disorder. So let's do one more. Let's go down the orange line. The projector isn't giving us a good yellow component, here. OK, so this is the orange line. And what's the difference between the orange line and the green line? The difference between the orange line and the green line is that in the orange line we're going to cool more slowly than we did on the green line. Because green means go. So that's the fast one, right? So here we are. We still get supercooled liquid, but we get down to a lower temperature before we have ceased to have higher and higher viscous liquid and have formed the solid. The knee in that curve occurs at a lower temperature. And that value is called the glass transition temperature. Why do we call it the glass transition temperature instead of just saying it's a solidification temperature? Because when I say solidification, before today you would say, OK, liquid became a solid. But after today, if someone says to you, solidification, you say, do mean ordered solid or disordered solid? So we distinguish. So every crystallization is a solidification, but every solidification is not a crystallization. So here we are. Look at this-- this has a smaller excess volume. Because if it was a slower cooling, that meant that's the equivalent of giving more time, more thermal energy, for things to find their crystalline position, which means there's less excess volume quenched in. OK. So let's just get that down, define these things. So this is a measure of disorder. Or glassiness, if you like. OK. So now let's define these two different temperatures so that we have the distinction. So we have, first of all, the classical one, which is crystallization. And crystallization represents the reactions of liquid. In both cases, we're going to convert a liquid to a solid, but a liquid goes to a crystalline solid. And that occurs at a unique temperature called the melting point. Imagine if I talked to you about the freezing point of water. Freezing point of water at atmospheric pressure is 0 degrees centigrade. If I asked you, well, if I cool the water really quickly or if I cool it slowly, does it make any difference to the freezing point of water? No. Why not? Why isn't all this happening? Because water is a tiny molecule and so cooling rate has no impact on the temperature of conversion. Those water molecules will always find their lattice sites. So this is a function only of composition. If I put something in the water and I make it impure, I know I can change its freezing point. If I put salt in water, I will depress its freezing point. But if I put pure water, it will free at 0 degrees centigrade. Later on, we'll learn that there are some pressure effects, but today that's not going to elucidate anything. It'll just be a distraction. So pure water, always the same thing. But now if you go to something like silica, which has the capability of forming a glass, the glass formation is a different reaction. Glass formation is written in this manner. We will start with supercooled liquid. It's liquid, but I'm already going to stipulate that it's liquid that's been cooled below the melting point. Supercooled means it's cooled below the normal melting point. So a supercooled liquid is going to form a glassy solid, and this occurs at tg. tg, which is the glass transition temperature. And that is very much a function of cooling rate. And, of course, the function of composition. Obviously, if we change the composition from SiO2, we're no longer comparing apples to apples. So composition is important in both instances, but only in the case of supercooled liquid do we form the glassy solid. And the degree of disorder is a function of the cooling rate. And how is it a function of the cooling rate? As the dt by dt, right? Change in temperature with time as dt by dt goes up, the degree of disorder goes up. So if I want to quench in the liquid state, can you imagine if I had liquid and I could instantaneously cool it? I would quench in all of the liquid disorder. If I quench it less rapidly, there will be some time for the atoms to strive for a degree of crystallinity. So all of this we've said with reference to silicate glasses. Now, there are other glasses. What are other glass forming oxides? Well, what do I have to look for? I should look for other compounds that form bridging oxygens, right? That's the key here. What's the unifying feature? It's bridging oxygens. Bridging oxygen leads to glass formation. So what are other compounds that could give us such bridging oxygens? Well, you could be lazy and say, well, if silica does it, then why don't I look up and down the column on a Periodic Table. If you go up to carbon that's no good because it forms gas, but if you go underneath you will form germania glasses. So this is germania, and the glasses are germinate glasses. You can also go to group three. B2O3. B2O3 forms sp2 hybrids. And the sp2 hybrids off of each boron, we have three oxygens. And that oxygen can bond to another boron. One, two, three. And this is all going to lie flat in a plane, isn't it? But this oxygen bridge is only specified in two dimensions, whereas the boron struts are specified in three dimensions, which means this oxygen bond between the two borons doesn't have to lie in the plane. It could tilt this up and if it does, this is going to have a greater volume. You can see this with the naked eye. I mean, I could have just taught the lessen by putting this slide up. If those are equivalent numbers of borons and oxygens, it's plainly obvious that on the right side you've got excess volume. This occupies a much larger volume than the image to the left. The image to the left is crystalline B2O3. The image to the right is the same stuff, except in many instances this boron-oxygen-boron bond doesn't lie in the plane. It tilts out of the plane and it causes all sorts of distortions and leads to excess volume. So this is called a borate glass. All right, so if we can do it with the borates, we can do it with anything that will form covalent bonds. So we can do it with P2O5, phosphate glasses. V2O5, vanadate glasses. As2O5, arsenate glasses. And Sb2O5, stibnite glasses. And these are used, actually, as additives to some of the glass that's put in computer screens so that they will gobble up excess oxygen on cooling and avoid bubble formation, which then makes the glass foggy. And a foggy computer screen is no fun to try to look through. So what are the properties of these oxide glasses? Well, first of all, they're chemically inert. Why are they chemically inert? Because they've got strong covalent bonds. So if you try to react something with them, it's going to take a very special compound that can trigger reactions. Which is why they're used for bottling beverages and packaging foods. You put vegetables and fruits in glass jars going back to ancient times. Until recent times, with the advent of the soda can, there were glass bottles and so. They're electrically insulating. Why are they electrically insulating? Electronic structure. These are all covalent bonds, high band gap, which is why the amorphous version is used in window glass. High band gap, which means light goes through. How do I think about whether something is transparent? Do a little finger demonstration. This is the band gap of the glass, and this is the band energy, the photon energy. If the photon energy is small it goes right through. That's called transparency, see. You can study quantum mechanics, but you know what? This is it. That's all you have to know. Now what happens if the band gap is small-- like around 2eV-- and here's the photon energy? That's called absorption and re-emission. Transparency, absorption. That's all it takes. You think I'm kidding. That's all you need to know. So now, the next thing. Mechanically brittle. Why are they mechanically brittle? Strong bonds. No opportunity for slip. Please don't tell me-- this is what students tell me every year and they get zero for the stupid answer-- that the reason glass is brittle is it's a distorted solid and therefore has no dislocations. Uh-uh. Dislocations take the stress required to cause slip and reduce it to a lower value. But you cannot cause this thing to slip once it is solidified. The only way to get this silicon to move relative to that silicon in a vertical shear-- let's say I want to make this silicon move up and this silicon move down-- there's only one way. It's called break that bond. When you break that covalent bond, that's called fracture. So there is no slip allowed because we have strong covalent bonds. If you want to reshape glass, what do you have to do? You have to heat it back up above its glass transition temperature. But you do not shape glass once it is solidified. You may have tried in the home, and then you end up with something called shards. And that's the reason. Optically transparent, we know that one already. OK. Last thing is the very high melting. Melting point of silica is over 1,500 degrees Celsius. You're saying, well, wait a minute. He's telling me we're going to make beverage containers, we're going to make food containers, we're going to make all sorts of useful objects, but that's going to cost a lot of energy to go away up there to melt this glass. But it does have desirable properties-- chemically inert, and so on. So what can we do? Back here. Glass transition temperate is a function of cooling rate and a function of composition. So next step is let's modify the composition of the silicate network in order to drop it's processing temperature so that we can make beverage containers at acceptable energy. So what's it going to take? I'm going to have to do something about those bonds. Because those bonds are what caused me to have to go to such high temperatures. So in order to reduce the processing temperature of silicate glasses via change in composition-- so this is material science-- change the composition in order to get desirable properties. And specifically, I want to lower the processing temperature. Even when I get these things liquid, they don't deform very well, because all these networks are entangled. So I'll give you another analogy. Suppose you're in the kitchen and you're going to cook some pasta. So you take some linguine and it's about a foot long. So you got this pound of linguine and you cook it in the boiling water and you pour in into a colander. And you may or may not rinse it with cold water. Whatever. Just leave it in the colander for a little while and what you'll find is the whole thing turns into one big mass. I'm not talking about the stuff got all gooey. I'm just saying it just hangs together. And if you're clever you can just gently slide it out. And it'll slide out of this one big mass. What's holding those strands of pasta together, by the way? Well, are they covalent bonds? Are they ionic bonds? Are they metallic bonds? I don't know, if you've got metallic pasta, you've got digestive problems. So what's the bonds? It's van der Waals bonds, right? Now, I can cause them to slip again, can't I? I can put a little water in there. And what does the water do? It goes in between and then it's got hydrogen bonds and they slide. I can put a little oil in there and then that slides. or. The other thing I could do is-- you know, have you ever seen some people, they break the pasta before they boil it? I did this experiment. So you take the pound, divide it two halves. By my math, that's two half pounds. So you cook the one half pound one foot in length and you cook the other half pound-- break them in three. So they're about four-inchers. And then you put them each in a separate colander. Which one do you think is going to be much more difficult to move around? It's the long stuff, right? The long stuff entangles. So we're going to do the same thing here with the processing. And basically, you can study and learn pretty much everything you need to know about polymeric networks in the kitchen with a pound of pasta. All you need to know. All right, so let's go and look at it. I want to reduce the amount, the length, of those chains. If I reduce the length of those chains, I can process at much, much lower temperatures. So what's my weapon, here? My weapon here is oxygen. Oxygen is going to go in there in the form of the oxide ion. And the oxide ion, O double minus, has a very, very high affinity for the bonding. And so what'll happen is oxide ion attacks the oxygen-silicon bond. It attacks the oxygen-silicon bond and breaks it. It breaks it in two and then incorporates itself into the network in the following way-- you see I have to silicons joined across an oxygen? This free oxide ion will come in here, interrupt that network in the following manner. So it's now broken the silicon chain. Now, I've got conservation of charge. This was 2 minus. I don't see any exposed charge here. I need 2 minus. This'll be minus 1, this'll be minus 1. I have conservation of charge, I have conservation of mass. And what I've done is I've broken the chain. This is called chain scission. Shorter chains, higher fluidity. Higher fluidity, which means lower processing temperature. Now, where am I going to get my-- I can't go to the lab and get a bottle of oxide anions. I have to have charged neutral species. And so what I'm going to look for is an oxide ion donor. I need an oxide ion donor. And where do I find an oxide ion donor? Well, better be something that's going to be a cation, isn't it? Because if I've got an oxide anion, I need a cation. And what kind of a cation? How do I make an oxide anion? I have to have electrons to give to the oxygen. So I need a good electron donor. Or to use a simple Anglo-Saxon word, a good metal. So a good metal oxide like, say, calcium oxide. So if I take calcium oxide and I dissolve calcium oxide in silica, it dissociates to give calcium cation plus oxide anion. And then the oxide anion migrates over here to our pal the oxygen bridge and results in two of these broken pieces. So we call this one a bridging oxygen, as I've been saying up until now, BO. And this one is called a terminal oxygen. This is bridging oxygen, this is terminal oxygen. Or some people, I don't know why-- they have no literary skills-- they call it non-bridging oxygen. I hate that because it tells you what it's not. So you might see non-bridging oxygen. So all of these up here, the silicates and so on, these compounds that have the ability to form oxygen bridges, these are called network formers. These are all network formers. Because they have the capacity to make oxygen bridges. And then compounds like calcium oxide that have the ability to donate oxide ions that break bridges, these are called network modifiers. So good examples-- any good metal. So I can look at lithium oxide, sodium oxide, these will all disassociate to give oxide anion. If calcium will work and you believe Mendeleyev, then you should probably think about magnesium oxide, calcium oxide, and anything in that series. A good ionic oxide like lanthanum oxide, yttrium oxide, these will be oxide ion donors. If you want to get yourself fired, use scandium oxide because it's frightfully expensive. And you can go to Group four. You can even use something like lead oxide or tin oxide. Even they will donate oxide ions. In fact, you can put a lot of lead oxide in, you'll modify this thing so much that it'll start to allow you to cut crystalline facets and this'll be called lead crystal. Why do we use lead? Well, number one, it modifies so I can cut it and I get a straight line instead of that conchoidal glass edge that if you've ever cut yourself you'll never do it again. The other thing is lead, because it's got so many electrons, has a very, very high index of refraction. So when you make your billion and you've got your crystal chandelier hanging, of course you want the cut crystal with a high index of refraction because it's going to make that candlelight look really elegant and romantic and so on. And if you make a super boatload of money, then you're not going to go with lead crystal. You're going to make diamond pendants, right? Because they've got a higher index. And we all want the best index, don't we? OK, so I think this is probably a good place to stay. So let's jump to the end, because we've got a few things here. So I said I was going to show you about metallic glass. Here's metallic glass. 1959, Pol Duwez, who was a professor at Caltech, reasoned that if he were able to cool liquid metal very quickly, at, say, a million degrees per second, he could freeze the random orientations of atoms in the liquid state and prevent them from finding even something, either simple cubic, body-centered cubic, face-centered cubic lattice. So he worked with gold silicon. Gold silicon has a very, very deep eutectic. Even though gold melts to over 1,000, mixed with silicon, it gets down to about 400 degrees Celsius. So he dropped this through a little orifice here. This is molten and it drops onto a water-cooled copper wheel spinning at a very high speed. And it makes a ribbon some tens of microns wide. And what came out here was disordered solid. It was metallic glass. This was the birth of rapid solidification, and this is metallic glass. This is metallic glass. This has no long-range order, no grain boundaries, no dislocations, but it's not transparent. Why? Because it's a metal. And what do you know about band gaps and metals? And it's different. It was a different Q-factor. It has different mechanical properties. It has no ductility in a classical sense. This is just for your reference. This is aluminum foil. And you know what aluminum is. It's got ductility. It's got no grain boundaries so the corrosion properties are different. It's got no magnetic wall boundaries. So that a transformer made of this stuff is half the mass of a transformer made of crystalline glass. It has some other uses. So it's used in magnetoelastic resonators for theft prevention-- oh, I'm sorry. I'm in Cambridge. I can't say theft prevention-- for inventory control. And so you see that it's 20% boron and all of this metal-- iron, chrome, and moly-- and they put this inside the object in the store. And then there's a permanent magnet there, iron-cobalt-chromium, that sets the metglass to a field. And then when you walk through the uprights, there's a 58-kilohertz signal that causes this thing to ring. So it excites and listens for the ring. And if you still have that little piece of metglass in there that hasn't been demagnetized-- Bing! And then, oh, jeez, I'm sorry. I really am. I meant to pay for this. Explain it to the police officer. All right. So that's inventory control.
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https://ocw.mit.edu/courses/5-95j-teaching-college-level-science-and-engineering-spring-2009/5.95j-spring-2009.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation, or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK today, teaching with blackboards and slides. And also several questions from the last time. And, related to that, a handout, which I've put online, which is on how to make a lesson plan. So I'll do all that through the questions. So now just a quick bit about lesson planning, because a couple of you pointed out that you'd have liked to know more about lesson planning. And reminded me that when I taught the chemistry TA workshops, I actually gave everyone a handout on how to plan a lesson. So I've put that handout online, and I'll just show you what it looks like here. So this is basically the sheet I use for planning any kind of lecture recitation. At the top, you need some kind of course, and date. And then the objective. So that's quite important. You'll see the exact sheet. And there's a PDF file, you can just use it yourself. There's a course objective for that particular session. And that relates to your overall goals for the course. So, for example, in this problem, my objective might be something like, to show that easy cases-- in fact, this was my objective-- I wanted to show that easy cases-- extreme cases-- are useful not just for checking formulas, but also for generating formulas. OK, so I write that on the top. So I remind myself, why am I here? What am I doing? Then there's a three-column table. Basically you just plan items in your lecture. The first column is minutes-- how long you expect each thing to take. And then the middle column, most of the space is, what's your goal in this local part of the lecture? And what are you going to do for that? So here's my overall objective. And my local goal may be, just this example-- getting that one third. And then here in the third column is props. So there's actually a piece here, too. So props are anything that I need to bring. So I just stick that in the margin. OK, bring the cones, or the pyramids, whatever they may be. And so then, when you're filling this piece in, about what to bring, you just scan down this last column, and just write everything here. OK, cones. Homework three, solution set two-- whatever it may be. So now, at a glance, before you go to your session, you review these two things. Make sure you have everything, your know why you're here. But then here, you're going to follow your script. Which is a loose script. It's not exactly a word-for-word script. It's, maybe, a few equations. Suppose there's a question. What is this constant? OK, so that's my overall, what I'm going to do for, say, ten minutes. And then I break it into sub-questions. OK. What is H and B? How many pyramids do I need? What's my goal shape? So these are questions that I ask students. And for each one, I write down the minutes. Maybe in parentheses, put the total up here. So now, when I've done my two or three pages-- and by the way I find, just at the level of detail I use, if I ever go beyond three pages, I never get to that material. So it's just sort of for a security blanket by the time I'm beyond three pages. It depends how detailed you write these things, but I've generally found that's where I am. And then, you estimate the time. And what you'll find is, the first time you do stuff, you'll massively underestimate the time. You'll be off by a factor of two. So, in other words, if you think it takes five minutes, it'll really take ten minutes. And that's true even after you take account of this rule. Sort of. OK, so then you put down all your minutes on this page, and the next page. You add them up, and you make sure that you're not over 50, or whatever the amount is. And then what you'll find is what you thought was 50 really is going to take you 130. So the next time, when you evaluate minutes, you'll have a better idea. And you'll find yourself actually tuning your time sense pretty well. So now I've got my time sense that I don't really need to put the minutes down, because I can just do it by number of pages. Three pages, too much. And so you'll find your own writing level, and detail level, how many pages to use. Now the other point is, where do these questions come from? OK, well this is a way you can turn any regular lecture into an interactive one. So suppose you have a long derivation that you're going to do. Or, for example, suppose the first way I planned this, was I wanted to show people that this was 1/3 over here, this is three here. And the way I was going to do it, was I was going to draw six pyramids, show them in a cube, and show that 6V, and do all this showing. Well, first draft, write it all out like that, with times and everything, without questions. Then, any time you come to something interesting-- so here. So before you have [? tel. ?] So now you look at your sheet, and you say, hmm, where did something interesting happen? Hopefully, there is at least one point. Because if there isn't, maybe you shouldn't be giving the lecture at all. So now, let's just say, by construction, you found some interesting things. OK, so that's interesting. Oh, it's interesting that it makes a cube-- it's not obvious. So just think, where does something require thought, and that you're short-circuiting the thought by telling. So then, what I do is I have a green pen, and I turn it into a question. So I'll circle it in green, and write, "ask" next to it. OK, or you can just rewrite your sheet. Telling something, then asking, and then continuing. OK so that's how you can turn any regular lecture into an interactive one. And that's your sheet that you walk into class with. And what you'll find, is that the first time you do it that way, there will be a bunch more spots you realize were actually subtle. Because, for example, at the end of class session, when you collect your feedback sheets, people will have questions about different parts. You'll realize, oh, there was actually something interesting that happened that wasn't obvious in one of these telling points. And you'll be able to turn that into a question, as well. The master sheet with all that formatted for you is online, on the course website. And you're welcome to use that in your own teaching, and distribute it to everyone you know Any questions about that? Yes. AUDIENCE: When you run over time, then you have to shift some of the things to the next lecture. So how do you do that? PROFESSOR: What happens when you run over time. The first time you teach the course, you basically find out there's just twice as much material in the course as there should be. And you'll always be running over time if you try to cover every single thing. So, the first time you run out of time-- there's two approaches to it. One is just to slow the entire pace down, and cover half of the amount of material. So that's quite a reasonable approach. The other is to sort of keep rhythm, but not cover every single thing. So my piano teacher, a while ago, she said, when you're sight-reading new piano music, the most important thing is, keep rhythm. So don't just stop, and then think for like ten seconds about one measure, and then continue in this herk and jerk. Just keep playing at speed, but skip some of the notes, and do whatever you have to do to keep rhythm. So you can try that approach too. And then, what you do is you say, OK look I'm not going to cover every single thing in lecture. A lot of stuff is in the book. And that's good to do anyway. So a mix of the two approaches is one way to deal with running over time. AUDIENCE: [INAUDIBLE] So each lecture you have some objective. And then, if you run over time, then in the first lecture, you only finish half of it. And then, the second one seems to start from the middle of [INAUDIBLE]. PROFESSOR: Right. OK. So that's one reason it's worth writing the objective first. Because then you know what the main goal is. So one way to do it, which I do like, is to make sure that-- right away-- the objective is reached. So you do it as sort of a layer cake. So the first example-- maybe the first two examples-- just from those, if everybody just doesn't do anything else but the first two examples, which you expect will only take 15 minutes-- but it takes 40 minutes, it's okay. Because the first two examples reach your objective. Now they don't reach the objective 100%, but they give you 80% of the objective. So you want to plan your lecture structure like a layer cake, or like JPEG. People know how JPEG compression works? So JPEG, the way you do it, is the low-frequency, quote, "most important" terms, come first. And then the higher-frequency, say, less important terms, come later. So sometimes you see things rendered in your browser step-by-step, and they just sort of take focus. And it gets better and better as you wait longer and longer. So you want to plan your lecture a bit that way. So that it's robust to time shortage. So that if, for example, there's a fire alarm halfway through your lecture, still the main point got across. Now it didn't get across in all the detail, and the beautiful glory of that you wanted, but the main point still got across. So then, if you run out of time, it's OK. You say, look, there's two more examples of this which we didn't do, but they further illustrate the main point. See the notes. And then you can keep to your plan. Assuming that your plan is a reasonable one. If your plan is that, I'm going to do f equals ma today, and I'm going to do rigid body rotation tomorrow, that's probably not a reasonable plan. And the fact that you ran out of time on day one is probably a good sign that you should actually spend a little more time on f equals ma. Does that answer your question? Other questions? Yes. AUDIENCE: I have a question about [INAUDIBLE]? A lot of people have told me over the years that it's good to write really big. And I know you're constrained by that the people have to be able to see. But, given that you're writing large enough for them to see, what's the advantage of writing any bigger than that? As opposed to having the [INAUDIBLE]? PROFESSOR: OK, so that's a good question. Actually, why don't we save that. We'll talk about that today. So the question was, how big should you write? Just enough so people can see, or bigger? So that you can keep the full story there. OK, so, any other questions on the lecture planning? And blackboard, and slides, we'll come to starting right now. Teaching with slides and blackboards. What are the advantages, and disadvantages of each. I'll tell you my bottom line, which is that my zero-th order term, so this is my objective-- if the lecture ends after the next two minutes, at least you'll have the zero-th order term. The zero-th order term is that, if you can't help yourself, use slides, but otherwise, use blackboard. For 90% of things, blackboard is much better for teaching than slides. There are some cases where slides are useful, and you can make an argument for it. There's some cases where maybe the slide is as good as a blackboard, few cases where it's better than a blackboard. But the default is, blackboards are better than slides. Now let me show you an example of why that's true. Suppose you're teaching the-- an example I've done with you before, slightly, is the Navier-Stokes equation. Let me write it down, to start with. Actually, I'll put it on a separate blackboard. So there's our topic. So we always know what our topic is, because it's up there, and it's just going to live there for a bit. And in fact, in some classrooms-- you can't do it so easily here, but my favorite classroom is 4-265 because it has pretty much this much blackboards in the front, and it has blackboards all around the side, and on the back. So I don't use the back ones so much, but the side ones are really, really useful. Because you can, for example, put the topic, like I've done here, on the side blackboard, and just let it stay there the entire time. You can put an outline of the lecture. OK, so let's say we have a lecture on Navier-Stokes equation. We want to look at them, try to understand them. So here are the equation. Well "are," "is," the plural is a bit ambiguous. This is actually three equations, one because it's a vector, vector, vector. Already there, you see something that's harder to do with slides. You'd have to work harder to do that in slides. You can just go back, and annotate as you go. And here, you can see the entire topic while you're looking at this. So, for example, the student has now got overwhelmed by all the little symbols here-- the del squared, the partial derivative, the v dot grad-- that part always scared me when I was a physics student. I knew what del dot v is, but what's v dot del? So you get scared by all that. And then people forget, they panic. Because their short-term term memory got filled with every single, little, chunk here. Their chunks are, remember, very small. And they forgot, what are they doing? Navier-Stokes equation fluid mechanics. OK. And now, you can even improve that first blackboard by talking about, what does this apply to? Drag, turbulence, river flow, airplanes. So here, you can have a reminder of why it's important. So now that just stays up there the whole time. And people can re-center themselves every time they get confused of what they're doing, just by that. And then, you can continue. You can say, OK, let me try to explain the meaning of these terms. So let's do the terms one at a time. What's this? So this term here is Dvdt. We've seen things like that-- that's sort of like an acceleration. You can annotate the equation right there. That's also hard to do on a slide. I mean, you can do it, you can work at it, but it's very natural on a chalkboard. Now, there's some things that are common to chalkboards and slides, which is the use of color. So now this, unfortunately, is blue, which is not the ideal color. Unfortunately, I don't have my-- orange is generally better, but anyway I have blue. Can you see the blue? So marginally? Maybe this orange, even though it's not as big, will be better than the blue. OK, well this shows you an example of color, and the importance of choosing the right color. There used to be a green one here, but they're isn't green anymore. Blue is not ideal, I won't use the blue. Let's try this. Is that better? OK. So let me use this orange one, and I'll try to write large, even though it's not as bold. OK, so the use of color. That's important in both slides and blackboard. What's the use of color here? Is it just because it's nice to have a pretty picture? No. The reason for using the color is that you get layering. So again, remember the students' dilemma. Pretty much all students' dilemma come from the following problem, which is that their chunk size is one symbol. You've been teaching the Navier-Stokes equation for 20 years, your chunk size is the entire equation. Right? So that's one chunk to you, you just write it down, like I just did. But for the student, every single, little thing is one symbol. So now, if you then write all of this stuff in the same color, as your rest of your equation, it just becomes more stuff. They don't know how to separate it into two different objects. But, if you write your labels in a different color, it pops out in a different layer. They know that lives in a different space, so you're already helping them chunk. So what is this? This is also a change in v. So it's sort of like an acceleration. Well, it's part of the acceleration, actually. So this is another acceleration. It's a change in v, but it's a change in v because you've moved to a new place. So this is a change in v because you waited-- you move to a new time. This is a change in v because a little piece of fluid moved to new place, where the fluid velocity is different. So these two things together are two different contributions to the change in v. So we'll actually call this whole thing D of vdt, which is actually its actual name. So this is an acceleration. These are the two pieces of it-- this is the acceleration. What's on the other side? Well, this guy, pressure-- p is pressure-- grading of pressure. So that's the differences in pressure. So that's going to do some kind of thing, like a force. And row is density. So you're taking a force-- I'll put it in quotes, because it's not quite a force, it's sort of force per area. But now you divide by density, and you get f over m. So f over m. So we have, here this guy-- acceleration is equal to f over m. Do you recognize that equation before? Yes, Newton's Second Law. So this whole thing, so far here, is just this is a equals f over m. So this is the pressure forces, and this must be the-- and this is the viscous forces divided by the mass. So these are the two contributors to the forces-- pressure forces, viscous forces-- and those produce the acceleration. So the Navier-Stokes equation comes out just as Newton's Second Law all over again. So now, this kind of drawing is quite difficult to do in the slide. Yes. AUDIENCE: So I actually would argue with you. You could prepare a set of slides with annotations, and then you could do everything. You could put all the information [INAUDIBLE] in a slide, or in multiple slides. [INAUDIBLE] animation. PROFESSOR: Yeah. So you could put them on multiple slides, and then flip to it, one by one That's true. So you can do it, which is why I said you can do it, but it's hard. Because it is a bit of a pain, and you can do it. OK but then the problem is, you want to now leave that on. So now you want to have not only this, but you also want to have this, and leave it on for the entire lecture. Because this is the meaning of the terms. And now you're going to leave it there, and talk about each term, one by one. Yes AUDIENCE: But you could also give them a handout, with all the slides there, so they would have everything in front of them. And if they ever want to go back, they can look back on the previous slide. [INAUDIBLE] but I think so far this is not a strong argument against the slides. Maybe the question's about interactivity. If somebody asks you a question, what about the third term? If you don't have a slide for that, then-- PROFESSOR: Right. And this is the point I'm coming to. So there's that. And I agree with you, you can do this with slides, you can give people a handout. Though the problem with the handout-- suppose you put all of this on the handout for them, so that now, when you erase this screen, and that screen, you go to another thing they still have something to refer back to you. The problem is that you want it all in one visual field. And you don't have that when it's on a handout. So you have a problem of split attention. So now the student's attention is switching between the screen, and the handout-- screen and handout. And when they look at the screen, they're trying to remember what they're confused about. And now they look at the handout. This is the analogy. When I was in Prague, I would look at a street sign, and I would read the word, and look at all the letters. [INAUDIBLE] or something like that. And I would look away. Two seconds later, I could not remember more than one letter that I saw on the street sign-- on the street name. The reason is, because each of the letters was random to me. So, yes, someone who's Czech, just says, oh yeah, that's Main Street, of course. No big deal. What's there to remember? It's Main Street, it's in the Main Square. Whereas me, as the foreigner, has no clue. And it's all new data to me. So it's the same way for the students. This is all new to them. So when they're looking at something else, you're talking about-- you're going to discuss, what's the dimensions of this term? They look at this thing, and now you have a big analysis of the dimensions of this, you want to check that all the dimensions are right, and your goal is eventually to work out the dimensions of this-- viscosity, let's say. They might be frightened of the partial. So they look at this, and they say, oh, yeah, where does the Dvdt come from? And then they look back at their handout, and they've already forgotten pretty much what this was. Because they are the foreigner. They are the tourist in this land. AUDIENCE: Yes. I was just going say that I would have said something very similar. The annotation part of it is to [INAUDIBLE] slides at all, because if I was making slides for this lecture, I would've done exactly what you did. But it would all be nicely spaced, and there would be no handwriting issues. So if anything, slides would be better for that. But I think that the argument for slides is [INAUDIBLE] having them linger on, but also pacing. Like, the fact that it doesn't flip up so quickly, because the [INAUDIBLE] that I've taken where they use slides and use [INAUDIBLE], it goes by so quickly that you have no chance to internalize. The fact that you have to write it out by hand forces you to slow down, in a way that is really helpful for the listener. And so that's something that's really important [INAUDIBLE] difference. PROFESSOR: Right. And that's a fundamentally important point about pacing. And this partly addresses your question about how big should you write. So one of the reasons for writing big, is that you're writing slower. And writing slower, generally forces your pace to match the pace that is absorbable by the students. So I'll come back to that point in just one second. So let me just finish the point about the visual field. The good thing about the blackboard is, you could call it cognitive offloading. In one glance, everything that the students need, can be there. So they don't really have to remember much. And the short-term memory isn't overfull. They're registers. They're a low-register machine. Or rather, they have the same number of registers as us, but they don't store as much information in their registers. So their registers don't get overfull, because everything's right there. Whereas there is a much bigger danger of register overfilling with slides. Then, as you point out, there's the issue of interactivity. So this is partly a performance question. Slides, not only are they much more pre-prepared-- I mean, you can prepare your blackboards, as well, but you actually do have to re-perform it each time. Whereas a slide, as you say, you just click to Return, or Next, and there you are, at the next slide. So there's much less performance in it. And it seems much more pre-scripted. Now, what does that signal to the students? The pre-scripting. It says to the students, look, this is all one, tight thing. It's sort of like a play on a stage, that you're only supposed to applaud at particular times. A modern-day play. Back in Elizabethan times, in Shakespeare's time, people cheered, and screamed, and cat-called, and it was a much more lively thing on the stage. But now, everyone sits really quiet, God help you if you cough, you try to wait, and you think about your coughing, and then you don't think about the play, and you try to prevent your coughing. So all that transfers a bit. That same mentality comes when you're using slides. People think, oh my God, this is a pre-scripted thing-- how dare I intervene, and break into that. So if your goal is to encourage students to question, it's harder to do that with slides. You still can do it. You can, for example, put questions on the slides-- and you definitely should do that you're going to use slides-- but the random, spontaneous character of your session-- the good side of that, is going to be minimized by the slides. So this one was just to continue saying, OK, now we're going to leave these two on, and we're going to do the dimensions of each term. I have one, two, three, four, let's say-- four terms I want to talk about. I'll do the dimensions of two terms on this board, dimensions of two terms on that board, now I'll have all the dimensions there. And now I can start, for example, doing dimensional analysis on this side of the room. And we'll eventually get to the Reynolds number, which is vr over nu. Once we know this guy's dimensions. And v, we know his dimensions, r the size of your sphere or something. So this is dimensionless. Now you've got the Reynolds number, you've also got, say, drag coefficient, which is f over 1/2 [INAUDIBLE] v squared times some area. So this is also dimensionless. So now, you have two dimensionless things. And then finally, the finale is-- So now, let's say by the end of the lecture, you get to the point where you've plotted the drag coefficient, a dimensionless thing, versus the Reynolds number, on a log-log scale. And then you can talk about this, you can go back and you say, OK, well what is this? This is really interesting. It looks like it was constant. And then what happened? Well, you can go back and talk about that. And as you talk about it, the entire graph is visible to everyone, as well as the source of all the pieces of the graph, which is all the-- why is everything dimensionless? Well, you have it on these two boards. And what are the terms that you used? Well, those are all on there. So all of that is visible in one glance. So that's an example of something that is hard to do in the slides. Now, it's not easy to get it right on the blackboard, either. You do have to plan. You have to plan by saying, what am I going to put on each board? And how am I going to structure the boards? But you can do that on your lesson plan, for example, on the back of the lesson plan. Say, OK, how many boards do I have? And what am I going to use? And then, if you want to take advantage of the fact that everything is visible all the time, you generally don't want to use the up-and-down nature of the board. You just want to use two boards on each of the regions. If you push this up, and start writing on this guy, you're now OK, still. But then, when you push this up, and you write on here, you've covered this guy up. And yes, you've lost some of that advantage. Again, you can plan it out, and do a trade off. But that is harder to do with slides. Now, your question about how big you should write so that everyone can see, versus telling the whole story. I think it is important to try to tell the whole story. It doesn't have to be the entire lecture sits on the set of blackboards, but a coherent chunk of the lecture should sit on the blackboards. You have two things you could control. One you can control is the writing size, and the other is what you write down. So this is a similarity between slides and blackboards, which is you don't want to write everything down. Because if you write everything down, then you say, well for example, did I write every single thing I said here? No, because I want the main ideas to come across. Every little detail that I say, that should be either in the notes, or in some book. But the main point should be on the blackboard at a level of detail that people can absorb. It's the same thing with slides. You don't want to put everything on the slides. So, you can still write big, if you just cut down how much you write. And generally, that is more suited to what students need. Because, again, they are living in the dirt of what is this, and what is this. And you want to try to bring them up higher. So you don't want to spend a whole bunch of time on all these little guys. You want to try to bring them up higher, and put some high-level chunks onto the little bits that they have. So use a big chalk, write, large, just don't write everything. And try to fit a coherent piece onto all the blackboards. It doesn't have to be a whole lecture. Although Feynman, at Caltech, when he gave the Feynman Lectures-- well, what later became the Feynman Lectures-- he did actually plan it so that he would start at one corner, and he would end all the way here. And I find that's too hard to do if you're also going to make the lecture at all interactive. For example, if you're going to have questions where you're going to have the students give answers, and you're going to write down the answers. Well, you're not necessarily going to keep all the answers for the rest of the class. So the goal of corner-to-corner, I think, is a bit too strict. But the general idea is sound. Any questions about that before we take a break, and then go onto slides, and slide design? Yes. AUDIENCE: I just wanted to comment that also, a lot of students, if there are slides, and they're available online, they will just print them out, and not come to lectures. PROFESSOR: That's true. A lot of students will just print the lecture slides, and then just think they can thumb through them, and understand the lecture. So that's partly-- you could call it the optimism of youth. That they don't realize how much they're actually not seeing, because they didn't actually come. They think if they just follow each slide, they've understood everything. But there's no way to put everything in a slide. If you packed a slide with everything you said that was worth remembering, or that was actually part of the structure of the argument, then it would become not a slide, but a book chapter. And so, yeah. I think there is a reasonable argument that if there's a book chapter-- maybe they don't want to come to lecture if all you're doing is reading the book chapter. But slide, and lecture, are so different. It's sort of like, if you just take a few photographs of the blackboard, could you replace the lecture? No, but you're right, students do think that. So actually I don't like giving them handouts of the slides, if I do happen to use slides. Because it's really not a substitute for a lecture. It's an aid to the lecture, just like the blackboard is an aide to the lecture. But what's important about the lecture is, what are you having them do in the lecture? And that's the reason they should come into lecture. Because in lecture, they're actually struggling with some stuff, and you've structured it so that they struggle, and learn. And somehow you have to convince them that that's the value. And it's hard, because of what you say. Other comments or questions? One second, in the back. AUDIENCE: [INAUDIBLE] But at the same time, I've probably taken at least half a dozen classes where every single word that the lecturer spoke, was present on the slide. And then it felt kind of worthless going, because you show up, you pick up the slide handouts, and you can read the slides during the lecture. You pay no attention to the lecture, because he's not saying anything that isn't there. PROFESSOR: Right. So that's sort of the worst of both worlds. So now, what they've done, is they've taken the constraints of the slide format-- because there's only so many words you can put on the slide, and make it even at all readable-- and often they don't make it readable, and I'll show you an example when we come back. And so then they've torqued the lecture to be basically reading out the slide. So now they messed up the lecture, and the slide still isn't really a good substitute. It's a huge problem. I think the answer to that, is that if you're going to use slides, they're really an aide, but the students really need to learn how to read a book, or your notes. Or something that's actually a coherent presentation of the material. Yes. AUDIENCE: So when you write something on the board to respond to a comment, and it [INAUDIBLE] some of your space that you could otherwise use to plan for the usual stuff. So when do you erase that comment? PROFESSOR: That's a good question. So how temporary should stuff be? One way to do it is to think about the number of minutes each thing deserves. So that-- I like things like that to be on the board the whole time, if possible. And now it may be that, halfway through, I might reclaim that board if I don't have a choice. But I'd like that to be there for a while, at least, because that's really the main reason we're doing all this stuff. And what are we doing. So ideally, I'd put that on the side blackboard, and just leave it there forever. Something like this, I would like it to stay there the whole time, so we remember. So this, I would mark as, "don't erase," if at all possible. So when I'm writing stuff down, I would know, how perishable is the material? That's the word I'm looking for. So suppose I ask people a question, like the wood blocks. OK, what's the frequency going to be? And we get a bunch of reasons. I think the reasons are quite perishable. After we write them down, and we talk about them, and then we reconcile everything, I'm perfectly happy to erase all the reasons, and just have the reconciled picture on the board. But while I'm reconciling, I want them all visible. So then, what I would do, is I would try to have all those reasons go, say, here. I might cover that up temporarily. What I would do, is I would write the reasons here, and then do the reconciliation here. And then-- so let's call this Reconciliation-- so this is actually the picture that I would use for the wood blocks, which is this. And I would draw a bunch of arrows, and stuff, dot, dot, dots, and spring [? bogs ?] OK. And now, I need to reclaim this board with perishable material on it. So now I reclaim the board. So this had reasons before. Lots of reasons. OK, and then, reclaim this. And I still have that reconciliation up there, and I have this board back. AUDIENCE: So you think it is a good idea to separate the perishable things on one board, and then the more permanent things on the other board? PROFESSOR: Yes. I think it is a good idea. That's a good point. I hadn't thought of it that way, but that is a good idea. Because that way you can reclaim the entire board back, and then replace it with more permanent things. Or, if something perishable is going to go back on it, put it here. And then, when it's done, to erase it, until a permanent thing goes on there. And then you move onto the next board. So it does require some board planning. But what you find, basically, is the first time you teach the course, you wish you'd planned it more in the first lecture. But then you start to be more automatic about it. Yes. AUDIENCE: I have a comment on the performance aspect of the board. So when you were writing these equations, and then as you went back in the vector symbols, and also when you're circling things, because if you're doing it right, the students are engaged. Their focus is [INAUDIBLE] chalk. PROFESSOR: That's a good point. AUDIENCE: It's as much of an interactive activity as they can with you writing on the board. Whereas with a slide, certainly [INAUDIBLE] a physical pointer, you might circle something, but then there's nothing there. A laser pointer's even worse, because you're not actually touching anything. Even if you have an animation drawing, you're still separate from it. So I think that's a big advantage of the board. If their attention is at your chalk tip, then that's some degree of engagement with that action. PROFESSOR: Yeah that's a really good point. So the point is that when I'm, for example, circling this, it directs people's attention, and then there's a permanent trace left. So you can, kind of, do it on a slide. If you knew you were going to do that ahead of time, you can make an animation that goes, circle. So you can do that, but it's hard. So one of the cures for that is this tablet, a kind of PC. So you can actually write on the tablet as you go. AUDIENCE: Yeah. But even so, the thing is that you're still over there [INAUDIBLE]. PROFESSOR: I know. AUDIENCE: [INAUDIBLE]. PROFESSOR: I think that's true. That, actually, we are together, doing this, on the object. AUDIENCE: Yeah that's exactly what I was not saying very well earlier, was you can do all this, what you're saying, on a slide, just as well. When I do [INAUDIBLE] presentations, I do actually animate the circle. I find that that's the most intuitive way for me to do something. But it's not the same as actually drawing it yourself. It's not the same level of activity. And obviously you kind of have to use slides for [INAUDIBLE], but in a classroom you don't have to. PROFESSOR: Right. OK, so that's what we'll talk about when we come back-- the difference being technical presentation, and teaching. And then we'll actually look at an example of slides, and we'll critique them, and I'll show you how to redraw them. OK so break for 10 minutes. We'll start at 10:18 sharp, by that clock. So you can set your clocks accordingly. And I will erase all this, and put up a slide that we're going to critique. And then we'll talk about technical presentations versus teaching. Because there's a lot of misconceptions that come from technical presentations. And therefore, people think we should use slides for teaching. OK, so see in 10 minutes, or if people have questions during the break. OK, so our next task is to figure out, if you are going to use slides, what should you do. And related to that point is, when are slides appropriate. People have the idea the slides are always appropriate for teaching, because they see them used all the time for seminars. So if you go, and give a talk at a conference, you go to a conference, most-- almost all the talks-- or at the department seminars, basically people now come with some form of slides. When I was an undergraduate, it was sort of transitional-- people came with overhead projector foils. And it was sort of shifting, from seminars used to be blackboard seminars, and then they shifted to overhead projectors, at least in physics. And now it's pretty much slides all the way through. But there's a fundamental difference between seminar presentation, and teaching. And seminar presentation-- basically you're talking to people who are sort of experts. They may not be experts in that particular area you've understood. Probably not, otherwise they would have invited one of them to talk, instead of you. So now, you're the expert locally, on that local area. But in the broader area, you're talking to basically people who are already interested in, and are professionals in the field. And maybe some apprentices in the field, who are going to soon be professionals, like graduate students. So your goal is not to slowly, necessarily, uncover one or two ideas, especially if it's a 10-minute conference presentation. Your goal is something like, here are the core, main ideas-- one or two points-- and here's something to show you that really interesting stuff is going on. And you should pay more attention to me is part of the goal of the seminar. Now that's very different in teaching. In teaching, your goal really is to kindle some new thoughts in the student. And who are very different from a seminar audience. They are not professionals. Maybe they would like to eventually be, but part of your job is to kindle that interest so they would like to become professionals. So you have a very, very different audience, and generally much different time constraint. In a seminar, maybe it's 10 minutes, maybe 15 minutes. Rarely, once you start getting well known, maybe you get invited to give a 50-minute seminar. But, teaching, you generally will have 50 minutes. So your time is different, your audience is different. So just because slides are used so much for seminars, does not mean they should be used so much for teaching. You should really use them when you don't really have a choice, and when it seems like it's the optimal thing to do. So for example, art history class. Now, an art history class, it would be fantastic to go to Sienna, and look at all the tapestries, and take the whole class with you. Now that's just too expensive. So, instead, you show slides. That's exactly why they're called slides-- they originally were 35 millimeter slides. 35 millimeter? Yeah, 35 millimeter slides. So for something like that, yes, slides are ideal. And drawing the tapestries of Sienna on the blackboard is not even close to showing the actual pictures. So, in that case, 100% go with slides. Other cases generally go towards blackboards, but there are cases where, for example, there are no blackboards in the room, and you have to teach with slides. There's no whiteboard, either. Whiteboards are not as good as blackboards, but they're a reasonable substitute. But there's neither, and it's just slides. Or, yes-- AUDIENCE: Why aren't whiteboards as good as blackboards? PROFESSOR: I find the markers always dry out, whereas chalk always works. So blackboard markers never quite work as well. And the other problem, just me personally, is that I'm sensitive to the chemicals in them. So I eventually just get a bit dizzy using them. So I don't like him for that reason. And there's quite a few people. I'm sort of the canary, so I'm more sensitive, maybe, than many people, because I have lots of allergies. But I think lots of people are bit sensitive to them. But, generally, I find the markers just don't work as well. And, also, there's very rare room has lots of whiteboards. Whereas many, many, classrooms have this many blackboards. When they replace them with whiteboards, usually it'll be one whiteboard. So then you're back to many of the disadvantages of slides, which is that you can't get a big field of view. But now let's say you are going to use slides. How do we make good slides? Well, one way to figure out what constitutes good slides, is to look at what constitutes bad slides. So this, here, is a horrid slide. So what we're going to do is, I'm going to show you how to re-write the slide. But first, let's figure out what are some of the problems with this slide. So, take a couple minutes with your neighbors. You can all see the slide. What are the things that just don't work about it? And there's many. It's a very competitive field. And I'll write some stuff down on here. By the way, one thing is that the JIT, that I just actually shortened. It actually was, "Just in Time Learning". I've just shortened it to improve the typesetting, assuming that the audience knows what it means. So the original slide didn't have that problem. So I see you found lots of trouble. Give me one thing that's wrong. Adrian. AUDIENCE: There's no message. PROFESSOR: What do you mean by message? AUDIENCE: What is the person trying to tell you here. You have to infer what the person is trying to tell you by analysing it yourself. PROFESSOR: OK, no message. So, the reader, the viewer, the student-- this was actually used in a class, never mind where, but it's not here. Yeah, what is the teacher trying to communicate? So Jean-Luc Doumont, if you saw him during IAP-- he has a great way of talking about what's a message, versus what's information. So the message-- so information, that's what that slide has. It has lots of information. That's the "what." Message is the "so what." His native language is not English. I wish I could come up with things like that in other languages. I think it's fantastic, because there's lots of "what." Well, 12% said Just in Time Learning-- JIT Learning-- was a bad idea. But your question is, so what? And there's no answer to that question. So there's no message. Yes. AUDIENCE: So a counterpoint to that would be, this is documentation, this isn't propaganda. Right? You don't want to say, this is what you should all think about this data, instead of, look at the data yourself, and make your own conclusion. Presenting the data in an accessible manner is a different issue. But I don't know if I agree with this, "so what" idea. PROFESSOR: And some of that is that, yeah, definitely in a seminar, you lean much more towards the "so what." And you would put in the "so what," really first. And in teaching, maybe you start with the "what," and then you lead to the "so what." So, for example, the wood blocks. Here's the "what--" what do you think is going to happen? And I haven't said, "so what," really. And then we talk about a whole bunch of ways of analyzing it, and then we finally lead up to the "so what.". But at some point, you do want a "so what." So, yeah, without seeing the entire lecture, it's hard to know if they ever gave a "so what." But I can tell you that the succeeding slides just went on to other topics. So it wasn't that they were using that to develop the point-- what do you think about Just in Time Learning, or how should we analyze Just in Time Learning. They just moved on. Yes. AUDIENCE: Going back to your art history example as a good use of slides, that doesn't have anything in it that gives you "so what." It gives the presentation of information, and then the discussion of it is what gives that. So I don't necessarily think there's something inherently wrong with giving information on a slide, and having not answered the discussion in the slide. PROFESSOR: It's not inherently wrong, and that's a good point. So the art history one, it's sort of a different use of the slide. So there, it's really like you're bringing in a prop. So the cones, these pyramids themselves, didn't have a "so what." They only got a "so what" as we used them. So it's the same way with the art history slide. It would get its "so what" in how you discussed the slide. But here, the slide is really the alternative to the lecture-- to the blackboard. If you've gone to slide talks, or slide teaching, which aren't art history. I think art history is really a separate example. It's, I'm telling you stuff, I'm telling you stuff, I'm just telling you in a way that you don't really know what I'm telling. Like, I may be telling it to you verbally, but the slide doesn't match what I'm telling you. Sorry, you're next. Go ahead first. AUDIENCE: You may have already said this, but is this slide supposed to be for the students, or is this for a faculty meeting? PROFESSOR: No, this was for a course about-- what was the course about? It was a computer science course somewhere. Maybe it was a discussion of a-- no, it was a computer science course, or a teaching course. I forget exactly what course it was, but it was from a particular course. So it's a slide used in teaching. Yes. AUDIENCE: That partially answers what I was going to say. This material doesn't belong in the lecture in any course, unless it's a course on teaching. The surveys are filled out, so they benefitted the lecturer so that they know what to do when planning and running the course. The students don't need to know what the results are. PROFESSOR: I think it wasn't for the course itself, it was for the students in a teaching course, is what I think it was for. And either way, whether it's for a seminar, or for a teaching course, it has the same problem, which is you don't even know what to think about it. You just have a lot of data. And there's no integration. OK, so that's the no message-- the "what," versus "so what." What else is wrong? Yeah, Wendy? AUDIENCE: The data is just terribly organized. [INAUDIBLE] One thing is that there's bullets under bullets. By the time I get to [INAUDIBLE], I'd rather skip, I've forgotten [INAUDIBLE], I don't remember what we were talking about. PROFESSOR: Right, so the organization is ghastly. AUDIENCE: And another good comment, which is that the numbers of the percentages are not organized in any way. They're not going from highest to lowest, or lowest to highest. So you've seen, maybe 52, you're like, I assume this is going in descending order, and then things get flipped around. PROFESSOR: Right. So the percentages are kind of random. It's sort of like the "what." To organize it towards the "so what," you would put some theme in it. But there isn't any, or none that's obvious. And the typesetting, yes, is related to the organization. The typesetting is a bit off. Yeah. AUDIENCE: So related to that, [INAUDIBLE] the numbers and the questions are given completely equal footing, because the numbers are organized, [INAUDIBLE]. And so the numbers just follow better to worse. I'm sorry, the answers to the questions are organized from better to worse. [INAUDIBLE] PROFESSOR: OK, so then it comes out. OK. AUDIENCE: And then it comes out, but that's not [INAUDIBLE]. PROFESSOR: So the Likert scale isn't obvious. The Likert scale is the better-to-worse scale of how much you agree with this statement. Right. So you're saying, also, the answers have equal prominence to the questions. AUDIENCE: Equal prominence to the percentages. You have no clue what's organizing the information. PROFESSOR: Right OK. So the emphasis is random. Yes. AUDIENCE: So [INAUDIBLE] so she suggested [INAUDIBLE] or [INAUDIBLE]. PROFESSOR: OK, so it's not visual. A pie chart, or a histogram, or something would help you see much more. So that's one of our principles, right, that your perceptual system is so much smarter than your symbolic processing system. So you're actually slowing everyone down, and making them much dumber, by forcing them to try to extract-- basically build a mental picture which you could build for them. Yes. AUDIENCE: I'd say there's too much. PROFESSOR: Too much. So what do you think is too much? AUDIENCE: All at once, you're getting lots of text, and it's all these different things. Too much [? given ?]. PROFESSOR: Yeah, it's too many chunks. It just overloads you. You don't relate it to the lack of messages. If there are messages in there, there are too many messages. You don't know what's important, and what to pay attention to. Other problems? Yes. AUDIENCE: Format switches midstream. PROFESSOR: Where? AUDIENCE: At the bottom, the last question. PROFESSOR: Oh yeah. You're right. OK So the format. AUDIENCE: [INAUDIBLE]. PROFESSOR: Right. It's sort of related to too much. Right, I tried to put too much on my slide, and I really tried, really hard. So the format changes at the end, right, so they could fit everything onto one slide. Yes. AUDIENCE: Extraneous use of bullets, colons, [INAUDIBLE]. Not helping. PROFESSOR: So a random use of colons and bullets. And it's not clear how those actually help. Other problems. Yeah. AUDIENCE: Is that all bold? PROFESSOR: Yeah. I think it is all bold. So everything's equally emphasized. So all bold. So again, it's presenting all the information, right? It's just you don't know what's important-- what's the "so what." Yep. AUDIENCE: The numbers are very far away from the text, so by the time you get over there, you don't know. PROFESSOR: Right, it's sort of like a table of contents. And that's OK, for a table of contents. It's not ideal, but there are ways of doing table of contents that don't have that problem. For a table of contents it's OK. You don't need to sort of get the entire table at a glance, right? You just want to know where chapter five begins, and you just scan across, and you go there. Whereas here, you can't match the things. You want to try to match it all together, and you can't, because the numbers are so far away from the text. AUDIENCE: I don't know what PI means. PROFESSOR: Sorry, that was my fault, too. I just tried to shift it. Peer instruction. Yeah, you're right, so there's a lot of use of acronyms. Yeah, so the JIT, and the PI were my contributions to it. So maybe I made it a little bit worse. Yeah. AUDIENCE: So how would you do this on the blackboard, instead, if you had a chance? PROFESSOR: So this one, actually, if I were going to present this, I would either draw the thing I'm about to show you, or I would use a slide for it. Either way. AUDIENCE: A slide with what? PROFESSOR: I'll show you what the redrawn slide is. OK, so now let's redraw the slide. So I think the slide that I redrew-- I'll show it to you-- basically answers most of these objections, though maybe not all. I think most of them. So what I'm going to show you is what Michael Alley calls the Assertion-Evidence design for slides. So Michael Alley, I'll write his name down. So Michael Alley, and his book is called The Craft of Scientific Presentation, I think. And this is to get around many of these problems. So the idea is, first of all, so related to what Jean-luc says, talking about messages versus information, each slide should have one message-- one assertion. So that's the assertion part of the slide. And the body of the slide is the evidence. And it should ideally be visual evidence, because of all the reasons we talked about for visual evidence. So here is an example of that. There. Is this the entire slide? No. This is just one, basically, third of it. But you can redraw the rest of the slide, with two more slides, this way. So this is one slide, with one message-- most students like in-class concept tests. If that's the point you're trying to make. The problem is, it's hard to redraw slides with no messages. Because you don't actually know what the message was, so you have to sort of infer it. But hopefully the author could actually infer it, and redraw it themself. So here you have to play author, and I guessed the message-- most students like in-class concept tests. The presentation is here, with a pie chart. And you notice all of the tags, the "I like them", the "please skip them," "I dislike them." They're all right next to their percentages. And then, what's color used for? There, color is used to emphasize the like. So the green, and the light green, are both categories of like. As asserted by the author. I mean, you could say, well, no "enjoy and learn from them" is the only category that matters, "like" isn't strong enough, or really you shouldn't lump them together. You can have those debates. But what the author-- the teacher-- is trying to say, is that most students like them. And you can see it just at a glance, because most of the pie chart is colored in. So this solves most of the problems. First of all, you don't have the ghastly typesetting that you have here. The typesetting here is just horrendous, this is sort of typical PowerPoint typesetting. The alignment is nonexistent. There's random alignment lines everywhere. If you look, for example, at the way top, the yellow bar doesn't align with anything else. The left edge of it just doesn't align with the bullets. The open bullets are just random. It's just a big mess, as you pointed out. So here, just get rid of all that. And you can do this with any program. You can make your figure however you want. You can put one assertion at the top. So now assertion, another synonym for that is "sentence headlines." So you want your title of your slide to be a sentence. If you're not writing a sentence, you're probably writing a topic. You're just mentioning something-- you're giving them "what." To give "so what," you really need a sentence. So fit a sentence of one or two lines there, and then, give visual evidence. Yes. AUDIENCE: To sort of play devil's advocate, I mean, I agree [INAUDIBLE], but the advantage of having all the information on one slide is you can see the different things, and see the different conclusions. PROFESSOR: If you could actually see them. The problem is, you can't actually see all the conclusions. I agree it's nice to see all the conclusions at once. So to do that, what I would do is, I would make-- I think there are really three messages here-- one for each of the questions. So I made a slide for the first one, which is about the concept test. You'd make another slide for the Just in Time learning, another one for, do lectures cover too much material. So now you have three. And now, basically you want an overall message of, overall these instructional changes were well liked. Then you could have one summary slide, which just lists those three points. And with maybe three small pie charts, to remind people of each pie chart, to fit it all together. But to try to pull it out of that, the people won't actually get the three messages. So you have to guide them towards it. Hopefully you can use this for your seminars as well, not just for your teaching. This works for teaching, as well seminars. Now, how would you actually put this into practice, in a teaching example? So that's what comes up next. I'll show you a mathematical example, which you could do on the blackboard, or I sort of optimized it so you could do it on slides too. And we can see how that goes. OK, so here is just the comparison, side by side, of the two slides. Just at a glance. I don't actually mean for people to be able to read the words, just to see that one has enough information that you can get a point, but doesn't overload you, and the other is just a rat's nest. OK, so then the mathematical example-- actually, a statistical mechanics example as well-- is the log of n factorial. So we're going to approximate the log of n factorial using pictures. Now why n factorial? Well, what is n factorial? 5 factorial is 5 times 4 times 3 times 2 times 1. Why do we care about it? It's the most important function in statistical mechanics when you're counting objects. And it shows up in probability theory all the time. And statistical mechanics has a lot of particles, so the n is big. So we'd like to approximate how big n factorial is, because it's so big we take the logarithm. OK, let's see if we can approximate the logarithm using pictures. So log of n factorial is the area of those rectangles. You can see rectangle for log 2, log 3, log 4, log 5, log 6, and log 7. And log of n factorial is just the sum of all of those. Say, 3 factorial is 3 times 2 times 1, so log of 3 factorial is log of 3, plus log of 2, plus log of 1. Log of 1 is 0, so all you have left is those rectangles to add up. OK, so now let's add them up, approximately. So the area under the curve is your first approximation. So the curve is just log k. So let's see what that is. Well, OK, that's just the integral of log k from 1 to n, from the lower limit to the upper limit. So that we can just do symbolically, you get n log n minus n plus 1. OK, now the error. Where does the error come from? Well, it comes from the protrusions beyond the log n curve. So there you have the protrusions. so we'd like to add up all those protrusions. Well, how are we going to do that? Each piece is almost a triangle. So let's just straighten out each little piece of log k curve, and make triangles. And now, it turns out they're much easier to add up if you double them, and make every triangle into a rectangle. OK, so now I'm going to stop here, and see if you can figure out how to add up all of those pieces. What's the sum of all of those shaded regions? Take a minute and check with your neighbor. OK, when you see it, raise your hand. Find someone around who sees it, if you haven't seen it, and check with them. Because it is fun to see. I don't want to spoil it unnecessarily. Here's a little hint. So what you do, is you hold your hand at the right page, and you whack all of the rectangles, and they slide across, and go psssssh, and stack, and form the last one. So that means the sum of all the corrections, times 2, is the last rectangle. So log 7, or log n, is twice your error, roughly. So now you just combine the integral, which is the piece under the curve, with the approximated protrusion, and you get that for log of n factorial. So let's see. So that is actually very close to Stirling's formula. Let's see how close it is. Well where's the error? Well, the error comes from when we straightened out the pieces of the log k curve, and made triangles instead of the funny-shaped region. So the error is those shaded guys, which is not very big. And most of the error happens low down, so if you just corrected for that, pretty much it would all go So here is the total error. Here's an example. The picture method, so we get 7 factorial, gives you 8.594. The exact answer for the logs of 7 factorial, 8.525. So it's an error only 0.07. So all of this just adds up to 0.07. Which ends up being a 7% error in seven factorial. So the moral of the story is that pictures can help you approximate log of n factorial, and here is your approximation. So one of the most important functions in statistical mechanics. 95% with pictures, and one tiny integral. OK, now. That's an example or something you could do if you wanted to teach it using slides. You could easily do that on the blackboard, and normally I teach it on the blackboard. But I made this as an exercise for myself to say, OK, well suppose I have to teach it with slides, what would I do? And you notice here. So when I first drafted it, this slide was missing. And I went from here to there directly. And then I thought, oh, actually-- so this is about the interactivity question-- oh, actually I'm depriving people of a chance to think about something, and realize it. On the board, it's really easy to pause. You just put the picture up, and you wait. And you say, OK, what do you think? And then you start drawing stuff. So I thought, OK, well you can actually simulate that on slides, too. So I put the intervening slide in there, so that people had a chance to stop here, and think about the result before we continued on to the solution, where it adds up to the last guy. So that's an example of basically somewhat interactive slide teaching of a mathematical idea. Yes. AUDIENCE: So why didn't you animate a slide [INAUDIBLE] with boxes to take advantage? PROFESSOR: That's a good question. Why didn't I? Actually, because, A, I didn't think of it, but it's a good idea. I should have probably done it. The problem is that I'm showing it as a PDF file, rather than through any program. So PDF animations-- I don't know if PDF has animations in it. It probably doesn't. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yes, that's true, it's sort of against the purpose of PDF, which is a stable format. Yes. AUDIENCE: I actually thought this entire presentation was really clear, and would be just as good as if you did it on a blackboard. Because it was interactive, slow, had pacing, the suspense of having us ask the questions. So I think one of the major reasons why I would be opposed [? to teaching ?] in this way, is just because it takes you more time to create the slide. From my own experience, it takes a long time to make really good slides, just fiddling around with [INAUDIBLE]. So I think if you could spend all the energy we're spending on slides instead on [INAUDIBLE]-- PROFESSOR: That's a good point. AUDIENCE: --or whatever, I think that might be [INAUDIBLE]. PROFESSOR: I think that is a really good point. And I hadn't thought of it until you mentioned it, but you're right. It took me a long time to make the slides right. To do slides well, you have to really think about the visual presentation, make everything visual, you're making a lot of diagrams. There's all kinds of little tweaks. Like, I made sure that all the diagrams are exactly the same size, so that when you flip from slide to slide, you don't get a shift. So all those things, you don't have to worry about on the chalkboard. And yes, I would advise people, just as you do, to spend that time thinking about your teaching, and where you're going to put the interactivity in. And do it on the blackboard. So this is something where, with a lot of effort, the slide reaches to the level of the blackboard. But with colored chalk, and lots of blackboard, the blackboard is just as good, and quicker. And the blackboard has the advantage that this could be two blackboards in your class. Say it's a class about visual representations, and you have two blackboards for this thing. The middle two blackboards for yet another example, the end two blackboards for yet another example. And you have all of them at once. Yes. AUDIENCE: Last time I taught a class where I used all slides, and when I put the amount of time that, things like you put into this, to make them all so they perfectly line up, and the animations were really helpful and everything, just creating the slides probably took half the overall time that I put into the class. But I think the one advantage was I was able to use printouts of those handouts as class notes. Which I handed out after the class. They didn't have them during class. So that's the only advantage, is you can reuse all of these [INAUDIBLE]. PROFESSOR: And that's exactly true. So all these nice figures, we did it for this class, and for a seminar on making slides. And then I used them verbatim in my Street-Fighting Mathematics textbook. So basically, to make good slides requires making something pretty much publication quality. And then, you do have something publication quality, but maybe the first time you teach it, that's not where you want to be spending your time. There was one other point that was raised in an email beforehand, which is, in Jean-luc's paper, he mentioned that he uses less popular alternatives to the Microsoft software for making slides. So, yeah, I highly do not recommend using Microsoft software, because it's not free software. And it's, I think, unethical to use non-free software. Free software meaning freely licensed, so you change the source code, view the source code, make your own versions, do whatever you want. And it's none of that. Second, it's all proprietary formats. So you're encouraging other people to use proprietary formats. No one has fully debugged, and decoded, all the formats that Microsoft uses. So you should use open formats at least. Now, you could use Open Office, but the problem is Open Office, the whole model of the way it does slides is kind of yuck, because it's just copied from the Microsoft way. So actually I prefer-- maybe it's because of my computer science math background-- what I prefer is to program all the slides. So let's see. [? Log.tech. ?] Where are the-- yeah. OK. So I actually write in [? Tech, ?] or version of [? Tech. ?] And here is your title of the slide. And I can change the formatting later. AUDIENCE: Maybe that's what took you so long. PROFESSOR: Maybe. Maybe. But the thing is, you can't make publication-quality stuff otherwise. So if you're going to actually do a good slide, basically you want to do that. Otherwise, I don't think people should bother. And mathematical typesetting is just ghastly unless you use [? Tech ?]. AUDIENCE: [INAUDIBLE]. PROFESSOR: Oh. And I don't think you can make it look as nice. And I'm speaking here about Open Office, and it's probably a clone of rubbish that it replaces. The mathematical typesetting is just not as good. And so for technical presentations, it's just not as good. So now, making the figures, I have the same philosophy. So the figures I actually program as well. Where is that fig [INAUDIBLE]. So here are the figures. So this is a function that draws logarithm graphs. And then I tell it which parts of the graph to put in. So that's how I do the-- you draw all the lines, you drew all the pieces. And then here I call it with different arguments, and I get different figures. So I'll put all the source code up for you guys, so you can see an alternative way of doing it. And some links, so you can try it yourself. And some templates. But it's not for everyone. Though it is a public format. It produces PDF, which is a public format. So I actually highly recommend it for those who have at all some programming affinity. AUDIENCE: [INAUDIBLE] programs that ends up as PDF format. PROFESSOR: That has the advantage of PDF, which is good. So I do recommend presenting things as PDF because you can view it on anything. It's just, you have the problem of, do you have good mathematical typesetting. And that's hard to find, except [? in Tech, ?] basically. That's the only thing that does it right. OK, so if you could just take one minute, fill out the sheets. I'll answer the questions next time. I'll have office hours as usual. We'll meet outside that door. And then we'll go to the cafe area, and have a chat at the tables, as long as people have questions. OK, so, the moral the story is use the blackboard if you can. Use the big visual field. But if you're going to use slides, Assertion-Evidence. Sentence headline. Visual evidence. One message per slide. Answers from lecture nine, to questions generated in lecture eight. Audio quality for this video is poor due to technical difficulties. PROFESSOR: [INAUDIBLE] is I'd like to go through the questions from before. There's one question from the time before, which is, could I get the reference for the works that said, there's a difference between telling people that they have intrinsic abilities-- saying, oh, you did that because you are really smart, versus, oh you did that because you tried really hard. And how those two ways of talking to people-- and children especially-- produced different interests, and confidence in trying new problems. I mentioned that that work was done a lot by a Professor at Stanford, in the Psychology Department. Could I get a reference for it? The next update of the website will have the reference, with the reference is Carol Dweck, D-W-E-C-K. And she's written a ton of books, and papers. But the one that [INAUDIBLE] popular, [INAUDIBLE] is called Mindset. Which is a really, really fascinating read. So that will be on the website. OK. First question was, you mentioned many times, problems that are PDF-- pass, D, or F, where P is made a decent effort, D is made an indecent effort, and F is didn't make an effort. That's great, but then how do you balance that with the fact that exams become high-stakes, hit-or-miss performances that completely determine grades. I don't use the exams that way. I give exams, they're balanced with the problem sets. The problem sets are PDF, and if you do P on all the problem sets, that contributes a large part of your grade, towards getting an A. So there's no necessity to make the exams all or nothing. Problem sets still account for quite a bit. [INAUDIBLE] It's just you're not forcing the students to get every last T, and dot every last I, which often means going through [INAUDIBLE] to try to do all that. Instead, you're saying, look, I want you to try to learn something big, important. If you take responsibility, then, you get a reward for that, by saying, OK, that'll help your grade, too. About slide talks. Were there any questions about slides talks versus teaching, versus blackboards. So one of the papers says that the best usage of slides is as a completed form that's understandable without the talk that comes with it. Is that for a research talk? Yes, for a research seminar, that's what you want. You want two channels. You want the auditory channel, and the slide channel, or the visual channel-- whatever's on the blackboard, however you do it. You want them to stand independently. So if people are deaf, and they just see the slides, they get something out of the talk. And if they're blind, and they just hear your speech, they get something out of the talk. That's, I would say, less true for teaching. The slides tend not to be so, message, message, message. You may develop a point, build up a bit of suspense, and then give a final message. So there's more interaction that happens from the lecture, generally, when you're teaching, than in a research seminar. So those are somewhat different. But generally, you do want your slides, or your blackboard work, to somehow stand independently. Although I've just violated that principle right there. This is meaningless unless you've heard what I said. But as I said, it's not an absolute rule. It's just that there's tendencies that are different between [INAUDIBLE] talks, and teaching. What do you do if you're assigned to a classroom with a bad blackboard situation. In other words, not like this. This is fantastic. It would be even more ideal if there was a slide [INAUDIBLE]. But what do you do if you go to a classroom with, for example, just one whiteboard, or just one or two blackboards? Then you have to work harder. Then you have to plan even more carefully, how are you going to use the blackboard. And be more telegraphic. Put coarser details on the boards, not every single detail. Use the notes more. So you optimize the constraints that you have. Should you only use the front half of the room. Suppose you're in a room which has blackboards all around. Should you only use the front half of the room? Yeah, I generally try to use only the front half of the room, because people can look at that, and glance out of the side of their eye at the side, and then, the front as well. The back forces people to turn around, so they're much less likely to use it. So it's sort of like putting something on the ceiling. [INAUDIBLE] never see it. So generally, yeah, I use the front of the room, and [INAUDIBLE] the side. Should all slides have a "so what" element, or is that only for research talks? Definitely for a research talk, all the slides should have a "so what," so people know what the point is. And ideally, for teaching, you should have two. Now, last time, I didn't mention one of the fundamental benefits of putting a "so what," an assertion, in every slide. So that was the idea. So the slide title is the "so what.". And then here is the evidence, maybe a picture, or a graph. Some kind of, ideally visual, evidence. So what's the advantage of doing that? It's not just for the audience-- it's for you, as well. What you do, after you make your series of slides, with your sentences, your "so what." You extract just the sentences, which is easy to do if you're using a program [? like Tech. ?] You just [INAUDIBLE], just the titles. However you do it. You make a list of sentences. And then you see whether the sentences flow. With a sentence, you can tell, very well, because whether your presentation, your teaching, your seminar is well organized, if it herks and jerks, if sentence one and sentence two seem to have nothing to do with each other, there's no transition between them, well, maybe you need a sentence one and a half, a slide that goes in between. Or maybe sentence two belongs down here, next to something that relates better. So you can see not just at the level of the individual slides, but at the level of the entire presentation, whether there's a thread that goes through. And if you remember, one of the principles of good teaching is having a good story. So this helps you make a good story. When you're teaching a class that has expectations for teaching lots of material, how do you reconcile your desire to teach the important things really well, and keep a reasonable pace? Some things aren't really reconcilable. Classes that, for example, say, OK, we're going to go through the 1,600-page textbook, there's really not much you can do to help that. And students are just hosed. But try to avoid that, mitigate it. Sometimes it seems like, just out of habit, everyone does that 1,600-page textbook. But you could find out, what are the actual requirements? What do people need for [INAUDIBLE]? If may be a lot less. It may be that a bunch of stuff was just put in there historically, and could be taken out. So you might have some freedom to reduce that. The other thing is you put more stuff in the notes, and you give the higher-level ideas in the lecture. Can I put my slides on the course web page? Yes, I'll do that, with the reference, on the next update. What about having slides that scroll rather than flip? This was a suggestion to avoid the problem I mentioned of signal overload, where you fill up the short-term memory with a slide, and then you've forgotten what's on the old slide. And so, [INAUDIBLE] too small to remember the whole slide. So what about having slides that scroll, instead of flip. And this may be a critique of the PDF format, which organizes things in [? pages. ?] I think it's actually not a critique of the PDF format, it's just a critique of the PDF viewers. So if you have a PDF viewer that can scroll, which many can, you could actually use that. The problem is that screens are not big enough. That's really the fundamental issue. So if this slide is taking up the entire screen space, when you scroll it, stuff goes off these screens, and gets replaced by new stuff. So that problem is really hard to work around. What you really need is multiple screens, and multiple projectors. Many questions asked, what do you do if you're in a room with a bad blackboard situation, where you just have one or two blackboards. Well, almost every room has a bad slide situation, because you only have one slide projector. So you only use a small part of the room. So that's the really fundamental problem. And if you could get around that, it wouldn't really matter so much whether the slides scrolled or flipped. You'd put one over there, one over there, maybe one over there. Of course now [INAUDIBLE]. My guess is that the t prepare is probably proportional to n cubed, or something like that. Depending on the number of slide projectors you have to deal with. Because you have to do all the coupling of [INAUDIBLE] n squared [INAUDIBLE]. You'd have to do all the coupling between all the slides, and it will probably take you twelve hours for every lecture that you give, of one hour. Which is kind of [INAUDIBLE]. Do I think the video camera for OCW impacts the style of my lectures, blackboards, pacing, et cetera? If anything, it probably improves it. Because I'm a little more conscious that I write clearly, and maybe it improves my dress code a bit, because I make sure to wear a tie. But other than that, it doesn't really affect, too much, what I do. And I usually forget that it's even there. It's much easier to write equations [? in Tech, ?] and then copy and paste them into a presentation, than doing the entire presentation [? in Tech. ?] Actually, I disagree. This is a case of start-up costs, versus running costs. So there's many examples of that, here at issue. It's actually a general principle worth knowing, because you can analyze lots of stuff. Another example of this principle is, suppose you see a book that you really like, that's available in print, from publisher n, and there's also a PDF file online. Well, you could just print it out and replace your printer. Or, you could buy the thing from the publisher. Now, unless the publisher's a total thief -- unfortunately, many of them are, but many of them aren't-- it's usually much better to buy it at the bookstore, because it's much cheaper, than to print it on your laser printer, get all the paper trimmed to the right size, and have it all bound. So the running cost for you to do one book-- you have no start-up cost, you just do it. There's no fixed amount you have to pay, no matter how many books you do. But there's a running cost, and your running cost is pretty high. Whereas the publisher, they have a big start-up cost. What they do, is they make all these plates, they take them to a PDF file, [INAUDIBLE], they make all these metal plates, and they ink the metal plates, stamp it onto the paper. So they do all that. That's a big start-up cost. But their running cost is much lower. They're much more efficient printing, per book. But it requires all that start-up cost. So there's a contrast between book publishers and you. So it's the same thing, actually, with doing [? Tech, ?] versus doing, say, regular word processing. With [? Tech, ?] there's a lot of start-up cost. No doubt about that. It's not what most people are used to, you end up programming your documents. And so, if you're only making one document, if you only write one letter in your life, no question it's faster to write the letter. Your have lower running costs. You don't have to worry about the start-up costs if you write it using a word processing program-- Open Office, for example. You just type, and you make it look like you want, and just spit it out. But, if you have to write 50 letters that all have a roughly similar format, it's much better to get the template. You pay the start-up cost once, learn [? Tech, ?] learn some kind of program template. And then, you just enter the blanks into the template. The text, dear who, the address, and all that. [INAUDIBLE] formatting, that's all dealt with once. So you pay start-up costs, in return for lower running costs. And that's basically true of lectures, too. Once you have all your templates ready, it's much faster to do everything [? in Tech, ?] or in some kind of programming, text-processing system, rather than in a word processor. The problem is that, because of the start-up costs, [INAUDIBLE] lecture the day before, or two days before. Because of the start-up costs, you always say, well, I could learn [? Tech, ?], and do this properly, and it would save me a lot of time later, but I don't have the time right now. And then you end up paying the higher running cost to do it some other way. But each time, before each lecture, it happens again, over and over and over again And you would have been better off [INAUDIBLE] actually paying the start-up cost. So, to that end, I will put my templates up, so other people can use them. So you don't have to pay so much start-up cost, and you can just take them, and use them for your own [INAUDIBLE]. So I think that was the main questions for this time. Any questions that have been created since then?
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https://ocw.mit.edu/courses/8-701-introduction-to-nuclear-and-particle-physics-fall-2020/8.701-fall-2020.zip
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MARKUS KLUTE: Welcome to 8.701. So in this lecture, we'll give you the first introduction to Feynman diagram. This is part 1 out of a few sections on Feynman diagrams. So this is really meant to introduce the topic such that we can use the same language to talk about Feynman diagrams before we then later on are able to use them as a tool to calculate interesting processes. This brings me right to the essence already. What is a Feynman diagrams and what can it be used for? They arise from pertubative calculations of amplitude for reactions. And that's exactly how we're going to use them later on. It turns out that the mathematical terms in the perturbation series can be represented as a diagram. And then you can turn this around and use the diagram in order to perform a calculation. So each of the diagrams then indicates a particular factor in the calculation. Again, and then you have a rule which allows you to, after drawing, you can then put the pieces together in order to perform [INAUDIBLE] calculation. The derivation of those tools or rules is beyond the content of this course. But I will teach you how to actually use diagrams in order to calculate things. So here's one example of a diagram. Let me just put this down here so you can see this. So this is an electron radiating a photon. You see components like those lines here. Those represent particles with energy and momentum also what to consider the spin. And they meet at a point. This point here is called a vertex. And this is where the interaction takes place. And in this example. The vertex is labeled with a q or e, representing the charge, the electric charge, which gives us the strength of the coupling. We already discussed when we talked about units that we can express the strength of the electromagnetic-- the coupling in QED with the electric charge. And that's shown below again. The amplitude then turns out to be proportional to the charge or to this coupling. And the diagrams with n vertices for n of those components here get a factor e, the charge, to the nth power in the amplitude, and e to the second. Because if we're going to calculate a probability, you have to square the amplitude. You get a factor of e to 2n. Again, don't get confused-- e is charge. So for n vertices, there will be a factor alpha to the nth power for the probability. And so since alpha is 1/137, you see that if I want to do a calculation, and diagrams which have n vertices will be suppressed, will not contribute much to our perturbation serious because alpha is much, much smaller than 1. So this is already an interesting finding. Can restrict yourself to calculating diagrams which have a couple of vertices or n vertices, but you don't have to calculate the entire series. You want to measure your calculation with experimental findings. Interesting here-- antiparticles. If you have a specific vertex and you calculated it, it can be reused. It can be reused for example by replacing a particle with an antiparticle or by re-labeling. One thing I haven't explained to you yet, you have to define when you write them which is the direction of time. We'll come to this direction. And so in this case here, you have a particle and an antiparticle unrelating to a photon. So far, so good so. This is again a good point to stop and just try to read the diagrams. Note that what happens in this discussion when you actually change the direction of time-- forward down. You want either directions. So now if you want to calculate the reaction, it's not sufficient to just use one word vertex. Why? Because a single vertex will not be able to give us a reaction. You can simply see this when you look at something like an electron plus electron photon. This is not really possible because of energy and momentum conservation in this diagram. So you need a couple of vertices in order to make a reaction. So this here is, again, we have potentially the time going this direction. There's a scattering between an electron and a muon through the exchange of a photon. Both particles have electric charge of e. And then you can just calculate what is the probability for a process like this to occur. We'll see how to do this technically later on. But hopefully you have a first impression. Again, let's label this now very quickly. So you have an incoming particle, a second incoming particle, outgoing particles, and an exchange particle. So this exchange particle is a photon. And there's two vertices in this diagram.
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https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/8.04-spring-2016.zip
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PROFESSOR: How about the expectation value of the Hamiltonian in a stationary state? You would imagine, somehow it has to do with energy ion states and energy. So let's see what happens. The expectation value of the Hamiltonian on this stationary state. That would be integral dx stationary state Hamiltonian stationary state. And we're going to see this statement that we made a few minutes ago become clear. Well what do we get here? dx psi star of x e to the i Et over h bar H e to the minus i Et over h bar psi of x. And H hat couldn't care less about the time dependence, that exponential is irrelevant to H hat. That exponential of time can be moved across and cancelled with this one. And therefore you get that this is equal to dx psi star of x H hat psi of x, which is a nice thing to notice. The expectation value of H on the full stationary state is equal to the expectation value of H on the spatial part of the stationary state. That's neat. I think it should be noted. So it's equal to the H of little psi of x. But this one, we can evaluate, because if we are in a stationary state, H hat psi of x is E times psi of x. So we get an E integral the x psi star of psi, which we already show that integral is equal to one, so we get the energy. So two interesting things. The expectation value of this quantity of H in the stationary state is the same as it's quotation value of H in the spatial part, and it's manually equal to the energy. By the way, you know, these states are energy eigenstates, these psi of x's, so you would expect zero uncertainty because they are energy eigenstates. So the zero uncertainty of the energy operator in an energy eigenstate. There's zero uncertainty even in the whole stationary state. If you have an H squared here, it would give you an E squared, and the expectation value of H is equal to E, so the expectation value of H squared minus the expectation value of H squared would be zero. Each one would be equal to E squared. Nothing would happen, no uncertainties whatsoever. So let me say once more, in general, being so important here is the comment that the expectation value of any time independent operator, so comments 1, the expectation value of any time-independent operator Q in a stationary state is time-independent. So how does that go? It's the same thing. Q hat on the psi of x and t is general, now it's integral dx capital Psi of x and t Q hat psi of x and t equals integral dx-- you have to start breaking the things now. Little psi star of x E to the i et over H bar. And I'll put the whole thing here. Q hat Psi of x E to the minus i et over H. So it's the same thing. Q doesn't care about time So this factor just moves across and cancels this factor. The time dependence completely disappears. And in this case, we just get-- this is equal to integral dx psi star Q psi, which is the expectation value of Q on little psi of x, which is clearly time-independent, because the state has no time anymore and the operator has no time. So everybody loves their time and we're in good shape. The second problem is kind of a peculiarity, but it's important to emphasize superposition. It's always true, but the superposition of two stationary states is or is not a stationary state? STUDENT: No. PROFESSOR: No, good. It's not a stationary state in general because it's not factorizing. You have two stationary states with different energies, each one has its own exponential, and therefore, the whole state is not factorized between space and time. One time-dependence has one space-dependence plus another time-dependence and another space-dependence, you cannot factor it. So it's not just a plain fact. So the superposition of two stationary states of different energy is not stationary. And it's more than just saying, OK, it's not stationary. What it means is that if you take the expectation value of a time-independent operator, it may have time-dependence, because you are not anymore guaranteed by the stationary state that the expectation value has no time-dependence. That's how, eventually, these things have time-dependence, because these things are not [INAUDIBLE] on stationary states. On stationary states, these things would have no time-dependence. And that's important, because it would be very boring, quantum mechanics, if expectation values of operators were always time-independent. So what's happening? Whatever you measure never changes, nothing moves, nothing changes. And the way it's solved is because you do have those stationary states that will give you lots of solutions. And then we combine them. And as we combine them, we can get time-dependence and we can get the most [INAUDIBLE] equation.
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https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/8.04-spring-2013.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Any questions from last lecture from before spring break? No questions, nothing? Nothing at all? We you all still on vacation? Thank you. Yeah. AUDIENCE: Could you talk about coherence states? PROFESSOR: Ah. OK, well, that's a great question. So coherence states-- actually, how many people have looked at the optional problems? Nice, OK, good. Good, so coherence states were a topic that we touched on the problem sets and on the optional problems, and the optional problems are mostly on the harmonic oscillator and nice problems revealing some of the structure of the harmonic oscillator, and it generalizes quite boldly. But here's the basic idea of coherence states. Let me just talk you through the basic ideas rather than do any calculations. So what's the ground state of a harmonic oscillator? What does its wave function look like? AUDIENCE: Gaussian. PROFESSOR: It's a Gaussian, exactly. It's minimum uncertainty wave packet. How does it evolve in time? AUDIENCE: Phase. PROFESSOR: Yeah, phase. It's an energy eigenstate, it's the ground state, so it just evolves in time with the phase. So if we look at the wave function for the ground state, phi naught, it's something like e to the minus x squared over 2a squared with some normalization coefficient, which I'm not going to worry about. So this is a minimal uncertainty wave packet. Its position distribution is time independent, because it's a stationary statement. It's momentum distribution, which is also a Gaussian, the Fourier transform, is time independent. And so this thing up to its rotation by the overall phase just sits there and remains a Gaussian. Now, here's a question. Suppose I take my harmonic oscillator potential, and I take my Gaussian, but I displace it a little bit. It's the same ground state, it's the same state but I've just displaced it over a little. What do you expect to happen? How will this state evolve in time? So, we know how to solve that problem. We take this state, it's a known wave function at time 0. We expand it in the basis of energy eigenstates, each energy eigenstate evolves in time with a phase, so we put in that phase and redo the sum, and recover the time evolution of the full state. And we've done this a number of time on the problem sets. Yeah. AUDIENCE: Would the Gaussian, the displaced Gaussian evolve the same way, keep its width, if it had any other initial width other than the one of the ground state? PROFESSOR: Let me come back to that, because it's a little more of a precise question. So we know how to solve this problem practically, algorithmically. But here's a nice fact. I'm not going to derive any equations, that's part of the point of the optional problems, but here's a nice fact about this state. So it's clearly a minimum uncertainty wave packet because at time 0, because it's just the same Gaussian just translated over a little bit. So what we'd expect, naively, from solving is we expand this in Fourier modes and then we let this the system evolve in time. We let each individual Fourier mode-- or, sorry, Fourier mode-- we let each individual energy eigenstate evolve in time, pick up a phase, and now what we get is a superposition. Instead of sum over n at time 0, we'd have sum over n cn phi n of x. Now, as a function of time, we get e minus I omega nt, and these phases are going to change the interference from summing up all these energy eigenstates, and so the system will change in time. Because the way that the various terms in the superposition interfere will change in time. So very naively, what you might expect, if you just took a random function-- for example, if I took a harmonic oscillator potential and I took some stupid function that did something like this. What would you expect it to do over time? Well, due to all the complicated interference effects, you'd expect this to turn into, basically, some schmutz. Just some crazy interference pattern. The thing that's nice about a coherent state, and this is where it gets its name, is the way that all of these interference effects conspire together to evolve the state is to leave it a Gaussian that does nothing but translates in time. A coherent state is coherent because it remains coherently a Gaussian as it moves along. It oscillates back and forth. And in fact, the peak, the center of this Gaussian wave packet, oscillates with precisely the frequency of the trap. It behaves just as a classical particle would have had you displaced it from the center of the harmonic oscillator trap, it oscillates back and forth. Now, on the other hand, you know that its momentum is changing in time, right. And at any given point, as you've shown on the problem set that's due tomorrow-- or you will have shown-- at any given moment in time, the momentum can be understood as the overall phase. You could just change that momentum by the overall phase, so it's the spatial rate of change of the phase, e to the ikx. So you know that the way the phase depends on position is changing over time. So it can't be quite so simple that the wave function, rather than the probability distribution, is remaining a Gaussian over time. It's not, it's got all sorts of complicated phases. But the upshot is that the probability distribution oscillates back and forth perfectly coherently. So that seems a little magical, and there's a nice way to understand how that magic arises, and it's to understand the following. If I take a state, phi 0, but I translate it x minus x0. OK. So we know that this state, phi 0 of x, was the ground state, what does it mean to be the ground state of the harmonic oscillator? How do I check if I'm the ground state of the harmonic oscillator? Look, I walk up to you, I'm like, hey, I'm the ground state of the harmonic oscillator. You're suspicious. What do you do? AUDIENCE: Annihilate. PROFESSOR: Annihilate. Exactly. You act with the annihilation operator. Curse you. So you act with the annihilation operator and you get 0, right. What happens if I act with the annihilation operator on this? Is this the ground state? No, it's been displaced by x0. And meanwhile, I told you that it oscillates back and forth. So what happens if you act with the annihilation operator? Should you get 0? No, what are you going to get? AUDIENCE: Something weird. PROFESSOR: Some random schmutz, right? If you just take a and you act on some stupid state, you'll just get some other stupid state. Except for Gaussians that have been displaced, you get a constant times the same wave function, phi 0 of x minus x0. Aha. It turns out that these displaced Gaussians are eigenstates of the annihilation operator. What does that mean? Well, it means they're coherent states. And so the optional problems are a working through of the study of the eigenfunctions of the annihilation operator, the coherent states. Is that cool? It's a state, it's a superposition. You can think about it like this. It's a superposition of the energy eigenstate. Any state is a superposition of energy eigenstates, and a coherent state is just some particularly special superposition of energy eigenstates. So you can think about it literally as c0 phi 0 plus c1 phi 1 plus c2 phi 2, blah blah blah. What does the annihilation operator do, what does the lowering operator do? It takes any state and then lowers it with some coefficient. So a coherent state, an eigenstate of the annihilation operator, must be a state such that if you take c 0 phi 0 plus c1 phi 1 plus c2 phi 2 and you hit it with the annihilation operator, which will give you see c1 phi 0 plus c2 phi 1 divided by root 2 plus dot, dot, dot-- or times root 2 plus dot, dot, dot. It gives you the same state back. So that gave you a relation between the coefficients c0 and c1, c1 and c2, c2 and c3. They've all got to be suitable multiples of each other. And what you can show-- and this is something you work through, not quite in this order, on the optional problems-- what you can show is that doing so it gives you a translated Gaussian. OK. So physically, that's what a coherent state is. Another way to say what a coherent state is, it's as close to a classical object as you're going to get by building a quantum mechanical wave function. It's something that behaves just like a classical particle in that potential would have behaved, in the harmonic oscillator potential. Did that answer your question? OK. Anything else? So, Matt works with these for a living. Matt, do you want to add anything? All right. I like these. They show up all over the place. This is like the spherical cow of wave functions because it's awesome. It's an eigenfunction, it's an annihilation operator, it behaves like a classical particle. AUDIENCE: Is there an easy way to define coherent states in potentials where you don't have nice operators like a? PROFESSOR: No, I mean, that's usually what we mean by a coherent state. So the term coherent state is often used interchangeably to mean many different things, which in the case of the harmonic oscillator, are identical. One is a state which is a Gaussian. OK. So in the harmonic oscillator system, that's a particularly nice state because it oscillates particularly nicely and it maintains its probability distribution as a function of time roughly by translating, possibly, some phases. But people often use the phrase coherent state even when you're not harmonic. And it's useful to keep in mind that we're often not harmonic, nothing's truly harmonic. You had another question a moment ago though, and your question was about the width. Yeah. AUDIENCE: It just translates without changing its width, with or without changing its shape at all. If the Gaussian were of a different width, would that still happen? PROFESSOR: Yeah, it does, although the details of how it does so are slightly different. That's called a squeeze state. The basic idea of a squeeze state is this. Suppose I have a harmonic oscillator-- and this is actually one way people build squeeze states in labs-- so suppose I have a harmonic oscillator and I put the system in a ground state. So there's its ground state. And its width is correlated with the frequency of the potential. Now, suppose I take this potential at some moment in time and I control the potential, the potential is created by some laser field, for example. And some annoying grad student walks over to the control panel and doubles the power of the laser. All of a sudden, I've squeezed the potential. But my system is already in a state which is a Gaussian, it's just the wrong Gaussian. So what does this guy do? Well, this is funny. The true ground state, once we've squeezed, the true ground state would be something that's much narrower in position space. I didn't draw that very well, but it would be much narrower in position space, and thus its distribution would be much broader in momentum space. So the state that we put the system, or we've left the system in, has too much position uncertainty and too little momentum uncertainty. It's been squeezed compared to the normal state in the delta x delta p plane. It's uncertainty relation is still extreme, it still saturates the uncertainty bound because it's a Gaussian, but it doesn't have the specific delta x and delta p associated with the true ground state of the squeezed potential. OK, so now you can ask what does this guy do in time. And that's one of the optional problems, too, and it has many of the nice properties of coherent states, it's periodic, it evolves much like a classical particle, but it's uncertainties are different and they change shape. It's an interesting story, that's the squeezed state. Yeah. AUDIENCE: Do other potentials have coherent states that are Gaussian? PROFESSOR: It sort of depends. So other potentials have states that behave classically. Yeah. There are generally-- so systems that aren't the harmonic oscillator do have very special states that behave like a classical particle. But they're not as simple as annihilations by the annihilation operator. Come to my office and I'll tell you about analogous toys for something called supersymmetric quantum models, where there's a nice story there. OK. I'm going to cut off coherent states for the moment and move on to where we are now. OK, so last time we talked about scattering of a particle, a quantum particle of mass m against a barrier. And we made a classical prediction, which I didn't quite phrase this way. But we made a classical prediction that if you took this particle of mass m and threw it against a barrier of height v0, that the probability that it will transmit across this barrier to infinity is basically 0. So this is the classical prediction. It will not transmit to infinity until the energy goes above the potential. And when the energy is above the potential, it will slow down, but it will transmit 100% of the time. So this is our classical prediction. And so we sought to solve this problem, we did. It was pretty straightforward. And the energy eigenstate took the form-- well, we know how to solve it out here, we know how to solve it out here, because these are just constant potentials so it's just plain waves. Let's take the case of the energy is less than the potential. So this guy. Then out here, it's in a classically disallowed state, so it's got to be a exponential. If we want it to be normalized, well, we need it to be a decaying exponential. Out here it's oscillatory, because it's a classically allowed region. And so the general form of the wave function of the energy eigenstate is a superposition of a wave moving this way with positive momentum, a wave with negative momentum, a contribution with minus k, and out here it had to be the decaying exponential. And then by matching, requiring that the wave function was oscillatory and then exponentially decaying out here, and requiring that it was continuous and differentiably continuous-- that its derivative was continuous-- we found matching conditions between the various coefficients, and this was the solution in general. OK. Now, in particular, this term, which corresponds to the component of the wave function moving towards the barrier on the left hand side, has amplitude 1. This term, which corresponds to a wave on the left with negative momentum, so moving to the left, has amplitude k minus i alpha over k plus i alpha, where h bar squared k squared upon 2m is equal to-- that's just the energy, e-- and h bar squared alpha squared over 2m-- that's on this side-- is equal to v0 minus e. But this, notably, is a pure phase. And we understood that, so we'll call this parameter r, because this is the reflected wave and this is an amplitude rather than a probability, so we'll call it little r. And we notice that the probability that we get out-- good. So if we want to ask the question, what is the transmission probability, the probability that I get from far out here to far out here, what is that probability? So if I consider a state that starts out as a localized wave packet way out here and I send it in but with energy below the barrier, what's the probability that I'll get arbitrarily far out here, that I will subsequently find the particle very far out here? 0, right, exactly. And you can see that because here's the probability amplitude, the norm squared is the probability density as a function of the position. And it goes like a constant times e to the minus alpha x. For large x, this exponential kills us. And we made that more precise by talking about the current. We said look, the current that gets transmitted is equal to-- that's a funny way to write things. H bar-- sorry. The current that gets transmitted, which equal to h bar upon 2mi psi complex conjugate dx psi minus psi dx psi complex conjugate on the right. So we'll put right, right, right, right. This is just equal to 0, and the easiest way to see that is something you already showed on a problem set. The current vanishes when the wave function can be made real up to a phase. And since this is some number, but in particular it's an overall constant complex number, and this is a real wave function, we know that this is 0. So the transmitted current out here is 0. The flux of particles moving out to the right is 0. So nothing gets out to the right. So that was for the energy less than the potential, and we've now re-derived this result, the classical prediction. So the classical prediction worked pretty well. Now, importantly, in general, we also wanted to define the transmission probability a little more carefully, and I'm going to define the transmission probability as the current of transmitted particles on the right divided by the current of probability that was incident. And similarly, we can define a probability that the particle reflects, which is the ratio of the reflected current, the current of the reflected beam, to the incident current. So this is going to be the transmission probability and the reflection probability. Yeah. AUDIENCE: Is there a square on t? PROFESSOR: Thank you. OK. So let's do a slightly different example. In this example, I want to study the same system, the wall, but I want to consider a ball incident from the left-- a ball. An object, a quantum particle incident from the left, with energy greater than v. Greater than v0. So again, classically, what do you expect in this case. You send in the particle, it loses a little bit of energy going up the potential barrier, but it's still got a positive kinetic energy, and so it just keeps rolling. Just like my car making it up the driveway, just barely makes it. OK. So here again, what's the form of the wave function. Phi e can be put in the form on the left. It's going to be e to the ikx plus b to the minus ikx, and that's over here on the left. And on the right it's going in the form ce to the ik-- in fact, let me call this k1x and k2x-- or sorry, k1x, which is on the left. H bar squared k1 squared upon 2m is equal to e. And on the right, we're going to have k2x because it's, again, a classically allowed region, it's going to be oscillatory but with a different momentum. e to the ik2x plus de to the minus ik2x where h bar squared k2 squared upon 2m is equal to e minus v0, which is a positive number, so that's good. So here's our wave function. Now, before we do anything else, I want to just interpret this quickly. So again, this is like a wave that has positive momentum, and I'm going to say that it's like a contribution, a term in the superposition, that is moving to the right, and the way to think of it as moving to the right, well, first off, it's got positive momentum. But more to the point, this is an energy eigenstate. So how does this state evolve in time? It's an energy eigenstate with energy, how does it evolve in time. AUDIENCE: [INAUDIBLE] PROFESSOR: It rotates by an overall phase, exactly. So we just get this times an e to the minus i omega t, where h bar omega is equal to e, the energy. And that's true on the left or right, because the energy is just a constant. So this term goes as e to the ikx minus i omega t, or kx minus omega t. And this is a point of constant phase in this move to the right. As t increases-- in order for the phase to be constant or for the phase, for example, to be 0-- as t increases, x must increase. By assumption here, k is a positive number. On the other hand, this guy-- that was a k1-- the second term is of the form e to the minus ik1x, and when I add the omega t, minus i omega t, this becomes kx plus omega t. So point of constant phase, for example, phase equals 0, in order for this to stay 0 then, as t increases, x must become negative. x must decrease. So that's why we say this corresponds to a component of the wave function, a term in the superposition which is moving to the left. And this is one which is moving to the right. Sorry, my right, your left. Sorry, this is moving to your right and this is moving to your left. OK? It's that little z2, it's harder than it seems. OK. In this set up, we could imagine two different kinds of scattering experiments. We can imagine a scattering experiment where we send in a particle from the left and ask what happens. So if you send in a particle from the left, what are the logical possibilities? AUDIENCE: It can go through. PROFESSOR: It could go through, or? AUDIENCE: Reflect back. PROFESSOR: Reflect back. Are you ever going to get a particle spontaneously coming from infinity out here? Not so much. So if you're sending a particle in from the left, what can you say about these coefficients? d equals 0 and a is not 0, right. Because d corresponds to a particle on the right hand side coming in this way. So that's a different scattering experiment. So in particular, coming in from the left means d equals 0 and a not equal to 0. Coming in from the right means a equals 0 and d not equal to 0 by exactly the same token. Cool? So I want to emphasize this. a and d, when you think of this as a scattering process, a and d are in guys. In, in. And c and b are out. This term corresponds to moving away from the barrier, this term corresponds to moving away from the barrier, just on the left or on the right. OK? Out and out. Everyone happy with that? Yeah. AUDIENCE: According to that graph, if d is-- no, that one. Yeah. If d is bigger than [INAUDIBLE] then the transplants will be 1. Doesn't that mean that d is also 0? PROFESSOR: So this is the classical prediction. I've written it, classical prediction. So let me rephrase. The question was basically, look, that classical prediction implies that b must be 0 on the other scattering process, and that doesn't sound right. So is that true? No, it's not true, and we'll see it again in a second. So very good intuition. That was good. OK, good. So in general, if we have a general wave function, general superposition of the two states with energy e with a,b,c, and d all non-0, that's fine. That just corresponds to sending some stuff in from the right and some stuff in-- sorry, some stuff in from the left and some stuff in from the right. Yeah? But it's, of course, going to be easier if we can just do a simpler experiment. If we send in stuff from the left or send in stuff from the right. And if we solve those problems independently, we can then just superpose the results to get the general solution. Yeah. So it will suffice to always either set d to 0 or a to 0, corresponding to sending things in from the left or sending things in from the right. And then we can just take a general superposition to get the general answer. Everyone happy with that? So let's do that in this set up. So here's our wave function. And you guys are now adept at solving the energy eigenvalue equation, it's just the matching conditions for a,b,c, and d. I'm not going to solve it for you, I'm just going to write down the results. So the results-- in fact, I'll use a new border. The results for this guy, now for e greater than v, are that-- doo doo doo, what just happened. Right, OK, good. So let's look at the case d equal 0 corresponding to coming in from the left. So in the case of in from the left, c is equal to 2k1 over k1 plus k2, and b is equal to k1 minus k2 over k1 plus k2. And this tells us, running through the definition of j, the currents, and t and r, gives us that, at the end of the day, the reflection coefficient is equal to k1 minus k2 upon k1 plus k2 [INAUDIBLE] squared, whereas the transmission amplitude or probability is equal to 4k1 k2 over k1 plus k2 squared. And a good exercise for yourself is to re-derive this, you're going to have to do that on the problem set as a warm up for a problem. So at this point we've got an answer, but it's not terribly satisfying because k1 and k2, what's the-- so let's put this in terms of a more easily interpretable form. k1 and k2 are nothing other than code for the energy and the energy minus the potential. Right, so we must be able to rewrite this purely in terms of the energy in the potential e and v. In fact, if we go through and divide both top and bottom by k1 squared for both r and t, this has a nice expression in terms of the ratio of the potential to the energy. And the nice expression is 1 minus the square root of 1 minus v0 over e over 1 plus the square root of 1 minus v0 over e norm squared. And suddenly, the transmission amplitude has the form 4 root 1 minus v0 over e over 1 plus the square root of 1 minus v0 over e squared. I don't remember which version of the notes I posted for this year's course. In the version from 2011 there is a typo that had here e over v, and in the version from 2012, that was corrected in the notes to being v0 over e. So I'll check, but let me just warn you about that. On one or two pages of the notes, at some point, these guys were inverted with respect to each other. But the reason to write this out is now we can plot the following. We can now plot the quantum version of this plot. We can plot the actual quantum transmission as a function of energy, and the classical prediction was that at energy is equal to v0, we should have a step function. But now you can see that this is not a step function, and that's related to the fact that b was not actually 0, as you pointed out. Hold on one sec. Sorry, was it quick? AUDIENCE: The notes are wrong. PROFESSOR: OK, the notes are wrong, good, thank you. That's great. OK, so the notes are wrong. It should be v0 over e, not e over v0 on the notes that are posted. We're in the process of teching them up, so eventually a beautiful set of nice notes will be available. Extra elbow grease. OK. So what do we actually see? Here we got for e less than v0, we got in d that the transmission probability was exactly 0. So that's a result for the quantum result. But when e is equal to v0, what do we get? Well, when e is equal to v0, this is 1. And we get square root of 1 minus 1, which is square root of 0, over square root of 1 plus square root of 1 minus 1, which is 0. That's 0 over-- OK, good. So that's not so bad. Except for this 1 over this e, this is a little bit worrying. If you actually plot this guy out, it does this, where it asymptotes to 1. And to see that it asymptotes to 1, to see that it asymptotes to 1, just take e0 gigantic. If you know it's gigantic, this becomes v0 over e, which goes to 0, and if e is much larger than the potential. So this becomes square root of 1. And in the denominator, this becomes square root of 1, 1 plus 1, that's 2 squared, 4, 4 divided by 4, that goes to 1. So for large e much larger than v0, this goes to 1. So we do recover the classical prediction if we look at energy scales very large compared to the potential height. Yeah? OK. So that's nice. Another thing that's nice to note is that if you take this reflection probability and this transmission probability, then they sum up to 1. This turns out to be a general fact and it's a necessary condition in order for those to be interpretable as reflection in transition probabilities. The probability that it transmits and the probability that it reflects had better add to 1, or something is eating your particles, which is probably not what you're looking for. So this turns out to be true in this case, just explicitly. But you can also, as you will in your problem set, prove that from that definition r and t, it's always true. A check on the sensibility of our definition. If it weren't true, it wouldn't tell you that quantum mechanics is wrong, it would tell you that we chose a stupid definition of the transmission and reflection probabilities. So in this case, we actually chose quite an enlightened one. OK, questions at this point? Yeah. AUDIENCE: Does this analysis even hold for classical particles? If we're talking about the difference between-- like if I threw a baseball and I happened to throw it at a potential that had height one millionth of a joule less than the energy of the baseball, would we observe this also? PROFESSOR: No. AUDIENCE: No? PROFESSOR: No, for the following reason. So here's my classical system. Classical system is literally some hill. So where I was growing up as a kid, there was a hill not far from our house. I'm not even going to go into it, but it was horrible. If you have energy just ever so slightly greater than the hill-- he said from some experience-- if you have energy that's ever so slightly greater-- OK. I'm going resist the temptation. So one of these days. If you have ever so slightly greater than the hill, and you start up here, what does that tell you? What does it mean to say you have energy just ever so slightly greater than the potential energy at that point? AUDIENCE: Up more slowly. PROFESSOR: Little tiny velocity. OK, and I'm going to say I have a little tiny velocity this way. OK. So what happens? We follow Newton's equations. They are totally unambiguous, and what they tell you is, with 100% certainty, this thing will roll and roll and roll, and then all hell will break loose. And then if you're very lucky, your car gets caught in the trees on the side of the cliff, which is later referred to by the policeman who helps you tow it out as nature's guard rail. I was young, it won't happen again. And So with 100% certainty, Newton's equations send you right off the cliff. And for style points, backward. So now I'm going to do the second thing. Newton's laws satisfy time reversal invariance. So the time reversal of this is, this thing has this much energy, and it shoots up the cliff. And ever so slowly, it just eventually goes up to the top, where everything is fine. OK, but does it ever reflect back downhill? No. And does it ever-- classically, when you roll the thing-- does it ever actually not go off the cliff and hit the trees, but instead reflect backwards. I wish the answer were yes. But sadly, the answer is no. Now, if that car had been quantum mechanically small, my insurance would have been much more manageable. But sadly, it wasn't. OK, so this is a pretty stark and vivid-- this is a pretty stark difference between the quantum mechanical prediction and the classical prediction. Everyone cool with that? Other questions? Yeah. AUDIENCE: I understand why it works out that r plus t equals 1, but I'm not sure I understand the motivation to use the square with the ratios. PROFESSOR: Excellent, OK. So the reason to do it, so great, excellent. How to say. Here's one way to think about it. The first thing we want is we want a-- so this is a very good question, let me repeat the question. The question is, look, why is it squared? Why isn't it linear? So let's think about that. Ah, ah, ah. The reason it's squared is because of a typo. So let's think about-- [LAUGHTER] PROFESSOR: Let's step back for second and let's think, OK, it's a very good question. So let's think about what it should be. OK. For the moment, put an arbitrary power there. OK. And let's think about what it should be. Thank you so much for this question. I owe you, like, a plate of cheese or something. That's high praise, guys. France, cheese. So should it be linear or should it be quadratic? What is that supposed to represent? What is T, the capital T, supposed to represent? AUDIENCE: The probability of transmission. PROFESSOR: The probability of transmission, exactly. The wave function. Is the wave function, the value of the wave function at some point, is that a probability? It's a-- AUDIENCE: Probability of density. No, it's the square root. PROFESSOR: Is it a probability of density? It is a probability amplitude. It is a thing whose norm squared is a probability. So we want something that's quadratic in the wave function, right. Meanwhile, some time before, we wanted a definition of how much stuff is moving past a point in the given moment in time. What's the probability density moving past a point at a given moment in time. And that's where we got the current j from the first place. j was the probability density moving past a point at a given moment in time. So notice that j is quadratic in the wave function, so it's how much stuff is moving past any particular direction-- we chose it to be the incident to the reflected bit-- at a moment in time. So it's a probability density-- it's actually a current-- and it's quadratic in the wave function. So it has the right units and the right structure to be a probability. If we squared that, we would be in trouble, because it wouldn't be a probability, it would be a probability squared. And in particular, it wouldn't normalize correctly. So that typo was actually a bad typo. So what happened, now thinking back to-- you have a video someday, so you can scroll back. What happened was I didn't put the squared on the first, I put the squared on the second, and then someone said squared on the first? Like, yes. But the correct answer is no squared on the second. So there are no squareds. Thank you for the question. Very good question. In particular, the thing that was so awesome about that question was it was motivated by physics. It was like, look, why is this thing squared the right thing? That's two factors of the wave function, it's already quadratic. It should be like a probability. Why are you squaring it? Very good question. Yeah. AUDIENCE: So in the transmission probability, why do we know it's 0 if the energy is less than v0? I mean, is there any rule? PROFESSOR: Yeah, OK, good. We'll come back to that question in a little bit, but let me just quickly say that we calculated it, and we see explicitly that it's 0. Now you might say well, look, the true wave function has a little bit of a tail. So there's some [INAUDIBLE] probability you'll find it here if you do a measurement. But the question we want to ask in transmission is, if you send it in from far away over here, how likely are you to catch it far away over here? And the answer is, you are not. That make sense? OK, good. OK, so so much for that one. Now, here's a fun fact, I'm not going to go through this in any detail. We could have done exactly the same calculation-- OK, we could have done the same calculation sending something in from the right. And sending something in from the right, in on the right, corresponds to d not equal to 0 and nothing coming in from the left, that's a equals 0. So we could have done the case a equals 0, which is in from the right. And if you do that calculus, what you find is c is equal to k1 minus k2 upon k1 plus k2 d. And b is equal to 2k1 upon k1 plus k2 d. And this says that plugging these guys in, the reflection and the transmission are the same. In particular, physically, what does that mean? That means the reflection and transmission are the same uphill as downhill. Downhill is uphill, they're the same both ways. But that's truly weird, right? What's the probability to transmit quantum mechanically if you have just a little bit of energy and you send the particle in? How likely are you to go off the cliff. If we have energy ever so slightly greater the potential, and the transmission amplitude is the same downhill as it was uphill, how likely are you to fall off the potential? AUDIENCE: Always. PROFESSOR: Never. Because if energy is just slightly greater than v0, then the transmission probability is very, very low. The transmission to go from here to here is extremely low if the energy is close to the height of the barrier. So had I only been on a quantum mechanical hill, I would have been just fine. This is a really striking result, but this thing, the fact that you're unlikely to scatter uphill, that's maybe not so shocking. But you're really unlikely to scatter downhill, that is surprising. So I'll leave it to you to check that, in fact, the transmission, when we do it in from the right, the transmission and reflection probabilities are the same. Recalling that transmission means going this way, reflection means bouncing back to the right. Cool? Yeah. AUDIENCE: [INAUDIBLE] PROFESSOR: Uh, because, again. Jeez, today is just a disaster. Because there's a typo. This should be times a. God. And the reason you know it should be times a, first off, is that these should have the appropriate dimensions. So there should be an appropriate power of a. And the second thing is that if you double this stuff in, you'd better double this stuff out. So if you double a, you'd better double c. So you know there had better be a factor of a here. And that's just a typo again. I'm sorry, today is a bad day at the chalkboard. Thank you for that question. Yeah. AUDIENCE: Is there a physical reason for why r and t are the same and so different from what you would classically expect, or is that just the way the math works? PROFESSOR: There is. I want you to ponder that, and either at the end of today or at the beginning of tomorrow, we'll talk about that more in detail when we've done a little more technology. It's a good question and you should have that tingling sensation in your belly that something is confusing and surprising and requires more explanation, and we'll get there. But I want you to just think about it in the background first. Yeah. AUDIENCE: Is there any physical experiment where r and t don't sum up to 1? PROFESSOR: Mm mm. Nope. So the question is, is there any experiment in which r and t don't sum up to 1? And there are two ways to answer that. The first way to answer that is, it turns out that r and t adding up to 1 follows from the definitions. So one of the things you'll do on the problem set is you'll show that using these definitions of the transmission and reflection of probabilities, they necessarily add up to 1, strictly. They have to. It follows from the Schrodinger equation. The second answer is, let's think about what it would mean if r and t didn't add up to 1. If r and t didn't add up to 1, then that would say, look, if you throw a particle at a barrier, at some feature in a potential, some wall, then the probability that it goes this way at the end versus the probability that it goes that way at the end if you wait long enough is not 1. So what could have happened? What could have gotten stuck in there? But that doesn't even conserve momentum, not even approximately. And we know that the expectation of momentum is time independent for a free particle. In between while it's actually in the potential, it's not actually time independent, there's a potential, there's a force. AUDIENCE: I mean, if it's whole particles, can't some of them get annihilated or something? PROFESSOR: Ah, well OK. So if you're talking about multiple particles and interactions amongst multiple particles, then it's a slightly more complicated question. The answer there is still yes, it has to be that the total probability in is the total probability out. But we're only going to talk about single particles here. But it's always true that for-- we're going to take this as an assumption, that things don't just disappear, that the number of particles or the total probability is conserved. There's a third way to say this, this really isn't independent from the first. Remember that these were defined in terms of the current? The current satisfies-- the current concept of probability conservation equation dt of rho is equal to the gradient minus the gradient of the current. So the probability is always conserved, the integral of probability density is always conserved, it's time independent. AUDIENCE: So it's by [INAUDIBLE]. PROFESSOR: Yeah, it's by construction. Exactly. Yeah? AUDIENCE: Looking at your expression for the transition probability, I'm having trouble seeing how that works out to 0 when e is less than v0? PROFESSOR: Oh, this expression is only defined when e is equal to v0. Because we derived this-- excellent question-- we derived this assuming that we had an energy greater than v0 and then that the wave function had this form. You cannot use this form of the wave function if the energy is less than v0. If the energy is less than v0, you've got use that form of the wave function. And in this form of the wave function, we derived that the transition amplitude is 0, because the current on the right, the transmitted current, is 0. So this calculation is appropriate when e is less than v0 and this calculation is appropriate when e is greater than v0. So you're absolutely correct. You can't use this one e is less than v0, it gives you not the same answer. In fact, it gives you a complex-- it's kind of confusing. The factors cancel. So it's not really a probability at all, and indeed, this is just not the right quantity to use. AUDIENCE: All right, cool. PROFESSOR: Cool? All right. So there are a bunch of nice things I want to deduce from what we've done so far. So the first is, look, I pointed out that this can be derived just explicitly and it gives the same results as before. That's not an accident. If you take this system and you just reverse the roles of k1 and k2, what happens? Well that's just replaces e by e minus v0, we can do that by doing this, replacing e by e minus v0 and e minus v0 by e, it swaps the role of a and d, and it gives you exactly the same things back. So if you're careful about that, you never have to do this calculation. You can just do the appropriate transformation on that calculation and it gives you the exactly the same thing. It's the same algebraic steps. But the other thing that's nice is that you can actually do the same thing from here. So as long as you set the d equals 0-- so there's another term here that we neglected, the plus delta x, we got rid of it. But you can analytically continue this calculation by noting that look, if we just set alpha, we want minus alpha equals ik2, then the algebra is all going to be the same. We just have ik2 instead of minus alpha. So we can just replace alpha with minus ik2 everywhere in our expressions, being careful about exactly how we do so, being careful to take care of factors of i and such correctly. And you derive the same results for both cases, which is a nice check on the calculation. So often, when you get a little bit of experience with these, you don't actually have to do the calculation. Again, you can just take what you know from a previous calculation and write down the correct answer. So it's a fun thing to play with, exactly how do you do that. So I invite you to think through that process while you're doing your problem set. Another thing is the reflection downhill thing which is pretty surprising. But here's the thing that I really want to emphasize. What this calculation shows you is not so much that-- it's not just that transmission downhill is highly unlikely when the energy is very close to the height of the potential barrier. That's true, but it's not the most interesting thing about this calculation. The most interesting thing about this calculation, to my mind, is the fact that from the detailed shape of the transmission as a function of energy, we can deduce what the potential is. Think about what that tells you. If you do an experiment, you have a barrier, and you want to know the shape of the barrier. Is it straight, is it wiggly, does it have some complicated shape. How do you measure that? Well, you might measure it by just looking. But imagine you can't, for some reason. For whatever, it's in a box or you can't look at it. Maybe it's just preposterously small. How can you deduce what the shape of that hill is? Well, one way to do it is to send in particles as a function of energy, more and more energy, and measure the probability that they transmit. OK. Now if you do so and you get this graph as a function of e, what do you deduce? You deduce that the barrier that you're scattering off of is a square step with this height v0. Are we cool with that? So apparently, just look at the transmission amplitudes, the transition probabilities, you can deduce at least something of the form of the potential. Which is kind of cool. If you didn't know the potential, you could figure out what it was. And this turns out to be a very general statement that you can deduce an enormous amount, and as we'll see, you can, in fact, deduce basically everything you want of the potential from knowing about the transmission probabilities as well as the phase shift, the transmission amplitudes. So this is the basic goal of scattering. And so the way I want you to think about it is imagine, for example, that someone hands you an object. A box. And the box has an in port and it has an out port. And they allow you to send in particles as a function of energy and measure transmission and reflection, you can measure transmission and reflection. Just like I'm measuring transmission off of your faces right now, from the light from above. So suppose that you do so, put it on your test stand and you measure transmission. You measure transmission as a function of energy, and you observe the following. The transmission as a function of energy is small up until some point, and then at some point, which may be the minimum energy you can meaningfully probe, you get something like this. So here's the transmission as a function of energy. So what can you say? If this is all the information you have about what's going on inside the box, what can you deduce about the thing inside the box? One thing you can deduce is that it looks kind of like a potential with height around v0. It looks kind of like a potential step with some height v0. This is asymptoting to 1. However, it's not, because it has these oscillations in it. So there's more to the potential than just a barrier of height v0. What I want to show you is that you can deduce everything about that potential, and that's the point of scattering. So let's do it. So the goal here, again, to say it differently, is what's v0? What is v of x? Not just the height, but the total potential. So another way to say this, let me set up a precise version of this question. I want to be able to do the following. I want to take a system that has a potential which is constant up to some point which I'll call 0, and then again from some point, which I'll call L, is constant again. And inside, I don't know what the potential is. So in here, there's some unknown potential, v of x, which is some crazy thing. It could be doing anything. It could be some crazy-- it could have horns and whatever. It could be awful. But the potential is constant if you go far enough away, and the potential is constant if you go far enough away. A good example of this is a hydrogen atom. It's neutral but there's a clearly and complicated potential inside because the proton and the electron are moving around in there in some quantum state, anyway, and if you send something at it, far away, it's as if it's not there. But close by, you know there are strong electrostatic forces. And so the question is what you learn about those forces, what can you learn about the potential by throwing things in from far away, from either side. Now one thing we know already is that out here, the wave function always-- because it's a constant potential-- always takes the form e to the ikx plus b e to the minus ikx. And out here it takes the form c e to the ikx plus d e to the minus ikx. And again, this corresponds to moving in. d is in from the right. c is out to the right. b is out to the left. And a is in from the left. So again, there are basically four kind of scattering experiments we can do. We can send things in from the right, which corresponds to setting a to 0. We can send things in from the left, which corresponds to setting t equal to 0. And all the information about what happens in v is going to be encoded in what's coming out, the b and c coefficients. And the way to make that sharp is just to notice that the transmission probability, if we compute for this system, assuming it's forming the wave function asymptotically away from the potential, the transmission amplitude is just c over a squared when you're sending in from the left. And the reflection is equal to b over a norm squared. This squared is a squared. [INAUDIBLE] function over amplitude squared, good. To learn about the transmission and reflection coefficients, it's enough-- suffices-- to compute, to know b and c as a function of a and d. All of the scattering information is in those coefficients b and c for a and d. And here I'm assuming that I'm sending in a monochromatic wave with a single, well defined energy. I'm sending in a beam of particles with energy e. I don't know where they are, but I sure know what their momentum is. So some well defined beam of particles with energy e. And these probabilities are going to contain all the data I want. So this is the basic project of scattering. Questions? AUDIENCE: So basically, it only depends on the transmission-- it only depends on the edges? PROFESSOR: That's a good question. The question is, does the transmission depend only on what you do with the edges. And here's the important thing. The transmission depends crucially on what happens in here. For example, if this is an infinitely high barrier, nothing's going across. So this transmission depends on what's in here. But the point is we can deduce just by looking far away, we can deduce the transmission probability and amplitude just by measuring b and c far away, b and c far away. OK. So the transmission amplitude is something you measure when very far away. You measure-- if I throw something in from very far out here to the left, how likely is it to get out here very far to the right? And in order to answer that, if someone hands you the answer to that, they must have solved for what's inside. That the point. So knowing the answer to that question encodes information about what happened in between. AUDIENCE: So I guess initially, your potential's going to-- so say it's stuff you did earlier. The potential drops down to the [INAUDIBLE] in the box. Is that going to be problematic? PROFESSOR: Yeah, for simplicity, I'm going to assume that the potential always goes to 0 when we're far away, because that's going to be useful for modeling things like hydrogen and, exactly. Carbon, we're going to do diamond later in the semester, that'll be useful. But we could repeat this analysis by adding an extra change in the asymptotic potential, it doesn't really change anything important. Yeah. AUDIENCE: It looks like even for just the simple step up [INAUDIBLE] you can't tell from just the probabilities why the step is going up or going down. PROFESSOR: Ah, excellent. Excellent, excellent. So good, thank you for that question, that's really great. So already, it seems like we can't uniquely identify the potential from the transmission probability if the transmission probability is the same for step up or step down. So what's missing? AUDIENCE: Maybe the energy [INAUDIBLE] PROFESSOR: But we're working with an energy [INAUDIBLE]. So the energy is just a global constant. We'll see what's missing, and what's missing is something called the phase shift. So very good question, yes. AUDIENCE: It looks like when we did the example for the step, that t equals [INAUDIBLE] over a squared. PROFESSOR: Yeah, it is, because it's the ratio. That's why I wrote t is c over here. So t is the ratio of j over ji. And in fact, jt here is equal to c squared times-- so this is the probability density, so it's the probability density times the velocity, what's the effect of velocity here? H bar k 2 over m, whereas j incident is equal to a squared-- probability density times the momentum there, the velocity there-- which is h bar k1 upon m. So here the ratio of j transmitted to j incident is norm c squared k2 over norm a squared k1. AUDIENCE: So because it's not level-- PROFESSOR: Exactly, it's because it's not level. So here, they happen to be level, so they're only [INAUDIBLE] factor, cancel, and the only thing that survives is the amplitude. Good question. Yeah. AUDIENCE: For the lead box example, is it sufficient just to know what's reflected back to solve the situation? PROFESSOR: So for precisely for this reason, it's not sufficient to know t. It's not sufficient to know t and r. But, of course, once you know r, you know t, so you're exactly right. Once you know the reflection probability, you know the transmission probability, but there's one more bit of information which we're going to also need in order to specify the potential, which is going to be the phase shift. But you're right, you don't need to independently compute r and t, you can just compute one. AUDIENCE: You need sensors on both sides of the box, to answer my question. PROFESSOR: You don't need sensors on both sides of the box, but you need to do more than just do the counting problem. We'll see that. OK. So let's work out a simple example, the simplest example of a barrier of this kind. We want constant potential, and then ending at 0, and we want a constant again from L going off to infinity. So what's the easiest possible thing we could do? Step, step. We're just doing what we've done before twice. So this is an example of this kind of potential. It's sort of ridiculously simple, but let's work it out. So we want scattering, let's start out, we could do either scattering from the left or scattering from the right. Let's start out scattering from the left, so d equals 0, and let's study this problem. So what we know-- and let's also note that we have a choice to make. We could either study energy below the height of the potential or we can study energy above the height of the potential. And so for simplicity, I'm also going to start with energy greater than the height of the potential, v0, and then we'll do e less than v0 afterwards. It'll be an easy extension of what we've done. OK, so this will be our first case to study. So we know the form of the wave function out here, it's a e to the ikx, b to the minus ikx. We know that for the potential out here it's c e to the ikx and d e to the minus ikx. The only thing we don't know is the form of the potential in here. And in here it's actually very simple. It's got to be something of the form-- I think I called it f and g, I did. f e to the ik prime x-- I'm calling this k, so I'll just call this k prime x-- plus g e to the minus ik prime x. And the reason I chose k prime is because we're working with energy greater than the potential, so this is a classically allowed region. It's an oscillatory domain but with a different k prime. So here, k squared, h bar squared over 2m is equal to e, and h bar squared k prime squared over 2m is equal to e minus v0, which is positive when e is greater than v0. This analysis will not obtain when e is less than v0, we'll have to treat it separately. So now what do we do? We do the same thing we did before, we just do it twice. We'll do the matching conditions here, the matching conditions here. That's going to give us 1, 2, 3, 4 matching conditions. We have 1, 2, 3, 4, 5, 6 unknown coefficients, so we'll have two independent ones. That's great. We set d equal to 0 to specify that it's coming in from the left and not from the right, that's 5. And then we have normalization, which is 6, so this should uniquely specify our wave function. Yeah. Once we've fixed e, we have enough conditions. So I'm not going to go through the derivation because it's just an extension of what we did for the first. It's just a whole bunch of algebra. And let me just emphasize this. The algebra is not interesting. It's just algebra. You have to be able to do it, you have to develop some familiarity with it, and it's easy to get good at this. You just practice. It's just algebra. But once you get the idea, don't ever do it again. Once you get reasonably quick at it, learn to use Mathematica, Maple, whatever package you want, and use computer algebra to check your analysis. And use your physics to check the answer you get from Mathematica or Maple or whatever you use. Always check against physical reasonability, but use Mathematica. So posted on the website are Mathematica files that walk through the computation of the transmission and reflection amplitudes and probabilities for this potential, and I think maybe another one, I don't remember exactly. But I encourage you strongly to use computer algebra tools, because it's just a waste of time to spend three hours doing an algebra calculation. In particular, on your problem set this week, you will do a scattering problem similar to the bound state probably you did last week, the quantum glue problem-- which you may be doing tonight, the one due tomorrow. Which is two delta function wells and find the bound states, so that involves a fair amount of algebra. The scattering problem will involve a similar amount of algebra. Do not do it. Use Mathematica or computer algebra just to simplify your life. So if we go through and compute the-- so what are we going to do, we're going to use the matching conditions here to determine f and g in terms of a and b, then we'll use the matching conditions here to determine c-- d is 0-- c in terms of f and g. So that's going to give us an effective constraint relating a and b, leaving us with an overall unknown coefficient a, which we'll use for normalization. The upshot of all of which is the answers are that, I'm not even going to write down-- they're in the notes. Should I write this down? I will skip. So the upshot is that the transmission amplitude, as a function of k and k prime-- the transmission probability, I should say, is 1 over-- actually, I'm going to need the whole amplitude. Shoot. The transmission probability is equal to-- and this is a horribly long expression-- the transmission probability, which is c over a norm squared, is equal to 4 k squared k prime squared cosine squared of k prime l plus k squared plus k prime squared sine squared of k prime l under 4k squared k prime squared. Seriously. So we can simplify this out, so you can do some algebra. This is just what you get when you just naively do the algebra. I want to do two things. First off, this is horrible. There's a cosine squared, there's a sine squared, surely we can all be friends and put it together. So let's use some trig. But the second thing, and the more important thing, is I want to put this in dimensionless form. This is horrible. Here we have ks and we have ls, and these all have dimensions, and they're inside the sines and the cosines it's kl. That's good, because this has units of one over length, this has units of length, so that's dimensions. Let's put everything in dimensionless form. And in particular, what are the parameters of my system? The parameters of my system are, well, there's a mass, there's an h bar, there's a v0, and then there's an energy, and there's a length l. So it's easy to make a dimensionless parameter out of these guys, and a ratio of energies-- a dimensionless ratio of energies-- out of these guys. So I'm going to do that, and the parameters I'm going to use are coming from here. I'm going to define the parameters g0 squared, which is a dimensionless measure of the depth of the potential. We've actually run into this guy before. 2m l squared v0 over h bar squared. So this is h bar squared, 1 over l squared is k squared over 2m, so that's an energy. So this is a ratio of the height of the potential to the characteristic energy corresponding to length scale l. So the width, there's an energy corresponding to it because you take a momentum which has 1 over that width. You can build an energy out of that h bar k squared over 2m. And we have an energy which is the height of the barrier, we take the ratio of those. So that's a dimensionless quantity, g0. And the other dimensionless quantity I want to consider is a ratio of the energy, e, to v0, which is what showed up before in our energy plot. Or in our transmission plot over there, e over v0. So when we take this and we do a little bit of algebra to simplify our life, again, use Mathematica, it's your friend. The result is much more palatable. It's t-- again, for the energy greater than v0-- is equal to, still long. But 1 plus 1 over 4 epsilon, epsilon minus 1 sine squared of g0 square root of epsilon minus 1. And upstairs is a one. So remember this is only valid for e greater than v0, or equivalently, epsilon greater than 1. And I guess we can put an equal in. So when you get an expression like this, this is as easy as you're going to make this expression. It's not going to get any easier. It's 1 over a sine times a function plus 1. There's really no great way to simplify this. So what you need when you get an expression like this is try to figure out what it's telling you. The useful thing to do is to plot it. So let's just look at this function and see what it's telling us. Let's plot this t as a function of epsilon and for some fixed g0. Keep in mind that this only makes sense for e greater than v0 or for epsilon greater than 1, so here's 1. And we're going to remain agnostic as to what happens below 1. And just for normalization, we know that the transmission probability can never be greater than 1, so it's got to be between 1 and 0. So here's 0 and 0. We're remaining agnostic about this for the moment. So let's start thinking about what this plot looks like. First off, what does it look like at 1? So when epsilon goes to 1, this is going to 0, that's bad because it's in a denominator. But upstairs, this is going to 0, and sine squared of something when it's becoming small goes like-- well, sine goes like that thing. So sine squared goes like this quantity squared. So sine squared goes like g0 squared, this goes like 1 over-- this is going like g0 squared epsilon minus 1 square root quantity squared, which is epsilon minus 1. So this is going to 0 and this, the denominator, is going to 0 exactly in the same way. Epsilon minus 1 from here, epsilon minus 1 downstairs from here. So the epsilon minus ones precisely cancel. From the sine squared we get a g0 squared, and from here we get a 4 epsilon. So we get 1 plus g0 squared over 4 epsilon. But 4 epsilon, what was epsilon here? 1. We're looking at epsilon goes to 1, so this is just g0 squared over 4. So the height, the value of t-- so we're not looking in here. But at energy is equal to v0 or epsilon is equal to 1, we know that the transmission amplitude is not 0, but it's also strictly smaller than 1. Which is good, because if it were 6, you'd be really worried. So g0 squared, if g0 squared is 0, what do we get? 1. Fantastic, there's no barrier. We just keep right on going through, perfect transmission. If g0 is not zero, however, the transmission is suppressed. Like 1 over g0 squared. What does this actually look like with some value. And what is this value, it's just 1 over 1 plus 1/4 g0 squared. Cool? OK. And now what happens, for example, for a very large epsilon? Well, when epsilon is gigantic, sine squared of this-- well, sine squared is oscillating rapidly, so that's kind of worrying. If this is very, very large, this is a rapidly oscillating function. However, it's being divided by roughly epsilon squared. So something that goes between 0 and 1 divided by epsilon squared, as epsilon gets large, becomes 0. And so we get 1 over 1 plus 0, we get 1. So for very large values it's asymptoting to 1. So naively, it's going to do something like this. However, there's this sine squared. And in fact, this is exactly what it would do if we just had the 1 over epsilon times some constant. But in fact, we have the sine squared, and the sine squared is making it wiggle. And the frequency of the wiggle is g0, except that it's not linear in epsilon, it's linear in square root of epsilon. So as epsilon gets larger, the square root of epsilon is getting larger less than linearly. So what that's telling you is if you looked at root epsilon, you would see it with even period. But we don't have root epsilon, we're plugging this as a function of epsilon. So it's not even period, it's getting wider and wider. Meanwhile, there's a nice fact about this. For special values of epsilon, what happens to sine squared? It goes to zero. And what happens when this is zero? Yeah, t is 1. So every time sine is 0, i.e., for sufficient values when root epsilon is a multiple of pi determined by g0, transmission goes to 1. It becomes perfect. So in fact, instead of doing-- wow. Instead of doing this, what it does is it does this. So let's check that I'm not lying to you. So let me draw it slightly different. To check, so it's going to 0, and the period of the 0 is getting further and further along. That's because this is square root, not squared. So epsilon has to get much larger to hit the next period. That's why it's getting larger and larger spacing. However, the amplitude goes down from 1, that's when this gets largest. Well, at large values of epsilon, this is suppressed. As epsilon goes larger and larger, this deviation become smaller and smaller. So that's what this plot is telling us. This plot is telling us a bunch of things. First off, we see that at large energies, we transmit perfectly. That makes sense. This was a finitely high barrier. Large energies, we don't even notice it. It tells you at low energies-- well, we don't know yet what happens at very low energies. But at reasonably low energies, the transmission is suppressed. And if you sort of squint, this roughly does what we'd expect from the step barrier. However, something really special is happening at special values of epsilon. We're getting perfect transmission. We get perfect transmission of these points, t is equal to 1. Perfect. Star. Happy with a big nose. I don't know. Perfect transmission at all these points. So this leaves us with a question of why. Why the perfect transmission, what's the mechanism making transmission perfect. But it also does something really lovely for us. Suppose you see a spectrum, you see a transmission amplitude or transmission probability that looks like this. You may get crappy data. You may see that it's smudged out and you see all sorts of messy stuff. But if you know that there's perfect transmission at some particular epsilon 1, there's more perfect transmission at another epsilon 2, more perfect transmission at epsilon 3, and they scale, they fit to this prediction. What do you know? That you've got a finite high barrier, and it's probably pretty well approximated by a square step. So in your problem set, you're going to get experimental data, and you're going to have to match to the experimental data. You're going to have to predict something about the potential that created a particular transmission probability distribution. And knowing where these resonances are, where these points of perfect transmission, is going to be very useful for you. Yeah. AUDIENCE: Wasn't there another potential that will also create periodic 1s? PROFESSOR: A very good question. So I'm going to turn that around to you. Can you orchestrate a potential that gives you the same thing? That's an interesting empirical question. So, on the problem set, you'll study a double well potential. So the question was, how do I know isn't another potential that gives me the same answer? At this point, we don't know. Maybe there is another potential. In fact, you can do all sorts of things to make a potential that's very arbitrarily close that gives you an arbitrarily similar profile. So if they're significantly different, how different do they look? So then the students say, well, I don't know. What about a double well potential? So, in fact, we'll be doing a double well potential on the problem set. A good question. And we certainly haven't proven anything like this is the only potentially that gives you this transmission amplitude. What we can say is if you get transmission amplitude that looks like this, it's probably pretty reasonable to say it's probably well modeled. We seem to be reproducing the data reasonably well. So it's not a proof of anything, but it's a nice model. And we're physicists, we build models. We don't tell you what's true, we tell you what are good models. If a theorist ever walks up on you and he says, here's the truth, punch him in the gut. That's not how it works. Experimentalists, on the other hand. We'll just punch them too, I guess. [LAUGHTER] PROFESSOR: We build models. OK, so this leaves us with an obvious question, which we've answer in the case of energy grade in the potential. What about the case of the energy less than the potential. So we haven't filled out that part of the graph. Let's fill out that part of the graph. What happens if the energy is less than the potential? And for this, I'm going to use the trick that I mentioned earlier that, look, if the energy is less than the potential, that just means that in the intervening regime, in here, instead of being oscillatory, it's going to be exponentially growing [INAUDIBLE] because we're in a classically disallowed region. So if the energy is less than v0, here we have instead of ikx, we have minus alpha x. And instead of minus ikx we have plus alpha x. So if you go through that whole analysis and plug in those values for the k prime, the answer you get is really quite nice. We find that t for energy less than v0 is equal to-- and it's just a direct analytic continuation of what you get here-- 1 plus 1 over 4 epsilon 1 minus epsilon. So epsilon is less than 1 now. 1 sinh squared of g0 1 minus epsilon under 1. OK, so now this lets us complete the plot. And you can check, it's pretty straightforward, that because sinh, again, goes like its argument at very small values of its argument, we get g0 squared 1 minus epsilon. Sorry, this should be square root 1 minus epsilon. Really? Is that a typo? Oh no, it's in there. OK, good. So again, we get g0 squared 1 minus epsilon, the 1 minus epsilon cancels, and we get 1 over 4 g0 squared. Which is good, because if they disagreed, we'd be in real trouble. So they agree, but the sinh squared is just a strictly-- or that function is just a strictly decreasing function as we approach epsilon goes to 0. It goes mostly to 0. So this is what we see. We see an exponential region and then we see oscillations with residences where the period is getting wider and wider. But this should trouble you a little bit. Here, what are we saying? We're saying look, if we have a barrier, and we send in a particle with energy way below the barrier, that's kind of troubling. If we send in a particle with energy way below the barrier, there's some probability that it gets out. It goes across. So for e less than v0, classically no transmission, we get transmission. Now, do we get resonances when e is less than v0? No, that's an interesting thing. We'll talk about resonances and where they come from, and we'll talk about the more generalized notion of a nest matrix in the next lecture. But for now I just want to say a couple of things about it. So I want to ask-- so this is called tunneling, this transmission across a disallowed barrier, a classically disallowed barrier-- so this transmission across a disallowed barrier is called tunneling. And just a quick thing to notice is that if we hold e fixed and we vary l, we vary the width of the barrier, how does the tunneling amplitude depend on the width of the barrier? At fixed energy, if we vary the width of the barrier-- which is only contained in g0-- how does the transmission amplitude vary? And we find that for large l-- for l much greater than 1 for a typical scale in the problem-- the tunneling amplitude goes like e to the minus 2 alpha l where alpha just depends on the energy in the potential. So what we see is that the probability of transmitting, of tunneling through a wall, depends on the width of that wall. And it depends on it exponentially. The wider the wall, the exponentially less likely you are to tunnel across it. And this fits the simulation we ran where we saw that tunnelling through a very thin wall was actually quite efficient. And we also saw that tunnelling across a wall can have resonant peaks. At special values, it transmits perfectly. We saw that in the simulations. So at this point I can't resist telling you a very short story. I have a very good friend who is not a physicist, but who is a professor at MIT. So a smart person, and very smart. She's one of the smartest people I know. She broke her ankle. She's always losing stuff, she's constantly losing stuff. And she broke her ankle, and she was going to the rehab place, and she parked her car right in front of the rehab place. She parked her car right in front because she's got the broken ankle and she doesn't want to have to walk a long way to get the car, she parked right in front. She went in, she did her two hours of rehab, and she came outside, and her car was gone. And she's always misplacing her car, she's always misplacing her keys, she's always losing everything. But this time, this time she knew where it was. And it wasn't there. So first she thought, oh crap, my car's been stolen. How annoying, I've got a broken ankle. But she looks around, and there's her car on the other side of the street pointed the opposite direction. And she's like, finally. Finally. She was explaining this to me and a friend of mine who is also a physicist, she said finally, I knew I had incontrovertible proof that quantum effects happen to macroscopic objects. My car tunnelled across the street. [LAUGHTER] PROFESSOR: At this point, of course, my friend and I are just dying of laughter. But she said my car tunnelled across the street. But here's the problem, I couldn't tell anyone. Because if I told anyone, they would know that I was crazy. Clearly I'm crazy, because I lose my stuff all the time, but surely it's just gotten misplaced. So I went home, and I got home and I was sitting down to dinner with my partner and our daughter and I was burning up inside because I wanted to tell them this crazy thing happened. I know quantum mechanics, it happens. I couldn't tell them anything, so I was just fidgeting and dying. And finally my daughter said, mom, didn't you notice that we moved the car across the street? [LAUGHTER] PROFESSOR: See you next time.
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https://ocw.mit.edu/courses/3-091sc-introduction-to-solid-state-chemistry-fall-2010/3.091sc-fall-2010.zip
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The following content is provided under a creative commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JOCELYN: Hi. Jocelyn here and we're going to go over fall 2009 exam three problem number three. As always, let's read the question first. Calcium ammonium phosphate dissolves in water according to the dissolution equation, for which the value of the solubility product, ksp, has been determined to be 4.4 times 10 to the -14. Calculate the solubility of the compound in water. Express your answers in units of molartiy, ie moles of the compound per liter of solution. So the first thing we want to do is write down what the question's asking. Part A-- how much calcium ammonium phosphate can dissolve in water? And we're going to call this cs-- or the saturation concentration. Next we need to figure out how to find this out, right? So it gives us that the ksp-- 4.4 times 10 to the -14. So we need to know something about the ksp and although it's called the solubility product, it's the same as other equilibrium constants. So equilibrium constants tells us about the concentrations at equilibrium and therefore the maximum solubility. Looking at the equation, we have calcium ammonium phosphate dissolving-- which is a solid-- dissolving into calcium ions, ammonium ions and phosphate ions. As with every equilibrium constant, we can write down an equation relating to the concentration of the species. So our ksp is the product of the concentration. And a common mistake made on this problem was that people included-- students included the calcium ammonium phosphate solid on the bottom as you normally would-- where you would normally put the reactant. However, remember that for equilibrium constants, we don't put solids in there because the activities don't change-- or the concentration doesn't change. It doesn't really make sense to add it in. So here we just look at the solvated ions and we see that we have a 1:1:1 molar ratio here. That's going to make our life a little bit easier. So now we need to figure out what the solubility of the compound is and to do that, we want to look at this chemical equation. We see that for one mole-- so one mole of calcium ammonium phosphate dissolved gives you one mole of the calcium ion, the sodium ion and the phosphate ion. Thus, we can say the amount of calcium ammonium phosphate dissolved, which we called Cs, is going to be equivalent to each of these concentrations. Because we're asked for how much of the compound is dissolved, but we're given the ksp, which is in terms of the concentrations of these solvated ions. I'm sorry. This doesn't equal the product. These are all equal. Now we can plug the Cs into our ksp equation that we had before. So instead of the concentration of calcium ions, we have-- it's equal to the concentration of the compound dissolved. The same goes for the ammonium and the phosphate, because again, we have 1:1:1:1 stoichiometric coefficients. And we can be a little more concise. Now it's just a matter of solving for the saturated concentration. So doing some algebra here. Then we plug in the numbers. And we get that the solubility of calcium ammonium phosphate is 3.53 times 10 to the -5 moles-- molar, sorry. So that means 3.53 times -5 moles per liter of water. That's not very much. So we can say generally or relatively this calcium ammonium phosphate is not that soluble in water. So let's move on to part B. Part B says, calculate the solubility of calcium ammonium phosphate in 2.2 molar calcium bromide. Express your answer in units molarity. Assume that in water, calcium bromide completely disassociates into calcium 2 plus cations and bromide anions. This is basically asking us for the same value. It's asking us for the solubility of calcium ammonium phosphate, but under slightly different conditions. So we want to write those conditions down. Now we have to ask ourselves why would the fact that we have a concentration of calcium bromide affect the solubility of calcium ammonium phosphate? And the thing to remember here is the common ion effect, right? 2 molar calcium bromide will-- when dissolved in water, the concentration of calcium from just the calcium bromide will be 2.2 molar, right? Because we're told that it completely disassociates. So if we look at the disassociation reaction, we see that for every mole of calcium bromide, we get one mole of calcium and two moles of bromide. So having calcium bromide will alter our answer from the previous problem because the ksp will always be the same. It's an equilibrium constant, right? So even though it has to do with calcium ammonium phosphate, we have to take into account that we already have calcium in the system. So before we start plugging in any numbers, we should think about this problem. Do we think that already having calcium in the system will increase or decrease the solubility of the calcium ammonium phosphate? Hopefully we can agree that it would probably decrease the solubility because you already have those calcium ions. So putting more calcium ions into the water is going to be harder and therefore less calcium ammonium phosphate will dissolve. Now that we know what kind of number we're looking for, we can start doing the actual calculation. So again, we'll start with our equation for the ksp, which for calcium ammonium phosphate has not changed. I'm just going to rewrite it on this board here. But now instead of having each of these be equal concentrations, we already have some calcium in the system. So we need to take that into account. And I'm going to call the saturation concentration Cs star because we're under different conditions, right? The calcium bromide does not contribute any ammonium or phosphate ions so those concentrations will just be determined by how much of the calcium ammonium phosphate dissolves. From before, we can use what we found in part A to simplify this a little bit. So in part A, we know that without anything else, without a common ion effect, that we have decided will decrease the solubility. We have 3.35 times 10 to the -5th molar solubility. Therefore, we can say with good certainty that our saturation concentration with the common ion effect will be much, much less than 2.2 molar because 2.2 molar is much greater than our pure solubility. Going back to our equation over here, that means we can have-- and now we have a fairly simple algebraic problem. Dividing by 2.2 and then taking the square root, we get that the saturation concentration under these conditions-- 2.2 molar calcium bromide-- will be 1.41 times 10 to the -7 molar. And going back to our answer from part A, we see that this is indeed a lower solubility. There's less calcium ammonium phosphate that can be dissolved and that makes sense from our previous thought about this problem. And so now that you've found the answer, box it and we're done.
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https://ocw.mit.edu/courses/5-95j-teaching-college-level-science-and-engineering-fall-2015/5.95j-fall-2015.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JANET RANKIN: A general comment is by and large, I thought you guys did a great job. Your learning outcomes were for the most part specific, measurable, and realistic. The biggest problem I think I had was with-- the biggest problem that you had on average was the idea of the learning outcomes really have to be measurable. So if you use a word like appreciate, it's not measurable. So that's something to remember. If you use words like discuss, it's potentially measurable. But you need to make sure that you set up a venue for students to discuss. You can't just say you'll be able to discuss that and assume that they'll be able to discuss it, That If that's in your learning outcome, you have to set up some venue for them to discuss and you to know what they're discussing. And then the two biggest things that come up, or the two most problematic words that come up-- and I compliment you all, because you did a great job. Nobody used understand. So that was awesome. But the other thing, there's words like explain and describe, which are perfectly good words to use in a learning outcome. But you have to make sure that when you say, students will explain the ideal gas law. They'll be able to explain the ideal gas law. Well do you know that they're really explaining it in the true sense of that word that they actually internalized what it means, and they're really explaining it? Or are they just spitting it back, spitting back, maybe, what they read in a book, a definition that they read in the book? So you have to make sure that you've set up a situation wherein they really are explaining. And they're not just remembering. So it's a great word to use. But you have to make sure that we're using in the right context. Same thing for describe. You can imagine that has the same issues. Describe da, da da, da, da. Well are you sure they're doing the describing? Or are they just remembering somebody else's description and writing it down? One good way around that is to tweak the parameters a little bit. To say, in a world where there is no oxygen, what do you expect this organism to look like, given that it has to run really fast or something? So come up with alternate realities that that actually force students to transfer and then re-explain. That can be helpful. That can be really, really helpful. I remember once in a kinetics class, we had been deriving the critical nucleus size for growth. And it's all based on a radius, a critical radius. And then on the exam, the guy said OK, they don't grow as sphere's. They grow as rectangular boxes. So then all of a sudden, we had to look at the ratio of the length to the height of the box or something. So that really wasn't just we were just remembering a solution or remembering a method. We had to actually think about what it meant and then reapply it. So just think about how you can get students to really describe, explain, derive even, because you can memorize a derivation. But if you really want them to derive it, what does that mean? How are you going to test that? So those are generally the biggest problems people have with learning outcomes. But by and large, you all did a great job. If you came in late, I'll give you your learning outcomes at the end to class. If you looked at the Wiki-- and this is a test to see who looked at the Wiki-- I had suggested you bring a web enabled device if you had one. And that's because I wanted to set up something called Back Channel The topic today is on active learning. And so this is just one method that might help your students, especially if you have a big class. So we don't really need it in this class, but I just wanted to give you the opportunity to play with it a little bit. It was designed for conferences actually. But it works pretty well, I think, in classes. So if you login to here, this is the URL. If you go to our Wiki page bookmarked, there's a link on the Wiki page, if you just want to link to it that way. And then at any point in the class, you can type a question in. And if you're on the page, you can see other people's questions. You can upvote and downvote their questions. And I will take a break in the middle, when you're doing some other things, and read what people have posted and then get back to you. So it's an opportunity for students to interact who may be a little more shy, who may not want to. I don't think we need that in this class. I don't think that there's any one of you that isn't willing to share your ideas, which is great. But this may be something that's of interest to you. Before we jump into the topic today, active learning, there were a couple of mud card questions that I just wanted to talk about. One was sometimes students may nod off during class. And if it's a big class, there's probably not a lot you can do about it. If you do enough active learning, though, that will keep them-- that will make them wake up. Especially pair activities or activities where they have to talk to one or two other people, or two or three other people, it's pretty hard to be asleep during those. And you'll see. We're going to do something called a lightning round later on. And it's virtually impossible to be asleep during that. So you want to make sure that it's worth the student's time to actually be in class and be awake for it. And active learning can play a huge role in that. Sometimes people are tired. They stayed up too late for whatever reason. And you can't control everybody in your class. But you want to make sure that you're doing everything you can do to keep them engaged. And if it's just one student every now and then, you probably don't want to take it personally. It's probably not you. If the whole class is falling asleep, then it's probably a good opportunity to rethink your strategies. Somebody asked if there were good times of day to have a lecture. And part of it is you. Part of is your schedule. Sometimes you're allowed. You can pick that. And sometimes you really don't have much control over when you choose to teach your class. People do say that students, undergrads, people between the ages of 18 and 21 are not particularly interested in being up at 8:00 or 9:00 in the morning. But again you may not have any control over that. Generally speaking, that's a pretty low time for people. So if you want to take that into consideration, you might. But it depends as much on you as the students. If you're going to be really tired at 4:00 but they're just waking up, then don't teach it at 4:00. Maybe don't teach at 4:00. And the last one was guiding readings. If you just put a lot of readings up, students might not know where to focus their attention until after the class. So there are reading questions, reading guides that you can post. Read such and such and try and think about how you might explain this, that, or the other thing. And then there's also pre-class questions you can ask that they might derive, that they may have to pull from the readings to answer. So the concept of just in time teaching, the idea that students do a reading and then they actually have to answer questions on the reading before they come to class and submit their answers. So there's a whole procedure around that. It can be very, very effective. You can do it for a large class. There's strategies and techniques to streamline the amount of time you put into it. But it does force them to do the reading. And it does guide, at some level, what they read. And that's a great great approach. And I'm happy to talk more about that at another time or later or offline. I did go to the Teaching with Technology conference, the Teaching Professor Technology Conference. And they were giving out little ribbons to stack on your badge. And this was a prominent one. It's in the syllabus, which I thought was just crazy timely because a couple of weeks ago. So if you want one of these to stick somewhere. And then they all said when I tweet, which was sort of weird. But anyway I brought those back. Anybody want some. Sometimes my computer doesn't wake up. So let's pretend that instead of the class active learning today, I decided I was going to teach you how to juggle. And I said, here's the equation for how to juggle. And here's the variables, F, D, V N, and H, which are ball in the air, ball in an hand, time hand is vacant, number of balls, number of hands. And this is this handy dandy equation, which if you balance it out, it describes the juggling process. And for those of you that are visual learners, I have a graph here, a yellow ball, a blue ball, and a pink ball. And then you can see that just shows the path of the balls. If you remember from physics, there's the equations of motion in the x direction in the y direction. And so I think I probably did a pretty good job of teaching you how to juggle. No, did a terrible job of teaching you how to juggle. I didn't even come close to teaching you how to juggle. And if I were to give you a test, let's say, and I were to measure whether or not you could juggle, I guarantee that some of you would be able to juggle. And I could say, I did a great job teaching how to juggle. But I didn't. If you knew how to juggle coming in, presumably you still know how to juggle. I did no harm, I think. However if you didn't know how to juggle coming in, you don't know how to juggle now. So why is this here now? Why did I bother to do this stupid thought experiment right now? AUDIENCE: Because this is clearly how most classes are taught. Obviously it would be easier to have a ball and show us how to juggle, active learning as opposed to just showing the theory behind it and ending there. JANET RANKIN: And I think a lot of times we really do, in science and engineering classes, we really put up a lot of equations. And then it's not like we expect somebody to do something physically with them. But we expect them to be able to manipulate the equations, transfer the equations, apply the equations. And we never really talked about that. We talked about the equations. So it's a real call in my mind, for us to think about what it is we really want students to be able to do. And then make sure that they're doing stuff to teach them that helps them learn how to do that. And I know you all read the readings and that they talk a lot about active learning. There's a few things about techniques for active learning. There's a few things about the why's of active learning. There's some data for why active learning works. But I'm guessing that there's a range of buy in at this point from you guys. Eh, I don't think it's worth it. Or I don't really know. So for the first 20 minutes, 25 minutes of class, what I'd like you to do is I'd like you to pretend that you were part of a department, the STEM department. Call it whatever field you want. And there is a proposal that your department should commit funds to train professors and develop resources, the resources necessary for the adoption and ongoing use of active learning techniques in all courses. And the target date is 2017. You've got a year to get all courses up to speed with active learning stuff introduced into these classes and with a plan for how to do that. So what I'm going to do is I'm going to just count off one, two, three, one, two, three. And I'm going to ask you to form a group. If you're a one, go with the one group. Two go with the two group. A three go with the three group. And you're going to have to physically move. So just remember your number. One. AUDIENCE: Two. AUDIENCE: Three. AUDIENCE: One. AUDIENCE: Two. AUDIENCE: Three. AUDIENCE: One. AUDIENCE: Two. AUDIENCE: Three. AUDIENCE: One. AUDIENCE: Two. AUDIENCE: Three. AUDIENCE: One. JANET RANKIN: It doesn't matter that they're not the same size. Can we get all the ones, let's say all the ones up here? All the twos here. And all the threes there. And I'll tell you what to do in just a sec. So if you're group number one, you have to adopt the position that you are pro. You are going to push for the adoption of this resolution. And you have to come up with a set of arguments and a set of reasons why it's a good thing. I don't care what you think personally. I don't care if you think active learning is the worst idea ever. For the purposes of this debate, you need to pretend you're totally into it. And you need to come up with some good arguments why. Two are against it. Again, I don't care how you feel personally. But come up with some good solid arguments why you should not do this. AUDIENCE: We are against it? JANET RANKIN: Against, you do you not want it. No. Group number three, you got a little bit of time. But get to know each other. And you're going to hear the arguments. You're going to hear a case presented by the pro people, the con people. And then you are going to deliberate and decide what's going to happen in your department. You guys are the deciders. It's an American political joke. But we'll let that go. And you guys get this lovely microphone. So I'm going to give you 10 minutes for this. So be ready to share your views. You've got about, maybe, two or three minutes of a spiel of proof. And then we'll do one exchange of debate. And then the deciders will decide. [SIDE CONVERSATIONS] JANET RANKIN: I'm going to ask someone from each of the groups to make a statement that puts forth the side that you're supposed to defend, if that is the side you're supposed to defend. So we'll start with the pros. So someone from your group ready to talk pros? All right, Michelle. AUDIENCE: So we are pro active learning. And part of the basis for thinking that active learning's important is that we believe that without active learning, the professor usually just resorts to lecturing. And we think that lectures are not effective. Students don't learn simply because the material is presented to them. Lecturing wastes resources, because you could be lecturing to 1,000. But only 15 students may be reached. And if you use active learning instead of lecturing, it can make sure the students are paying attention. You can check that they're understanding. You can know what they are getting and what they're not getting. We also feel that active learning is a life skill. And that if students learn to learn by do things and they learn by talking to each other and by going through a process and solving problems, that they can take this skill to outside of the classroom later in society. Then that the society that you have students who are used to being active learners and who know how to solve problems that way. And we also think that the end, the big point that we have is that you remember things that you use. If you learn something, if it goes in one ear and out the other, if you don't actually use it, you're not going to remember it. If you're not forced to actually work through a problem and really think about it and use it in a classroom, that you're much less likely to actually hold on to that information. So these are all reasons why we think that we should promote active learning in our department. JANET RANKIN: OK, thank you very much. The con, or I guess I should say the not whatever. Anyway, group two. AUDIENCE: Do you want to do it? AUDIENCE: OK. We start with a basic question. The question is, why do we want to change what is working? We are training our students. They are getting job. The professor are doing well. They are collecting grants. The original [INAUDIBLE] is OK. So why do you want to change what is working? So that is the first question. Now the first thing is that, if you want to go about this, it's going to consume a lot of money. It's going to take a lot of money. And then it cuts into your funds. The one funds for other such, want to buy equipment for the laboratory, want to go hire scientists, want to go hire post doc. So why do we want to change? Where do we get these funds? We divert funds for the training of professors. And they're going to be little bit of no funds for all this such work. And secondly, the efforts to train professors is likely to be very huge. Professors are used to this method of teaching. And as you saw, they will not want to change. There's going to be a lot of time changing them. And we're not even sure, if we try to change them, whether they're going to be able to do it at the end of the day. Thirdly, we are not sure that the stakeholder would be happy. What about the alumni that are very much used to the method of teaching? What about the parents? are they now going to go against us? Will they not see that the professor no longer, just come to the class, give them group work like this without teaching, without writing? Then the professor is no longer teaching. And they are just going to collect their salary. And then, there's not to be constrained on time to cover syllabus. When I saw I spend a lot of money being in group work, individual, or this and that. So how do you complete your syllabus? You have about 20 topics to cover this semester. And by the time you start active learning, how are going you cover about 50% of them? Then you push all of that up [INAUDIBLE]. So it's not going to work. We should just drop the idea. What have we value very last class? You have your class of 500. How do you want active learning to be effective? How do you want to treat them? It's good here, because we are just about 15 or less than 20. By the time you have 500 students in a class, it's going to be very difficult for you to have active learning, difficult for the professor to control classes, and of course, difficult for you to really get something concrete out of the whole thing. So my final guys and professors, I think it's not necessary. the. System is working. Let it continue to work this way. Thank you. JANET RANKIN: So now what I'm going to do is I'm going to let each group ask one question or a set of clustered questions to the other group. The other group can answer them. And then the deciders can ask a couple questions. And then we'll let them deliberate. So group number one, like to ask some questions to group number two or make some comments? Go ahead. AUDIENCE: Yeah, I listened to you speak. And I want to ask you. You are saying that the system is changing. It is changing because the status quo isn't delivering. So that is why we are adopting active learning. And this is proven from a research perspective. There is actually letting more students learn more materials faster and more effectively. So what do you think about that? AUDIENCE: I'll answer it by saying the different research shows us that the student that is good is good. [INAUDIBLE] So why can't we just concentrate on getting those students? And what your presentation didn't say about funds, was the whole idea is to fund the proposal. And if we divert funds to this new program, and that's a very big issue. You have to concentrate on the question that is the diversion of funds away from research. So if we can get the new students and they can learn by [INAUDIBLE], and it's working great-- go ahead. AUDIENCE: So you want to throw away the baby with the bath water? If you have 20 students in a class, and you have five active learners, and then you want to discover 15 that wouldn't. This program is geared towards each one of the students in the class to participate. And I think it's more cost effective. If you could get all the 20 students to learn and to learn actively, I think that would pay itself off at the end of the-- AUDIENCE: No, the problem there is that you have to face the proposal. This is 2017. It's a lot of change in a very short time. The proposal itself is already against-- AUDIENCE: Our department could be three people big. We don't know how much the department is. [INTERPOSING VOICES] AUDIENCE: But then the issue there is that you're going to retrain professors. Professors are really stuck in their ways. And you spend all this money. And it's very short time. There's lots of people from the labs, and money from the labs in this very short time to train them on something we don't know if they're ready to buy into it. AUDIENCE: So I think your assumption is that-- you're like, we're deferring funds from labs and resort. But you never once said you're diverting funds from anything related to the students. You're thinking of it from the professor perspective. But really professors are here to teach their students. And you're like, oh, we're taking money away from the research, from the lab equipment, from all these different things that sound great to the professor but have literally nothing to do with the students learning experience or the student's experience on campus. And from the student's perspective, that's really frustrating, when that's all the professors care about. And that's what makes a difference between a good professor and a crap professor is one who cares. And I think, if you want a professor to care, oftentimes they're willing to make the effort that is needed to change their teaching style or do whatever it is that's necessary to have their kid interested and engaged in the same material that this professor loves. So you keep talking about diverting funds. But you never once mentioned diverting these funds away from this one teaching initiative to a different teaching initiative. So just keep that in mind . AUDIENCE: So the problem is that when we are looking at the proposal itself, the time, 2017, this is already the end of 2015. So that means you are going to do this, by 2017, that means you're going to do it by 2016. We're not against active learning in the [INAUDIBLE]. [LAUGHTER] [INTERPOSING VOICES] AUDIENCE: --the proposal. We are against-- we need these two group and against this proposal. So for this proposal, we are against this proposal, because you know how much it's about the funds and the resources. And that is why we are against the proposal. And also you pick whatever [INAUDIBLE], we're not even sure that it is going to work. JANET RANKIN: I'm going to take, because you guys actually did ask a question, so I'm going to count that to you Did you have another separate one? AJ, you look like you might have some questions. AUDIENCE: Yeah actually I was just curious, so if this initiative is really great, it looks like a lot of this research has been done since '80s. Why isn't it adopted? Why are we having this conversation now in 2015? AUDIENCE: I think we should look back at this issue of the lecturing system that we are using now. It's static. When there was no printing, lecturers then, they get the material. They reach the students. So lecturing is a Latin word that means reading. Now since then, now we are printing. We have information readily available that people can get on at their home and their own and develop to something. Too So there is a need for us to change to a new system, because the old system that more facilitated lecturing has changed. So then we have to change. JANET RANKIN: I'm going to turn it over to the deciders who now have to discuss. They've been resting for many minutes. So now they have to. So you have questions for each of the groups? AUDIENCE: Yeah. AUDIENCE: Yeah, thank you. Great arguments on both sides. I found it interesting, when you talked about the larger classes, so my first question is for this group. So how would active learning apply to a large class? The proposal is that it's for all courses. And so are there courses where this doesn't lend itself to? AUDIENCE: Even in the classes with 700, 800 students here, they still do active learning for the freshmen, i.e. Clicker questions. Oftentimes professors have to talk to your partners. So they'll pose a question. Everyone will click on their answer. And then they'll be like, now go back and talk to the person sitting next to you. And then you'll vote again in two minutes. So I think that's just one example of how you can engage a class of a lot of students. AUDIENCE: And to add to that, actually it's also been found out that when peers speak to themselves, they even learn the material better than when the expert, who may have lost touch to what it could be like to learn from a student's perspective. So peers speak to themselves and teach themselves even faster than the expert who is in front of them. AUDIENCE: Yeah I have comment for the against group. You mentioned that you don't require to create or active learning, because some students are very smart. They get it right. But don't you think that education is a right of every child? And they deserve to have a good shot at it. So why do you want to pull one child out of here, because he's not smart enough? AUDIENCE: I just to answer, I think we are pushing for the best education. We believe that is the traditional way. We think that active learning can happen outside of the classroom. This is the way it's worked for 300 plus years over at Harvard, one of the best universities, MIT for 100 plus years. We think that essentially it's working. And the reason we're having this conversation right now is because it has worked. So we want to continue that legacy. We think that every system is going to have problems. And if you analyze any system, you're going to find issues. And there are certainly some smaller inefficiencies within the lecture. But by and large, it's trusted. And it works. AUDIENCE: I want to add on what adding. So I think we are not pushing those less smart kids away, because they can actually learn from each other outside of class. We just want to say that doing the lecture is more efficient if the professor lecturing at them, giving all the information. As a footnote to that-- sorry-- if professors come to the class and just give theory and go away, and the smart kids get it, then how do you measure the professor's performance? How do you think about, say, something like students information on educational quality? How do you measure that? AUDIENCE: You don't really need the professor to be extremely active and do active teaching in the class. The full student can actually get what they need. You have a bunch of homeworks. You have a bunch of test questions, exam questions, that student, whether they like it or not, they will surely be forced to read those things. And by the time they come back, they come back learning. Lot of graduate from MIT, they are l really doing well. About 25% start up their own business. Probably like 70% get jobs when they finish. What are you talking about? The system is working. And That's what we're saying. And their value is a great. The evaluation is done on the type of textbook, the mode of lecture, the [INAUDIBLE] lecture. So these are the things that I try to evaluate in the evaluation process. And it has been great. And we want it to continue. AUDIENCE: So we addressed the measuring impact. And that goes for both. But for the active learning, how do we quantify how much active learning needs to go into these different classrooms in the department? So how will we ensure the appropriate amount has been implemented in different classrooms? Basically, how much active learning needs to go into a classroom to create change or to strengthen our student learning? AUDIENCE: How much has to go in or how you're going to make sure that the professors are doing it? AUDIENCE: Both, I guess, is the question. If there's a professor that's more on the lecture side that might throw one thing pair share in and call it a day. So how do we ensure the appropriate active learning techniques and the quantity are there for the classrooms? AUDIENCE: I think we'd have to go back to the research and see how much, whether it's 30% of the time needs to be spent on active learning activities, or to see what has been proven, because active learning has been proven to be effective. But we would have to go back and make sure that we set a standard that, if it is 30% of the time or 70% of the time. And I think oversight from the department would be important, in that the leadership in the department will have to sit in on classes. And as part of the training of the professors, there has to be training. But then there has to be someone to observe them in the class, give feedback. The training has to be ongoing. It's not just a three hour seminar on this is with active learning is. And this is what you should do. I think it's going to be a long term investment in the professor training that someone's going to have to check up and give them feedback. And then they can adjust from there. JANET RANKIN: I'm sure that we could [INAUDIBLE] all day. But I'm going to give the deciders two and a half minutes, three minutes, to talk to each other and come up with a decision. All right, the committee has met. And now they're going to determine your fate in this department. AUDIENCE: Does this mic work? Is this on? Do I need to use the mic. You can hear? Well thank you again for your arguments. Certainly the panel felt that the pro arguments in terms of the benefits of active learning are great. And we definitely support that. However, given the time frame, and this proposal is very aggressive, it would be extremely disruptive. Also the amount of funds that we think that it would take to implement this would be very high without any proven way to measure success. So what we are willing to do is commit funds to a pilot program, where we can then measure what the success will be and then make it a decision next year as to whether or not we invest for the rest of the departments. [INTERPOSING VOICES] JANET RANKIN: So I hope you found this exercise useful. I just want to debrief on the exercise itself, why we bothered to do it. It took probably 45 minutes of time to do it. But I made a decision that would be useful. So what did we get out of this exercise? What was the point? AUDIENCE: Whether you think that's it's worth it or not, it's really important time. Also, it seems this confrontation is happening a lot in the department, especially [INAUDIBLE] as they go through the program. I assumed this test would be debated quite a bit, internally and externally. JANET RANKIN: Other reasons we engaged in this? AUDIENCE: I think it gives everybody the opportunity to think deeply about the subject matter. And it also, even if you are against or for something, you have to have a deeper knowledge to be able to bring out the good points, certain issues that might actually make see [INAUDIBLE]. JANET RANKIN: [INAUDIBLE], you want to add to that? AUDIENCE: The topic is active learning. And what we just demonstrated is active learning in practice. JANET RANKIN: Yes, exactly. And the idea to Gordon's point, the idea that it can be very, very useful for you to have students take an arbitrary position and argue it, because they will learn things, they will think about things, that they would not have thought about otherwise. And it's usually not a good idea to say OK everybody that thinks this, go over here. Everybody that thinks this, go over here, because all that does is reinforce what people think already. And maybe it's not completely correct. So this is often a very good technique, when there's two clear viewpoints or two clear sides. And you want students to really engage with that. And exactly, we could have had a discussion. Or I could have had PowerPoint slides that said, here are the good things. Here are the pros with active learning, bum, bum, bum, bum, bum. Here are the cons of active learning, bum, bum, bum, bum, bum. And that would've taken-- just to come out of my neutral position here-- that would have taken probably four minutes. Instead we took 35 or so, 40 minutes. But I think you probably engaged with some of the ideas and concepts a little bit deeper, because you had to come up with it yourself. So yes, all those reasons. And it does come up often. People are discussing it. And it's not so clear cut. There's not a right answer. And you're going to want to say, OK, well we'll do it by 2019. Or we'll do it. But we won't do that. There's going to be deals and all those kinds of things. And that's OK. It's OK that there's not a right answer. But that was the point of this exercise. So that served as our discussion for active learning. But you guys did it. I didn't do anything. AUDIENCE: And that buttresses the point that active learning is the way to go. JANET RANKIN: Thank you. Yes, this is very meta. This is very meta here. Yeah and I do have to feel a little bit bad when you're like, what? Then we're going to do all the work. And the lecturer's just going to sit back and not do anything. I took it personally. But no, no, it's OK. One thing I wanted to bring up is if you haven't read the Scott Freeman article, it's in the reading list for today. It shows-- let me just bump to it-- it shows essentially these results. He did a meta analysis of 225 studies on active learning or on classes that that used active learning. And they showed a 12% decrease in the failure rate in classes that used active learning. And so I was telling the pro group that what say in the paper is that if it had been a clinical drug trial, and 12% of the people on the drug were having a better outcome than the people on the placebo, they would have had-- or the people on the control-- they would have had to stop the trial and give everyone the drug. So in the sense of why we're still lecturing, that's to me one of the biggest compelling reasons for why we should be doing more active learning. 12% of the students are going to pass the class, on average, normalized. And they actually considered both experienced lecturers, who were just lecturing, people that were considered to be good lecturers that were just lecturing, as well as graduate students who were first time teachers but were using active learning. So they looked at all sorts of-- they mixed up all sorts of outcomes. And active learning appears to, the data's pretty conclusive. And this isn't the only study that shows it. So there are definitely issues. There are definitely implementation issues. They're definitely sticking points. But it's something that I think we can't ignore. AUDIENCE: I just want to add that it's not only that we have 12% decrease in the failure rate. We also have big shifts in the letter grade students get. For instance, more people shift from grade C to grade B or from B to A. And B to see a test like that. So that's really significant also. JANET RANKIN: So I just want not have learning outcomes. But I did want to have the discussion before I even introduced them. So I think you'll be able to explain the impact of active learning exercises in the classroom. You've already started to do that. And then you'll be able to identify and develop active learning exercises that you'll want to use. And you'll be able to think about how you might use them. I think, pretty attainable learning outcomes. I'm going to show the slide a lot throughout the rest of the time. But remember, it's constructive alignment. We've seen this diagram before. But the beauty of some of these active learning techniques is that their formative assessments of students' understanding, of students' learning. And so they are actually activities that happen in the class. But they're also measures, informative, formative measures of whether students are learning or not. So if somebody had come up with a really terrible argument, or it was clear that nobody had read any of the papers on the active learning research, then I would know that from this debate. I know what you've read and what you haven't read, and what you're thinking about and what you're not thinking about. And I get a good sense for whether or not it you know it and how much and how well you know it. And so do you. And we'll see that a little bit more in these other situations, where if you get a clicker question wrong or you get a multiple choice question wrong or that you can't participate in the lightning round, then you know that you've got to step up to the plate. Or you need to try something different. Or you're not really getting it. So you get information. The learner gets information. And the teacher gets information. AUDIENCE: I just thought that was really powerful in the readings, as well. They mentioned that, that students who, when they get in the group sections, if it's clear they really are getting a lot of things wrong, they know they need to step up. And that's in a different way, I think, than getting a bad grade on a paper. And you know the average distribution or whatever. Or the professor says you're wrong. But when you know that the people sitting right next to you are getting it and you're not, that really, I think, will push the student forward to do something. JANET RANKIN: It's news you can use. It's information. And then you're not being penalized. You're not in trouble. It's not affecting your grade if you act on it, if you take the information and use it. So that's why, I think that's the beauty of it. There are many, many active learning techniques. And if you looked at the website from University of South Florida has just, I think, it's close to 200 techniques for what you might do. I'm doing a bit of a-- I call it the foie gras method. You know how they make foie gras? They force feed the goose. So I'm going to give a lot. And we're going to do a lot of different active learning things in this session. It's a little bit comical. I know it's kind of funny how many we're trying to cram into two hours. But I'm going to try to do that. So we'll do some of these. And we can talk about any one of them in particular. But a nice one to get students in the frame of your class is to they come in. And they have five minutes before you do anything to write down what they thought was the most interesting part of the reading or answer a certain question related to the reading or solve a problem or whatever it is. You give them a prompt. You sit down. You focus them. And then they're off. I do want to just focus on the word active for just a second. It may help you to think about active and interactive learning. There's all sorts of ways to chop it up. And at some level, it's semantics. But at some level, it isn't. So students can be active if they're sitting by themselves thinking about something and writing something down or solving a problem. That is active learning. They don't have to be talking to each other. You can do active learning in a class of 3,000 people. Stop and think. Answer this question. That's active learning. That's way more than happens in a lot of classes. Interactive learning generally implies that students are talking to each other. That's the most general categorizations you can get, active, interactive. So we're going to do a bunch of stuff with clickers or plickers, these things. But there's some other techniques you can do. You can put up a graph or an image. And you could say, OK, somebody come up and show me the green boundary in this picture. Show me a triple junction. Show me where the building, the crack in the building started. Show me whatever. And you pass the pointer. And then you have the student who has the pointer come up with another question and pass it to somebody else. So that's another nice technique that takes the responsibility off of the teacher. And so you're not picking on students. We've been doing mud cards, think, pair, share. We're going to do a lightning round. I'll talk a little bit more about jigsaw later. And if time allows, I'll show you a demo I did on diffusion using the students as the atoms. Sometimes if you're trying to ease in active learning, slowly into your class-- you don't want to do too much-- you want to think about the short time things that don't require huge disruptions. If you want to go back to your seats, you can. Maybe I should let you guys do that. So you know what we'll do? While they're taking a break, I'm going to demo the beach ball. So this is nice. This is the beach ball technique. So when you catch the ball, I'd like you to tell me if you've had any experiences with active learning in classes that you've taken. And you can't count this one. AUDIENCE: I had active learning in my electricity and magnetism class. And that was called TEAL. And it's very polarized. The people for and against it, especially the students. So they basically tables of nine. And then within that table, you had groups of three. And each of those would be a problem solving group. So you'd have four problems to solve per two hour class period. And that would be going up to the board look and doing it by themselves. And then the teacher at the end would reveal the answers and tell us how to go through it. And, for my group at least, there was always-- or my table-- there was always this one person who would write every single time-- JANET RANKIN: 3,000 People. Stop and think. Answer this question. That's active learning. That's way more that happens in a lot of classes. Interactive learning generally implies that students are talking to each other. And that's the most general categorizations you can get active, interactive. So we're going to do a bunch of stuff with clickers, or plickers, these things. But there's some other techniques you can do. You can put up a graph or an image. And you could say, OK, somebody come up and show me the green boundary in this picture. Show me a triple junction. Show me where the building, the crack in the building started. Show me whatever. And you pass the pointer. And then you have the student who has the pointer come up with another question and pass it to somebody else. So that's another nice technique that takes the responsibility off of the teacher. And so you're not picking on students. We've been doing mud cards, think, pair, share. We're going to do a lightning round. I'll talk a little bit more about jigsaw later on. And if time allows, I'll show you a demo I did on diffusion using the students as the atoms. Sometimes if you're trying to ease in active learning, slowly into your class-- you don't want to do too much-- you want to think about the short time things that don't require huge disruptions. If you want to go back to your seats, you can. Maybe I should let you guys do that. So you know what we'll do? While they're taking a break, I'm going to demo the beach ball. So this is nice. This is the beach ball technique. So when you catch the ball, I'd like you to tell me if you've had any experiences with active learning in classes that you've taken. And you can't count this one. AUDIENCE: I had active learning in my electricity and magnetism class. And that was called TEAL. It's very polarized, the people for and against it, especially the students. So we basically had tables of nine. And then within that table, we had groups of three. And each of those would be a problem solving group. So you'd have four problems to solve per two hour class period. And that would be going up to the board and doing it by themselves. and then the teacher at the end would reveal the answer and telling us how to go through it. And for my group at least-- or my table-- there was always this one who would write every single time. And I feel like that detracted from active learning. I think it worked better for other people. JANET RANKIN: It's definitely polarized. Students really are not-- I think we talked a little bit about this when we did some of the learning theories-- but students really are polarized about whether they like it or not. The data that there's been some pretty good analysis of learning, measuring how much students learn, and then how much they retain and all that kind of stuff. And when you go through that, students seem to learn more and retain more in the active learning format, even though they really at least half of them don't like it at all. So it's an interesting commentary That's a good great story. So why don't you give the ball a toss? There you go. No, you have to answer a question. Have you ever experienced active learning in a classroom besides this class? And how did it work or not work? AUDIENCE: I would say yes to that. It's definitely the traditional method, sitting in front, face to face lecture. JANET RANKIN: Fair enough. Give a toss, toss to anyone. Are we going to get the same answer from all of you? If it's just-- AUDIENCE: Is it the same question? JANET RANKIN: Yeah, same question. AUDIENCE: Yeah, I have. Because I recently studied in Sweden. And that's the way, most of the time that way, they teach in Sweden. So for some classes, generally in higher level classes, according with d-level classes, they use active learning. JANET RANKIN: Great and students buy into it? They're OK with it? They work at it? AUDIENCE: Culturally, they tend to veer towards active learning. JANET RANKIN: And that's a very interesting point, the idea that it can be a cultural thing. If everybody else is doing it, or it's expected, then that's what you do right. So some of these changes, there's an incubation period before they actually take off and they're widely accepted. AUDIENCE: In fact in Denmark, I think that's the only way they each. JANET RANKIN: It's pretty active. I have a colleague who's in, or friend who works at the University of Copenhagen, and all new faculty have to take a yearlong course on teaching, no matter what department, physics, math, a year long, the faculty. And here you could never force the faculty to do anything. So it's pretty interesting that they've been doing it for so long. AUDIENCE: I have a question about that. Is it the way that professors, their role as actual teachers is viewed? [INAUDIBLE] But seriously, people [INAUDIBLE] or something, I think definitely trying to enforce a teaching class would be a lot more receptive than if you tried to enforce that here, where teaching isn't necessarily the focus. JANET RANKIN: But there, it's a good research university. But on the teaching side, that's the norm. That's the expectation. AUDIENCE: But what about requiring-- I know it's hard when you have professors that have been doing what they're doing for 25, 30, 40 years. It's hard to get them to adopt new techniques. But I think that requiring things-- the faculty that you can require things of are the new faculty, who have no power yet. JANET RANKIN: Yes and no. If the tenured system, the requirements for tenure don't change, then asking them to take on new-- oh, you guys should try active learning. Revamp this course. And then they don't get tenure. So it really has to be a whole systemic change, where the department buys in. The school buys in. The university buys in and says, OK, these are the criteria for tenure. And in that is you have to use active learning. Or you have to-- that's where it would have to get change. Otherwise you can't expect people that aren't tenured to do it. You shouldn't. Let's give the ball one more toss. AUDIENCE: So it's been a long time since I've been in a classroom. I would say, in a regular classroom, I don't that I've been in a class that does active learning. But I have taken courses that are largely lab courses, like a mouse embryology course, where you have 15 minutes of lecture. And then you go into the lab And you do it. And the professor will walk around and help you with the technique. You're actually there with the mice. And you're actually doing something. So does that count? JANET RANKIN: Lab courses can be their own genre. But there has been this movement to take that's a little bit of what TEAL does, at a more simple level. And there's a little bit of lecture. And then students are supposed to answer some questions or do something. And then they come back together. And Studio Labs at WPI has that same format, where the classrooms are built with lab benches. And students sit at lab benches and see a lecture. And then they do some experiments. They're mini experiments. It's not whole mouse dissection. But they do a mini experiment. And then they report back. And there's an interchange like that. So there are some blending of the two. Often what people do in labs is different. And usually a good lab is fabulous. A good lab is fabulous, which is another argument for why active learning is good, because everybody thinks back about, oh, that lab course was so great. I learned so much and did so much. And yet that's the lab course. And then we have the lecture. So it's an interesting comment. So that was the beach ball. And you see you guys got to throw the ball at other students. So I didn't have to pick on every-- I didn't have to call on people or pick on people. I had a little demo here about why it's important, just if you ask a question or you ask students if there are any questions or what questions they have. And that's generally a better phrase. What questions do you have? It's more welcoming, opposed to saying OK, who's stupid enough. Who's not following me? Say, OK, what are the questions? It's a more welcoming phrase. But the idea of waiting. So here, if I click on this little guy, you should hear 30 seconds of music. So imagine you ask a question, 30 seconds for people to think about it and formulate an answer. This is what 30 seconds feels like of dead time. Well, there'll be music. [MUSIC PLAYING] That's not part of it. So did that feel like a long time? I know we had a little break there. But that feels like a lot. Especially when you're standing up here, it feels like a long time. But 30 seconds, give them 30 seconds. The point is, if you want to give students 30 seconds, you're going to have to time it. You totally have to time it, because you will not estimate 30 seconds effectively. You will totally underestimate it. And then another technique that you can piggyback on that is saying, OK, take 30 seconds. I'm not going to take any answers for 30 seconds. So you may have an answer right away. But I don't want to hear it for 30 seconds, because there's always that guy in the front. I had this guy in my high school math class. And he used to sit on his-- this little short guy-- and he used to sit on his foot on the chair. He's on his chair. And he's sitting on his foot. And he had his hand like this. And he was in the front row. And whenever the teacher asked the question, he would go. And he would come up off of it. And so he was always the guy that answer the questions. But that left some of us who took a little more time not having the opportunity, kind of like having the one person write all the time in the TEAL group. So that's a good way to go about that. We talk a little bit about using clicker questions to solicit feedback. And we've used the plickers in this class. We'll use them again. You can get dedicated clickers, these dedicated punch in things. They use them for TEAL, I think. Students may have to purchase them. That may cost money. The department may have to purchase them. That's another source of money. AUDIENCE: They make us buy everything. JANET RANKIN: So when I was at Brown, actually, the department would buy them. And they'd leave them in the library. And then you could check them out with your student ID. And then you got that. So you got it for the semester. But it was linked to your ID. And if you didn't bring it back, they'd charge you for it. But that's another way to do it, a fairer way to do it. We'll look at something called Socrative later on. I'll use it later on. But it's a text. It's an online format. So students need a smartphone or a laptop. But they don't need a separate clicker. There are issues with that, because not everybody is going to have a smartphone. And are you biasing your class against people that don't happen to have the technology? This is why I like plickers, because you don't need to buy anything. I give you the card. And I need to have the smartphone. But you don't. And then you could always use index cards or do the whole, put one finger up if you think it's A and two if it's B and three if it's C. That gets a little tough in a bigger class, because it's hard to see. It's hard to count. I like the fact that the clickers, plickers, and the Socrative or Poll Everywhere give you a histogram. So you get a really quick idea about who's getting it and who isn't getting it. So we're going to do a couple of these. Remember-- this is from the label of the free body diagram-- that these misconceptions always linger in a lecture class. We talked about this free diagram the other day. So let's just try this one here, give you another little warm up on the picker's. So you have the card. And I'm just going to ask you to look at the question. And when you're ready to answer, just hold up your card. And if you can keep your finger away from the pattern, that would be great, because it makes it hard to read it. And then however you want to-- AUDIENCE: Does this work for larger classrooms? JANET RANKIN: It does. I have a colleague who uses it at Brown. And he just started teaching this semester. And he showed me his-- or a friend-- he showed me his plickers. And he printed them out on 8 by 10, like a sheet of paper, cardstock. And they're enormous. They're like this big. AUDIENCE: How do you not get [INAUDIBLE]? Well Well they have to hold them up. You have to. And you can walk around a little bit. But I'll show you this. So I have the histogram. And I measured everybody. So now I know what you think about this question. I know who's got it right. I don't know who's got it right. But I know that most of you have chosen the correct answer. But there were a few of you picked up one of the other answers. And so now I know that it's a non-trivial number of you that picked an incorrect answer. So now I know exactly what I need to do. It's given me incredibly good information that there is some out there that have a pretty strong misunderstanding of the literature. And that's that some of you picked A. There's a lot of data to support that active learning, active technique supports student learning. I call your attention to the Freeman paper that's in the reading for today, also to the [? Haik ?] article, which was referenced. So those are two of the more compelling readings. But if you go through, you'll see quite a few studies that unambiguously show that active learning has an impact. So the incorrect answer is that it always involves group work. That's not true. It doesn't always involve work. As we said, you could just ask students to sit and think and then do something with their thoughts. And that would be active. So Rachel, you're smiling. Is there a-- OK. AUDIENCE: Sometimes my face just does that. JANET RANKIN: I didn't know whether it was a smirk or whatever. And did I show you what this looks like? So you can see with the histogram looks like. I did? [INTERPOSING VOICES] You weren't here, though, when I did. The time I used it before. I'll show you offline. So the histogram is really, really useful. So it really can ferret out some misunderstanding. We don't have to go through all these. But this is another way to check for misunderstanding. I can ask this kind of question. It's a concept question. It's a conceptual question. And then I can ask you to pick an answer. And then I can see where your misconceptions lie about this topic. So it can be very, very-- it should be very, very focused. It's not just a busywork question it's a focused question with focus distractors. The distractors are the incorrect answers. AUDIENCE: Can we do this? I want to see what the spread of answers is. JANET RANKIN: Sure. Let's do it. AUDIENCE: They asked this question-- I laughed, because they asked a very similar question at Harvard graduation. JANET RANKIN: Exactly. What was it called, a world of their own? AUDIENCE: Yeah one of those. JANET RANKIN: Two times, check. I've got to clear my responses. Let's go. So I'm getting you folks in the back without a problem. I got one. I'm missing one of you. Wow. So there's any number of things I could ask. We're going to do another pair, share around this kind of thing. But maybe not for this one. But this is amazing. Basically it's A got three votes. B got three votes. C got four votes. And D got three votes. So it's pretty amazing. AUDIENCE: What is the right answer? JANET RANKIN: Biologists? AUDIENCE: B. JANET RANKIN: The answer is B. And it's the most-- maybe not surprising-- but the idea that it's taking in things from the air. They're taking in gases. But there's chemical reactions which manifest themselves in a weight gain. [INTERPOSING VOICES] It's not a biology class, which is why-- so that's just super telling. That would be incredible. If this were a biology class and that was the spread, that would just be really, really informative for the instructor. AUDIENCE: Really worrying. JANET RANKIN: Well yeah, that too. But it's biology class. So don't worry. But you do want to think very, very carefully about the wrong answers, because you want the wrong answers to tell you something about what students understand or don't understand. Generally speaking in the adoption of clickers, plickers, whatever, takes a three step process. One is that people generally start with these simple fact questions, and then more challenging conceptual questions, which I would put the one we just did at. That students, they might-- arguing with each other could be useful. And then the third is you've just totally revamped the lecture. And it's all centered around these questions. And depending on how students answer, that's how you navigate through the lecture or through the course, class period. I would never recommend that anybody start with this one. I think that we're all capable of starting with two actually. And I think starting with three may be a waste of time, unless you really just want to check and see if people did the reading or something. It's a big class. And you just want to see if they've done the reading. But in general, I would say put your energy into coming up with some really good questions. And maybe don't ask as many. And then as I keep saying, I hope you can start to see how that's really, really linked. It's a measure of whether they get it. But it's also an activity to help them get it. So I've been minoring the Back Channel. Nobody's saying anything except for Dave. And I answered him. And because I gave you a break before, I may scoot on, continue on. So if you want take a second and login, that's fine. But I may not-- AUDIENCE: That's the link for last year's. AUDIENCE: I entered it, and I put entered into last year's. And I got really confused. JANET RANKIN: Uh oh. So the one for this year is back here. Hold on. AUDIENCE: It's like four something, 4125 instead of 3969. JANET RANKIN: Oh OK, thank you. So while you're doing that, I'm going to set up something which is called lightning round. And we're doing these two. So I'm going to just, sitting in your seat, if you can think about this question, just the question on the left. Based on research at NASA, what was the approximate net global change in temperature between 1880 and 1975? It's not a plicker question. It's just a question. Try to think about what you think the answer is. AUDIENCE: What is Celsius to Fahrenheit? JANET RANKIN: To get from Celsius to Fahrenheit? So you have to-- well I always just double it and add 30. So it's 8/5 plus 32. [INTERPOSING VOICES] 8/56 plus 32 is the real equation. Is it 8/5 or 9/5? AUDIENCE: 9. JANET RANKIN: 9/5 That's why I just double it. Close enough. So what I'm going to ask is that-- let's see-- David, Gordon, and Julie. And if you can move down, down, just to the end. And then if Adam, [INAUDIBLE], Dave can come around? Just keep going. Keep going. Keep going. That's fine. One, two, three, four-- no, no. You have to stand up, guys. No relaxing. Yes, but that's not-- I can tell a story. Please stand up and move over. Move over. And just keep coming around. Everybody's up. Everybody's doing this. Michelle, so I need six, Michelle come on over here next to [INAUDIBLE]. That's fine. Now [INAUDIBLE], you come around here please, and go opposite David, please. And everybody else just file in and go opposite someone. Sarah, do you want to participate? So here's what happens. I'm going to tell you to go. And you have two minutes. The person in this line takes some amount of time and says what their answer is and why they chose it. And the person in that line can ask questions or whatever, and then say what they thought and say why they thought that. So you're going to have a really quick conversation about your choices. And after two minutes, I'm going to tell you to please stop. And then you have to get a new person. And I'll show you how to do that. Everybody understand what we're doing? You're having a two minute exchange of ideas. And it's going to be loud. Believe me. And it's the person opposite you. Are you ready? Everyone ready? On your mark, get set, go. [INTERPOSING VOICES] Let me ask a question. Did you encounter people that had different choices than you? AUDIENCE: Yes. JANET RANKIN: Yes. So that generally means it's a pretty good question, because you get to hear lots of different perspectives. Now if this were the central part of the class, this topic, you might let the pairings go even longer, even more pairings. So did anybody change their mind based on what they heard a partner say? OK, a couple of you. Generally, what happens then is you would want to have a de-brief, where we talk about why you picked the answer you picked and what you picked. And most of the time, the choices are a little been informative. If you found, as the instructor, that everybody picked a negative number, well you're in a very different place than if people are arguing over how much positive it is. So you're going to use that information to find out about where the misconceptions are. What's the advantage of this method? Gordon. AUDIENCE: It's very interesting, I thought. One I found out is that the students talk to each other. We never talk to each other. And it's easier for them as to [INAUDIBLE]. So at the beginning, they are aware of conflicts. But as we rotated, there was a convergence. It was very interesting. JANET RANKIN: [INAUDIBLE] have your hand up. AUDIENCE: Yes, one time I was laughing just because it was very effective. There was so much [INAUDIBLE]. And when we started, one resisted some passionated idea about this is how it should be. And by the time it went to all the people, people can get all the idea. You try to look for the superior idea. Then you decide, oh, this is better. When we try to switch again, you listen to a better argument. So from which one decides without [INAUDIBLE] the first one? JANET RANKIN: A couple of points. One is that it's impossible not to-- when you pair people up, it's pretty hard for somebody not to participate. So if you have somebody that's resistant, they might still be resistant. But it's virtually impossible for them not to participate. And also it's a little less scary. One person, it's really noisy. You know you only have to talk to them for a minute or two. So it's a little less scary. Everybody's doing it. The other thing is that you definitely want to have some sort of de-brief. You can't just stop and go, OK, let's go on. You're going to have to find out what people's answers are. And you're going to have to address it or at least reveal the right answer, because there's many places many, many places for misconception. And also, then, if you do it enough, you can maybe-- students can find arguments or people's explanations that resonate with them. And then you get exposed to different kinds of arguments or different kinds of ways of explaining the problem. So in this case, the answer is actually positive 0.4 degrees. So B is the correct answer, just so you know. AUDIENCE: Yes, I just want to make a comment that sometimes this method could be confusing as well. For instance, if you listen to two superior arguments, one against. For instance, I talked to someone who was actually in support of negative. And she brought an argument and so forth like that. And somebody thought of positive. And there was a strong argument also for that. Sometimes if you just be sitting on the fence, I don't know where to get off. JANET RANKIN: Well I think it's OK to be confused for a while. That's fine. As I said, if this were a course on global warming, climate change, whatever it is, we would have to make sure that we heard all of those explanations or a sampling of those explanations and said, oh, well that's a good. I can see why you think that. But you've forgotten about this. Or I can see why, so that you can process it all so that people don't leave confused. But it's good to hear the wrong answer, to be convinced of the wrong answer, maybe, but then to see why it's wrong. That's an important learning activity. AUDIENCE: But this method is only applicable for where you have cases where you can debate. Because if you have very clear cases, for example, if I ask, what is the color of the sky? It's a definite. But when it's fuzzy, then this can be-- JANET RANKIN: Yes, or when it's a concept you think people might have difficulty with. Yeah, it's not going to work if it's a real clear cut. I want to demo one more method. This is think, pair, share. And I know we've been doing those forever. But I'm going to couple it with a clicker question. And I think we can do it in seven minutes. So I have two blocks here, because this is just too fun, really. I have two blocks. They're the same width and the same. Length but they are different thickness. And I have this thing. And I'm going to tap on the thin block. Makes more or less-- it's not really a note. But it's some sound. And I'm going to ask you, when I do the same thing, the same force, the same place on the thick block, will the note be higher, lower, or the same as the thin block? And this should be A, B, and C, not one two and three. So but if it's A, B, and C go ahead and vote with your plickers. Do you know that's what they auto correct to? That's what it, if you're typing and you write plickers, it autocorrects to pluckers, which can be pretty embarrassing. AUDIENCE: So high is for A. You've not clicked on the thicker one. JANET RANKIN: I know. That's why you have to tell me what you think it is. AUDIENCE: Is this a A, B, and C? JANET RANKIN: A B and C. So there's no D. There's a reason for this. If I told you the answer, then there wouldn't be much fun in voting. And I have everybody. So I have a distribution. I'm not going to tell you the distribution. I'm not going to share that with you I'd like you to take-- just cluster. It could be groups of three or four, whenever you have nearby, and talk about why you picked the answer you picked. And then I'm going to give you the opportunity to vote again. So in the interest of time, two minutes to discuss this. [SIDE CONVERSATIONS] I wanted to just call attention to a couple things I did. I walked around and tried to eavesdrop on your conversations so that I could tell what you were thinking about, but without putting you on the spot, without asking you to say, in front of everybody, what you thought. So that's one thing to do. Now I have a sense for your mental models, how you're thinking about the problem. So I have some really good information about why you think the answers you think. What I'm going to do is I'm going to give you the chance to re-vote. If you want to change your vote, you may. Or you don't have to. Some of the clicker software actually lets you save consecutive histograms, which can be useful. You can compare if there's change. This doesn't let you do it. So I just have to-- AUDIENCE: Voting twice there? AUDIENCE: I'm voting for [INAUDIBLE]. He's not present, so-- JANET RANKIN: Voting in Chicago, early and often. We have exactly the same distribution, which flies in the face of some other data I was going to show you. But that's OK. I'll just in the interest of-- so one thing I could do is I could call on you. And I could say, OK, if you said it was lower, tell me why you thought it was lower. And I could get some arguments for why you thought it was lower on the board. So some people think of cellos make lower sounds than violins. And some people think that doors on mansions have a low sound. And other people talked about xylophones and the length and the wavelength of the xylophone, and the longer wavelengths for the longer pieces. So I know a lot about how you're thinking about this problem. The other thing I know is that I am not telling you the answer yet. So there's a little bit of suspense in here. There's a little bit to try to keep you interested and wanting to know, wanting to understand the answer. AUDIENCE: Janet? JANET RANKIN: Yes? AUDIENCE: So you said we had the same distribution. But can we have a show of hands for how many people changed their answer? JANET RANKIN: That's a great question. Yes. Raise your hand if you changed your answer. It could've been amazing. AUDIENCE: So sometimes we've done that in my physics class. When they ask, who changed their answer? And everyone raises their hand. It's the same exact distribution. Or everyone goes from A and B to C and D. And even C and D originally only had three people, you'll see the entire shift. And it turns out A and B was right. When we vote the second time, we all picked another answer. JANET RANKIN: That's very funny. That could have happened. Sometimes a nice follow up question is how confident are you in your answer. So then that's a really good-- because you may have 80% of the people h the right answer but if you say how confident and hardly anybody is very confident then, you do need to talk about it some more. So it gives you some really, really good information. So here we go. Here's this one. It's higher. Exact same material. AUDIENCE: So when you play the marimba and the xylophone, you have bigger bars. That they get really thin when they're on the bass. The bass notes are really thin. And the really small high notes, they're really, really thick. JANET RANKIN: So that's one thing. The other thing is that on a xylophone, it's a standing wave, because you fixed the endpoints. And the wave just vibrates based on the length of the pins, the separation of the pins or the mounts. So in this case, it's a free vibration. So that's another different misconception. And then if you think-- so the easiest way to think about it is a diving board. If you had a thin diving board, and you went and jumped on it, it would go way down and come back up again. That's a long wavelength. And that's a low note. And then if you think about a stiffer diving board that you'd be kind of like, mmm. That's a shorter wavelength. So it's different. You have different-- and then any kind of a cavity, like a cello or a door of a house, if it's a big cavity, it's a bigger standing wave. It's still a standing wave. So those are very different. They are different models. So this technique is really great, because it really gives you incredible insight into what students' misconceptions are. And then you can really make sure to address those misconceptions to deal with their faulty models. So for the post-class assignment, I've asked you to think about what active learning techniques you might adopt in your own teaching, in your own classes. And then think about how you would adapt them, anticipate any issues or whatever. So hopefully you've gotten a good smattering. And I'll post the slides, which have some links and things in there if you want to look at videos. And also check out that list from University of South Florida. That has some amazing techniques that you might find useful. I'm going to send everybody off. But if you'd like to comment on the mud cards-- I know we're over time-- I have them here, if you want to make a comment. But otherwise, I will see you next time. And I'd like the plickers back.
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https://ocw.mit.edu/courses/8-962-general-relativity-spring-2020/8.962-spring-2020.zip
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[SQUEAKING] SCOTT HUGHES: Good afternoon. So we spent our last lecture laying out some of the basic foundations, making a couple of definitions. I want to quickly recap the most important concepts and definitions. And then, let me be blunt, I kind of want to get through these definitions, which I think it's important to do them precisely, but there's nothing significantly challenging about them. We just need to make sure they are defined very precisely. So now that you've kind of seen the style of these things, I would like to sort of move through the next batch of these definitions quickly enough that we can start to move into more interesting material. So a quick recap-- and my apologies, my daughter has some kind of a virus that I am desperately trying to make sure I do not catch. And so I will be hydrating during this lecture. So I just want to recap some of the most important concepts we went over. So this whole class is essentially a study in spacetime. Later, we're going to connect spacetime to gravity. And general relativity is going to become the relativistic theory of gravity. So we began with a fairly mathematical definition. Spacetime is a manifold of events that is endowed with a metric. Manifold, for our purposes, is essentially just a set in which we understand how different members of the set are connected to each other. Events are really just when and where something happens. We haven't precisely defined metric yet. We will soon. But intuitively, just regard it as some kind of a mathematical object that gives me a notion of distance between these events. I tried and I will continue to try to be very careful to make a distinction between geometric objects that live in the manifold, the events themselves with things like a displacement vector between two events, other vectors, which I will recap the definition of in just a moment. They have an existence and sort of a reality to them that is deeper and more fundamental than the representation of that object. So when I say that the displacement vector is delta t, delta x, delta y, delta z, that is according to observer O. And when I write that down-- I will be very careful as much as possible. I will occasionally screw this up-- but I will try to write this down without using an equal sign. Equal sign implies a degree of reality that I do not want to impart to that representation. So an equal dot with a little o here, that's my personal notation for delta x is represented by these components according to observer O. And for shorthand, I was sometimes just write this by the collection of indices delta x alpha. A different observer O bar, they will represent this this factor using the exact same geometric object. They all agree that it's this displacement between two physical events in spacetime. But they assign different coordinates to it. They give a different representation to it. And we find that representation using a Lorenz transformation. And I'm not going to write out explicitly the Lorenz transformation matrix lambda. I gave it in the last lecture. And I'm assuming you're all experts in special relativity, and I don't need to go over that. So I will then using this as sort of the prototype, the general notion of a vector in spacetime, which we'll often call a 4-vector for the obvious reason that it has 4 components, I'm going to treat that as any quartet of numbers that has transformation properties just like the displacement factor I just went over. So if there's some quantity A that has time like an x or y and z component, as long as a different observer, for reasons having to do with the underlying physics or whatever the heck this A is, as long as that different observer relates their components to observer O's components via the Lorenz transformation, you got yourself a 4-vector. Any random set of four numbers, that ain't a 4-vector. You need to have some physics associated with it. And the physics has to tell it that it's a thing that's related by a Lorenz transformation. So I want to pick up this discussion by introducing four particularly important and special vectors, which to be honest, we're not going to use too much beyond some of the first couple of weeks or so of the course, but they're very useful. And one should often bear in mind, even later in the course when they've sort of disappeared, that they're coming along for the ride secretly. And these are basis vectors. So if I go into frame O-- I'm just going to pick some particular reference frame-- I can immediately write down four special vectors. So remember, this is a Cartesian type of coordinate system. So I'm going to introduce e0. I'm going to represent this by just the number 1 in the time slot and 0 everywhere else. e1, or ex if you prefer, I'm going to write that down like so. And you can kind of see where I'm going with this. The analogy, if you've all seen unit vectors in other classes, hopefully this is fairly obvious what I'm doing. I'm just picking out a set of, you know, little dimensionless simple quantities that point along the preferred directions that I've set up in this inertial reference frame. A compact way of writing this-- so notice, I have four of these vectors. And these vectors each have four components. And so what I can say is that the beta component of unit vector e alpha is representative, according to observer O, by delta alpha beta, the Kronecker delta. If you're not familiar with this one, I'm usually reluctant to send you to Wikipedia. But in this case, I'm going to send you to Wikipedia, Wiki via Kronecker delta. Yeah, so what that does is it just emphasizes that at least in-- by the way, I should put little o's underneath all of these things, because I have chosen this according to observer o's representation. So this is just saying that a coin to observer O, these are four very special vectors. The utility of these things-- up high for everyone in back to be able to see-- the utility of doing this is that if I now want to write the vector A as a geometric object, I can combine the components that observer O uses with the basis vectors that observer O uses. And I can sum them all together. I can put them together. And then I've got the-- it's not a representation. That's a damn vector. OK? I put it all together using sort of an internally consistent set of numbers and basis vectors. And so I am free to say this, where I actually use an actual equal sign. You might stop and think, well, but those aren't the components of the observer O bar would use. And you're right. Observer O bar would not use those components. They would also not use those basis factors. We're going to talk about how those things change. But observer O bar does agree that if they were handed O's components and O's basis vectors, this would give me a complete representation of what the vector is. OK? So again, I'm really harping on this sort of distinction between the geometric object and the representation. This is the geometric object. And we can take advantage of this. The fact that that combination of things is the geometric object is a tool that I'm going to now-- is a fact that I'm going to exploit in order to figure out how my basis vectors transform when I change reference frames. So let me just repeat what I wrote of there. But-- oops-- good point for just a slight editorial comment. When you're talking and writing and there are millions of little sub scripts and indices, sometimes the brain and the appendages get out of sync with one another. I caught that one. I don't always. OK? So if you see something like that and you kind of go, um, why did that alpha magically turn into a beta, it's probably a mistake. Please call it out. OK? All right, so this is how I build this geometric object, using the components and the basis vectors as measured by O, as used by observer O. Let's now write out what they would be if they were measured by a different observer, an O bar observer. I know how to get components, the barred components from the unbarred components. I don't yet know how to get the barred basis vectors from the unbarred basis vectors. But I know that they exist. And that once I know what they are, this equation is true. OK? These are just two different ways of writing this geometric object, which every observer agrees has its existence that transcends the representation. So let's rewrite what I've got on the right-hand side of the rightmost equal sign here. I'm going to write this as lambda mu bar beta A beta e mu bar. And then now I'm going to use a trick, which is what the-- it's not even so much a trick, but I'm just going to use a fact that is great when you're working in index notation. So often when students first encounter this kind of notation, your instinct is to try to write everything out using matrices and things like row vectors and column vectors. It's a natural thing to do. I urge you to get over that. If I get some bandwidth for that, I'm going to write up a set of notes this weekend showing how one can translate at least 2 by 2 objects-- two index objects and one index objects in a consistent way between matrices and row vectors and column vectors. But we're rapidly going to start getting into objects that are bigger than that, for which trying to represent them in matrix-like form gets untenable. In a little while we're going to have a three index object. And since we don't have three-dimensional chalkboards, making this sort of matrix representation of that is challenging. Soon after that, we'll have four index objects. And we'll occasionally need to take derivatives of that four index object, giving us a five index object. At that point, the ability to sort of treat them like matrices is hopeless. So really, you should just be thinking of this as ordinary multiplication of the numbers that are represented by these components as written out here. And an ordinary multiplication like this, I can just go ahead and swap the order of multiplication very easily here. So what I'm going to do is move the-- hang on a second. Yeah, yeah, yeah, now I see what I'm doing. Sorry about that. I misread my notes. I'm just going to move the A onto the other side of my lambda. And then I'm going to use the fact that beta on my right-hand side is a dummy index. So in that final expression I wrote down over there, beta is a dummy index. And I'm free to adjust it to put it into a form that's more convenient for me. So let's begin. Let me just write where I've got this equation right now over there. So A alpha E alpha equals-- and what I've got then is component A beta lambda mu bar beta-- keep mu bar. Sorry about that. So I'm going to remap my dummy index. The reason I did that is I can now move this to the other side and factor out the A alpha. OK, everyone can see that I hope. So see what I did was I arranged this so that I've isolated essentially only the transformation of the basis vector. So this equation has to hold no matter what vector I am working with. It's got to hold for an arbitrary A alpha. So the only way that can happen, this means that my transformation of basis vectors obeys a law that looks like this. Now, on first inspection, you're going to go, ah, that's exactly what we got for the components of the 4-vector. Caution-- I'm going to remind you, it's actually on the board over there. OK? The barred component is playing a different role than the barred basis vector here. If you want to get the barred basis vector from the unbarred basis vector, you need to work with the inverse of this matrix. That's what this is telling us here. OK? All that being said, if you're just working through this and you've got your components set up and you're sort of hacking through it, the algorithm for you to follow is actually simple. Really, all we're doing is lining up the indices. We're summing over the ones that are repeated and requiring that those that are on both the left-hand side and the right-hand side appear and equal one another. Or as an old professor of mine liked to say about 12 times a lecture, line up the indices. So that's essentially what we're doing. In this case, I line up-- if I have my Lorenz transformation matrix with the barred index up top and the unbarred down below. Boom, I line up the indices. And that tells me what the unbarred basis components are from the barred ones, and vice versa for the components. So what basically, as I just said, that tells me if I actually want to get this guy given this guy, I need to work with the inverse. Put this up high so that everyone in the back can see it. So this, again, is one of those places where you might be tempted to sort of write out a matrix and do a matrix inversion. But before you do that, remember this is physics. OK? The inverse is going to be-- the inverse matrix is going to be the one that does, at least for certain quantities-- what the Lorenz transformation does is that it relates objects, if I'm at rest, if I consider myself at rest, it tells me about things according to a frame that it's moving with respect to me. The inverse matrix does the opposite. It would say to that person being at rest, what is the matrix that tells them about things according to me, and they see me moving with the exact same velocity, but in the opposite direction. So to get the inverse matrix, to get the inverse Lorenz transformation, what we end up doing is we just reverse the velocity. So I've just wrote down for you-- and it's worth bearing in mind, every one of these lambdas that is there, they are really functions. And so my e alpha that I wrote down over there, I'm going to use an under tilde to denote a 3 vector. In one of your textbooks, it's written with a boldface. But that's hard to do on a blackboard. So if I want to go the other direction, well, I just need to have the inverse transformation. And I have that. Bear in mind, I mean, those are really the exact same matrices. OK? In terms of the function that I'm working with here, I'm just flipping around the direction in order to get these things out. OK? So, as I said, you might be tempted just to go ahead and do the matrix inversion. Let we just do a quick calculation to show you that that would work. And the reason I'm doing this is I just want to quickly step through a particular step, which is, again, sort of in the spirit of swatting mosquitoes with sledgehammers, it's the kind of analysis that you're going to sort of have to do off to the side now and again. So given that I know e alpha is-- let's see, let's use beta-- e beta bar be e beta bar. But I now know that I could write this guy as lambda-- let's use a gamma-- beta bar minus v in the gamma direction. Now, you look at that, and you go, ooh, look, I'm actually summing over the betas there. Let's gather my terms a little bit differently. So notice, I have unbarred basis vectors over here on the left, unbarred ones over here on the right, a bunch of junk in these braces here in the middle. The only way for that to work is if after summing over beta, that bunch of junk is, in matrix language, we'd say, it's the identity matrix. In component notation, we're going to call it the Kronecker delta. Just as little aside, if I started with the barred on the left side and the unbarred in the right-hand side and did a similar analysis, it would take you at this point now that you are all fully expert in this kind of index manipulation, it should take you no more than about a minute to demonstrate to yourself that you can get a Kronecker delta on the barred indices in a similar way, just by using them in a slightly different order. OK? All right, so far all still basically just formulas. So I'm going to start now doing a little bit of formalism that will very quickly segue into physics. We can all take a deep sigh of relief and go, ah, OK, something you can imagine measuring. So that's nice. So I have introduced 4-vectors. I haven't introduced the basis vectors and their components. I haven't really done anything with them yet. So before we start doing things with them, let's think about some operations that we can do with 4-vectors. So the first one which I'd like to introduce is a scalar product. To motivate the scalar product that I'm going to define in the same way that I defined 4-vectors as a quartet of numbers whose transformation properties are based on the transformation properties of the displacement, I'm going to motivate a general scalar product between 4-vectors by a similar kind of quantity that is constructed from the displacement. So let me recall a result that I hope is familiar from special relativity. So working in units where the speed of light is 1, you are hopefully all familiar with the fact that this quantity is something that is in variance to all observers. I did not use a represented by symbol there, because no matter whose delta t, delta x, delta y, delta z I use there, they will all agree on the delta s squared that comes out of this. If I want to-- actually, let me write of a few things done before I say anything more. So this is an invariant. This is the same-- let me actually write this in a slightly different way-- it is the same in all Lorenz frames. So pick some observer, get their delta t, delta x, et cetera, assemble delta s squared, pick a different observer, do the Lorenz transformation, assemble their delta t prime, delta x prime, et cetera, make that, boom, they all agree. So we're going to use this to say, you know what, I'm going to call that the inner product of the displacement vector with itself. So I'm going to call this delta x dotted into delta x. And so what this means is that built into my scalar product-- so if I write this as a particular observer would compute it-- this is the scalar product that I'm going to define with respect to the displacement vector. And this is usually the point where somebody in the class is thinking, why is there a minus sign in front of the timelike piece? I can't answer that. All I can say is that appears to be-- more than appears to be. There's a whole frickin' butt load of evidence-- that's not how nature is assembled. OK, it's connected to the fact-- so the fact that all of my spatial directions are sort of entered with the same sign, but my timelike direction, it has a different sign. It's connected to the fact that I can easily move left and right, front and back. It's a little bit of effort. I can move up and down. But I cannot say, oh, crap, I left my phone at home I'll go back 15 minutes and pick it up. You cannot move back and forth in time. Time actually, which is the timelike component of this thing, it enters into the geometry in a fundamentally different way from the spatial things. And that's reflected in that minus sign. Anything deeper than that, let's just say that we're probably not likely to get very far with that conversation. Depending on what kind of muscle relaxants you enjoy using on the weekend, you might have some fun conversations about it. But it is not something that you're really going to get very far with. You just kind of have to accept that it's part of the built-in geometry of nature. OK, so this I am defining as the inner products of the displacement vector with itself. I define vectors as having the same transformation properties as the displacement vector. We can similarly define an inner product of a 4-vector with itself. OK, now, I'll put this on another board. So A dot A, I'm going to define this-- or rather I will say it is represented according to observer O as minus A0 squared plus A1 squared plus A2 squared plus A3 squared. I realize there can be a little bit of ambiguity in the way I'm writing it here. You just have to-- if you're ever confused, just ask for clarification about whether I'm writing something to a power or whether it's an index label. Context usually makes it clear. Handwriting sometimes obscures context though. The reason why I'm doing this and a real benefit of this is that whatever this quantity is, because A has the same transformation properties of the displacement, this must be a Lorenz invariant as well. The underlying mathematics of Lorenz transformation doesn't care that I wrote a zero instead of delta x here. It doesn't care that I wrote A1 instead of delta x1, et cetera. It just knows that it's a thing that goes into that slot of the Lorenz transformation. So all observers-- this is how I represent it according to observer O. But all observers agree on that form. And as a consequence, this is going to be something that we exploit a lot. Even when we move beyond the simple geometry of special relativity, a generalization of this will be extremely important for, not an exaggeration to say, everything. So a little bit of terminology-- so if A dot A is negative, and depending upon which is bigger, A0 squared or the sum of the other ones-- it could very well be negative-- we say that A is a time like vector. This traces back to the fact that if A were the displacement vector, if the displacement, the invariant interval, were negative that would tell me that the two events, which are the beginning and the end of the interval, I could find a frame at which they're at the exact same location and are only separated in time. In the same way, this is basically saying that I can find a frame when this vector points parallel to some observer's time axis. If this is positive, A is spacelike. Everything I just said about timelike, lather, rinse, repeat, but replace time with space. OK? And if A dot A equals 0, we say A is-- there's two words that are commonly used-- lightlike or null. Null just traces back obviously to the zero. Lightlike is because this is a vector that could lie tangent to the trajectory that a light beam follows in spacetime. OK? OK, let me get some clean chalk. So, so far, I've only talked about this inner product, this scalar product. Oh, and, by the way, I'll use inner product and scalar products somewhat interchangeably. But this allows me to reiterate a point I made in Tuesday's lecture. When I say scalar, scalar refers to a quantity which is-- you know, it doesn't have any components associated with it. So in that sense, it's familiar from your eager intuition of, you know, not to vector. But it has a deeper meaning in this course, because I also want it to be something that is invariant between reference frames. So A dot A is the scalar product. It gives me a quantity that all observers agree on. Now, I've only done scalar products of vectors, A and the displacement vector, with themselves. So a more general notion, if I have vectors A and B, then I will define the inner product between them, as observed by O, as constructed by observer O, rather, like so. It's not hard to convince yourself, given everything we've done so far, that this quantity must also be invariant. I'll sketch a really quick proof. Let's define-- let's say we have two 4-vectors, A and B. Their sum, by the linearity rules that apply to these vectors, must also be a vector. And so if I compute, C dot C, this is an invariant. With a little bit of labor, that basically boils down to middle school algebra. You can show that C dot C is A dot A plus B dot B plus twice A dot B. This is invariant. This is invariant. This is invariant. And so this must be invariant. So this is really useful for us, because we now have a way-- I've introduced these objects, these geometric objects, these 4-vectors. We are going to use them in this class to describe quantities that are of interest to the physicist who wants to make measurements in spacetime. We've now learned one of the things when you're doing stuff in relativity is you have to be careful who is measuring what. What are the components that 4-vector as seen by this observer? What about their friend who's jogging through the room at 3/4 quarters speed of light? What about their friend who's driving 2/3 the speed of light in the other direction? You have all these really annoying calculations that you can and sometimes have to do. This gives us a way to get certain things that are invariant out of the situation that everyone is going to agree on. Invariants are our friends. So earlier today, earlier in today's lecture, I talked about how I can write my 4-vectors using the basis factors. So another way of writing this-- so what's sort of annoying is every time I've actually written out the inner product, I have used the represented by symbol. I don't want that. I want to have equal symbols in there, dammit. So let's take advantage of the fact that A dot B, I know how to expand A and B using components and basis vectors. And again, using the index notation, I can just pull everything out and rearrange this a little bit. Whenever you get down to a point like this, we now get to do what every mathematician loves to do-- give something a name. I'm going to define the inner product of basis vector A with basis vector B to be a two index tensor-- eta alpha beta. What's lovely about this, this is a totally frame invariant quantity. We know that. And so I've now found a way to write this using the components as something that gives me a result that is totally frame invariant. Now, when you hack through a little bit the algebra of this, what you'll find is that the components of this metric-- oh, shoot, I didn't want to actually say it out loud-- the components of this tensor, which pretend you didn't hear me say that-- the components of this tensor has the following components-- I just said something circular-- has the following values. This is, as I unfortunately gave away the plot, this is, in fact, the metric that I said at the beginning is the quantity that I must associate with spacetime in order for there to be a notion of distance between events. I haven't really said what a tensor is carefully yet. I'm going to make a more formal definition of this in just a moment. But this is your first example of one. And so the way in which this actually gives me a notion of distance is through this that I wrote down right here. If I have two events in spacetime that are separated by a displacement delta x, the delta s squared, which I obtained from this thing, is fundamentally the notion of distance between those two events that I use. And notice, it's a little less normal of a distance than you're used to when you do sort of ordinary Euclidean geometry. This is a distance whose square can be negative. What we like to say is that when you're working in special relativity, it's not necessarily positive-- the distance between two events is not necessarily-- the distance squared is not necessarily positive definite. If it's negative, though, that just means it's sort of dominated by the time interval between them. If it's positive, you know it's dominated by the space interval between them. If it's zero, well, you actually know-- it's actually a little bit confusing at that point. They could be, in fact, you know, very widely separated in both space and time, but in such a way that a light beam could connect them. So there's a lot of information encoded in that. Now, as we move forward-- hang on a second-- as we move forward, we're going to upgrade this. So right now, our metric is just this simple matrix of minus 1s, 0s, and 1s. One of things that we're going to do is sort of a warm-up exercise to the more complicated things we're going to do later is we're going to move away from special relativity and Cartesian coordinates. We're going to look at it in polar coordinates. That's going to be kind of like a warm-up zone. And so when we do that, we're always going to reserve eta for the metric of special relativity when I'm working in Cartesian coordinates. It's just a great symbol to have for that. And it's a useful thing to always have that definition in mind. I can continue to do special relativity, but then working in coordinates that are, you know, spherical-like or polar-like or something like that, then this is going to become a function. And what that is going to mean is that things like my little basis vector is going to have more complicated behavior. A little later in the course, we will then show that when gravity enters into the picture, essentially the essence of gravity is going to be encoded in this thing as well in a way where, again, it's going to be a function. It's going to vary as a function of space and time. And the dynamics of gravity will be buried in that. It's sort of funny that it really does just sort of start out-- I mean, if you take that thing and you set delta t equals 0, this is just the bloody Pythagorean theorem. That is all this is. Put time back in, and it sort of is the generalization of Pythagoras to spacetime. And, in fact, we're going to take advantage of that and sort of define a geometry that looks like this as being flat in the same way that a board is flat, and the Pythagorean theorem works perfectly on it. Then, we're going to start think about what happens when it becomes curved. And you start thinking about things like, what is the geometry on the surface of the sphere look like? That's just sort of pointing ahead. So I just throw that at you, so that you get ready for some of the concepts that we'll be talking about soon. So let me write this actually in terms of differentials. It's sort of useful for what I want to say next. So a little differential, if I have two events in spacetime that are very close to one another, I can write them like so. And what I've essentially done here is written dx as dx alpha e alpha. Before I get into some more sort of a couple of important, fairly important 4-vectors, the reason I did this is I want to make an important point about some notation and terminology that is used. If it is the case that the displacement vector is related to the differentials of your coordinates like so, we say that e alpha is a coordinate basis vector. What it does is it transforms a differential of your coordinate into a differential vector in spacetime. Now, you, may be thinking to yourself, OK, well, what other kind can there be? Well, this is where my little spiel there a second ago about how we're going to start looking at more complicated things, it's going to become important. So when we're working in a Cartesian-like coordinate system, the fact that this is what we call a coordinate basis vector isn't very interesting. Suppose I was working in some kind of a curvilinear coordinate system, OK? Spherical coordinates. Now, let's just focus on 3-space for a second. So if I write a sort of analogous equation in curvilinear coordinates-- OK, so here's the 3-space version of that. Now let's imagine that i equals 1 corresponds to radius, i equals 2 is theta, i equals 3 is phi. Then, this would be dr er plus d theta e theta plus d phi e phi. Does that disturb you at all? OK, this has dimensions of length. These have the dimensions of angle. In order for this to work, er must be dimensionless. e theta must have the dimensions of length. e phi must have the dimensions of length. This is what a coordinate basis looks like when I am dealing with-- well, we're going to use is a lot in this thing. I introduce this right now because you are all probably looking at that, and some small part of you inside is weeping, because what you want me to write down is this. Ah, isn't that better? OK, this looks like something you're used to. So I throw this out here right now just because I want to make sure you're aware that there are some equations and some foundational stuff you guys have been doing over the years, particularly, this shows up a lot when you've done E&M out of a textbook like Purcell or Griffiths or Jackson, because there's some derivative operators, which are assuming that your basis vectors are what we call orthonormal. So my e i hat here, it is an orthonormal basis. And orthonormal basis is defined such that the dot product of any two members of this thing gives me back the chronic or delta. That is not necessarily the case when I work with a coordinate basis. Our basis has er dot er equals 1. Yay, that one's nice. But when I do e theta dot e theta, I get r squared. e phi dot e phi will be r squared sine squared theta. And what I'm going to do when I start generalizing these things, I'm going to change my-- this thing which I defined up here-- do I still have it on the board? Yeah, yeah, right here. So when I set eta alpha beta is e alpha dot e beta and I made it this thing, I'm going to generalize this and say at the dot product of any two basis vectors it gives me a more general notion of a metric tensor. And the values in the metric tensor maybe functions like this. Right now, throw that out there, you know, this might be sort of just like a peak of the horrors that lie ahead. OK, we're not going to worry about this too much just yet. But I want you to be prepared for this. In particular, it's really useful to have this notion of a coordinate basis versus an orthonormal basis in your head. We're going to start defining some derivative operations soon. In fact, probably won't get to them today, but they will be present when we start doing Lecture 3. And there's a couple of results that come up where everyone's sort of like, wait, I knew that the divergence had a factor of r on that derivative there. Where'd it go? It's because we're not working in an orthonormal basis. All right, I'm a little sick of math. So let's do a little physics. So, so far, the actual only physical 4-vector that I've introduced is the displacement vector. From the displacement vector, it's really easy to make the probably the first and simplest important 4-vector, which is known as the 4-velocity. This tells me the rate of displacement of an observer as this person moves through spacetime per unit-- and we're going to be careful about this in this class-- d tau is the time interval as measured along the trajectory of the observer with 4-velocity u. In other words-- that's a very long winded way of saying it-- it's an interval of proper time. In English, the word proper time sounds very like, woo, I don't want to use improper time. I better use that. But this actually I think it comes from French. It just refers to the fact that it's one's own time. Apparently in German people say eigenzeit. So, you know, there's a couple of different words for it. But proper time is what we use. In special relativity, if we see someone going by with constant velocity, a particular observer who sees, you know-- we're here in the room. Someone comes through. Their 4-velocity is u. We would see their 4-velocity to have the components gamma gamma v, where gamma, I'll remind you, is the special relativistic Lorenz factor. And I'll remind you again we've set speed of light to 1. A very useful thing, which we're actually going to take advantage of quite a bit, is that in the rest frame of u-- pardon me for a second-- in the rest frame of u, or I should say of the observer whose 4-velocity is u, they just have 1 in atomic direction, C, if you want to put your factors back into their. And that's basically just saying that the person is standing still, but moving through time, because you are always moving through time. All right, from the 4-velocity for an observer who has-- or for an object, I should say, who has some mass, we can easily define the 4-momentum, where this m is known as the rest mass of this object. It's worth a bit of description here. You will often see, particularly in some older textbooks that discuss special relativity, people like to talk about the relativistic mass. And that comes from the fact that if I write out what this thing looks like according to some particular observer, you have this gamma m entering into both of the components. And so older textbooks often called gamma m the relativistic mass. That's not really the way people have-- over the course of the past couple decades, they've moved away from that. And it's just more useful to focus on the rest mass as the only really meaningful mass, because, as we'll see in a moment, it's a Lorenz invariant. We'll see how that is in literally about 3 minutes. And so what we're instead going to say is that as seen by some particular observer, this has a timelike component that is the energy that that observer would measure and a set of spacelike components that are the momentum that that observer will measure. So where we get a bit of important physics out of all this stuff is by coupling these two 4-vectors to the scalar products that we made up. So the first one, if you do u dot u, according to any observer, that's just going to be minus gamma squared plus gamma squared v squared. And with about 20 seconds worth of analysis, you can find that this is always equal to minus 1. Actually, there's an even trickier way to do this. Suppose I evaluate this in the rest frame of the observer whose 4-velocity is u? Well, in the rest frame, v is 0, and gamma is 1, and I get minus 1. And this is an invariant. So whatever I get in that particular frame must be obtained in all frames. That's a trick we're going to use over and over and over again. Sometimes you can identify-- you know, you get some kind of God awful expression that just makes you want to vomit. But then you go, wait a minute, what would this look like in frame blah, blah, blah? And you sort of think about some particular frame. And in that frame, it may simplify. And if it does and it's a frame invariant quantity, mazel tov, you have just basically won the lottery. You've got this all taken care of. Go on with your life. So the 4-velocity has a scalar product of itself that is always minus 1. OK? How about the 4-momentum? Well, the 4-momentum is just 4-velocity times mass. So that's just minus m squared. But we also know it's related to these two other quantities, which are important in physics. This is related to the energy and to the momentum. So this is also equal to minus e squared plus-- so this is the ordinary 3-- the magnitude of the 3-vector part of this thing as measured by the observer who breaks up the 4-momentum in this way. So what this means is I can manipulate this guy around a little bit here. Anybody who works in particle physics is presumably familiar with this equation. Sometimes it appears with the p squared moved onto the other side. If it looks a little bit unfamiliar to you, let me put some factors of C back in this. So remember, we have set C equal to 1. When you put it back in, that's what this is. So it drops out of this in a very, very simple way. One of the uses of this-- and many of you have done exercises presumably in some previous study that does this. And there'll be one exercise on the p set that was just posted where you exploit this. So a key bit of physics, the reason why we care about 4-momentum is it's in one mathematical object allows us to combine conservation energy and conservation of momentum. So conservation of 4-momentum puts both conservation of energy and conservation of momentum into one mathematical object. So if I have n particles that interact, then the total 4-momentum is conserved in the interaction. OK? So, yeah, we talk a little bit more about this and then just sort of quickly move on. So combining this with the fact that we are free to change our reference frames often it gives us a trick that allows us to really simplify a lot of analysis. So if I have n particles that are sort of swarming around and doing some horrible bit of business that I need to study and I need to have a good understanding of, we can often vastly simplify our algebra by choosing a special and very convenient frame of reference in which to do our analysis-- choose the center of momentum frame. So this is the time frame in which that that p tote-- so C-O-M, center of momentum-- has zero spatial momentum. In that frame, you have just as much momentum going to the left as going to the right, just as much going forward as going backwards, as much going up as going down. And so this turns out to be-- so the classic example of where this is really useful is when you are studying particle collisions and you're looking at things like the production of new particles. So imagine you've got particle A with some 4-momentum PA coming in like this. Particle B's got some 4-momentum coming in like this. These guys collide. And they do so-- I work in the center of momentum frame. I might want to just calculate the energy at which they just happen to produce some new pair of particles at rest. I would like to find the threshold for this particular creation process. OK? So you're going to play with one problem on the p-set that's kind of like that. Let's see, what do I have time to do? I think I will do-- yeah, I think I can do two more things. So all the dot products that I have been talking about so far have been a dot product of a 4-vector with itself. I did u dotted into u. I did p dotted into p. I invented a frame in which p has a particularly simple form. And then when you actually do some of analysis, you would probably take that p tote and dot it into itself. I haven't done anything that looks at the crossing between these two things, dotting one into the other. So let me to go through a very useful result that follows by combining p with u. There's a very specific notion of p and a very specific notion of u. Let's let p be the 4-momentum of a particle. I call it a. Let's let u be the 4-velocity not of a, but the 4-velocity of observer O. So particle a might be a muon that was created the upper atmosphere and is crashing through our room right now. Observer O might be your hyperactive friend who is jogging through the room at half the speed of light. The question I want to ask is, what does O measure as the energy of particle a. So the naive way to do this, which I will emphasize is not wrong, what you might do is sort of go, OK, well, we're sitting here. This room is our laboratory. I've measure this thing in my lab. So I know p as I measure it. I can see O jogging by. So I know O's 4-velocity as I measure it. So what I should do is figure out the Lorenz transformation that takes me into the rest frame of O. Once I have that Lorenz transformation, I'll apply that Lorenz transformation to the 4-vector p. Boom, that will give me that energy. That will work. That will absolutely work. But there's an easier way to do it. So one thing you should note is that everybody represents that 4-velocity as an energy-- excuse me, they represent the 4-momentum as an energy and a 3-momentum. In particular, though, they represent it as the energy that they would measure and the 3-momentum that they would measure. So that means p as seen by O is e according to O and p according to O. That are acquainted o is what we want. And I just told you a moment ago, you know, if you have p in your own reference frame, and you have u in your own reference frame, you can do this whole math with Lorenz transformations and get it out. But you also know that in O's own reference frame, O represents their 4-velocity as 1 in the timelike direction, 0 in the spatial direction. So what this means is if I go into O's reference frame, if I go into their inertial reference frame, notice that if I take the dot product of p and u, I get e times 1 and p times 0. So that is just negative. It's exactly what I want-- modulo minus sign. And so you go, OK, well, I'll flip my minus sign around. And you think, OK, great, but I did that using those quantities as written down in O's reference frame. And then you go, holy crap, that's the invariant scalar product. I'm done. Mic drop. Leave the room. What this means is you start with p as you measure it, u as you measure it. Take the scalar product between the two of them. Boom, the answer you want pops out. No nonsense of Lorenz transformations. None of that garbage needs to happen. You just take that inner product and you've got it. So that sort of says in words, no matter what representation you choose to write down p and u in, take the dot product between the two of them, throw in a minus sign, you've got the energy of the particle with p as measured by the observer with u. It's sort of late in the hour, or the hour and a half I should say. So let me just sort of emphasize, there are occasional moments in this class where if you're dozing off a little bit, I suggest you pop up and tattoo this into a neuron somewhere. This is one of those moments. OK? This is a result that we're going to use over and over and over again, because this holds-- this isn't just in special relativity. When we start talking about the behavior of things near black holes, there's going to a place where I basically at that point to say, well, I'm going to use the fact that the observer measures an energy that is given by-- and I'm going to write down that. The dot product that's involved is a little bit more complicated, because my metric is hairier. But it's the exact same physical concept. OK? Let me just do one more. And then I'll go talk, without getting into the math, about what I will start with on Tuesday. Last 4-velocity, which is probably useful for us to quickly talk about is-- so we've talked a lot about 4-velocity. That is just one piece-- when we're talking about sort of the kinematics of bodies moving in spacetime, you need more information than just velocity. Sometimes things are moving around. There's additional forces acting on them. And so we also care about the 4-acceleration. And so this is what I get when I take the derivative with respect to proper time of the 4-velocity. So will there be some homework exercises that use this. The main thing which I want to emphasize to sort of conclude our calculations for today is that when I talk about a 4-velocity of acceleration, this has an extremely important property. It is always the case that a dotted into u equals 0. If you're used to sort of 3-dimensional intuition, that may seem weird. OK? Anytime you see a car accelerate from a stop, that's a case in which its acceleration is clearly not orthogonal to its velocity. But the issue here is, this is not a spatial dot product. This is a space time dot product. And some of your intuition has to go out the window because of that. It's very simple to prove this. Remember, u dot u equals minus 1. So d of u dot u, d tau, which is just 2, u dot a is the derivative of minus 1, which is 0. OK? So this is something that we will exploit. If you want to describe the relativistic kinetics of an accelerating body, this is a great thing that we can use to exploit. You often need a little bit more information. We have to give you as a bit of additional information some knowledge about what the orientation of the acceleration is and things like that. So whenever you are given-- I'll get to you in just a sec-- whenever you're given any kind of differential quantity like this, it's not enough to know-- it's like the acceleration of loss, you have to also have boundary conditions. And that sort of tells you what the initial direction is. Question? AUDIENCE: Is that time still the proper time? SCOTT HUGHES: That time is the proper time, yes. AUDIENCE: So it's not accelerating? SCOTT HUGHES: That's right. So you can still define a proper time for an accelerating observer. It will not relate-- hold that thought. You're going to play this a little bit more in a future problem set. I mean, the key thing is that the way-- if you have an d observer, an interval of proper time as compared to an interval of time for, you know, someone in a rest frame that sees this person accelerate away, the conversion between the intervals of time, the two measure, it evolves. So, you know, let's say you're in this room with me. In fact, it turns out that if you accelerate at 1G for a year, you get to very close to the speed of light. So let's say that you were in a rocket ship right now that launched with an acceleration of G. Initially, you and I synchronize our watches. And so an interval of a second to me is the same as a second to you. Half a year later, you're moving at something like half the speed of light. And I will see a noticeable time delay. An interval of a second as you measure it looks long compared to me. Six months later, you're actually quite close to the speed of light, and it gets dilated even more. So last thing, which I'm going to say, and I'm not going to get into too much detail with this yet, is we're going to begin next time by making a little bit more formal some of the notions that go around. So we've increased some physics and some vectors. I've given you guys one tensor so far, the metric tensor. And so I'm going to give you-- in fact, I will write down a very precise definition of this right now, and we'll pick it up from there on Tuesday. So the basic idea-- so you guys have see-- the one tensor you've seen so far is the metric tensor. And what the metric is is it's sort of a mathematical object that I put in a pair of 4-vectors. And it spits out a quantity that is a Lorenz invariant scalar that characterizes what we call the inner product of those two 4-vectors. More generally, I'm going to define a tensor of type 0 N as a function or mapping of N 4-vectors into Lorenz invariant scalars, which is linear in each of its N arguments. So I will pick it up here on Tuesday. And let me just say in words, the metric is a 0 2 tensor. I put in two 4-vectors. It spits out a Lorenz invariant scalar. We're going to before too long come up with a couple of things that involve three real vectors-- excuse me three 4-vectors, too many numbers here, a trio of 4-vectors, which it then maps to a Lorenz invariant scalar. Some of them will take in four 4-vectors and produce a Lorenz invariant scalar. Notice I wrote this in sort of funny way. The 0 N sort of begs for there to be sort of an N 0. OK? To do that I have to introduce an object that is sort of dual to a vector. We're going to talk about that. Those are objects called one-forms, which actually happen to be a species of vector. We're actually going to then learn that the vector is itself a tensor. And so we will make a very general classification of these things. And we'll see that vectors are just a subset of these tensors. And at last, we'll sort of have all the mathematics in place. We can sort of lose some of these distinctions and just life goes on, and we can start actually doing some physics with these. All right, I will pick it up there on Tuesday.
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https://ocw.mit.edu/courses/5-111sc-principles-of-chemical-science-fall-2014/5.111sc-fall-2014.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality, educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. OK. So we have-- you can show the answer-- 57%, 1 and 3. So this is where we were at the end of last lecture, so if you didn't get this written down in your notes, you want to look at these definitions. So here in this, we have NH3 with its lone pair, and it is acting as the Lewis base, and BF3 is acting as the Lewis acid. So a Lewis base donates its lone pair electrons, and the acid accepts them. So what was the other definition? We had the Bronsted-Lowry. Does someone want to tell me what that definition was? Can we have a volunteer tell us the Bronsted-Lowry definition? Pull out your notes. There you go. AUDIENCE: So a Bronsted-Lowry-- CATHERINE DRENNAN: I'm not sure it's on. It is on? OK. AUDIENCE: Bronsted-Lowry base is something that accepts H plus or H3O plus. CATHERINE DRENNAN: Yeah. So the base accepts the hydrogen ion or the proton, and the acid donates. So is this definition now incompatible, or are they completely opposite, or could they be worked together? Does anyone have an opinion on that? So how many people think that these could be consistent definitions in some way? OK. Oh, quite a few people. One could make that argument. So if you know now that you can make the argument, does someone want to try to make it? We have a bag. All right. Our mic went away, but can we get up over there? So where did our mic runner go? I should have explained they were needed twice. All right. All right, who said they might try this? AUDIENCE: All Bronsted-Lowry bases are Lewis spaces, but not all Lewis bases are Bronsted-Lowry bases because you can act as a proxy with no hydrogen ion if it connects directly. But if a hydrogen ion bonds, an example H3O, to the lone pair of oxygen, then it's the same thing as a Lewis base. CATHERINE DRENNAN: OK. so that's right. The Lewis acid-Lewis base definition works if you don't have a hydrogen ion involved or a proton involved, but the other still applies. So remember that when you're accepting H plus, you're taking it without an electron, so the base is taking H plus, so it's donating its electrons to form the bond. And the acid, when it gives up H plus, it's keeping all of the electrons. It's accepting the electrons, so when the acid gives up H plus, it's accepting, and when the base takes H plus, it's using its electrons. So the Lewis definitions are lone-pair or electron-centric definitions, whereas, Bronsted-Lowry is thinking about the proton, but they both work together. So that's a good review of the definitions we talked about last time, and now I'll try to get my mic back on. All right. So today we're going to continue talking about acid base, and we're going to continue talking about acid base quite a bit. So we have a number more lectures on acid base. So we're going to start today with the relationship between pKw, pH, and pOH, we're going to talk about the strengths of acids and bases, and then we're going to start doing equilibrium acid-base problems. And we're going to do a bunch of those. And students who are doing this and learning it for the first time are telling me there is an infinite number of different kinds of problems. It's really not true. There are only five different types, and today hopefully, we'll get through two of them. So first we need to think about water because water is a really important solvent when we're talking about acid base equilibrium. And we mentioned last time that water can act as an acid and a base. Does anyone remember what that's called, something that can act as an acid and a base? AUDIENCE: Amphoteric. CATHERINE DRENNAN: Amphoteric. So here we can think about two waters going to H3O plus and OH. We could also write it out instead of just saying two waters, one water that's acting as the acid and another water is acting as a base. So the water that's acting as an acid gives up its hydrogen ion, or proton to the base. The base accepts it forming a conjugate acid, and this water that lost H plus becomes OH minus. So water plus water, H2O plus H2O, forms an acid and a base. So we can ask the question, then how much water is in a glass of water, or how much H2O is actually in this water container that I have here? So when I'm asking that question, how much H2O in a container of water and if I know the delta G0 of that process, I'm asking about what is the relationship between our products here, our ionized acid and our base over here, the ionized products of water, compared to water at equilibrium. Because this container, the water, is in equilibrium. So if I'm asking about the ratio of products to react as an equilibrium, what am I asking about? Just yell it out. AUDIENCE: Equilibrium. CATHERINE DRENNAN: I'm asking about the equilibrium constant. Exactly. I'm asking what is K. What is the equilibrium constant. So let's just briefly review the relationship. We're given here a delta G0, and we're asking about K. What is the relationship, how do you calculate K if you know delta G0? And we have some equations that will be given to you. Sorry. Not quite yet. There's going to be a clicker question coming. We have some equations here. So we have delta G0 equals minus RT natural log of K, which we can rearrange to solve for K. And let's just have a little reminder of what these terms are. So we'll put this up here. T is temperature, and we're going to be talking about room temperature. This isn't room temperature now. And in fact, almost all acid-based problems are going to be at room temperature. There might be a few differences, but we have a lot at room temperature. We have a constant, R, and so 8.314 joules per Kelvin per mole and delta G. So we were already told the value for delta G here. Now, thinking about this room temperature, this constant, and this value for delta G, do you expect a large or small value for K if you have this delta G0 value? And that's our clicker question. All right. Let's just do 10 more seconds. Yep. So K, you would expect to be a small value less than 1. And you could think about that mathematically from these expressions, or you could think about it in terms of the value of for delta G. So if we solve this and we put in our value for delta G0, which is a positive value-- so the forward direction of the reaction is non-spontaneous. So we put in our positive delta G0 and our other values here, and this is, again, joules per Kelvin per mole. And we have to make sure that we convert kilojoules to jewels and our units will cancel out. And if we do that, we get this value for natural log of K and this value for K of 1.0 times 10 to the minus 14th at room temperature. And you will find that you end up memorizing this value if you do enough acid-base problems. In fact, you might have it memorized already. So this is, in fact, a small value. And that indicates that only a small proportion of water molecules are ionized. About 1 molecule in every 200 million are ionized, so it's a very, very small value. So there is a lot of H2O in a glass of water. So in here it's mostly H2O. Very few molecules are ionized in that. Now, part of the reason why you're likely to have this value of K memorized is because it has a special name. This K is called Kw for water. So that's easy to remember. And Kw, the equilibrium constant for water, is equal to the hydronium ion concentration times the hydroxide ion concentration. And since this is an equilibrium constant, if you're at constant temperature at room temperature, this will always equal 1.0 times 10 to the minus 14th. This product is always going to be this at room temperature. And that is why it's going to turn out to be important because you're going to use that information in solving a lot of problems. All right. So if this is an equilibrium constant, we want to think again about expressions for equilibrium and just make a note that we did not include water squared on the bottom of this expression. So Kw equals hydronium ion concentration times hydroxide ion concentration. You don't put it over water. And the reason for this is because this is a pretty pure solvent, and you don't include pure solvents in your equilibrium expression. You also don't include solids. So we talked about that when we were talking about solubility. So you have to remember that solvents like water is nearly pure, doesn't go in. Pure other liquids, pure other solids are not included in your equilibrium expression. All right. So we now know about water. We know about Kw. Let's do a few more definitions and then think about how all of these terms are related to each other. So a definition that many people already know, probably if you've taken any kind of science before, you're probably aware that pH equals minus log of the hydronium ion concentration of H3O plus. If you don't know it, it will be on your equation sheet, so it doesn't really matter, but there it is. And there's also pOH. pOH equals minus log of the hydroxide ion concentration, OH minus. Now, let's think about the relationships of pKw, pH, and pOH. So relationships. So I just told you and maybe you've already memorized that Kw is going to be equal to the hydronium ion concentration times the hydroxide ion concentration, and that's going to be equal to 1.0 times 10 to the minus 14th at room temperature. So now, if we take this expression and we take the log of all of these terms and multiply by a negative value, then we will get this expression. So minus the log of Kw is pKw. And minus log of the hydronium ion concentration is what? pH minus log of the hydroxide ion concentration is what? AUDIENCE: pOH. CATHERINE DRENNAN: pOH. So pKw equals pH plus pOH, and that is equal to 14.00 at room temperature. And this is a very useful expression, the fact that pH and pOH are equal to 14. If you know one of these, then the other one by doing a simple subtraction. So you will find yourself using this expression quite often in the problem sets. And so I'll say that one, the problem set that that's due on Friday has just a couple of acid-based problems at the end. The next problem set will be 100% acid-base problems, so there's a lot to look forward to. So let's talk about the strengths of acids and bases. So if you have a pH of pure water, it should be equal to minus log times 1.0 times 10 to the minus 7, or the pH equals 7, which is a neutral pH. So if we have a scale here, pH minus 1 to pH 14, 7 is a neutral value. If we're talking about things that are acidic, the pH of an acidic solution is less than 7, and so this is the acidic part over here. And the pH of a basic solution is greater than 7, so that is down in this part. The EPA defines waste as corrosive if the pH is lower than 3, so that would be corrosive, or higher than 12.5, also corrosive. So living creatures like to be more in the neutral range. When you get to things that are very low pH or very high pH, that is less good. So let's now check out a few pH's, and I'd like to have five TAs come down to help me with this. And I'll just show you what we're going to do. So we'll have some volunteers. We have pH paper. The pH paper has an indicator on. You put the paper in the solution and then try to see which thing matches best. So this is a very quick and dirty way to estimate pH. There's pH meters and indicator dyes that work better, but we'll have five people measure five things. We have ammonia, a cleaner. We talked about the importance of cleaning bathrooms last lecture. We have soda. We have vinegar, which is often used in salad dressings. Before I came here I just went to a random sink and got some MIT water to measure. And then this, this is special. This is a prescription medicine. So this is a solution of iron 2 sulfate. And this is prescribed if have an iron deficiency. An iron deficiency is pretty bad, and a lot of kids have iron deficiencies. It's really common in infants. So my daughter had an iron deficiency, and this was her medicine. In fact, this box is completely full of this medicine. So kids cannot swallow pills very easily. How many of you actually cannot swallow a pill? A lot of adults are pill challenged. How many are happy swallowing pills? Let's do a more positive thing. OK. So swallowing pills is not everyone's favorite. When you're a kid, it's really impossible to tell someone to put something in their throat and go back and try to get it down. They choke, spit up, it's awful. So kids have to take medicine in solution. So this medicine is 300 milligrams in 5 milliliters. It's very concentrated, so my daughter had a pretty severe iron deficiency. So when she tried taking it, and I tasted it, it was horrific beyond belief. And after she had taken it a couple of times, she started to have sores on her tongue and in her mouth. And I as a chemist said, let's measure the pH. So you will measure the pH of this prescription medicine for a four-year-old and tell me whether the EPA would prove it or not. Anyway, so here are the TAs, and they're going to come around. Raise your hand if you're willing to measure a pH. TAs, just grab a solution. I should have had you pour it, but OK, we're doing it. And we'll go around, and we'll measure, and tell me what you find, and I'll write it on the board. Yeah, OK. All right. So we have pH. All right. Do we have an answer? Did we get one? What do you have? AUDIENCE: Ammonia. CATHERINE DRENNAN: Ammonia is 12. So almost corrosive, but not quite. You can still have clean, but if you want an excuse not to clean, you're like, sorry, too close. Too Corrosive Do we have another one? AUDIENCE: MIT water is 7. CATHERINE DRENNAN: MIT water is 7? AUDIENCE: MIT water is 7. CATHERINE DRENNAN: Awesome. So often, actually, water that you get from a tap is not 7 because often there are ions that are dissolved in it. But a little bit around 7, a little bit of ions never hurt anyone. We have lots of ions in our body. Soda? AUDIENCE: 3 and 1/2. CATHERINE DRENNAN: Soda at 3 and 1/2. So soda is definitely getting into our corrosive range. AUDIENCE: Vinegar at 2. CATHERINE DRENNAN: And we have vinegar at 2. And that is corrosive. People often use pure vinegar to clean out a coffee pot because it is pretty corrosive. It really does a good job of cleaning it. When you're having it in salad dressing, you don't want to have it pure vinegar. That would not be good to drink. And what about medicine? AUDIENCE: Medicine has a pH of 2. CATHERINE DRENNAN: 2. So we also have in the corrosive category the medicine that my four-year-old daughter was asked to take. So when she started getting these sores in her mouth from taking her medicine, I brought some pH paper home and measured the pH and discovered that what she was taking was corrosive. So I went to the doctor and said, can you prescribe something different. This medicine is corrosive. And the doctor looked at me and it's like the pH, the pH is like 2 or lower. And she looked at me, and she's like, what do you mean pH? This is why I spend many lectures on pH in this class and acids and bases. Some of you will be doctors, and some of you will not encourage parents to buy this much corrosive thing to feed their child when they're iron deficient. So this is what I'm talking about. And for those of you who are not going to be doctors, many of you will probably have children with an iron deficiency. It's super common. So you will know that you need to measure the pH of the solution before giving it to your child. So what did I do? So what I did was I bought adult pills, same stuff, iron sulfate, and I made sure that this told me the number of milligrams that were in there. I looked at what was prescribed. So I got the pills with the correct number of milligrams, and then I crushed them, and that stuff also tastes really nasty, but a little tip for the future, there is one and only one that I discovered, taste that can cover the taste of iron 2 sulfate, and that is Nutella. So don't give your child corrosive medicine. Give them crushed pills in Nutella. pH is important. pH is important. So let's talk about the strengths of acids. And you might want to reconsider huge soda consumption too. I don't really know, but that's something you might want to think about. Strength of acids. So here we have some acetic acid, CH3COOH aqueous, dissolved in our solvent, which is liquid water. And this is an acid, so it will give up a hydrogen ion or a proton to the water, which acts as a base and forms our hydronium ions, H3O plus, and also the conjugate base of the acid. So it's the acid missing H plus, CH3C00 minus. So if we're talking about the strength of the acid, what we really want to know is how much of that acid forms ions, how much of it ionizes. That is going to determine how strong it is because the amount that ionizes is equal to the amount of the hydronium ions you get, and pH equals minus the concentration of hydronium ions. So the pH depends on the extent to which it ionizes. And again, we're talking about at equilibrium, so we're talking about a equilibrium constant, and this one also gets a fancy name. In acids and bases, all the equilibrium constants get their own little names, and this is the acid ionization constant, Ka. So it's the equilibrium constant for an acid, Ka. So that's at least very easy to remember. So now based on what you know about equilibrium constants, why don't you tell me the answer to this? All right 10 more seconds. All right. So this one is wrong. It's always products over reactants for an equilibrium expression. Water is a solvent and is nearly pure, so it should not be there. This should be in the expression because it's aqueous. If it were solid, it wouldn't be included, but it's aqueous, so its concentration is going to change, so it is in our expression. See answer to 3 for that one. Enough information, and it's not correct. So we can write the expression for Ka. So that's equal to the products, our hydronium ion concentration, our conjugate base over are conjugate acid. And so we can look at what the value is for this expression, and we can look it up. There's lots and lots of tables of these in your book. And it is 1.76 times 10 to the minus 5 or minus 3. I don't have my glasses, whatever it says there. I think it's 5 at room temperature, which is small. So when Ka is small, it means that very little of it is ionizing. And so only a small number of our acetic acid molecules are donating their proton when they're dissolved in water. So that's the definition of a weak acid. When it has a very small Ka, it's not ionizing very much. Now, in doing these units, a lot of people get hung up on the names of the different acids and one of the first steps in solving a problem is to write the equilibrium expression out, but people get hung up with that. So don't get hung up with that. You can use generic expressions for acids and water. If you don't want to write out the whole name of the acid, you can always just say HA, aqueous, plus the solvent, water, goes to hydronium ions plus A minus, the conjugate of the weak acid. So this is an acid in water, and the acid is HA here. And an acid expression should be forming hydronium ions. If it's acidic, you should have acidic pH, and so you need to have some H3O plus around. You can also write this expression as BH plus plus water. BH plus can give up the H plus and become B and also generate hydronium ions. So here are the acid is BHA. And often when you're looking at these kinds of problems, a weak acid will be HA, but sometimes you have a problem involving the conjugate acid of a weak base, and that's often expressed as BH plus. So both of these expressions are valid for an acid in water. So now let's think about strong acids versus weaker acids, and here's our definition in this class. A strong acid has a Ka greater than 1, a lot of strong acids have a Ka really, really, really greater than 1, and the acid ionizes almost completely. So if you say it's a strong acid in water, whatever concentration of that acid you put in is going to be equal to the concentration of hydronium ions. It basically goes straight. You form lots and lots and lots of products at equilibrium. A weak acid has a Ka of less than 1, and a weak acid does not for many ionized species in solution. Not very much H3O plus is formed. So what about pKa? pKa's are really important in organic chemistry, in biology, in a lot of areas. And every year the organic chemistry faculty in 512 talk about pKa's in organic, and the students there say, no, we never learned about that. They're like, you took freshman chem, right? It's like a GIR or you pass the test. You should know about pKa's, and they're like, no. So they contact me, and I'm like they did not take 5.111 if they do not know what a pKa is. There are other courses that sometimes people decide to take that are not 5.111, but in 5.111, you know about pKa's. So what I want you to do is later in life when you're in a class and pKa's come up and everyone else in the class is like, I don't know. You're like I took 5.111. I can answer that. That makes me very happy. So pKa's. so pKa is minus the log of the Ka. That's easy to remember. We already talked about the relationship of Ka with strong acids or weak acids. The lower the value of the Ka the higher the value of the pKa. That's just out of that expression. So the higher the pKa means what about the acid? Think about what you know about Ka to answer this question. All right. 10 more seconds. All right. So that is correct. So we have a weaker acid. So if we have a low Ka, that means that it is a weak acid, and then the higher value of Ka. So the higher the Ka, that's going to mean the lower the value or the weaker of the acid. So just remember the relationship from the equation, and think about it, and you can think about the value of Ka, products over reactants. You might have a lot of products. That means a lot of ionization and a stronger acid and fewer products, that means it's weaker. So let's look at some tables, and I just put the values that are with the arrows in your notes. I didn't put the entire table. The tables are in the book, and this is page one of the table, so there's a lot of values here. So this has the acid. It tells you what the HA term is, what the A minus, so what the weak acid is, what the conjugate base is. It gives you the Ka and the pKa. And so you can see the relationship between Ka and pKa. So up here, HI is at the top of this table. So that probably means it's the strongest or the weakest. Which do you think? Is this the strongest or the weakest? You just yell it out. AUDIENCE: Strongest. CATHERINE DRENNAN: It is the strongest. Right. So it has Ka value that is much, much, much, much, much, much, much, much, much, much, much greater than 1. And it has a very, very, very negative tiny, tiny pKa value. So this is a super strong acid. In fact, do not use that. There's no reason really you would ever want to be using it. It is not a good thing to play with. All right. So HCl is used more often in pH-ing things. Is this a strong or weak acid? AUDIENCE: Strong. CATHERINE DRENNAN: It's also strong. It's K or Ka is also greater than 1, but not as strong as this. We have 10 to the seventh, and that one, gosh, I should bring my glasses, 10 to the 11th. That's super strong. That's still pretty strong. All right. So let's look down here now. Tell me, is this a strong or weak acid? AUDIENCE: Weak. CATHERINE DRENNAN: That is weak acid. The Ka is less than 1 here, and I think that's a minus 2. And the pKa value, now we're out of the negative number, so it's on the bigger side. And down here, is this strong or weak? AUDIENCE: [INAUDIBLE]. CATHERINE DRENNAN: Also weak. And it's even weaker than this one. It's like 10 to the minus 4, and pKa value is 3.75. So you see you can look at the Ka values and think about whether it's a strong or weak acid. You can also look at the pKa values. This is really, really small negative value. This is a bigger value down here. And this is not been the weakest acid. There's a whole other page of tables for acids, lots and lots of acids. Less bases. There are a few, and you will see this space quite often, NH3. So we have a base in water. The base accepts a hydrogen ion or proton from the water, forming NH4 plus. And then after water loses its hydrogen ion, it forms OH minus. So the equilibrium expression or the K for this base in water problem is called the base ionization constant, or Kb. And we can write that expression here. We have the ammonium ion concentration times the hydroxide ion concentration over the concentration of ammonia. Again, water is not in the expression. It stays pretty much pure throughout this entire thing. It's the solvent. And so this is our expression for Kb. And if you're writing an expression for Kb, you should always make sure, check yourself that you're putting hydroxide ion concentration in there. If it is a base in water, it should be forming hydroxide ion concentration. If you start writing a Kb and you have hydronium ion concentration in there, you want to stop and rethink what you're doing. So in this case, we also have a weak base. 1.8 times 10 to the minus 5 is a small number for Kb. It's a small equilibrium number, so that means you're not a lot of products. Not a lot ionized when you put this weak base in water. So only a tiny amount of NH3 ionizes to NH4 plus and OH in solution, so this is what they call moderately weak base. So as we saw before, you can have generic expressions for a base in water. We can write B aqueous plus water goes to BH plus. The base accepts a hydrogen ion or proton, and the water loses one forming OH minus. We can also write the expression that often exists from the conjugate of a weak acid, and this is a weak base. So A minus plus water goes to HA plus hydroxide ion concentration. So either of these are generic expressions that you can write for a base in water. So in terms of the definitions, it's the same. You would say a strong base is something that ionizes almost completely to give OH. There aren't a lot of examples of strong bases. Most of the ones you'll see in your class are things like sodium hydroxide. Yes, that is hydroxide. That's a strong base, and if you put it in water, you should definitely form a lot of hydroxide. So they're not a whole lot of examples there. So they're not all those tables. It's not like the acids. But there are some terms that you still need to know. So pKb equals minus log of the Kb. And again, the larger the value for Kb, the stronger the base. Again, it tells you you have a lot that have ionized. And because of this relationship of this equation, the larger the pKb the weaker the base. Because if you have a big number here, you're going to have a small number there. So now, let's talk about conjugate acids and their bases. This is super important for buffers that, hopefully, we'll get to in class on Friday. So the stronger the acid, the weaker its conjugate base. The stronger the base, the weaker its conjugate acid. And here we have HCl, which we determined was a strong acid a little while ago, giving up its hydrogen ion to water, forming hydronium ion concentration and Cl minus. Cl minus is not a very good conjugate base. It's not a good base. So if we look at this table when we have a strong acid, its conjugate it is going to be ineffective as a base. You're basically not pushing that equilibrium back at all, the other direction. It's completely ionized, and it stays that way. But if you have a moderately weak acid, you're going to have a very weak base, a very weak acid, you'll have a moderately weak conjugate base, a strong base and you'll have something that's ineffective as an acid. So they are inversely related to each other. So let's think about why this would be true. And this relationship holds Ka times Kb equals Kw. And we know that Kw is 1.0 times 10 to the minus 14th at room temperature. And if we put logs and then minus logs by everything, we can derive this expression, which is the pKa plus the pKb equals the pKw equals 14.00. So you can't have an acid and its conjugate base both be strong. You can't have a base in its conjugate acid both be strong. The pK's need to add up to 14. So if you have something that's a good acid, then its conjugate base is not going to be particularly good. The pKa and the pKb must add up to 14. If you have something that's a good base, its conjugate acid is not going to be that good. So these are connected to each other. So we can think about, then, the strong acids and the strong bases again. So we talk about equilibrium, but really this is pushing the equilibrium pretty much to completion. If it's a strong acid in water, however much strong acid you put in is how much hydronium ion concentration you get out. A strong base is going to give you that amount of hydroxide ion concentration. Strong acids and bases push the equilibrium pretty much completely toward ionization. We have these numbers of 10 to the 11th, 10 to the seventh. These are hugely over here. We're forming almost ionizing completely. Now, for weak acids and bases, it's very different. Here the equilibrium you are going back and forth. You have this dynamic equilibrium. The acid in water is forming hydronium ions and an A minus, but a minus is also a somewhat good, weak base pushing the equilibrium back the other way. So in this case, we have a back and forth between the forward and the reverse. So if you have very weak, you have moderate, you have moderate, you have very weak, and this is what's important for forming buffers, and buffers are really important. Let's see if we can get to one of these types of problems because they're five we need to get to in this unit. And there are only five. And people will say salt and water is another, but salt and water just breaks down into weak acid and weak bases. So everyone can learn how to do five types of problems, so this is very doable. All right. So let's just look at equilibrium of weak acids and we'll save weak bases for last time. So a weak acid is vitamin C, and I brought some vitamin C here. I tried to do a demo once and discovered that vitamin C, it's really coated well, so it does not dissolve until it hits the acid of your stomach, so I could not dissolve it. So I'll just hold this up right here. But if we had a 500 milligram tablet, which I think is what this is and if I had 100 mils of water and if this wasn't coated so there's no way you can dissolve it, then I could calculate the pH. So how am I going to do this? I'm given the Ka, I'm told the number of milligrams, and I'm told the volume of water. So the first thing you want to do is calculate molarity. So you want convert grams to moles or milligrams to grams to moles. And then you want to use your volume, and you have 100 mils, but we're going to do molarity. So we want to convert milliliters to liters. And we can calculate 0.0284 molar moles per liter. Then we can write our expression, and if you want to, you can just write HA here and A minus. You don't have to write out the whole thing if you don't want to. All right. So here is the expression again. Now, just as we saw with chemical equilibrium, we can write this table talking about the initial molarity. So we have a weak acid we have no hydronium ion concentration. We have no weak conjugate base over here, so the change then is going to be the constant, the molarity that we had, and these are change in molarities, so don't put moles or other things here, molarity here, minus x, that's what you have in equilibrium, plus x plus x. So we can write our expression now for Ka. We're given the value Ka, and we know how to write an expression. We have our hydronium ion concentration, our conjugate base over here over our conjugate acid. This is equal to x squared 0.0284 minus x. Now, if you want to with these problems, you can make an assumption that x is going to be small and then not have to use the quadratic equation, but you do need to check your assumption. But I'll show you how to do it with an assumption, so we're going to assume x is small, and this value, the x, will drop out. So we can rewrite this expression as x squared over just the concentration we started with. And then we can solve for x. And x, in that case, is point 0.00151, but really two significant figures. Our Ka is just 2. Now, we can check the assumption. And we're going to check the assumption that this is, in fact, small. And if it's less than 5%, then we can use that. So here this is the x we calculated. Here is the amount times 100% gives us 5.3%, which is more than 5. 5 is the magic number for the course, and so then we have to use the quadratic equation. Now, a lot of you probably have calculators that can work with quadratic equation, so you don't really care, but if you want to check the assumption, you can, and sometimes we'll need to check it. So this value, this percent here, is sometimes called percent ionization because it's x over the amount you started with. It's the percent that ionized or percent deprotonated. If this is an acid, it's the amount of hydronium, the amount that deprotonated over what you started with. So sometimes you're asked to do this even if you're not checking any assumptions. All right. So we got to end with if a quicker question. We get here this is really two significant figures. So why don't you tell me how many significant figures are in your answer for pH. All right. 10 more seconds. So let's take a look at the answer here. So there's the answer. So this had two significant figures, which means you get two significant figures after the decimal point. So maybe we'll have one of these in the clicker competition on Friday, and we'll do our weak bases next time. So again, we've been studying acid base. And there are only five types of acid-base problems, weak acid in water, which we already did, check, weak basin water, which we're going to do right now, and as soon as we're done doing week base in water, I'm going to explain to you how salt and water are just weak acid in water or weak base in water problems. So you already know how to do them as soon as they teach you about weak bases. And then we're going to move on and do buffers. And then next week we're going to do strong acids and strong bases, and then you'll have all five. You already have enough that after today's lecture, you can do, I think, all the problems set 7 except for the last two questions, I believe, or at least a large fraction of problem set 7. I will warn you that acid-base problems take a lot of writing and a lot of time to work. One problem often has five or six parts to it, and they're all time-consuming part pretty much. So don't leave problem set 7 to the last minute. That would be a mistake on this particular problem set. Weak bases. Acid-base type problem 2. So in this example, we have ammonia, NH3, in water, and we measured the pH of some solutions before and saw they were basic. We had a pH of, I think, 12 last time. And so this conjugate base goes to a conjugate acid. So the base accepts a hydrogen ion or proton from water, becomes NH4 plus, and the water that lost its hydrogen ion becomes OH minus, and that is a base in water problem because you're forming hydroxide ions. For a base in water, you're talking about the equilibrium constant for that base in water, which has a special name KB, B for base. And it's important to remember the A's and B's on the equilibrium constant because it's always checking your work. Is this a base in water problem. People sometimes try to apply Ka's when they should be doing Kb's. So pay attention to basin water. We're thinking about Kb. So here if we're going to calculate the pH of a solution, we have molarity of 0.15, and room temperature, pretty much everything is at room temperature. So whenever you're asked to calculate the pH of a weak base in water, you want to think about at equilibrium, what's the condition now, what will be the condition, what's the change, and what will be the condition at equilibrium. And so you can make this table. You can write out the expression. Again, you can forget about water. Water is our solvent. It's not going to be showing up. And then put in our initial molarity, 0.15. And we put zeros in the other categories. And why don't you try the rest? Is it slowing down? SAM: [INAUDIBLE] answers are really quick. CATHERINE DRENNAN: All right. Let's just do 10 more seconds. Yup. Excellent. Sam told me everyone responded really quickly. So that is, in fact, what you're going to do. We'll put that up here as well. So we lose some of this minus x. So at equilibrium, we have point 1.5 minus x. And we get plus x here and plus x there. Now, we can write our Kb for our base in water, its products, the conjugate acid, NH4 plus, times the concentration of hydroxide ions over NH3. Again, water is the solvent. It's not in the equation. And then we can write it out-- this is x. That's x, so we have x squared. And then the change in the amount of the weak base, when it's in water, is 0.15 minus x. So at this point, you can either use the quadratic equation, or you can make an assumption that x is going to be small compared to 0.5. We don't want to make the assumption up here. We want to get to this point before we try the assumption, and you always need to check your assumptions to make sure they are correct. But if we use the assumption here just to simplify the math. And so the assumption is that this x is small, so we can drop that x minus x out, and then we can calculate that x is 0.00164. And then we can check the assumption, and that is another clicker question. All right. 10 more seconds. All right. It is number 1. So here you're checking the assumption and the way that you check the assumption is that you put x that you've calculated by simplifying it, divide it by the number that you're asking is it smaller than, and then times 100 because our rule is 5%. So if this value is greater than 5% of this value, you need to use the quadratic. If it's less than 5%, then you can go ahead and use your assumption. And so this is the assumption we're talking about, that this number is under 5%. And for weak acids and bases, it very often is less than 5% because they're weak, so there's not a lot of ionization. And this is often referred to as percentage ionization as well. All right. So our checking assumption, again, that x 0.00164 divided by 0.15 times 100% is 1%, 1.1, that's less than 5, so the assumption is OK. And then we can calculate pOH because this is a weak base in water, so the x that we calculated is the amount of hydroxide ion in solution. So we calculate pOH, and we get 2.79. And that's two significant figures, everything we had after the decimal because everything that we had was two significant figures. And so this is two significant figures, two significant figures after the decimal point. And then you need to calculate pH because the problem asked for pH. So this is, again, at room temperature. So we can use 14 minus 2.79 is 11.21. That's our pH. And don't make the mistake of stopping here forgetting that x is hydroxide instead of hydronium ion and tell me that the pH of my weak base in solution is pH 2. So one of the things that's really good about this unit is that there are some good checks that you've got the right answer at the end. If you're talking about a base in solution, you should have a pH above what value? AUDIENCE: 7. CATHERINE DRENNAN: About 7. So it should be on the basic side of neutral. pH 7 is neutral, so it should be above 7. And so if this was your answer for pH, that wouldn't make a lot of sense. So always check at the end. And it's fine if you get to an exam and you get to the end and you did something wrong and you realize that your answer doesn't make any sense, if you write, this answer doesn't make any sense, but I don't have time to figure out what I gave wrong, you will get points for recognizing that that is not a valid answer. Because it's a weak acid problem and you have a basic pH or vice versa. So keep that in mind. There's lots of partial credit. Grading acid-base problems on exams is a whole new fun adventure that the TAs have no idea what is going to be happening. But I'll be right with you there for 10 hours of grading, and we'll have a good time. Write neatly, please.
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https://ocw.mit.edu/courses/8-421-atomic-and-optical-physics-i-spring-2014/8.421-spring-2014.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So over the last lecture, we have talked about coherence within an atom, coherence between two levels, coherence between three levels. And today, in the last class, we want to talk about coherence between atoms. So this is now, I think, for the first time in this course that we really have more than one atom. Well, maybe we discussed some collision or broadening, or we discussed [INAUDIBLE] interaction between two atoms. But usually, atomic physics is one atom at a time. But now we want to understand one important phenomenon which happens when we have many atoms. And the phenomenon is called superradiance. So I left something good for the end. And superradiance has in common the word super with superconductivity and superfluidity. And it really represents that the atom, many atoms act together. And the word super also means coherence among atoms-- superfluidity and superconductivity have macroscopic wave function where all the atoms, the meta waves are coherent. The phenomenon of superradiance, as we will see, has not so much to do with coherent atoms. It has more to do with coherent photons. So it's more-- some people regard superradiance as a laser without mirrors. But you'll see where the story leads us to. So just to set the stage-- for many atoms, we should first talk about single atoms. And all that is described in this landmark paper by [INAUDIBLE] in 1954, which is posted on our website. So if you have a single atom prepared in the excited state, it decays to the ground state, and we want to characterize the system by emission rate as a function of time. So the emission rate initially is gamma, the natural language of the excited state. And then, of course, the emission rate decays because we don't have any atoms left. Similarly, the probability to be in the ground state is zero initially, and then with an exponential approach, it eventually goes to unity after a while, after we have only atoms in the gamma state. So this is rather straightforward. But now we want to bring in a second atom. And I'm asking, what happens when we have not the one atom, but two atoms? One is in the ground state. One is excited. So pretty much what we have added to the original situation with one excited atom was we've brought in one ground state atom, which naively you would think does nothing. But that's not the case. What happens is-- and I assume just for review-- we will drop the assumption later, but we assume for now that all the atoms are within one optical wavelength. What we then realize is for two atoms-- and I will show you that in its full beauty-- that the initial rate of light which comes out of the system is the same. So the extra ground state atom does not change the initial emission rate, but it goes down faster. And if we ask what is the probability that the atom is in the ground state, we find that it's only one half. So in other words, normalized [INAUDIBLE] system, we have a ground and excited state atom, and what comes out is only half a photon. Half of the atoms do not decay. So it's not the same rate and the same decay. Something profoundly has happened. And this is what you want to understand. So let me give you the correct answer. The rate of emission is a function of time for this situation. We start out with gamma, but then the emission decays, not with gamma but with 2 gamma. And the probability that both atoms are in the ground state-- or that the second atom, so to speak, is in the ground state-- will only asymptotically go to 1/2. And it does so exponentially-- but again, with the time constant, which is two times faster than for the single-atom system. So we have the same initial emission rate, but only probability of 1/2 to emit at all. So in order to understand it, we have to look at an atom in the excited state and atom in the ground state. And we want to write down the wave function as a superposition of a symmetrized and antisymmetrized wave function. I should tell you, I'm going very slowly for two atoms. And then once I've introduced the concept for two atoms, with a few pen strokes, we can immediately discuss in atoms. So all the phsyics, all the understanding what goes on in superradiance is already displayed for two atoms. So we want to have a superposition of symmetric and antisymmetric wave function. The symmetric one is a normalized wave function which is ge plus eg. And we call that the superradiant wave function, for reasons which will become clear in a moment. And if you have a minus sign here, the antisymmetric combination, we call this subradiant wave function. Now, what happens is, we have to consider-- so we have symmetrized the wave function. Well, I didn't really tell you why, but it's always good to symmetrize. Symmetry is, if you can use it, something good. And the reason why I symmetrized it is because I want look at the interaction Hamiltonian. And if I look at the interaction Hamiltonian-- the one we have seen many, many times but now for two atoms-- we will immediately realize that this interaction Hamiltonian is symmetric. So therefore, if the Hamiltonian is symmetric, it's a really good starting point to have the wave function for the atoms expanded in a symmetric basis. And since I want to emphasize that the whole story I'm telling you today has nothing to do with the kind of second quantization-- it is about spontaneous emission, but it's not involving any subtlety of spontaneous emission and field quantization-- I want to write down the interaction Hamiltonian both in a classical and a quantum mechanical way. In the classical way, we have the dipole moment d1. We have the dipole moment d2. And the atoms talk to the electric field at position RNT. And now you realize where some of the assumptions are important, since the atoms are localized to within a wavelength, they rarely talk to the same electric field. There are no phase factors. In about 55 minutes or so, we introduce phase factors for extended samples. But for now, we don't. And therefore, what the atoms couple with is with a dipole moment, which is a sum of the two dipole moments. So this is classical or semi-classical. So what enters in the Hamiltonian is only the sum of the operators for the two atoms. And the same happens in the QED Hamiltonian. And actually, I will get a little bit more mileage out of the QED Hamiltonian, as you will see in a moment. Because with the QED Hamiltonian we describe the atomic system-- so at first atom one-- with the raising and lowering operator with the atoms interacting with a and a [INAUDIBLE]. And then I have to add the term where the index one and two are exchanged. So we are introducing here-- that's convenient for two-level atom-- the spin notation sigma plus and sigma minus are the raising and lowering operator which flip the atom from the ground to the excited state and vice versa. But the important part now is-- and this is where, actually, everything comes from in superradiance-- that the coupling involves not the individual spins, little sigma plus, sigma 1 and sigma 2-- it only involves the sum of the individuals. i equals 122, and later we extend the sum to n. So therefore, what matters for the interaction of the atoms with the electromagnetic field is the sum of all the atomic spin operators. And the sum is, of course, symmetric against exchange. So therefore, when we are asking what is the coupling of the state which I called the superradiant state, the one where we had symmetrized eg plus ge, or we ask, what is the coupling of the subradiant state to-- well, the state where both atoms are in the ground state. Well, now we can use symmetry. The left-hand side is symmetric. The operator is symmetric. And now only the symmetric state will couple. The antisymmetric state will not couple. So therefore, the subradiant state, eg minus ge, cannot decay. That's why we call it subradiant. I think a better word would be non-radiant, but non is definitely subradiant. And for the matrix element eg and ge, we find that we have actually an enhancement of the coupling by a factor of square root 2. So now we pretty much know what we have to do. You want to use the symmetry of-- let's assume we consider ground and excited state of each atom as spin 1/2. But now we want to look at the total spin, the total pseudo angular momentum of the two atoms, and later we extend it to n atoms. So we want to use now the power of the angular momentum description. And that goes like follows. We have four states of two atoms. And this is gg, ge, eg, and ee. And if I denote with ground state spin down, excited state spin up, I'm talking about 2 spin 1/2 states. And 2 spin 1/2 states can couple to s equals 1, total s equals 1, and total spin s equals 0. And that's what I've done here. I've arranged the states ee, the symmetric superradiant state, the ground state, and the subradiant state. A variation energy level diagram-- here we have 0 excitation energy, here we have 1 excitation energy, and here we have two excitations energies of the atom. But I've also labeled now the spin labels for the combined system. Those symmetrized states correspond to a spin equals 1. It's a triplet letter with three different magnetic quantum numbers. m equals plus 1 means everything is highly excited. m equals minus 1 means we are in the total ground state. And here we have the simulate state, which is the antisymmetric state or the subradiant state. And our interaction Hamiltonian is the total spin plus minus. It is the raising and the lowering operator. And you know that the raising and lowering operator for the spin is only making transitions within a manifold of total s. It just changes the end quantum number by plus minus 1. So the Hamiltonian cannot do anything to the simulate state, because there is no other simulate state to couple. But within the triplet manifold, the sigma plus sigma minus operator is creating transitions between the different end states. And the coupling constant, which for an individual atom was little g is now factor of square root 2 enhanced. And we will see in a few minutes, that for n atoms, it's square root n enhanced. And if any speak, that's where the word super in superradiance comes from. Yeah, actually let me just quickly add the diagram for the single atom. The single atom has only an excited state, a ground state. It corresponds to s equals 1/2. And we have magnetic quantum numbers off plus 1/2 and minus 1/2. And the coupling due to the light atom interaction goes with the coupling constant g. So the key message we have learned here is that when we have several atoms within an optical wavelength, we should use for their description symmetrized and antisymmetrized states. Or when we generalize to more than two atoms, we should just add the total angular momenta by treating each atom as pseudo spin 1/2. And it is this angular classification which tells us how the radiation proceeds. Because the coupling to the electromagnetic field is only involving the lowering and raising operators for the total spin. And this only acts on a manifold where the total spin s is conserved. And what we get is transitions with delta in plus minus 1. So the question, have those effects been observed? Yes, they have, actually. And they're important for a lot of research. But just for two atoms, the simplest observation is when you take two atoms-- let's say two sodium atoms-- bring them very close, and you form a sodium 2 molecule. And to some extent, in four states where the molecule are binding is not completely changing the electronic structure, we can regard the sodium 2 molecule as consisting of two sodium atoms. And indeed, if you do spectroscopy of the sodium 2 molecule, you find some molecular states which are very long lived, like the subradiant states, which do not radiate at all, but then you find states which have a spontaneous emission rate which is about two times faster than the spontaneous emission rate. So you find that you can understand some radiative properties of molecules by assuming that they are related to the sub and superradiant state of the two atoms which form this molecule. So an example here is sodium 2 molecule. A state where the gamma molecule is approximately 2 times gamma sodium, or other states where it's very small. OK. Now we understand the basic four of superradiance in two atoms. And therefore, we can now generalize it to end particles. But before I use the spin algebra to describe end particles, I want to glean some intuition where we just consider-- and this takes us back to the beginning of the course-- where we consider end spins in a magnetic field. And I really invite you to think now completely classically. We'll describe it quantum mechanically in a moment. But I've often said in this course, if in doubt, if you have a classical description and a quantum mechanical and they seem to contradict, usually there is more truth in the classical description. It's so much easier to fool yourself with the formalism of quantum mechanics. So let's take end spins in a magnetic field and ask what happens. So we have end spins. So these are now real spins. They have a real magnetic moment. These are tiny little bar magnets. And we do pi over 2 pulse. And after we've done a pi over 2 products, the spins are aligned like this. Let's assume we had our magnetic field. And now what happens is these spins will precess at the line of frequency. So now you have your end spins. They precess together. And if you have a magnetic moment which oscillates, the classical equation of electromagnetism tells you that you have now a system which radiates. But compared to a single atom, the dipole moment is now n times the single atom dipole moment. So therefore, what do we expect for the radiated power? Well, if the electromagnetic radiation by an oscillating electric or an oscillating magnetic dipole moment scales with a dipole moment squared, therefore, we would expect that the power radiated is proportional to n squared. And that means I have to take the perfect of n [INAUDIBLE]. This means this is n times higher than if you assume you have n individual particles, and each of them emits electromagnetic radiation. what I'm telling you is if you scatter n spins through your laboratory, you excite them. Pi over 2 pulse, they radiate. They radiate a power which is proportion to n. But if you put them all together, localize them better than the wavelengths, their radiated power is proportional to n square, which is an n times enhancement. So the way how I put it for n spins-- and this is a situation of nuclear magnetic resonance-- this is the completely natural picture. But if I would have asked you the question-- let's take n atoms which are excited and put them close together, you say, well, each atom does spontaneous emission, and if you have n atoms, we get n times c intensity you would have gotten a different result. So we are so accustomed to look at spins in NMR as a coherent system, look that all the spins add up to one giant antenna, to one giant oscillating dipole moment, whereas for atoms, we are so much used to saying each atom is its own particle and thus its own thing. So for n excited atoms, they are usually regarded as independent. However-- and this is the message of today-- there shouldn't be a difference. All 2 level systems are equivalent. Side remark-- for NMR in spins, it is much, much easier to observe the effect, because the condition that all the spins are localized within one wavelength is always fulfilled if the wavelength is meter or kilometers. But if you have atoms which radiate at the optical wavelengths, this condition becomes nontrivial. That is partially responsible for the misconception that you treat the two-level system which is a spin in your head differently from the two-level system which is an atom. So the important difference here is lambda. And we have to compare it with a sample size. And usually, the sample size is much larger in the optical domain, and is much, much smaller in the NMR domain. However-- and that's what we'll see during the remainder of this class-- some of the dramatic consequences of superradiance will even survive under suitable conditions in the extended samples. So when we have samples of excited atoms much, much larger than the optical wavelengths, we can still observe superradiance. So therefore, for pedagogical reasons, I first complete the focus on the case that everything is tightly localized. We derive some interesting equations, and then we see how they are modified when we go to extended samples. But I want to say, the intuition from spin systems, the intuition from classical precession and nuclear magnetic resonance, will help us what happens for electronically excited atoms. So we want to use this other spin 1/2 system as a powerful analogy to guide us. So before I start with the angular momentum formalism, I want to emphasize that what are the ingredients here. Well, we're talking about coherence-- coherent radiation, coherence between atoms-- and we'll talk about radiation. And the important part here is the following. That when we talk about radiation, we have the situation that all atoms interact with a common radiation field. In other words, all the spins, all the atoms have to emit their photons into the same mode of the electromagnetic field. And therefore, you may be right in some limit that the atoms are independent, but not the photons they emit. They go into the same mode. And therefore, the emitted photons cannot be treated independently. And that's why the classical picture is so powerful for that. Because in the classic picture, we do a coherent summation of the field amplitudes. So we have constructive interference. The superposition principle of field amplitudes build into our equations and deeply engraved in our brains. And that's why when we use classical arguments, we automatically account for that the photons interfere, that the photons are emitted into the same mode of the electromagnetic field. And eventually, this leads to the phenomenon that we have coherence and enhancement when we look at spontaneous emission for n atoms which are sufficiently localized. So let me also discuss what we have assumed here. Number one is, we have assumed we have a localization of the sample smaller than the optical wavelengths. The other thing-- and this is really important-- we are talking here about a collective phenomenon where n atoms act together and do something. They develop the phenomenon of superradiance. They decay much, much faster than any individual atom could do by itself. But nevertheless, we have not assumed-- or we have actually excluded in our description-- that there is any direct interaction between the atoms. The atoms have no [INAUDIBLE] interaction. They're not forming molecules. They're not part of a solid with shared electrons. The atoms are, in that sense, non-interacting. And therefore, in a way, as long as they are just atoms, independent. Finally-- and I want you to think about it-- you can think about already for two atoms before we generalize it to n atoms. Think about it. What was really the assumption about the atoms? Do the atoms have to be bosons to be in this symmetric state? Can they be fermions? Or can they be even distinguishable particles? If the two atoms where one would be a sodium atom and one would be a rubidium atom-- but let's just say we live in a world where sodium and rubidium atoms emit exactly the same color of light. Would we have been sub and superradiant state for two atoms, one of which is sodium and one of which is rubidium? Yes? AUDIENCE: [INAUDIBLE]. PROFESSOR: Exactly. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yes. And that confuses many people. It is the indistinguishability of the photons they have emitted. It is the common mode where the photons are emitted. The atoms can be distinguished. Also, we've made the assumption that the atoms are localized to within an area which is much smaller than lambda. But you could imagine you have a solid state matrix and you have one atom here, one atom there, and you can go with a microscope and distinguish them. So therefore, the moment you can distinguish them because they are pinned down in a lattice-- or if you don't like a lattice, take two microscopic ion traps a few nanometers apart, and you tightly hold onto two ions-- it doesn't matter whether they are bosons or fermions. It only matters whether your bosons or fermions when the atomic wave functions overlap and you have to symmatrize it. As long as you have two atoms which are spatially separated, it doesn't matter whether they are bosons or fermions. And that also means they can be completely different atoms. You can already call the Boson A, Boson B. Now you can call it sodium and rubidium, and they can have different numbers of nucleus. They can be different numbers of neutrons in the nucleus. It could be different isotopes of the same atom. The whole collective phenomenon comes when they emit a photon into the same mode. OK. So now we want to treat to a formula treatment for end particles. So we have now the individual pseudo spins one half. We perform now with some overall end particles. We get the total spin s. The total spin s quantum number has to be smaller or equal to n over 2, because we have m spin 1/2 systems. The end quantum number is 1/2 times the difference of the atoms which are in the excited state minus the atoms which are in the ground state. And this is, of course, trivial. Trivially must be smaller than s. Because m is this z component of s. And we are now describing the system by the eigenstates s and n of the collective spin. So that means we have the following situation. We have a manifold-- we want to show now all the energy levels. We have a manifold which has a maximum spin n over 2. The next manifold has n over 2 minus 1. And the last one has-- let's assume we have an odd number of particles-- x equals 1/2. So here, we have now n energy levels. We can go from all the n atoms excited to all the n atoms being de-excited. In the following manifold, we have s is one less. And therefore, we have a letter of states which is a little bit shorter. And eventually, for s equals 1/2, we have only two components plus 1/2 and minus 1/2. So those levels interact with the electromagnetic field. The operator of the electromagnetic field, we have already derived that, involves the sum of all of the little sigma pluses, sigma i pluses, and we call the sum of all of them s plus and s minus. And the matrix element is now for spontaneous emission. You have a state with sm. s minus is the lowering operator for the n particle system, so it goes from a state with a certain number of atomic excitations to one excitation less, and that means this is the act of emitting spontaneously one photon. The operator s minus stays within the s manifold, so we state in the same letter, which is characterized by the quantum number s. But we lower the end quantum number by one. The end quantum number is a measure of the number of excitations. And we know from general spin algebra that this matrix element is s minus m plus 1 times s plus n. There are, of course, pre-factors like the dipole matrix element of a single atom. But I always want to normalize things to a single atom. And by just using the square root, if you have a single particle, which is in the s to m equals 1/2 state, then you see that it this square root is just 1. So therefore, for when I discuss now the relative strengths of the transitions between those eigenstates, I've always normalized to a single particle. For a single particle, the transition matrix element is 1. OK. So therefore, what we want to discuss now is, we want to discuss the rate, which is the matrix element squared, or the intensity of the observed radiation relative to a single particle. So the intensity-- and this is what we are talking about-- is now the square root of the square root, or the square of the square root, which is s minus 1 plus m times s plus m. Pretty much, this is the complete description of superradiance for strongly localized atoms. It's all in this one formula. Once we learned how to classify the states, we can just borrow all the results from angular momentum, addition, and angular momentum operators. So I want to use this formula for the intensity and look at which is the most superradiant state, the state where all the particles are symmetric. And this is a state with s equals n where the spin is n over 2. So I'm looking now at the letter of m states, and I want to figure out what happens. So the maximum m state is m equals s, all the atoms are excited. And now the first photon gets emitted. So just put s equals m equals n half into the formula for the intensity, and you find that the intensity gives us this expression is just n. So we have n excited atoms, and they initially emit with an intensity which is n. And this is the same as for n completely independent atoms. So nothing really special to write home about. But now we should go further down the ladder, and let's look at the state which has m equals zero. Well, then the intensity of the matrix element squared for the transition, which goes to m equals zero, has an intensity which is n over 2. I will just look. s is n over 2, and m is zero. So we have the question whether we have odd or even number of particles, but it doesn't really matter. What dominates is always the big factor n over 2. So what we find out is that we have an enhancement, huge enhancement over independent atom, because this intensity goes with n squared, and this proportionality to n squared, this is a hallmark of superradiance. So this is what is characteristic for superradiance. We have an n times enhancement relative to the a singular atom. So this is one important aspect. Now, in the classical picture, that should come very naturally. If you have all the spins aligned and they start the [INAUDIBLE] procession, there is not a lot of oscillating dipole moment. But when half of the spins are de-excited, they are now in the XY plane. Now you have this giant antenna which oscillates and radiates. So it's clear that at the beginning, the effect is less pronounced, and if you're halfway down the Bloch sphere, then you would expect this n times enhancement. But now let's go further down the ladder and ask what happens when we arrive at the end. So I'm asking now, what is the intensity when the last photon gets emitted. There is only one excitation in the system. And the answer is, it's not one like an independent atom. If you inspect the square root expression, you find it's n. So we have one excitation in the system, but it's completely symmetrized. And therefore, we have an n times enhancement. And I want to show you where it comes about. So there's only one particle excited. And here, we have an n times enhancement. By the way, the states with the classification s and m are called the [INAUDIBLE] state. And this state here, which has a single excitation but it radiates n times faster than a single atom, is a very special [INAUDIBLE] state. And there is currently an effort in Professor [INAUDIBLE] lab to realize in a very well-controlled way this special [INAUDIBLE] state in the laboratory. There are non-classical states, because they're not behaving as you would maybe naturally assume a system with a single excitation to behave. So let's maybe try to shed some light on it. One way how you can intuitively understand superradiance is really with a classical antenna picture that you have n spins which form a microscopic dipole moment which oscillates. And this is a very nice picture to understand the n times enhancement when we have half of the atoms excited and the other half de-excited. Let me now give you a nice argument which explains why a single excitation in this system now leads to an n times enhanced decay. The situation is that the initial state for this last photon is, we have an excited atom, and all the atoms are in the ground state. However, we could also have in this nomenclature the second atom excited. Or we could have the third one excited, and so on. So therefore, what we have is-- because we are in the left-most [INAUDIBLE] which has the maximum s spin quantum number of n over 2, that means everything is fully symmetrized. So therefore, we have to fully symmetrize by summing over the n possibilities. And our final state is, of course, the fully symmetrized [INAUDIBLE] state. And now you realize that you have a coherent summation over-- you have n contributions. So therefore, the matrix element has n contributions compared to single atom. The normalization is only square root n. So therefore, the matrix element is square root n times larger than for an individual atom. So by simply having one atom excited and n minus 1 atom not excited, but if you now have the fully symmetrized state, you don't know for fundamental reasons which atom is excited. You have a superposition state where the excitation can be with either of the atom. This state, which has a similar quantum of excitation, radiates n times faster than a single atom would. Let me make a side remark. Maybe some of you remember when we went to [INAUDIBLE] QED, we had just proudly quantized the electromagnetic field, and we discussed the vacuum Rabi splitting. And I told you that the vacuum Rabi splitting is if the cavity is not empty but is filled with n atoms, because of the matrix element of the a dagger operator, you get an enhancement of the vacuum Rabi splitting, which is square root n, then photon number. But then I showed you the important first observation, the pioneering research at Cal Tech by JF Kimball and Gerhard Rempe, and they didn't vary the photon number. They varied the atom number. And when they had more flux in the atomic beam, the Rabi splitting became larger and larger. Well, we've just learned that when we have n atoms, that the matrix element for emitting the photon is square root n times enhanced. So if you put n atoms in a cavity filled with little n photons, the Rabi splitting between the two modes has square root n plus 1 in the photon number and square root big n in the atom number. So the effect I've shown you in the demonstration of the vacuum Rabi splitting is this scaling with the atom number. This actually can be understood as a superradiant effect. OK. So that's pretty much what I wanted to tell you about the basic phenomenon of superradiance. Now I want to discuss two more things. One is superradiance in an extended sample. Ad we have time for that. But I also want to discuss with you the question. Let's assume we have the same system, and we just convinced ourselves, yes, it's superradiant. Photons are emitted n times faster. Now, what would you think will happen when we are not looking for spontaneous emission, but we shine a laser light on it, and we are asking for induced emission? Or the other way around-- we ask-- and you know that from Einstein's treatment, it's completely recipocal-- where we are asking the question, what happens to the absorption process? So is a stimulated emission process or an absorption process, are they also enhanced n times? I don't know. Do you have any opinions about that? AUDIENCE: [INAUDIBLE]. PROFESSOR: Yes. It's a subtle way of counting. I've shown you that certain matrix element-- especially the matrix element when the spin is in the middle, is at 90 degrees-- that we have matrix elements which are n times enhanced. And of course, if you ask for absorption or stimulated emission, we are talking to a system which has an n times enhanced matrix element. And you would say, things go n times faster. Why don't you hold this thought for a moment? Now let me just wear another hat and let me say that we have assumed that we have n independent spins that are closely next to each other, but they're not interacting. And now I take these n spins, and for stimulated emission and absorption, we can just use a picture of Rabi oscillation. On the first Rabi cycle, we emit. On the next Rabi cycle, we absorb. So if I take now-- and why don't you think not about these pseudo spins electronically, atoms with [INAUDIBLE] excitation-- just think of real spins which have a magnetic moment, and you drive them to the magnetic field. So now you have your n little spins. You apply a magnetic field to them, time-dependent magnetic field, and the time-dependent magnetic field is now driving the spin in Rabi oscillations. And the external drive field talks to one spin, talks to the next, talks to all of them. But each of the spins does exactly the same Rabi oscillation it would do if all the other atoms where not present. So the picture is, you have an external field. All the atoms couple to the external field. But the coupling of each atom to this external field is exactly the same as of a single atom coupling to the external field. And the Rabi frequency for each atom is exactly the Rabi frequency you would get for a single atom. So therefore, based on this picture, I would expect that I have my end spins, and these can now be real spins with a magnetic moment, or can be atoms in the electronically excited state. I coherently drive them with a drive field, and they will do Rabi oscillations. But the frequency or Rabi oscillations will be the same as for a single atom. OK. We've just held the thought that their matrix elements in the Dickey states, which are square root n times larger, and the Dickey state seem to suggest something to us which would say there should be an enhancement, whereas the analysis in independent atoms, which are driven by an external field, also seems compelling. So now we have to reconcile the two approaches. Is the question clear? We want to figure out-- we have matrix elements in the Dickey state which suggest enhancement, but the simple picture of n independent atoms driven by an external field would say there is no enhancement. OK. So let me just write down more formally what I explained. When we have an initial state, which is all the atoms are in the ground state to the power n, and the state n would evolve when it is driven in a state phi of t, so now the exact wave function for our n particles is nothing else than the time-dependent solution of Schrodinger's equation for the single particle. So this is single particle to the power n. So therefore, this is pretty much a mathematical proof unless I've made a mistake, which I haven't. So it takes exactly half a single atom Rabi period to completely invert the population exactly s for a single atom. So that's the result. However, if you describe the system by Dickie states, you have matrix element which are matrix elements which are proportional to n. However, if you want to-- I've described it just as a two-level system. Each atom does Rabi oscillation. I've said, OK, the system of n atoms is just n individual systems. But if you insist to describe it as a collective spin, then we have the Dickie states. Then we have the n times enhancement of the matrix element. But then we also have to go through n states. So we have n steps in the Dickie letter ladder. And one can say now-- and this is the exact argument-- you have n steps. You take each step n times faster, but the total time is the same. So n times 1 over n is 1. OK. But now when we talk about spontaneous emission, we are not driving the system with an external field. It's really driven by the system itself, which emits photons into the empty mode. Spontaneous emission, each step is proportional to the matrix element squared, because we're talking about [INAUDIBLE] spontaneous emission. So this is proportional to n square. And if you say that we have n steps, well, then we have n squared over n. Then we have a speed up. Each step is n squared faster. We divide by n, and we get the superradiant speed of which is n. So superradiance is something that you observe in spontaneous emission, but you cannot absorb it in a driven system. Because in a driven system, you can say you have a classical external field. And this external talks to one atom, to n atoms exactly in the same way. It is really the interference of spontaneously emitted photons which is at the heart of superradiance. As a side remark, we are talking here about coherent effect, which is n times enhanced. And you can actually regard that as a kind of bosonic enhancement in the emission of photons, because the photons are bosons. When Bose-Einstein condensation was discovered and people were thinking about basic experiments, of course, one thing which was on our mind is, we wanted to show that there are processes in the Bose-Einstein condensate we are n times enhanced. For fermions, they would be suppressed. This is the flip side. Big enhancements for bosons, complete suppression for fermions. And we found that, for instance, the formation of the condensate had an n times enhancement. There was a stimulation factor. But we also thought you should actually-- there may be ways where you can observe suppression of light scattering or enhancement of light scattering. But we thought about it with two laser beam [INAUDIBLE] scattering, and the idea seemed compelling. And then we said, no, wait a moment. If you use laser beams, everything is stimulated. You can observe bosonic enhancement and fermeonic suppression only when you have spontaneous events. If you drive it in a unitary time evolution, you will not be able to see quantum statistical suppression or enhancement. And the same thing as we have seen here-- when you have a stimulated system, everything is undergoing a unitary time evolution, and the unitary time evolution for n atoms is the same as for a single atom. You need the element of spontaneous emission. So I'm not proving it to you, but I'm just making as a remark-- what we have seen here, that the superradiance only shows up in spontaneous emission and not when we drive the system-- a driven system is a unitary evolution. And the same conclusion which we just got for superradiance also applies if you want to observe fermionic suppression or pulsonic enhancement in quantum gases. It needs an element of spontaneous scattering or spontaneous emission. Yes? AUDIENCE: If we think of it in terms of interference of photons, how does that tie in here? Because if the stimulated [INAUDIBLE] photons are still interfering, then you can get emission [INAUDIBLE]. PROFFESOR: The quick answer is, you have a classical field which you use for-- you have a laser field for stimulate emission for absorption. There are so many photons in the laser field that the few photons which your system emits do not matter. They are really talking to a classical field, and it doesn't matter whether the other n minus 1 atoms have emitted a photon, because you have zillions of photons in your laser field, and they determine the dynamics of the system. OK. Super radiance would not be as important as it is if it could not observed in extended samples. So now I want to use the last 10 minutes to show you what is kept and what has to be dropped when we talk about extended samples. So let's-- it doesn't really matter, but for pedagogical reasons, let's assume we have an extended sample which is much, much longer than the optical wavelengths along the long axis of the cigar and smaller along the short axis. The second condition that the cigar is smaller than along the long axis does not really matter. No, I'm not making this assumption. So it's a cigar, and let's just assume that everything is-- just saw a contradiction in my notes. So anyway, we have now a system which is-- let's have a Cuban cigar, a really thick cigar. And this is now our extended sample. And what I need is, I need the cross section of the sample A. And let's assume the length is l. This is diameter d. It's a cigar much, much larger than here. And yes, we are talking about superradiance, we are talking about spontaneous emission. But if you see a long cigar with excited atoms, you think immediately about lasing action. The photon is emitted is amplified along the path. And of course, the preferred direction where you would expect the maximum effect to happen is when the light is emitted along the long axis of the cigar. So you want to consider now preferential modes along the x-axis. So if you now assume that you have many atoms, and they emit light, if an atom here and here would emit light in this direction, it may constructively interfere. But in another direction, it will destructively interfere. But let us now consider what is the solid angle into which all of the atoms can coherently emit. Well, you know that from classical optics, the emission into a solid angle of lambda square over a can be coherent. Sort of similar to when you have a double slit and you ask, over what angle do the two slits emit inface. You get a bright fringe, and you get a dark fringe, you get the next bright fringe. The coherence, the angle over which the pass lengths differences do not add up to more than lambda. It's the deflection-limited angle which for a beam of size d is lambda over d. And if you take it to the second dimension, the solid angle is lambda squared over d squared. So that's what I'm talking about. So if you would give all the atoms in your assemble just the right phase that they are coherent to emit into the x-axis, they will also coherently emit into a small, solid angle, and the solid angle is given by this number. So the just of it is-- and I will not completely prove it to you, but I just want to give you a taste-- is that therefore, we still have a superradiant enhancement. We know the superradiant enhancement previously when we had the localized system was n. But now we have the n atoms act together, but they're not acting together for emission into 4 pi. They are acting together for emission into the solid angle. And if I write the big n as density n times l times a squared, the a squared cancels out and I get n lambda square l. And if you remember that the cross section of an atom was lambda square for absorption, if the atom is excited, the cross section for amplification of light for stimulated emission is also lambda squared. So lambda squared is the gain cross section. And what we find now as a superradiant enhancement factor is nothing else like something which reminds us of a laser, which reminds us of optical gain. And actually, the lasing phenomenon in superradiance in extended samples has a lot of analogies. In some limits, it's even identical. When we are talking about spontaneous emission, we are not talking about stimulated emission. But if you have a system which is in some excited state superradiant Dickey states, and we are asking what are the spontaneously emitted photons coming out, to say different atoms emit into the same mode, and now you have to add up the feeds coherently, this is a language we have used so far. Or if you use a language atom emits a photon and this photon gets amplified on its way out, those two language strongly overlap or in some limits are even identical. So the amplification of a photon on its way out, this is behind superradiance. But when the localize the atoms to lessen the wavelengths, well, the atoms pretty much emit as a whole and there is no pass lengths of the size of the optical wavelengths where you can say the photon propagates, gets amplified. So we have looked at just what comes out of it. But in extended sample, you could even address this situation. How do the photons get amplified, magnified, augmented when they travel from the center to the edge? So you could actually ask, what is the light intensity as a function of the position within the cigar? For localized samples, you can't. So let me just write that down. So this is analogous to optical amplification in an elongated, inverted medium. OK. So you can formally describe that. You can now define new Dickey states with respect to the preferred mode. And the preferred mode is the mode in the x direction. So what I've done is-- remember, we have those atoms. Those atoms are now sitting at different positions, x1 and x2. And if I define Dickey states which have phase factors into the ik x1, into the ik x2, if now this atom emits a photon and this atom emits a photon, well, the second photon is x2 minus x1 ahead of this photon if you think of those atoms sitting aligned in a string. But the phase vector is exactly canceling the propagation phase for the first atom in such a way that if you are now coupling these states to the electromagnetic field, the phase factors of the electromagnetic field in the mode cancel with those phase factors, and you again have the situation that each state here has an equal amplitude for emission. So now you have n possible contributions, and the normalization is 1 over square root n, and everything falls into place. And you can define that for two excited atoms with two phase factors and so on. So you can use immediately the same formalism. And what happens is those phase factors for the interaction Hamiltonian-- and our interaction Hamiltonian is now different. It is di. And now in an extended sample, we have to keep track of the precision of the atom. So for the coupling to the mode in the x direction, we have those phase factors. So all the phase factors cancel. And actually, I'm not telling you whether this is plus or minus in order to cancel. You pick the sign that they all cancel, and then you have superradiance. You have fully constructive interference. Yes. So all this looks now the same as superradiance, but there are also things which are different, and this is the following. If the atom would emit now photons, not in the preferred mode, then-- remember, we had the Dickey ladder. We had the most superradiant ladder, little bit less superradiant ladder, and eventually we had the subradiant ladder in order of smaller and smaller total quantum number s. Emission in the preferred mode stays in each letter, and we have the superradiant cascade. But emission into other modes is now coupling different s states. Because the operator or the phase factor into the IKR has broken to complete permutation symmetry between the sides. We have changed the symmetry. We have not the completely symmetric sum. We have a symmetric summation with phase factors. So therefore, the phenomenon is somewhat different. But we still have superradiant cascade for the preferred mode. And the result is that we have an enhancement for the most symmetric for the superradiant states, which is given by that. And this is nothing else than the resonant optical density of your center. So in experiments-- many of them go on in [INAUDIBLE] lab, where he uses collective spin and the storage of single photons in n atoms, the figure of [INAUDIBLE] of his samples is always the optical density, the number of atoms times lambda squared times the lengths. Finally-- and sorry for keeping you three more minutes-- the form of superradiance which is very important is Raman superradiance. We don't have an excited state where we put a lot of excitations on, because the excited state would be very short lived. So what we instead do is, we have Rabi frequency omega 1. We have a large [INAUDIBLE] delta. And then the spontaneous emission with the coupling constant g takes us down to the excited state. In the case that the Rabi frequency is much, much smaller than delta, we can eliminate the excited state from the description. And what we obtain is now a system which has an excited state. The widths of this excited state-- this is pretty much the virtual state here-- is the scattering rate which is the probability to excite the atom is Rabi frequencied over detuning squared. That's just perturbation theory. And then we multiply with gamma or gamma over 2. So this is the rate of spontaneous emission out of the virtual state. And from this virtual state, we go now to the ground state. And the Rabi frequency, or the coupling for this virtual state, is the original coupling g between ground and excited state, but now pro-rated by the amplitude that we have mixed the excited state into the virtual state. So therefore, we have now obtained a superradiant system. And for instance, we did experiments which became classic now because they are conceptually so clear-- we took a Bose-Einstein condensate, we switched on one strong of resonant laser beam, and then we had a system which was 100% inverted, because we had no atoms into the final state. The final state is a Bose-Einstein condensate but with a recoil kick. So by just having a Bose-Einstein condensation and shining this laser light on it, we had now in this picture a 100% inverted system, which is the ideal realization of a fully inverted Dicky state. Everything is completely symmetric, and then we observed superradiant emission of light pulses. OK. So this has been the important realize-- so important experiments have been done via BECs in my group and with cold atoms in with laser code samples in [INAUDIBLE]. So why is superradiance so important? And this my last statement for this class and for the semester. So if you have extended sample superradiance, those samples are no longer coupling to the electromagnetic field with the coupling constant g. The coupling constant g is now multiplied by the optical density of your sample. And there is a lot of interest for current research for quantum computation, manipulation of photon states, and all that, to do cavity QED. And in cavity QED, we try to have very good mirrors, very small mode volume to have a very, very large g. But this large g which we achieve in a cavity, if you put many atoms in it, gets enhanced by the optical density. So the cavity enhancement and the superradiant enhancement is multiplicative. And often, it's very favorable for single photon manipulation if you do Bose. You getting enhancement form the cavity and enhancement due to superradiance. And the person who has really pioneered work along this direction is [INAUDIBLE] here at MIT. Anyway, yes, with five minutes delay, I finish the chapter on superradiance. Well, that's the end of this course. Let me thank you for your active participation. Sometimes as a lecturer, you learn as much as the students. And I think partially based on your questions and discussions, this is really true. I've learned a few new aspects of atomic physics. I hope you have learned something, too. And good luck in the future.
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https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/8.04-spring-2016.zip
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PROFESSOR: So we go back to the integral. We think of k. We'll write it as k naught plus k tilde. And then we have psi of x0 equal 1 over square root of 2pi e to the ik naught x-- that part goes out-- integral dk tilde phi of k naught plus k tilde e to the ik tilde x dk. OK. So we're doing this integral. And now we're focusing on the integration near k naught, where the contribution is large. So we write k as k naught plus a little fluctuation. dk will be dk tilde. Wherever you see a k, you must put k naught plus k tilde. And that's it. And why do we have to worry? Well, we basically have now this peak over here, k naught. And we're going to be integrating k tilde, which is the fluctuation, all over the width of this profile. So the relevant region of integration for k tilde is the range from delta k over 2 to minus delta k over 2. So maybe I'll make this picture a little bigger. Here is k naught. And here we're going to be going and integrate in this region. And since this is delta k, the relevant region of integration-- integration-- for k tilde is from minus delta k over 2 to delta k over 2. That's where it's going to range. So all the integral has to be localized in the hump. Otherwise, you don't get any contribution. So the relevant region of integration for the only variable that is there is just that one. Now as you vary this k tilde, you're going to vary the phase. And as the phase changes, well, there's some effect [? on ?] [? it. ?] But if x is equal to 0, the phase is stationary, because k tilde is going to very, but x is equal to 0. No phase is stationary. And therefore, you will get a substantial answer. And that's what we know already. For x is going to 0 or x equal to 0, we're going to get a substantial answer. But now think of the phase in general. So for any x that you choose, the phase will range over some value. So for any x different from 0, the face in the integral will range over minus delta k over 2x and to delta k over 2x. You see, x is here. The phase is k tilde x. Whatever x is, since k tilde is going run in this range, the phase is going to run in that range multiplied by x. So as you do the integral-- now think you're doing this integral. You have a nice, real, smooth function here. And now you have a running phase that you don't manage to make it stationary. Because when x is different from 0, this is not going to be stationary. It's going to vary. But it's going to vary from this value to that value. So the total, as you integrate over that peak, your phase excursion is going to be delta k times x-- total phase excursion is delta k times x. But then that tells you what can happen. As long as this total phase excursion is very small-- so if x is such that delta k times x is significantly less than 1-- or, in fact, I could say less than 1-- there will be a good contribution if x is such that-- then you will get a contribution. And the reason is because the phase is not changing much. You are doing your integral, and the phase is not killing it. On the other hand, if delta k times x-- delta k times x is much bigger than 1, then as you range over the peak, the phase has done many, many cycles and is going to kill the integral. So if k of x is greater than 1, the contribution goes to 0. So let's then just extract the final conclusion from this thing. So psi of x 0 will be sizable in an interval x belonging from minus x0 to x0. So it's some value here minus x0 to x0. If, even for values as long as x0, this product is still about 1-- if for delta k times x0, roughly say of value 1, we have this. And therefore the uncertainty in x would be given by 2x0. So x0 or 2x0, this x0 is basically the uncertainty in x. And you would get that delta k times delta x is roughly equal to 1-- so delta k delta x roughly equal to 1. So I'm dropping factors of 2. In principle here, I should push a 2. But the 2s, or 1s, or pi's at this moment are completely unreliable. But we got to the end of this argument. We have a relation of uncertainties is equal to 1. And the thing that comes to mind immediately is, why didn't Fourier invent the uncertainty principle? Where did we use quantum mechanics here? The answer is nowhere. We didn't use quantum mechanics. We found the relation between wave packets, known to Fourier, known to electrical engineers. The place where quantum mechanics comes about is when you realize that these waves in quantum mechanics, e to the ikx represent states with some values of momentum. So while this is fine and it's a very important intuition, the step that you can follow with is-- it's interesting. And you say that, well, since p, the momentum, is equal to h bar k and that's quantum mechanical-- it involves h bar. It's the whole discussion about these waves of matter particles carrying momentum. You can say-- you can multiply or take a delta here. And you would say, delta p is equal to h bar delta k. So multiplying this equation by an h bar, you would find that delta p, delta x is roughly h bar. And that's quantum mechanical. Now we will make the definitions of delta p and delta x precise and rigorous with precise definitions. Then there is a precise result, which is very neat, which is that delta x times delta p is always greater than or equal than h bar over 2. So this is really exact. But for that, we need to define precisely what we mean by uncertainties, which we will do soon, but not today. So I think it's probably a good idea to do an example, a simple example, to illustrate these relations. And here is one example. You have a phi of k of the form of a step that goes from delta k over 2 to minus delta k over 2, and height 1 over square root of delta k. That's phi of k. It's 0 otherwise-- 0 here, 0 there. Here is 0. Here is a function of k. What do you think? Is this psi of x, the psi x corresponding to this phi of k-- is it going to be a real function or not? Anybody? AUDIENCE: This equation [? is ?] [? true, ?] [? but-- ?] PROFESSOR: Is it true or not? AUDIENCE: I think it is. PROFESSOR: OK. Yes, you're right. It is true. This phi of k is real. And whenever you have a value at some k, there is the same value at minus k. And therefore the star doesn't matter, because it's real. So phi is completely real. So phi of k is equal to phi of minus k. And that should give you a real psi of x-- correct. So some psi of x-- have to do the integral-- psi of x0 is 1 over square root of 2 pi minus delta k over 2 to delta k over 2. The function, which is 1 over delta k in here-- that's the whole function. And the integral was supposed to be from minus infinity to infinity. But since the function only extends from minus delta k over 2 to plus delta k over 2, you restrict the integral to those values. So we've already got the phi of k and then e to the ikx dx. Well, the constants go out-- 2 pi delta k. And we have the integral is an integral over x-- no, I'm sorry. It's an integral over k. What I'm writing here-- dk, of course. And that gives you e to the ikx over ix, evaluated between delta k over 2 and minus delta k over 2. OK, a little simplification gives the final answer. It's delta k over 2pi sine of delta kx over 2 over delta kx over 2. So it's a sine of x over x type function. It's a familiar looking curve. It goes like this. It has some value-- it goes down, up, down, up like that-- symmetric. And here is psi of x and 0. Here is 2 pi over delta k, and minus 2 pi over delta k here. Sine of x over x looks like that. So this function already was defined with the delta k. And what is the delta x here? Well, the delta x is roughly 2 pi over delta k. No, it's-- you could say it's this much or half of that. I took [? it half ?] of that. It doesn't matter. It's approximate that at any rate now. So delta x is this. And therefore the product delta x, delta k, delta x is about 2 pi.
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https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/8.04-spring-2016.zip
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PROFESSOR: What is a Rydberg atom? Well, it's a-- an atom can be a Rydberg atom if the outermost electron is in a very high principal quantum number. That's a definition of a Rydberg atom. The last electron is an n-- little n, very large. Now that is very interesting because when that happens, Rydberg atoms-- when that happens, you have a nucleus with charge ze and you have lots of electrons. And the last electron-- suppose the last electrode is in an orbit further out. You see the orbits do become further out in general. And suppose the last electron is in some large value of little n, so it's somewhat out. So here is the last electron. This electron sees a nucleus with charge z, but it also sees z minus 1 electrons, all the other electrons. The last electron is outside the nucleus and the cloud of the other electrons. So the last electron sees a charge plus 1 z from here and z minus 1 of the electrons. So the last electron sees charge 1. So in some sense, to a good approximation, the last electron says, oh, there's a hydrogen atom here. I'm part of a hydrogen atom. I don't see my friends. The other electrons are too close to the nucleus. And I'm out there going around as if I were hydrogen. So that's a very nice application of hydrogen atoms. So the first question that we want to understand, that it's, again conceptually, is what is the size. If n is large what is the size of the atom? Now I must say, I myself, when I look at these things after a few years that I don't teach quantum mechanics, I look at here and say, OK, this is the solution. Well, maybe the size is na0. Unfortunately, that's completely wrong. And we're going to try to explain what was wrong in looking there. You know, you see your hydrogen atom wave function. There's nothing like really the size, there's no such precise definition as the size. But you could say, what's the expected value of the radius. That's a reasonable definition of size. And you know from this wave function, is going to come out to a0, a0 over 2, 2a0, maybe pi over 2a0. Something like that. So from here you would say, well, it's going to come out to pi and na0 or something like that. That is not true. So how do we see that it's not true? We'll take a little time in a few minutes. But how can we get the size of that atom correctly in an intuitive way? Again, we want to just understand a few results about hydrogen atom that become part of your intuition. So the important result here is the virial theorem. Now whenever I think of the virial theorem I say, oh, there was a factor of 2 or a 1/2 there. How did it go? It takes me a few seconds to try to reconstruct that it's this. For the hydrogen system, for any 1 over r potential, there is this relation within the expectation value of the kinetic energy, the potential energy. You've seen that result probably a couple of times already in this course. Because it's a real important result. It looks like OK, just a theorem about these things. But the intuition is really important. So how does one remember or picture that? One way is the following. And that's the way I like it. I imagine the energy line here. And here is 0. And the one thing I remember is that, yes, there is a kinetic energy here. And actually the bound state energy is exactly the same, but the opposite. That's a way to remember that. That's really what is going on. That's the key thing. The kinetic energy and the bound state energy are just of opposite signs. Why do we see that? Because then we have that V, from this equation, is minus 2T. And if V is minus 2T, means that T plus V is T minus 2T. And T plus V is the total energy. So T plus V is equal to the total energy E of the bound state. For any stationary state, this is true. And this thing, given this condition, is minus T. So that's the way we think of it. So we then have a nice result because the expectation value of the potential would then be minus 2 times the expectation value of T, which is itself minus the energy. So you have a minus 2 times minus the energy. And it's equal to the energy. And this is correct. The expectation value of V is the energy and the expectation value of Eb is negative, and the energy's negative. So what is this? This is the expectation value of minus e squared over r. In fact, of-- yeah, I'll leave it like that. So I'm back to z equal 1, because we're talking about Rydberg atoms. Oh, 2Eb, I'm sorry, here. Thank you. OK, so expectation value of the potential we have here. And 2 times the energy-- yeah, I would have gotten this wrong. Thanks for correcting it before it did damage to the derivation. We have it there, e squared over 2a0, 1 over n squared. So what can we cancel? Well, the 2's cancel. The e squares cancel. The signs cancel. And we get expectation value of 1 over r is equal to 1 over n squared a0, which is suggesting very clearly that the typical radius is not na0. It's n square a0. So this is exact. The virial theorem is exact. The energy is exact. This is exact. That's not quite the expectation value of r. The expectation value of r is not the inverse of the expectation value of 1 over r. It's somewhat related. But there's no theorem, because it would be false, that the expectation value of 1 over a random variable is the expectation value of-- is 1 over the expectation value of the random variable. It's just not true. This nice result for 1 over r is exactly true. And it's l independent. On the other hand, the expectation value of r can be calculated with a bit more effort-- a lot more effort. And it's equal to this. It's just the same thing and 1, with a little correction, which is 1/2 1 minus l times l plus 1 over n squared. So actually, the expectation value of r in the hydrogen atom is l dependent. Not terribly strongly l dependent, but somewhat l dependent. To get an idea, this is equal to n squared a0 times 3/2 for l equals 0. When l is equal to 0, you see this whole bracket becomes 3/2. And for the maximum l, l equals n minus 1, this is roughly 1. This is roughly 0, to a good approximation but not exactly. It becomes n squared a0 with corrections that are very, very small. It's pretty accurate. All right. So, first thing we've learned is that we got a radius, expected radius, that goes back n squared a0. So it's kind of interesting to see what went wrong if you would have thought with a form of the solution. So psi nlm equal Ar to the l Wnl, a polynomial, e to the minus r over na0 Ylm of theta and phi or fnl of r times Ylm of the solid angle. And what do we know about this polynomial? It's of degree n minus l plus 1. That's it. And it depends on r. So what we're looking at is what was the error in thinking that the typical r was na0. And you see, when you make a mistake like this-- the mistake I made of saying oh, na0 must be right-- and you find that it's wrong, it's very important to go back and learn why did you get the wrong answer. That's what we're doing now. So the one question I can ask to begin with is what is a probability density to find the electron between some radius r and a radius r plus dr. So, you know, this is a probability. And it depends on theta and phi. And it's very complicated. How about giving me a probability along r that they can integrate along r and visualize how this is dependant on r? So the probability to find the electron in this shell must be equal to the value of the wave function squared times the volume element. And the volume element here is psi squared r squared dr times-- you would say 4 pi, but it's not spherically symmetric. So you have to integrate over solid angle. That is the volume element. And since I have to integrate over solid angle, I must have psi squared here. So that's the right equation. If you want to make things look perfect, put the d cubed x before the psi squared here. And the problem is that the d cubed x is big enough that it, in some sense, has partial integrals. The So notation is not perfect, but somehow you must imagine this whole volume element that is still infinitesimal but involves some integral already. So you have this. And you get then r squared dr fnl squared and you have the integral, the omega of this Y star lm Ylm. And that integral is exactly 1. Spherical harmonics are normalized. So now I can cancel the dr. And I get that the radial probability distribution, which is a nice concept, is really r squared fnl of r squared. Radial probability. So our mistake-- my mistake must have been that I didn't include all that was relevant. The exponential is one part. But there is the polynomial. And the polynomial must be causing the trouble. Indeed, that's what is happening. Let's look at fnl from the top blackboard. That includes r to the l times that polynomial. And it's a polynomial of that degree. So it begins like a0 plus up to coefficient a prime r to the n minus l plus-- well, minus l minus 1. And then I have the exponential. OK. Now I cannot do this-- I don't want to do this calculation exactly. It's too complicated. So let's ignore the lower part of the polynomial. And we're thinking the radius is going to be reasonably big. So it's a reasonable idea to keep the power of the polynomial that is the largest. So what is the largest? And here you see a nice thing, actually r to the l times this polynomial is a polynomial that begins with r to the l and finishes with r to the n minus 1. So it has like equal number of terms as reaches a value of n minus 1. The last term in the polynomial, when you multiply it in, this begins like r to the l and the last term is r to the n minus 1. So let's take this to be proportional to r to the n minus 1 equal to minus r na0. And here is the fight that actually changes the answer. Because this is fnl. So actually I can write p of r is proportional up to a constant of normalization, and the approximation they have then r squared times this polynomial squared. But I'm taking the last term of the polynomial, so we get r to the 2n e to the minus 2r over na0. That's a probability distribution. And with this probability distribution, then you see what's happening is that there is a fight between an exponential that has a typical length where it decays to half its value, that is related to na0 but the maximum is delayed because it's multiplied by a function that the higher the value of n, the slower it is to take off. x squared or r squared takes off slower than r, takes of slower than r to the 4th. So here you have a thing that just grows like that, but it takes forever to take off. And the result of this is a function that just picks up at some point over here. And we want to find the maximum. So the maximum comes from taking a derivative. So the maximum of p of r is determined by setting the derivative of this equal to 0. So you get 2n over r times the same r to the 2n e to the minus 22 over na0. So the first term, the derivative is 2r and r to the 2n minus 1, which is divided by r and the same thing. The second derivative gives you minus 2 over na0. And that's it. So from these two you get that n over r is equal to 1 over na0. So r equals n squared a0, as we had predicted. So it's the interplay of the polynomial with the other thing that makes the radius go very high. So what about this Rydberg atoms in nature? Well, they've been observed in interstellar gases. You see, there was this phenomenon of recombination when protons captured electrons as the universe cooled off, and formed hydrogen atoms. And that recombination sometimes works in such a way that the proton can capture an electron. And it captures it in a very high quantum number. And it keeps happening as we observe this electron. So people have observed in astrophysics n equal 350. And [INAUDIBLE] here the electron being captured, just not in the usual size but almost a million times bigger. So what happens for this thing r is equal 0.53 times 10 to the minus 10 meters. That's a0. And then you have 350 squared. And that gives you 6.5 microns. That's actually pretty big. A blood cell is 8 microns. A red blood cell is 8 microns. They are not stable because eventually they spiral in. But if you have an atom that, for example, has an n equal to 5, it jumps to n equals to 1 in 10 to the minus 7 seconds. These atoms are rather stable. Instead of lasting 1/10 of a millionth of a second, they can last a millisecond, 1/10 of a second, sometimes even 1 second. It takes a long time to go down that spiral. The energy levels-- if you have energies that go like 1 over n squared, the energy levels, the separation between them, goes like 1 over n cubed. So lots and lots of states there. And it takes a long time for it to decay from one to another. So they observe them in the lab in different ways. They create them with lasers now. You have the outermost electron, you kick it to another orbit with one-- they use three lasers. In the lab at MIT, three lasers. One and two to kick it to an n equal 10, and then the third one to kick it to n equals 60. And they detect those atoms by ionization. A normal atom, you can't ionize. You would need millions of volts per centimeter to ionize it with an electric field. These atoms you can ionize very easily. So they can see that they've been created that way. So also in terms of sizes, the diameter of hair is about 50 microns. Hair is very thin. But you see it. So you're about a factor of 5 or 10 to being able to see that atom with your naked eye. It's pretty impressive. Incredible, in fact. So a nice laboratory, those are almost semi-classical atoms. All what Bohr was doing of calculating, you can derive this law by assuming that the transitions between orbits in the hydrogen atom emit photons of the right frequency. It's all kinds of fun things.
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https://ocw.mit.edu/courses/5-310-laboratory-chemistry-fall-2019/5.310-fall-2019.zip
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PROFESSOR: Take a moment to look around you. Look at the person on your right. Now look at the person on your left. Turn around and look behind you and in front of you. If you have a question on mathematics, or you want to know something about the potential new elements in the cosmic dust of space, or perhaps you want to know something about the architecture of the most beautiful bridge in the world, the Millau Bridge in southwestern France, chances are there is a good possibility that someone in this class may be able to answer your question. There are 11 different majors in this small class. This presents a great opportunity for you to actually mingle with these majors and broaden your ideas. Welcome to 5.310. 5.310 is a non-major sequence in the chemistry department. In chemistry, in MIT lingo actually, this is a 2-8-2 course. 12 credits. In MIT policies, one credit unit is 14 hours of work per semester. So if we look at this, we've actually got 28 hours of lecture, 112 hours of lab, and 28 hours of outside work for a total of 168 hours. But you really only have-- you're really only going to be in lecture for 17 of those hours. And there are five labs. Each lab is four days. That's 20 days times 4 hours is 80 hours. One of the labs is five days so that's 84 hours of lab. This outside work is on the low side so I'm going to fix it. I'm going to increase that to 50. Does that make you feel good? We're moving some hours around here. And there's one more thing. 17 hours of lecture but three of the labs on day four you get out right after the quiz. You can finish the experiment in three days. I know. You're all excited about that, right? So we're really only talking 74 hours in lab here. So I'm going to increase this to 60. So we end up with 151 hours. If you subtract those, you have a surplus of 17 hours. So I just want you to see this so that you feel good about it and you know you're not-- if you look at this and do the calculation, you're going to be spending between 10 and 11 hours a week in this course. And on the long side, if you use these surplus hours, you'll be at the 12 credit units. Now take a look at this slide for a moment. If you look at the slide, you can see that 5.310 ties in to 5.35. 5.35, six, seven, and eight are the URIECA modules, Undergraduate Research Inspired Experimental Chemistry Alternatives. I got it all out. OK. Good. So we've got 12 labs here. Each lab is approximately four units. And what happens is 5.310 ties into those labs. Those are the labs that the chem majors take. So what that means is if you're a freshman, if you're a sophomore or a junior, and you suddenly decide that you like chemistry and you want to change your major, or you want to double major, I'll give you full credit for 5.35 after taking 5.310. So you'll eliminate three of those modules from your program. It's a good deal. Now let's talk a little bit about the course. So is this course useful to my future? It is an introductory laboratory chemistry course. But you will learn basic skills that you'll be able to carry away with you for the rest of your tenure at MIT and beyond MIT. What you're going to learn in this course-- you'll learn about small scale synthesis. You'll learn about inert atmosphere techniques, thin layer chromatography and column chromatography. You'll do an atmospheric distillation and a vacuum distillation. You'll also operate a variety of instruments. And these are the most modern instruments that you're going to find in pharma companies, in chemical companies, and industry when you go out. You'll be operating things like polarimeters, refractometers, density meters. You'll operate a tabletop nuclear magnetic resonance spectrometer, 60 megahertz just for you. You'll operate a robotic GC, an IR spectrometer, UV spectrometer, and a mass spectrometer. You'll also get to put your samples in and watch an inductively coupled plasma mass spectrometer running. There's only two at MIT. And you also will go to the X-ray lab here at MIT. And you'll see the most modern X-ray diffraction machine from Germany running some of your samples from the essential oils lab. In one part of the lab, you'll go on a field trip to the Charles River. Can take your lunch. You're going to bring back water samples. You'll be testing those water samples for dissolved oxygen and phosphate levels. Above all these things, you're going to learn organizational skills that you'll have with you for the rest of your career wherever you go. Can you give me any hot tips? Students always ask for a hot tip. And I can tell you one thing that actually works for me. And that is sometimes we all get frustrated and discouraged. And when that happens, I usually find a quiet place and I talk to my brain. And I tell my brain everything is OK. Everything is going to be fine. And time and time again, this works for me. I can get myself out of it. And if you read Oliver Sachs, in one of his books, he says that if you can imagine things, they actually can turn into reality. So are there any brain and cognitive science folks? Oh good. Look at this. I've got four of them. Can any of you offer any comments on how this helps me? Is there something there? Anyone? No takers. Anyone else? So I guess the thinking about, you must become happier and clear your thoughts out. And if you become happier and clear your thoughts, you become smarter too. Right? And I actually brought a brain that I keep in my refrigerator, a beautiful brain. And I'm just going to set it here so that you can admire it. So the brain is very important. Some of you in this course are very, very smart. You're all very smart because you got into MIT or at the top of your class. You're valedictorians. I recall several years ago I had a freshman advisee who came to me. And she actually tested out of every subject. She sat down and I could not find one subject to put this student in. So I was kind of beside myself. I went out of my office and I bumped into the head of the physics department walking down the hall and said, I've got this advisee and I don't know what to do. Did you ever have one like this? He said once every 10 years. So I went back and I put her into all the most advanced classes, some with graduate students. And then I ask her how she managed to do all this because she came from an island in the Pacific. And she said to me OCW, OpenCourseWare. I watched the lectures that MIT puts out and I learned it all myself. But there are geniuses sprinkled throughout MIT. And there are probably geniuses in this class. Should you still take 5.310? Yes, because geniuses need practical skills. And in 5.310 can help you to bind those basic skills and make your genius become reality. Now there are a couple other key things to help you succeed in this course. One, if you feel like you need help and you don't know what to do, you're lost, come to see us. And we'll get you back on track. The second thing, which is important, is you need to be able to accept your mistakes. If you make a mistake, don't start crying and say it's all over. I blew this lab. I just messed everything up. I'm done for here. I don't know what to do. Just accept the mistake and ask yourself, what can I get out of this mistake that I made. And when you do that, you really can help yourself for the long term. I mean it's not anything embarrassing if a student works for three days and then, by accident, they drop their product. It happens. So don't feel bad about it. Just come to us and we'll work something out. But don't get stressed out over it. The last thing that could really help you here is, in this class, it's all about lab reports. So you've got to write these four lab reports and give an oral report at the end. My advice to you is don't wait until the last day of the last weekend when these reports are due to actually try to write this report out. Start early. When you're in the lab, a four hour lab, and a lot of those labs you'll get done at four so you'll have an hour. We have beautiful write up areas in the new undergraduate chemistry lab. You can sit there. And that's the point where you can actually write out a paragraph about what you did that day, what you actually did, what you saw, and what you found. And that's your procedure and observations for your lab report. And while it's fresh in your mind, you do it. And the procedure and observations typically is about a page and a half, no more than two pages, of your lab report. You could also work on the background of the report because you're talking about why you're doing this experiment, what you're going to get out of it, something about the history of the experiment, what was done before. So you can work on these sections along the way rather than wait until the very end and do a marathon session, try to trying to get the lab report done. So with those things, I'd like to talk a little bit about the course. And I'm going to cover seven broad areas. One is academic integrity. The second thing is the lab policies and then, most importantly, grading, how this course will be graded. And then we'll look a little bit at safety, because you do have a safety lecture you're going to. And we'll look a little bit at the lab notebooks that are required for the course. And we'll talk a little bit about waste management. And finally, we'll spend a little bit of time talking about calibration of instruments. So let's start with academic honesty. MIT has one of the best integrity programs of any school that I know. They have a website devoted to it. They also have printed material that you can pick up and you can read through it. And I think probably to summarize it in just one sentence, you don't want to present as your work the efforts and product of another person. And the penalties can be quite severe. You could actually fail the assignment. Could fail the course. You could be suspended from the Institute. You could even forfeit your degree. So it's pretty serious business. In 5.310, I think there are two areas that you need to actually look at. One of them is there are a lot of lab reports out there. They're in the dorms. And they're in the sororities and fraternities. And they call them bibles. So you don't want to actually go out and take pieces out of those and put them in your own reports. The reason for that is it's not right to do that. And the second reason is we could have an electronic copy of one of those reports on file. That would not be good. The other thing is I guess the innocent thing is when students sit down together and they're talking about the lab. You might talk with your lab partner. You can talk and you're writing things down. But what you don't want to end up happening is you don't want to have the same sentences in both lab reports. So you've got to put things in your own words even when you talk with each other and you're writing your reports up. That's pretty much all I have to say about academic integrity. S the undergraduate lab policies-- you're actually picking to work on either Monday/Wednesday or Tuesday/Thursday. And depending on the safety lecture that you elected, some of you went yesterday to the safety lecture and you enrolled in the Monday/Wednesday section. Today the safety lecture is at 1 o'clock right after this class. And Tuesday/Thursday people will go there, attend the safety lecture, and then you'll go up to lab to check in your lockers. If there are any Tuesday/Thursday people who would like to switch to Monday/Wednesday, Monday/Wednesday will have fewer students. So there'll be much better TA interaction with you. So if you can, just let Sarah know. She'll be down front at 1 o'clock registering you for your lockers. And if you'd like to do Monday/Wednesday, you can check in today. We'll just give you a Monday/Wednesday locker. The lab itself opens at 1 o'clock every day. And the TAs will give a pre-lab lecture at 1:05. It's about a 20 minute lecture. It's going to cover exactly what you're doing in that four hour period. And they will also demonstrate some of the glassware, maybe the pipettes, and anything you're going to be using in that lab. And they'll also show you the instruments that you might be running that day. So it's pretty important, the pre-lab lecture. And it will also firm up what you heard in the lecture and possibly didn't understand something, your TAs can really help in that respect. On the fourth lab day of each experiment, there's a quiz. And that will cover-- you've already done the three days of the experiment so you should be in pretty good shape for the quiz. You should understand what you're doing and know. You shouldn't have any problem with that. The laboratory, we try to clean up about quarter to 5:00. I can't recall in 5.310 any time where students have to stay after 5 o'clock. That's a good thing because you want to go home. Some of you have sport practice. And we understand that. And the last thing is we've indicated some select Fridays that are make-up labs. And there will be one Friday at the end of each four day lab period. So if for some reason you've missed something because you're sick and you couldn't do it on day four, then we have that option. And those are scheduled in the lab syllabus in your packets. Safety goggles. You have to buy these at the VWR stockroom, the basement of building 56. And I mean, you put these on and they actually really-- they hug your face. They don't look too glamorous, but you're not going to a beauty contest. You're going to the lab. But you've got to buy a pair of these. And you've got to wear them at all times. The other thing is your lab coat is a fire-resistant lab coat. These are top of the line lab coats. We issue you one of these when you come. We'll also give you a baggy like this. And you can write your name on the bag and hang it in your locker and leave it there. You never take it out of the lab. You leave it there. And first thing you do when you come in, at either entrance, either from building 13 or building 16 when you enter building 12, you'll have lockers. And you can grab your lab coat and goggles, put them on, and you'll be ready for the lab. We've got plenty of gloves. We use nitrile gloves. And generally, those gloves do not give students any problems. But if you have an issue with that, just let us know. The attire. So the lab coats have to be worn at all times. And you can't wear-- you can't come in with open-toed shoes, low cut jeans, t-shirts. You have to be really covered. So my suggestion is that you bring a little bag, a change of clothes, and you keep it in the locker. Students did this last semester. It worked really well. And then you can just change out and change out again when you're leaving. And that way you can wear your sandals, and your shorts, and anything you want, but you can't wear that in the lab. Cell phones, radios, iPods. You have to keep those in your backpacks. And this is what the lockers look like at the entryway. So you just grab one of those. And that'll become your locker. Once you hang your lab coat in there, no one else is going to put anything in there. And we've never had any problems with the lockers in the undergraduate lab. If you want to put a lock on your locker for the semester, that's fine with me. Obviously, no eating and drinking in the lab. You can't bring beverages and food into the lab. There are chemicals on the countertops and around that you just-- chemicals and food just don't mix. This is very important. Report any accidents or injuries promptly. So if you do get cut or you spill a chemical on yourself, you should tell us. If you leave the lab and then a couple of days later come back and you've broken out in a red rash on your legs or arms, it's much harder for us to go back and try to figure out what happened. But if you tell us right away, then we can track it down. We know what chemical it was. We know how to treat it. And MIT Health will help with that. If you need special accommodations or you've got medical conditions that you would like to talk about, you can come and see me on that. And you should do it this week just so we're aware of it. And I mean we don't want-- we had a student about three years ago who just passed out in the lab. I mean literally down on the floor. She was out. And if we know ahead of time if there is a condition or something that we should be aware of, we can know better how to treat that. Grading. This is the grading scale. Pretty traditional. It's not inflated and it's not curved. So it's a traditional grading scale. And this is what we use in 5.310. So the grading is all based on five labs. Each lab is worth 100 points. Total number of points is 500 for the course. And this will make you happy. There's no final exam. There was a final exam when I took over this course 10 years ago. But I got rid of it. It was not nice. So you've got 500 points. And how is that broken down? So of that 100 points, 20 points is your quiz that you'll take. So that incorporates into that. Then you have your pre-lab notebook and your post-lab notebook. That's 10 points. Then there are 5 points which are noted as discretionary points for the TA. And that would be like does this student show up on time. Does the student clean up their area at the end of the lab? Does the student wear their safety glasses or do they walk around with the glasses up in the air? So that's important. And then your lab report, either written or oral, would be 65 points. I'll show you this is the course textbook. This is the book that we recommend that you get. It's Mohrig. and it's Laboratory Techniques in Organic Chemistry. Very good book. It has chapters on all of the instrumentation and a lot of the techniques that you'll be actually using in the course. Well worth to buy this book. The other book that is not required but it's on reserve and we have copies in the lab is this ACS style guide. This will help you in terms of writing up your lab reports, writing up the reference section, knowing how correctly to put things together. Attendance. Attendance is not mandatory for this class. But there is an attendance sheet going around. And each lecture day we will send it around. If you don't attend lecture, your grade isn't going to be penalized. But if you do attend all of the lectures, at the end of the semester, if your grade somehow is within a half a point of a higher grade, 89.5, technically you'd get a B plus, we'll look at the lecture attendance and that grade could be pushed up to an A minus. So that's how the lecture attendance works into this system. And if you have any questions as I'm going along, just ask. So to get a passing grade in 5.310, you need to turn in the four written reports and you need to deliver the oral report. There are some penalties. So for a late lab, we use three times n minus 1 plus 2, where n equals the number of days the lab is late. Also, with each lab, you need to attach a cover sheet to your written lab. No cover sheet is minus 2 points. There are also some late points on the oral report. If you're late for an oral report, it's minus five points. If it has to be rescheduled, it's minus 10 points, just so you're aware of the late points. You don't want to use these if you don't have to. You really want to try to turn your labs in on time. And that two points or five points could make a big difference at the end. So this is the hard copy of the course manual. And you'll be allowed to bring this in with you to lab. You don't have to write up the pre-lab step by step that you're going to do. You can bring this in and follow it. So we're saving you a couple hours a week here. I hope you appreciate that. Students used to have to write the pre-lab out step by step in their own words. So this is a big time saver for you. I want to keep this course under that 12 credit limits. You will need a lab notebook. And the lab notebooks are available either at the MIT Coop or they have them at the VWR stock room. And they're around $15. And the notebook should be like 100 duplicate pages with carbonless paper because what's going to happen is each day you go to lab, the TA will initial your pre-lab. And then at the end of lab, you'll go to the TA with your notebook. And the TA will initial each page of your post-lab notes. Then you'll tear them all out and hand them to your TA. You keep the original. We get the copy. The reason for that is when it comes down to the final lab report, you cannot have anything in the lab report that's not in your post-lab notes in your notebook. So let's just take a quick look and see. For the pre-lab, this is pretty simple. You don't have to write out the whole procedure, but you do need to have your title, date, the name of the experiment, an introduction, a couple of sentences what this experiment is about, what you hope to get out of it, maybe a couple of sentences on safety issues that you spotted in the experiment, and then any pre-lab equations you think you'll need. The first pre-lab is due Monday and Tuesday next week. That's when the lab actually will commence. Tables are very good to put in your lab notebook and in your lab reports as well as drawings that might help you to actually visualize what you're setting up. For the post-lab, this is probably the most detailed part because you've got to write everything down that you're doing in lab, all your observations, everything that's happening. When you're over at the distillation, you're going to record the temperature that the distillate started to come over. You're going to record the drops per minute that's coming over. All of these notes go into your lab book while you're working in the lab. So let me show you a couple examples of lab notebooks, what students have done. This is one example where a student had-- sorry-- where a student had put their pre-lab notes on one side and left the other side for their post-lab. This isn't necessarily the best way to do it because you don't know how much space to leave for your post-lab. So another way to do it would be just to write out your pre-lab on one page, and then stop, and then start your post-lab on the next page. The other thing that comes up is these are four to five day labs. So when you read the whole experiment, one way to do it is write your lab up for the whole experiment all in one time. Then you don't have to write it up each time. That works for a lot of students. But we only require you to write the pre-lab for the day that you're actually doing. So I want to talk just a bit about some questions that come up a lot, like what is the stability of my chemicals. You're going to be working with some very air sensitive chemicals. So you might go over to the balance area and mass out your chemical for the first lab. You have iron chloride tetrahydrate. You mass that out but then you walk away. If you don't put the cover on the chemical, the green chemical will turn brown. And it's not going to be any good for the other students who are doing the lab because the iron gets oxidized. And you need iron-2 to do the experiment. So be aware of that. And then how do I get what I need from a stock solution? So you're going to see chemicals like this labeled stock solutions for the lab. The one thing you don't want to do is open this up and go into a stock solution with a dirty pipette, because if you do that, you'll contaminate the solution for the whole class. And some of the experiments are really quantitative and very sensitive, especially the Charles River where we're looking at the phosphate levels. So what you want to do is you want to make sure that-- and the TA should do this-- have these chemicals poured out in labeled beakers so that they're ready for the students to actually go into and draw out what they need. If you don't see that, always ask your TA. Never go directly into the stock solution. And then the order in which you add chemicals is very important. You always want to-- you always want to go from the concentrated to the less concentrated. You would never want to take water and add it to acid because it would splash right back in your face. You always want to go the reverse. Add the acid to the water. Concentrated to less concentrated. And if you spill chemicals while working, let us know. It's very important. Don't leave them. A couple or few years ago, we had a class during IAP. And one of the freshmen spilled a half a bottle of urea on the floor and left it. And the TA was beside himself. He was saying who did this. Nobody would own up to it. But we just happened to be filming this class. And there is a video online that you can go watch. And you can see-- you'll know who it is when you watch the video. So you always want to let us know. And we always want to clean up if there is a spill. And if it's a spill, you have to let us know right away. We have to determine how bad it is so we don't have to evacuate the lab or something. And then if you get chemicals on yourself, you know you've got to let us know because chemicals on yourself, we can take care of it right on the spot, get you washed off. If we have to, we'll walk you over to MIT Health to have them look at it. And chemical waste. There are a lot of different kinds of waste containers in the lab. We've got boxes here for glass and plastic. And you've got to look at the box before you throw something in. Broken glass would go in the glass box. Plastic pipette tips go in the plastic box. And make sure that you empty them of your chemical first into the waste container before you throw the pipette's tips into the box. There are bottles for liquid waste. And everything is labeled. You can see the red labels here. A couple of things you should know. Don't pour into the top of this. This is a lid so that it actually opens up and then you pour it in. And then you always have to close this container after you've added your waste. If you leave a waste container open during an active lab and we're inspected, MIT can be fined thousands of dollars. So you've got to keep the waste containers closed. The other thing is you've got to be aware of what you're putting in. So read the container. Make sure it's the chemical that you're using that's going in. You never want to put acetone in to anything except a bottle that says acetone, because we're going to be using hydrogen peroxide in this class. And hydrogen peroxide and acetone and you have an explosion. So what you're making is you're actually creating these acetone peroxy compounds. This is the dimer. It's called acetone peroxide explosive apex. And then there's also a trimer. And this is the triacetone peroxide. So what happens with these is they form a white crystalline powder. And it smells like bleach. And when that forms, just any movement could trigger it. This is the same stuff that the shoe bomber had in his shoe if some of you remember 10 years ago on the plane. And so it's very frightening. So you have to be mindful of what you're putting in the containers. The acetone we're going to recycle. So it will have containers just for acetone. And that's where you're going to dump it. The other thing you want to be aware of is nitric acid. We're using nitric acid in the Charles River lab. You don't want to mix this with any kind of organic solvents. If we put this, say, with ethanol, then you're making this C2H5 ethyl nitrate. And this is also a primary explosive. The first thing that happens is you'll see orange smoke billowing out. And then you'll hear this whining noise. And it gets louder and louder. And then the hood goes. The ceiling goes. So if you see smoke or you hear noises, get away from the hood. But don't put nitric acid with anything except in the container that says nitric acid. There are also some containers for solid waste by the scales. So as you're massing your chemicals out, if you have a little bit left over, they go in the can with you with your weigh boats. Don't put gloves into the cans. Gloves can go in the trash. We have a box for needles here, which is great. And don't put needles in the trash, because the cleaners grab those trash bags at night and they could get poked. And that would not be good. So the needles, we're going to try to count them out. And the TA will be judicious in coming around with the needle box and collect them at the end. Calibration of melting point. I just want to show you one thing here. If you go to the melting point and you take a melting point of your sample, an uncalibrated melting temperature might be 64 to 66 degrees. Once it's calibrated, you're up to 79 to 80.9 degrees. So we have four standards. And each standard has a range where it starts to melt and ends melting. So you put your standard in. Then you look through the scope. And you record your observed where it's melting, when it started, when it stopped. And you have two points for each standard. So you can do a linear regression and draw a straight line and get an equation for the line. And then every time that you use that melting temperature, you can calculate-- you put in your experimental and calculate what it should be from the equation. So we've got we've got a couple of minutes left. And this is a chemistry class. So I'd like to do a demonstration just to end this class. And I need a volunteer. Yes. Come on. And what is your name? AUDIENCE: Autumn. PROFESSOR: Autumn. OK, Autumn. So Autumn, I've got a cup here and some water. So I'm going to pour some water into the cup. And you tell me when to stop. OK? AUDIENCE: Stop. PROFESSOR: OK. So what I'm going to do is I'm going to cover this up. OK. And then I'm going to carefully turn this over. So far so good, right? You know where I'm going with this, Autumn? I'm going to put it on your head. Face your fellow students. And we'll go up here very gently. And there we go. Now I'm going to let you hold it, Autumn, because I don't want to be responsible. OK. OK. You OK so far? AUDIENCE: Yep. PROFESSOR: OK. All right. Autumn, what does this say? AUDIENCE: Do not remove this card. PROFESSOR: Whoops. OK. Let me take this off from Autumn. AUDIENCE: What's on my head? PROFESSOR: What happened to the water, Autumn? AUDIENCE: Is there something inside that absorbed it? PROFESSOR: She said is there something inside that absorbed it? OK. Let's see what we got here. We made a polymer. AUDIENCE: Cool. PROFESSOR: So unbeknownst to Autumn, I had a little bit of powder in her cup. And when I put the water in, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1. AUDIENCE: What kind of polymer is it? PROFESSOR: I knew you were going to ask that. What kind of polymer is this? AUDIENCE: I'm Course three. PROFESSOR: OK. So you'll all know this. Hold that for a minute, Autumn. This is the diaper polymer. This is sodium polyacrylate. And I actually took one of these diapers. And I cut it open. And I got 4 grams of sodium polyacrylate out of the diaper. So this is how they work. These polymers, you can see this polyacrylate is a monomer repeating unit. And what happens is if you look inside, it has a lot of sodiums. But when we put the water around it, it starts to send sodium out and pull water in by osmosis to balance the sodium atoms. So all of a sudden it swells up. And you've got this system here. Now we're going to do a little experiment. We're going to take this and put those on now. OK. So I'm going to take this and put it into a little baggy. Hold that bag. AUDIENCE: Add salt [INAUDIBLE]. PROFESSOR: I'm going to add some salt to this. And now we're doing the reverse of what we just did. We had added water to the polymer before but now we're putting salt in. So just zip that bag up. And this is the fun part, Autumn. You get to squish it around. So now there are sodium ions on the outside of the polymer. And it's panicking. It's starting to pull them in and send some of the water back out. So the polymer is actually going back to a liquid. So that's how this works. Great. Give her a hand. She did a great job. Thank you so much, Autumn. Well, thank you all. And the next lecture is Tuesday. Don't forget to go to the safety lecture. And that is downstairs, two flights down.
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https://ocw.mit.edu/courses/8-03sc-physics-iii-vibrations-and-waves-fall-2016/8.03sc-fall-2016.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. YEN-JIE LEE: OK, so welcome back everybody. Happy to see you again. So today, we are going to continue our exploration and understand single harmonic oscillator. And this is actually a list of our goals. And we would like to learn how to translate physical situations into mathematics so that we can actually solve the physical problem, so we actually have and predict what is going to happen afterward. And we also sort of started this course by solving really simple examples, single harmonic oscillators. And as a function of time, you will see that, for our next class, the next lecture, we are going to bring in more and more objects. And of course, more objects means more excitement also, in terms of phenomena, but also more complication on the mathematics. So we will see how things go. And then after that, we are going to go through infinite number of oscillators to see what will happen. Of course, we will produce waves. That's very exciting. Then we'll do all kinds of different tricks to do those waves. So what we have we learned last time? So last time we went over example, a simple harmonic oscillator. It says you have a rod fixed on the wall, and you can actually go back and forth. And we also introduced a model of the drag force, or drag torque, and that's actually proportionate to the velocity of the motion of that single particle. And the interesting thing we learned last time is that we have three completely different behaviors if we actually turn on the drag force. The first one is on that damped. Damping is actually very small. Then we have the solution in this form. It's oscillating. The amplitude is decaying exponentially. As we make the drag force larger and larger, you will pass a critical point, which actually give you a solution, which you don't have oscillation anymore. The cosine disappeared. Finally, if you actually put the whole system into water or introduce something really dramatic-- a very, very big drag force-- then you have overdamp situation. And there you see that the solution is actually a sum of two exponential functions. So this is actually the one equation which actually works for all the damped situation we discussed up to now. And this is actually the map. Basically, if gamma goes to zero-- gamma actually controls the size of the drag force. Then we got no damping. Then you have a pure, simple harmonic motion. And as we increase the gamma, then you get see that the behavior is changing as we increase the gamma. So you can see that we can use a quantity, which is called Q. Q is actually defined as a ratio of omega at zero, which is basically the natural angular frequency of the system. And gamma is a measure of how big the drag force is. If we make a ratio of this to quantity, then you'll see that, at Q equal to 0.5, it reaches a critical point, which actually the behavior of the whole system changed. And you can see that the oscillation completely disappeared. So that is actually what we have learned last time. So what are we going to do today? We have been really doing experiments really with our hands, hands-on, right? So basically we will prepare the system. Then we release it. Then we don't touch it again and see how this system actually evolves as a function of time. So that's what we have been doing. So today, what we are going to do is to start to drive this system. We can introduce some kind of driving force and see how the system will respond to this external force. So that is actually what we are going to do. And that will bring us to the situation of damped driven harmonic oscillator. So let's immediately get started. So we will use the example which we went through last time as a starting point. So set example from the last lecture is a rod, which is fixed on the wall. And the lens of this rod is over two. And I define a counter-clockwise direction to be positive. And I measure the position of the rod by this theta, which is the angle between the vertical direction and the pointing direction of the rod. And we have went through with the math, and we got the equation of motion without external force, which is already shown on the blackboard. So now, as I mentioned at the beginning, I would like to add a driving force, or driving torque, tau drive. This is equal to d0 cosine omega d t. So I am adding a driving torque. The amplitude of the torque is actually d0. And there's actually also a harmonic oscillating force, or torque, and the angle frequency of this torque is omega d. And that means our total torque, tau of t, will be equal to tau g t-- this is actually coming from the gravitational force-- plus tau drag, which is to account for the drag force. So this time we are adding in a tau drag. So I'm not going to go over all the calculations on how did the right from the beginning to the end. But I will just continue from what we actually started the last time. So if I have additional driving torque there, that means my equation of motion will be slightly modified. This time, my equation of motion will become theta double dot plus gamma theta dot plus omega 0 squared theta, and that is equal to d0 divided by I. This is actually divided by I because, in order to get the acceleration, I'll need to divide my torque by a movement of inertia of this system and cosine omega d t. This is actually the oscillating frequency of the driving torque. And just a reminder, gamma is defined to be equal to 3b m l squared. And the omega 0 is actually defined to be square root of 3g over 2l. So as I mentioned in the beginning, this is actually giving you a sense of the size of the drag force. And the right hand side, the omega 0 is actually the natural angular frequency. So, of course, we can actually simplify this by replacing this term, or this constant, by symbol. So the symbol I'm choosing is f0. And this is defined to be d0 divided by I. Therefore, I arrive at my final equation of motion-- theta double dot plus gamma theta dot plus omega 0 squared theta, and that is equal to f0 cosine omega d t. So I hope this looks pretty straightforward to you. So this is our equation of motion you can see from this slide. So we have three terms in addition to the theta double dot. The first one is actually related drag force, or drag torque. The second one is actually related to so-called spring force. So that is actually be related to the spring constant or, because of the restoring force of the gravitational force. The third one is actually what we just add in as a driving force. So one question which I would like to ask you is-- so now I bring one more complication to this system. So now I am driving this system with a different frequency, which is omega d. The question is, what would be the resulting oscillation frequency of this driven harmonic oscillator? What is going to happen? Well, this system actually follows the original damped oscillator frequency, omega, which is actually close to omega 0. Or what this system actually follows the driving force frequency. Finally, maybe this system chooses to do something in between. We don't know what is going to happen. So our job today is to solve these equation of motion and to see what we can learn from the mathematics. Then we can actually check those results to see if that agrees with the experimental result, which is through those demos, OK? So as usual, I have this equation of motion here. So one trick, which I have been using, is to go to complex notation, right? Therefore, I can now re-write this thing to be Z double dot plus gamma Z dot plus omega 0 squared Z, and that is equal to f0 exponential i omega d t. So basically, I just go to the complex notation. And we would like to solve this equation. So in order to solve this equation, I make a guess, a test function. I guess Z of t has this functional form. This is equal to A-- some kind of amplitude-- exponential i omega d t minus delta. Delta is actually some kind of angle, which is actually to account for the possible delay of the system. So if I start to try-- for example, this is a system, which I am interested-- I start to drive this system, it may take some time for the system to react to your driving force. So that's actually accounted for by this delta constant. And the amplitude is actually what we were wondering, what would be the amplitude. Therefore, you have some kind of a constant in front of the exponential function. And you can see that this exponential is actually having angular frequency, omega d. And that is actually designed to cancel this exponential i omega d t here in the drag force. So now we can as you calculate what z dot of t would be equal to i omega d Z. Z double dot t will be equal to minus omega d squared Z. With those, we can now plug that back into this equation of motion and see what we can actually learn from there. So basically, what I'm going to do is to insert all those things back into the equation of motion. And that is actually going to be like this. Basically, the first term, the double dot, you get a minus omega d squared out of it. The second term, gamma Z dot-- I have Z dots here. Basically, I would get plus i omega d gamma out of the second term. That's third term, I get omega 0 squared out of it. And that is actually multiplied by Z. And this is equal to f0 exponential i omega d t. All right, and we also know from this expression Z is equal to A exponential i omega d t minus delta. That's the test function. So this is actually equal to A exponential i omega d t minus delta. So now what I can do is-- I have some constant in the front. Multiply it by exponential i omega d t. And now I can actually cancel this exponential i omega d t with the right hand side term. Very good. The whole equation is actually exponential free. Now I don't have any exponential function left. And exponential i delta is actually just a constant. So now this equation is actually independent of time. So what I getting is like this-- basically if I multiply the both sides by exponential i delta, then I get minus omega d squared plus i omega d t plus omega 0 squared A. And this is going to be equal to f exponential i delta because I multiply both sides by exponential i delta. And this is equal to f cosine delta plus i f sine delta. Just your last equation. Any questions so far? So look at what I have been doing. So I have this equation of motion. As usual, I go to complex notation. Then I guess Z equal to A exponential i omega d t minus delta because my friends from the math department already solved this, and I'm just following it. Then I can calculate all those terms, plug in e, and basically, you will arrive at this equation. This equation is a complex equation. So what does that mean? That means one equation is equal to two equations because you have the real part, you have the imaginary part. Therefore, that's very nice because I have two unknowns. The first one is A, a constant. And the second one is delta. Now I have two equations I can solve what would be the functional form for A and the delta. And let me go immediately solve this equation. So if I take the real part from this equation, basically what I'm going to get is omega 0 squared-- this is real-- minus omega d squared-- this is also real-- times A. A is actually some real number. This is actually equal to f cosine-- f0. Sorry, I missed a zero here. So that zero I missed. This should be f0. f0 cosine delta. And I can also collect all the terms, which is imaginary terms. Then I get only the second term from the left hand side is with i in front of it. Therefore, I get omega d gamma A from the left hand side. And from the right hand side, there's only one imaginary term. Therefore, I get-- this is equal to f0 sine delta. So now I have two equations. I have two unknowns. Therefore, I can easily solve A and delta. So I call this equation number one. I call this equation number two. So now I can-- sounding in quadrature the two equations-- in quadrature. And the left-hand side will give you A squared omega 0 squared minus omega d squared squared plus omega d squared gamma squared. That is actually coming from the second equation. That gives you the left hand side. It's a square of the sum the first and second equation. And the right hand side will become f0 square cosine delta cosine squared delta plus sine squared delta. And this is equal to 1. So that's actually the trick to get rid of delta. Then I can get what will be the resulting A. A is actually a function of omega d. Omega d is given to you. It's actually determined by you-- how fast do you want to oscillate this system. And this is equal to f0 divided by square root of this whole thing. So this will give you omega 0 squared minus omega d squared squared plus omega d squared gamma squared. Then we can also calculate what would be the delta. The trick is to take a ratio between equation number two and the equation number one-- 2 divided by 1. Basically, you will get tangent delta. This is sine divided by cosine. f0 actually cancel. This is equal to what? Equal to the ratio of these two terms. After you take the ratio, A drops out. Basically, what you get is gamma omega d divided by omega 0 squared minus omega d squared. So we have solved A and the delta through this exercise. So what does that mean? Originally, I assume my solution to be A exponential i omega d t minus delta. Therefore, I would like to go back to the real world, which is actually theta. So basically, if I take the real part, I would get theta of t, which is actually the real part of Z. And that will give you A omega d cosine omega d t minus delta is also a function of omega d. So we have done this exercise. And you can see that the first thing which we see here is that there's no free parameter from this solution. A is decided by omega d. And delta is also decided by omega d. There's a lot of math, but actually we have overcome those and that we have a solution. But it is actually clear to you that this cannot be the full story. Because you have a second-order differential equation, you need to have two free parameters in the solution. What is actually missing? Anybody can tell me what is missing. AUDIENCE: The homogeneous solution. YEN-JIE LEE: Very good. The homogeneous solution is missing. So that's actually why we actually have no free parameter here. Once is the omega d is determined, once the f0 is given, then you have the functional form which decides what is actually theta. So what in actually the full solution? A full solution should be, as you said, a combination of homogeneous solution and the particular solution which we actually got here. So if I prepare the system to be in a situation of, for example, underdamped situation. Then what I'm going to do is actually pretty simple. What I am going to do is to just copy the underdamped solution from last lecture and combine that with my particular solution, which I obtained here. So that actually to see what actually the full solution looks like. I have A omega d cosine omega d t minus delta is a function of omega d. This is actually so-called steady-state solution. And, of course, as you mentioned, I need to also add the homogeneous solution and basically the no -- this actually -- according to what I wrote there, I have a functional form of exponential A exponential minus gamma over 2t equals sine omega t plus alpha. So I changed A to B because I already have the A there just to avoid confusion. Then basically, you get B exponential minus gamma over 2t cosine omega t plus alpha. Basically, they are two free parameters, B and alpha. Those two free parameters can be determined by initial conditions. So, for example, initially I actually release the rod at some fixed angle of theta initial. And also the initial velocity is 0. Then I can actually practice solution A using those initial conditions to solve B and alpha. Any questions so far? Yes. AUDIENCE: Are we assuming it's underdamped? YEN-JIE LEE: Yeah, I'm assuming it's underdamped, the situation. So that's the assumption. So it depends on the size of gamma and the omega 0. Then you have actually four different kinds of solution. If gamma is equal to 0, then what you are going to plug in is the solution from no damping as a your homogeneous solution. And if you prepare this system underwater, damping is colossal, it's huge, then you actually plug in the overdamped solution to be your homogeneous part of the solution. Any other questions? Very good question. So now maybe you got confused a bit. I have now omega d. I have also omega. And there's another one we just called omega 0. What are those? So omega 0 is the natural angular frequency without given the drag force. If you remove everything just like without considering any drag force, et cetera, and that is actually the natural frequency of the system. And what is omega? Omega, according to the function, omega is defined to be omega 0 squared minus gamma squared over 4 square root of that. That is actually the oscillation frequency, which we actually discussed last lecture, after you add drag force into it again. Finally, omega d is how fast you actually drive this system. So that is actually the definition of these three omegas. So you can see that, if I prepare my solution to be underdamped situation, then basically you will see that this is actually so-called a steady-state solution because A omega d is a constant. So it's going to be there forever. And the second term is actually B exponential minus gamma over 2t. It's decaying as a function of time. So if you are patient enough, you wait, then this will be gone. So that is actually how we actually understand this mathematical result. And now, of course, you can actually take a look at this. This is actually just assuming some kind of initial condition and plug in the solution and plot it as a function of time. And you can see that this function looks really weird, looks a bit surprising. What does that mean? It looks really strange. But at some point, this superposition of these two functions-- because one of the functions actually dies out, disappears-- then you will see that, if you wait long enough, then you actually only see a very simple structure, which is oscillation frequency of omega d. And that means a large t. In the beginning, the system will not like it. You drive it, and the system don't like it. Like if I go and shake you, in the beginning, you would not like it-- maybe. And if I shake you long enough, and you say, come on, OK, fine. I accept that. So that is actually what is going to happen to the system. So now I would like to go through a short demonstration, which is actually the air cart, which you seen before. There's a mass and there are two springs in the front and the back of this cart. And, of course, as usual, I would turn on the air so that I make the friction smaller, but there's still some residual friction. And you will see that this mass is actually oscillating back and forth. And the amplitude can become smaller and smaller as a function of time. Now in the right hand side, I have a motor, which actually can drive this-- I can actually shorten or increase the length of the right hand side string. Then I actually introduce a driving force by the right hand side motor. If I turn it down, this is what is going to happen. So we can see now this motor is actually going back and forth. And it has a slightly higher frequency compared to the natural frequency. So the frequency of the motor is higher. And you can see that this cart is actually oscillating. But you can see that sometimes it pulls and sometimes it moves faster. So you can see that it's actually moving. And it stops a little bit because they are all superposition of two different kinds of oscillating functions come into play. You can see that now. It got slowed down, and it can become faster and slower and faster. But eventually, if you wait long enough, what is going to happen? What is going to happen? If we wait long enough-- AUDIENCE: [INAUDIBLE]. YEN-JIE LEE: Exactly. So basically, if you wait long enough, as you said, you will actually just oscillate at the frequency of the driving force. You can see that this motion looks really bizarre, right? Sometimes it stops, and sometimes it actually continues to move. And are you surprised? Probably you are not surprised anymore because we know math is the language to describe nature. And indeed it predicts this kind of behavior. That's really pretty cool. In order to help you to learn a bit how to actually translate a physical situation into mathematics, what I am going to do is to introduce you another example so that actually we can actually solve it together. So now I would like to drive a pendulum. So I prepare a pendulum at time equal to 0. This is a string attached to a ball with mass equal to m. And the length of the string is equal to l. And the angle between the vertical direction and the direction of the string is theta. And, of course, I can actually give you initial condition X initial, which is actually measured with respect to the vertical direction, and time equal to t. This is actually the original vertical direction, the same as this dashed line. And I can actually move the top of the string back and forth to some position. And, of course, this string is connected to the ball. And this system is actually driven from the top by the engine's hand, so the engine is actually shaking this system from the top. And I do it really nicely. So basically, I define that the displacement, d, as a function of time, is equal to delta sine omega d t. So that is actually what I'm going to do. OK, and I would like to see what is going to happen to this pendulum. So, as usual, the first step towards solving this problem is to define a coordinate system. So what is actually the coordinate system I'm going to use? So now I define pointing upward to be y. I define the horizontal direction pointing to the right hand side of the board to be x. So that's not good enough. I still need the origin, right? So now I also define my origin to be the original position of the ball which is actually completely addressed before I do the experiment. So that this is actually the equilibrium position of this system actually. Then I define here to be 0, 0. So once I have that defined, I can now express the position of this mass of this ball to be xt and yt. And see what we are going to get. Of course, as usual, we are going to analyze the force actually acting on this ball. So therefore, as usual, we will draw a force diagram. So basically, you have the little mass here, and you have actually two forces acting on this little mass, or little ball. This is Fg pointing downward. It's a gravitational force. And now this is actually equal to minus mg y. And there's also a string tension, T. Since we have this definition of theta here, basically I have a T which is actually pointing to the upper left direction of the board. Oh, don't forget-- actually there's a third force, which is actually the F drag. F drag is actually equal to minus bx dot in the x direction. Now I would like to write down the expression for also the string tension, T. The T is actually equal to minus T sine theta in the x direction because the T is pointing to upper left direction. So therefore, the position to x direction will be minus sine theta and plus T cosine theta in the y direction because the tension is actually pointing upper left. As usual, this is actually pretty complicated to solve. I have this cosine. I have this sine there, right? So what I'm going to do is to assume that this angle theta is very small, as usual. So I will take small angle approximation. Then basically, you have sine theta is roughly to be equal to theta, and that is actually equal to what? Equal to-- here. Basically, you can actually calculate what will be the theta. The sine theta, or theta, would be equal to x minus d divided by l. And of course, taking a small angle approximation will bring cosine theta to be 1. Then after this approximation, my T will become minus T x minus d divided by l in the x direction plus T in the y direction because sine theta is replaced by this approximated value because sine theta is actually replaced by 1. Any questions? Yes. AUDIENCE: Is that a constant or is that the change in sine? YEN-JIE LEE: It's a constant. Yeah, I was going too fast. So this is actually a constant of my amplitude. AUDIENCE: And the drag force is only in the x direction? YEN-JIE LEE: Yes, it's only in the x direction. So I'm only trying to actually move this point back and forth horizontally. So now I have all the components T, Fg, f drag. And, of course, you can see that I already ignored the drag force in the y direction from that formula because I am only considering the system to be moving in the x direction. Therefore, I can now collect all the terms in the x direction. Basically, you will have m x double dot. This is equal to minus b x dot, which is actually coming from the drag force, minus T x minus d divided by l. This is actually coming from this term in the x direction. Let's look at those in the y direction. And y double dot would be equal to minus mg plus T. The minus mg is from the gravitational force. And this T is coming from the y component of the string tension. And of course, since we are taking a very small angle approximation, there will be no vertical motion. Yes. AUDIENCE: Why did we use the small approximation for sine theta when we're going to use x minus d over l, which represents psi instead of just theta? YEN-JIE LEE: Yeah, so also in this case, they happen to be exactly the same. And why I care is actually the cosine theta. Otherwise, I would have to deal with cosine theta. And also, this y double dot would not be equal to 0, which is what I'm going to assume here. Good question. So the question was, why do I need to take an approximation? Because I want to get rid of cosine theta. So now from this y direction, I can solve T will be equal to mg because I assume that there's no y direction motion. And I can conclude that-- originally, I don't know what is actually the string tension. It's denoted by T. Now, from this second equation, I can conclude that T will be equal to mg, which is the gravitational force. Then, once I have that, I can go back to x direction. Basically, I get m x double dot. This is equal to minus b x dot minus mg over l x minus d. Everything is working very well. And I just have to really write down the d function explicitly. What is d? d is just a reminder, delta sine omega d t. So I will plug that into that equation. And also I will bring all the terms related to x to the left hand side just to match my convention. All right, so now I will be able to get the result, m x double dot plus b x dot plus mg over l x. And that is actually equal to mg over l d. And this is equal to mg over l delta sine omega dt. So basically, I collect all the terms, put it to the left hand side. And I write down T explicitly, which is this. Then I can divide everything by m. Then I get m x double dot plus b over m x dot plus g over l x. And that would be equal to g over l delta sine omega dt. Now, of course, as usual, I will define this to be gamma, define this to be omega 0 squared. And I would define this to be f0, which is equal to omega 0 squared delta. It happened to be equal like that. And then this actually becomes x double dot plus gamma x dot plus omega 0 squared x equal to f0 sine omega dt. Am I going too fast? OK, everybody is following. So we see that ha! This equation-- I know that. I know this equation, right? Because we have just solved that a few minutes ago. Therefore, I know immediately what will be the solution. The solution is here already. I have A omega d and the tangent delta, the function of force there. Therefore, I can now write down what will be the A. So A is actually just equal to f0 divided by square root of omega 0 squared minus omega d squared squared plus omega d squared gamma squared. So now the question is-- what does the result actually mean. I have this function. I have that function, tangent delta. It's solved. It's actually the amplitude of the steady-state solution and also the phase difference between the drag force phase and the steady-state oscillation phase. So that's actually the amount of lag and the size of the amplitude. But through this equation it is very difficult to understand. So what I'm going to do is to take some limit so that actually we can help you to understand what is going on. So suppose I assume that omega d goes to 0. So what does that mean? This is the engine's hand and is moving really slowly and see what is going to happen. If I do this, then you will find that A omega d-- since omega goes to 0, this is gone, this is gone. Therefore, you will see that omega A will be equal to omega 0 divided by-- I'm sorry-- of f0 divided by omega 0 squared. And that is actually equal to g delta over l divded by g over l. And that will give you delta. So what does that mean? This means that, if I drive this thing really slowly, then the amplitude of the mass will be equal to how much I actually move, which is delta. OK. Do you get it? In addition to that, tangent delta-- since I am taking the limit omega d goes to 0. Therefore, tangent delta will be equal to 0, and that means delta will be equal 0. Any questions? So that means there will be no phase difference. The system has enough time to keep up with my speed. The second limit, obviously, omega d goes to infinity. What does that mean? That means I'm going to hold this as a string and shake it like crazy really fast and see what will happen, OK? So in that case, you will get A omega d, and that one goes to 0, because omega d goes to infinity. This one goes to 0. And also, tangent delta will go to infinity. Therefore, delta will go to pi. So that means they will be out of phase. Any questions so far in these two limits? OK, so what I'm going to do now is to take a small toy, which I made for my son, who is one-year-old now. Because I would like him to learn wavelength vibration before he goes to quantum, right? Hey? So I made this toy. And he looked at it. So you can see now, I can demonstrate what is going to happen when omega is approaching to 0. OK? I am already doing it. Can you see it? No? It's a very exciting experiment. Can you see that? You see that this is the origin vertical direction. If I do it really, really, really slowly, you can see that the amplitude of the ball is actually exactly the same as the displacement I introduced. So that's kind of obvious. So now, let's see what is going to happen if I drive this system like crazy. OK, not going up and down. Eeeee-- that's the maximum speed I can do. Maybe you can do it faster. But you can see that nothing happened. So amplitude is close to 0, because what you have been doing is-- disappear, you sort cancelling each other. And it's actually not going to contribute to the motion of this ball. So now, you can see that I can also test, what is actually the natural frequency? And what I am going to do is to oscillate at around the natural frequency to see what is going to happen. Let's see what is going to happen. You can see that the delta is really small, right? Can you see the delta. It's really small-- very small-- very small. But you can see that the amplitude, the A, is huge. What does that tell us? What does that tell us? Yeah? AUDIENCE: Well, we're experiencing resonance. YEN-JIE LEE: Yes, we are experiencing resonance. And also, that also tells you that the system is under-damped very much. The Q value is very big. So if I calculate the amplitude, A-- now, I can calculate the amplitude, A, at natural frequency. What I'm going to get is-- now, I can actually plug in omega d equal to omega 0. So if I plug in omega d-- omega d equal to omega 0-- then what is going to happen? So this term is working. So you have A is equal to f-0 divided by omega 0 gamma. Omega-d is now equal to omega 0. And that is going to give you-- so f-0 is actually omega 0 square delta divided by omega 0 gamma. And the one omega 0 actually cancels. Then, basically, you will get Q times delta. What is Q? Just a reminder, it's actually the ratio of omega 0 and the gamma. When the Q is very large, what does that mean? That means it's so close to an idealized situation that direct force is very small. You can see that in the example which I have been doing. So you can see that, ah, it is really the case. So you can see that if my delta is something like 1 centimeter, but the amplitude is actually at the order of 1 meter, maybe. What does that mean? That means that Q is actually, roughly, 100. So you can actually even get a Q out of this experiment. Any questions so far? OK, that's very good. So now, we can go ahead and take a look at the structure of the A and the delta. As we demonstrated before, we make sense of those three different kinds of situations-- omega d goes to 0, omega d goes to infinity. And of course, I would like to know the force structure of A and delta. Therefore, what I'm going to do is to plug A omega d as a function of omega d. So what I'm going to get is this will be equal to delta when omega d goes to 0. We just demonstrated that. And this will increase to a large value and drop down to 0, when omega d goes to infinity. And you can see that this is around omega 0. And you are going to get a huge amplitude at around omega 0. But not quite. The maxima is actually slightly smaller than omega 0. You can actually calculate that as part of the homework. So that makes sense. Now, I can also plug the delta, which is the phase difference-- and you can see that this phase difference will be, originally, 0, when the omega d is very small. And this is actually the omega 0. I hope you can see it. And this will be increasing rapidly here and approaching to pi. So that means, when you are shaking this system like crazy-- very high frequency-- then the system cannot keep up with the speed. The amplitude will be very small. And also, the amplitude will be out of phase completely. So let's actually do us another demonstration, using this little device here. This is actually what you see before, the ball with a Mexican hat. And you can see that there is a spring attached to this system. And on the top, what I am going to do is to use this motor to drive this system up and down, as a direct force. So now, what am I going to do is to come from a very low-frequency oscillation. So you can see that the natural frequency is sort of like this. And you can see that, now, I am driving this system really slowly. You can see, this is actually going up and down really slowly. And you see that-- huh-- the amplitude is actually pretty small. There's no excitement for the moment. All right, so what I'm going to do is, now, I increase the speed of the motor and see what will happen. So you can see, now, it's actually driving it with higher and higher frequency. You see that-- huh-- something is happening. You can see the amplitude is getting larger and larger. I'm still increasing the frequency-- increasing, increasing-- until something-- something happened! Right? Did you see that? It starts to oscillate up and down. Because right now, you can see that-- look at the top. The frequency of the motor is now really close to the natural frequency of this system. So a resonance behavior will happen. And what you are going to get is that, OK, omega d around omega 0. Then you are going to get large time amplitude. So now, what I am going to do is to continue to increase the driving frequency to a very large value. OK, now it's actually doing the "mmmmm"-- doing it really fast. You can see on the top-- very fast. OK, I even get it even faster. You see that-- huh-- indeed, this system is now oscillating at a larger frequency. It's trying to keep up with the driving force. But you can see that the amplitude is actually much smaller than what had happened before. So before the class, you may actually think that, OK, drive it really fast. Maybe we'll increase the amplitude. But in reality, actually, it will give you a very small amplitude. Another thing, which is interesting to know, is that you can see that, when the driving force is actually at the maximum. And actually the position of this mass is actually at the minimum. So they are actually out of phase. I hope you can see it. It's like this. OK, so what you can see is that, when I understand the system and I try to drive it with the natural frequency, what is going to happen is that I'm exciting this system to a state of resonance. So basically, you'll get some resonance behavior. So I have shown you that this works for driven mechanical oscillator. It also works for the spring-mass system. And there are many other things which also work, which is around you. For example, if you happen to be my office hour, you would notice that the air-condition in my office is actually creating a resonance behavior. You'll see low frequency sound-- "mm mm mm"-- something like that. And that is because the pipe actually happens to have the frequency match with the resonance frequen-- OK, the airflow actually happened to excite the pipe, so that it's actually oscillating up and down at that frequency. So what I did was I tried to turn it down to low and see what happened. But unfortunately, it actually excited another resonance. I see, now, not a low-frequency sound, but a very high-frequency sound. I will post a video, actually online. So my life is hard, right? But I'm a physicist. Is So I choose to use the median. Then I actually stay between the two resonances. Then I don't hear the additional sound, which bothers me. Another example is that, when I was in Taiwan as a undergrad, I was living outside in a apartment. And with my flat-mate, we owned a very old washing machine. So in the middle of the night, the washing machine would started to walk around, like my flat-mate. And we are not scared. That is because the oscillation frequency-- actually, the rotation-- happened to match with the frequency of the washing machine. Therefore, when we started to wash our clothes, it start to walk around in the room. So as a physicist, what we have decided is to make the spin slightly slower, or even faster. Then, actually, you can see that, when you do that, then you get rid of the resonance behavior. So it's not walking around any more. We can control it. Another thing which is interesting is that the resonance behavior is not only in the physical objects, which we actually deal with these days also. But either you learn quantum mechanics and upon the field theory, you will find that there are resonance also in a mass wave function. So basically, you can see that these are examples of the Z boson resonance peak. So if you scatter a electron and positron then, basically, you'll see that the cross-section have a resonance peak at around 90 GeV. And that is actually another very interesting example of a resonance in particle physics. Finally, the last example, which I am going to go through is an example involving a glass. We have prepared a very high-quality glass here. Maybe you have seen this glass before. They are pretty nice. And I usually use it to enjoy my red wines, which you cannot, enjoy, probably now. So you can see that this is the glass. And if I put a little bit of water on my hand and I rub it-- [VIBRATING TONE] --carefully, I can actually excite one of the resonance frequencies. So you can see that we have all everything working on a single particle. And that will give you one resonance frequency. If I work on two particles, which you will see that in the next lecture, I would get two resonance frequencies. And this glass is made of infinite number of particles. Therefore, I will have infinite number of resonance frequencies. When I'm rubbing it, I'm actually giving input of all kinds of different frequencies. But the glass will be smart enough so that it will pick up the one it likes the most, which is the resonance frequency. So you can see that the sound is actually, roughly, 683 Hertz. And you can actually measure it with your phone. So on the TV commercial, you may have seen that there's a lady singing. And she's singing so loudly such that the glass-- "bragh!"-- breaks. Can we get a volunteer today to sing in front of us? Oh-- singing. Can you sing it-- high frequencies? "Ahhhh." [LAUGHTER & APPLAUSE] Very good try. But it didn't work. OK, I guess it's really difficult to perform that in front of so many people unprepared. But fortunately, we are MIT. So we have designed a device, which actually can help us to achieve this. So this device actually contain a amplifier here. And I can now control the frequency of the sound through this scope. And this amplifier will actually amplify the signal and produce a sound wave and try to actually isolate this glass. So we are going to do this experiment. So we will need to change the loud setting a bit. Because the sound is going to be, probably, too loud. Just for safety-- some of you may not survive. [LAUGHTER] So I'm handing out these. OK, who is closer? OK, maybe you. AUDIENCE: [INAUDIBLE] YEN-JIE LEE: Oh. Oh, sorry. [LAUGHTER] I'm so sorry. What? I don't need that. OK. So just for safety, I will put this on. And what I am going to do is also put these glass on. OK, maybe I'll do this first. AUDIENCE: [INAUDIBLE] YEN-JIE LEE: OK, so what I am going to do now is to start producing sound wave. [LOUD TONE] So through the camera, you should be able to see what is actually shown on the screen. So you can see that, if this glass is actually moving, the wood inside would also vibrate. So you can see that, clearly, we don't have resonance yet. So what I am going to do is to increase the frequency and see what happens. So now, it's actually 643. It's actually still below the resonance frequency. Now, I have measured the frequency. It should be 684. So now, it's actually 653-- 663 Hertz-- 673 Hertz. Can you see the movement? You cannot see the movement yet. 683-- you see? You see, now-- AUDIENCE: Yes. YEN-JIE LEE: --the frequency of the sound is actually matching with one of the natural frequencies of the glass. Apparently, the glass likes it. And now, you can see that it is still vibrating. And the next step, which we are going to do, is to try to increase the amplitude, increase the volume of the sound, and see what happens. Maybe you want to cover your ears, just for safety. [LOUD TONE] OK, then the glass may break, if we are lucky. Let us see what is going to happen. [INCREASINGLY LOUDER TONE] Oh! [APPLAUSE] TECH SUPPORT: Good job. YEN-JIE LEE: Very good. TECH SUPPORT: Perfect. That's the quickest one we've had. YEN-JIE LEE: Yeah, thank you very much. Thank you, glass. [LAUGHTER] So you can see the power of resonance. So if I tune down the frequency slightly more, then you will be where? You'll be here. Then you will not have enough amplitude to break the glass. And also, as we discussed before, the quality of the glass should be really, really high, such that the resulting amplitude will be very large. Then you can actually break it with a external sound wave. And if we go above the resonance frequency, then you would not also move a bit. Because if you go to very large omega d, then amplitude will be pretty small. OK, let me try to switch back to my presentation. I think we did. Sure. So this is actually what we have learned today. We have learned the behavior of a damped driven oscillator. We have learned the transient behavior. So what is actually transient behavior is a mixture of steady state solution, which was coming from the driving force, and the homogeneous solution. If you wait long enough, this will decay and disappear. Is And we have learned resonance. So an IOC circuit, which you actually solved that in your P-set, in pendulum, which I just show you, which helped my son to learn wavelength vibrations-- and air condition, washing machine, glass-- particle physics. We can see damped os-- driven oscillator or resonance almost everywhere. So I hope that you enjoyed the lecture today. And what we are going to do next time is to put multiple objects together so that you see the interaction between one particle to the other particle and see how we can actually make sense of this kind of system. Thank you very much. And I will be here if you have additional questions related to the lecture.
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https://ocw.mit.edu/courses/7-012-introduction-to-biology-fall-2004/7.012-fall-2004.zip
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As I'm going to argue repeatedly today, biology has become a science over the last 50 years. And, as a consequence, we can talk about some basic principles. We can talk about some laws and then begin to apply them to very interesting biological problems. And so our general strategy this semester, as it has been in the past, is to spend roughly the first half of the semester talking about the basic laws and rules that govern all forms of biological life on this planet. And you can see some of the specific kinds of problems, including the problem of cancer, how cancer cells begin to grow abnormally, how viruses proliferate, how the immune system functions, how the nervous system functions, stem cells and how they work and their impact on modern biology, molecular medicine, and finally perhaps the future of biology and even certain aspects of evolution. The fact of the matter is that we now understand lots of these things in ways that were inconceivable 50 years ago. And now we could begin to talk about things that 50 years ago people could not have dreamt of. When I took this course, and I did take it in 1961, we didn't know about 80% of what we now know. You cannot say that about mechanics in physics, you cannot say that about circuit theory in electronics, and you cannot say that, obviously, about chemistry. And I'm mentioning that to you simply because this field has changed enormously over the ensuing four decades. I won't tell you what grade I got in 7. 1 because if I would, and you might pry it out of me later in the semester, you probably would never show up again in lecture. But in any case, please know that this has been an area of enormous ferment. And the reason it's been in such enormous ferment is of the discovery in 1953 by Watson and Crick of the structure of the DNA double helix. Last year I said that we were so close to this discovery that both Watson and Crick are alive and with us and metabolically active, and more than 50 years, well, exactly 50 years after the discovery. Sadly, several months ago one of the two characters, Francis Crick died well into his eighties, and so he is no longer with us. But I want to impress on you the notion that 200 years from now, we will talk about Watson and Crick the same way that people talk about Isaac Newton in terms of physics. And that will be so because we are only beginning to perceive the ramifications of this enormous revolution that was triggered by their discovery. That is the field of molecular biology and genetics and biochemistry which has totally changed our perceptions of how life on Earth is actually organized. Much of the biology to which you may have been exposed until now has been a highly descriptive science. That is you may have had courses in high school where you had to memorize the names of different organisms, where you had to understand how evolutionary phylogenies were organized, where you had to learn the names of different organelles, and that biology was, for you, a field of memorization. And one point we would like, hopefully successfully, to drive home this semester is the notion that biology has now achieved a logical and rational coherence that allows us to articulate a whole set of rules that explain how all life forms on this planet are organized. It's no longer just a collection of jumbled facts. Indeed, if one masters these molecular and genetic principles, one can understand in principle a large number of processes that exist in the biosphere and begin to apply one's molecular biology to solving new problems in this arena. One of the important ideas that we'll refer to repeatedly this semester is the fact that many of the biological attributes that we posses now were already developed a very long time ago early in the inception of life on this planet. So if we look at the history of Earth, here the history of Earth is given as 5 billion years, this is in thousands obviously. The Earth is probably not that old. It's probably 4.5 or 4 or 3 billion years but, anyhow, that's when the planet first aggregated, as far as we know. One believes that no life existed for perhaps the first half billion years, but after half a billion years, which is a lot of time to be sure, there already begins to be traces of life forms on the surface of this planet. And that, itself, is an extraordinary testimonial, a testimonial to how evolutionary processes occur. We don't know how many planets there are in the universe where similar things happened. And we don't know whether the solution that were arrived at by other life systems in other places in the universe, which we may or may not ever discover, were the similar solutions to the ones that have been arrived at here. It's clear, for example, that to the extent that Darwinian Evolution governs the development of life forms on this planet that is not an artifact of the Earth. Darwinian Evolution is a logic which is applicable to all life forms and all biosystems that may exist in the universe, even the ones we have not discovered. However, there are specific solutions that were arrived at during the development of life on Earth which may be peculiar to Earth. The structure of the DNA double helix. The use of ribose in deoxyribose. The choice of amino acids to make proteins. And those specific solutions may not be universal. Whether they're universal in the sense of existing in all life forms across the planet, the fact is that many of the biochemical and molecular solutions that are represented in our own cells today, these solutions, these problems were solved already 2 and 3 billion years ago. And once they were solved they were kept and conserved almost unchanged for the intervening 2 or 3 billion years. And that strong degree of conservation means that we can begin to figure out what these principles were early on in evolution of life on this planet and begin to apply them to all modern life forms. From the point of view of evolution, almost all animals are identical in terms of their biochemistry and in terms of their physiology. The molecular biology of all eukaryotic cells, that is all cells that have nuclei in them, is almost the same. And, therefore, we're not going to focus much in this course this semester on specific species but rather focus on general principles that would allow us to understand the cells and the tissues and the physiological processes that are applicable to all species on the surface of the planet. Let's just look here and get us some perspective on this because, the fact of the matter is, is that multicellular life forms, like ourselves, we have, the average human being has roughly three or four or five times ten to the thirteenth cells in the body. That's an interesting figure. The average human being goes through roughly ten to the sixteenth cell divisions in a lifetime, i.e. ten to the sixteenth times in your body there will be cells that divide, grow and divide. Every day in your body there are roughly ten to the eleventh cells that grow and divide. Think of that, ten to the eleventh. And you can divide that by the number of minutes in a day and come up with an astounding degree of cellular replication going on. All of these processes can be traceable back to solutions that were arrived at very early in the evolution of life on this planet, perhaps 550, 600 million years ago when the first multicellular life forms began to evolve. Before that time, that is to say before 500 to 600 million years ago, there were single-cell organisms. For example, many of them survive to this day. There were yeast-like organisms. And there were bacteria. And we make one large and major distinction between the two major life forms on the planet in terms of cells. One are the prokaryotic cells. And these are the cells of bacteria, I'll show you an image of them shortly, which lack nuclei. And the eukaryotic cells which poses nuclei and indeed have a highly complex cytoplasm and overall cellular architecture. We think that the prokaryotic life forms on this planet evolved first probably on the order of 3 billion years ago, maybe 3. billion years ago, and that about 1. billion years ago cells evolved that contained nuclei. Again, I'll show them to you shortly. And these nucleated cells, the eukaryotes then existed in single-cell form for perhaps the next 700 or 800 million years until multi-cellular aggregates of eukaryotic cells first assembled to become the ancestors of the multi-cellular plants and the multi-cellular animals that exist on the surface of the Earth today. To put that in perspective, our species has only been on the planet for about 150, 00 years. So we've all been here for that period of time. And a 150,000 sounds like a long time, in one sense, but it's just a blink in the eye of the Lord as one says in terms of the history of life on this planet, and obviously the history of the universe which is somewhere between 13 and 15 billion years old. You can begin to see that the appearance of humans represents a very small segment of the entire history of life on this planet. And here you can roughly see the way that life has developed during this period of time from the fossil record. You see that many plants actually go back a reasonable length of time, but not more than maybe 300 or 400 million years. Here are the Metazoa. And this represents -- Well, can you hear me? Wow, 614 came in handy. OK. So if we talk about another major division, we talk about protozoa and metazoa. The suffix zoa refers to animals, as in a zoo. And the protozoa represents single-cell organisms. The metazoa represent multi-cellular organisms. And we're going to be focusing largely on the biology of metazoan cells this semester, and we're going to be spending almost no time on plants. It's not that plants aren't important. It's just that we don't have time to cover everything. And, indeed, the molecular biology that you learn this semester will ultimately enable you to understand much about the physiology of multi-cellular plants which happen to be called metaphyta, a term you may never hear again in your entire life after today. That reminds me, by the way, that both Dr. Lander and I sometimes use big words. And people come up to me afterwards each semester each year and say Professor Weinberg, why don't you talk simple, why don't you talk the way we heard things in high school? And please understand that if I use big words sometimes it's to broaden your vocabulary so you can learn big words. One of the things you should be able, one of the big take-home lessons of this course should be that your vocabulary is expanded. Not just your scientific vocabulary but your general working English vocabulary. Perhaps the biggest goal of this course, by the way, is not that you learn the names of all the organelles and cells but that you learn how to think in a scientific and rational way. Not just because of this course but that this course helps you to do so. And as such, we don't place that much emphasis on memorization but to be able to think logically about scientific problems. Here we can begin to see the different kinds of metazoa, the animals. Here are the metaphyta and here are the protozoa, different words for all of these. And here we see our own phylum, the chordates. And, again, keep in mind that this line right down here is about 550 to 600 million years ago, just to give you a time scale for what's been going on, on this planet. One point we'll return to repeatedly throughout the semester is that all life forms on this planet are related to one another. It's not as if life was invented multiple times on this planet and that there are multiple independent inventions to the extent that life arose more than once on this planet, and it may have. The other alternative or competing life forms were soon wiped out by our ancestors, our single-cellular ancestors 3 billion years ago. And, therefore, everything that exists today on this planet represents the descendents of that successful group of cells that existed a very long time ago. Here we have all this family tree of the different metazoan forms that have been created by the florid hand of evolution. And we're not going to study those phylogenies simply because we want to understand principles that explain all of them. Not just how this or that particular organism is able to digest its food or is able to reproduce. Here's another thing we're not going to talk about. We're not going to talk about complicated life forms. We're not going to talk very much, in fact hardly at all, about ecology. This is just one such thing, the way that a parasite is able to, a tapeworm is able to infect people. This is, again, I'm showing you this not to say this is what we're going to talk about, we're not going to talk about that. We're not going to talk about that. There's a wealth of detail that's known about the way life exists in the biosphere that we're simply going to turn our backs on by focusing on some basic principles. We're also not going to talk about anatomy. Here in quick order are some of the anatomies you may have learned about in high school, and I'm giving them to you each with a three-second minute, a three-second showing to say we're not going to do all this. And rather just to reinforce our focus, we're going to limit ourselves to a very finite part of the biosphere. And here is one way of depicting the biosphere. It's obviously an arbitrary way of doing so but it's quite illustrative. Here we start from molecules. And, in fact, we will occasionally go down to submolecular atoms. And here's the next dimension of complexity, organelles. That is these specialized little organs within cells. We're going to focus on them as well. We're going to focus on cells. And when we start getting to tissues, we're going to start not talking so much about them. And we're not going to talk about organisms and organs or entire organisms or higher complex ecological communities. And the reason we're doing that is that for 40 years in this department, and increasingly in the rest of the world there is the acceptance of the notion that if we understand what goes on down here in these first three steps, we can understand almost everything else in principle. Of course, in practice we may not be able to apply those principles to how an organism works or to how the human brain works yet. Maybe we never will. But, in general, if one begins to understand these principles down here, one can understand much about how organismic embryologic develop occurs, one can understand a lot about a whole variety of disease processes, one can understand how one inherits disease susceptibilities, and one can understand why many organisms look the way they do, i.e. the process of developmental biology. And so, keep in mind that if you came to hear about all of these things, we're going to let you down. That's not what this is going to be about. This also dictates the dimensions of the universe that we're going to talk about because we're going to limit ourselves to the very, very small and not to the microscopic. On some occasions we'll limit ourselves to items that are so small you cannot see them in the light microscope. On other occasions we may widen our gaze to look at things that are as large as a millimeter, but basically we're staying very, very small. Again, because we view, correctly or not, the fact that the big processes can be understood by delving into the molecular details of what happens invisibly and cannot be seen by most ways of visualizing things, including the light and often even the electron microscope. Keep in mind that 50 years ago we didn't know any of this, for all practical purposes, or very little of this. And keep in mind that we're so close to this revolution that we don't really understand its ramifications. I imagine it will be another 50 years before we really begin to appreciate the fallout, the long-term consequences of this revolution in biology which began 51 years ago. And so you're part of that and you're going to experience it much more than my generation did. And indeed one of the reasons why MIT decided about 10 or 12 years ago that every MIT undergraduate needed to have at least one semester of biology is that biology, in the same way as physics and chemistry and math, has become an integral part of every educated person's knowledge-base in terms of their ability to deal with the world in a rational way. In terms of public policy, in terms of all kinds of ethical issues, they need to understand what's really going on. Many of the issues that one talks about today about bioethics are articulated by people who haven't the vaguest idea about what we're talking about this semester. You will know much more than they will, and hopefully some time down the road, when you become more and more influential voices in society, you'll be able to contribute what you understood here, what you learned here to that discussion. Right now much of bioethical discussion is fueled by people who haven't the vaguest idea what a ribosome or mitochondrion or even a gene is, and therefore is often a discussion of mutually shared ignorance which you can diffuse by learning some basics, by learning some of the essentials. Here is the complexity of the cell we're going to focus on largely this semester, which is to say the eukaryotic rather than the prokaryotic cell. And this is just to give you a feeling for the overall dimensions of the cell and refer to many of the landmarks that will repeatedly be brought up during the course of this semester. Here is the nucleus. The term karion comes from the Greek meaning a seed or a kernel. And the nucleus is what gives the eukaryotic cell its name. Within the nucleus, although not shown here, are the chromosomes which carry DNA. You may have learned that a long time ago. Outside of the nucleus is this entire vast array of organelles that goes from the nuclear membrane, and I'm point to it right here, all the way out to the outside of the cell. The outside limiting membrane, the outer membrane of the cell is called the plasma membrane. And between the nucleus and the plasma membrane there is an enormous amount of biological and biochemical activity taking place. Here are, for example, the mitochondria. And the mitochondria, as one has learned, are the sources of energy production in the cell. And, therefore, we'll touch on them very briefly. This is an artist's conception of what a mitochondrion looks like. Almost always artists' conceptions of these things have only vague resemblance to the reality. But, in any case, you can begin to get a feeling for what one thinks about their appearance. Here are mitochondria sliced open by the hand of the artist. And, interestingly, mitochondria have their own DNA in them. One now accepts the fact that mitochondria are the descendents of bacteria which insinuated themselves into the cytoplasms of larger cells, roughly 1.5 billion years ago, and began to do a specialized job which increasingly became the job of energy production within cells. To this day, mitochondria retain some vestigial attributes of the bacterial ancestors which initially colonized or parasitized the cytoplasm of the cell. When I say parasitized, you might imagine that the mitochondria are taking advantage of the cell. But, in fact, the mitochondria represent the essential sources of energy production in the cell. Without our mitochondria, as you might learn by taking cyanide, for example, you don't live for very many minutes. And the vestiges of bacterial origins of mitochondria are still apparent in the fact that mitochondria still have their own DNA molecule, their own chromosome. They still have their own ribosomes and protein synthetic apparatus, even though the vast majority of the proteins inside mitochondria are imported from the cytoplasm, i.e., these vestigial bacteria now rely on proteins made by the cell at large that are imported into the mitochondrion to supplement the small number of vestigial bacterial proteins which are still made here inside the mitochondrion and used for essential function in energy production. Here is the Golgi apparatus. And the Golgi apparatus up here is used for the production of membranes. As one will learn throughout the semester, the membranes of a cell are in constant flux and are being pulled in and remodeled and regenerated. The Golgi apparatus is very important for that. Here's the rough endoplasmic reticulum. That's important for the synthesis of proteins which are going to be displayed on the surface of cells, you don't see them depicted here, or are going to be secreted into the extracellular space. Here are the ribosomes, which I might have mentioned briefly before. And these ribosomes are the factories where proteins are made. Again, we're going to talk a lot about them. And, finally, several other aspects, the cytoskeleton. The physical integrity, the architecture of the cell is maintained by a complex network of proteins which together are considered to be the cytoskeleton. And they enable the cell to have some rigidity, to resist tensile forces, and actually to move. Cells can actually move from one place to the other. They have motile properties. They're able to move from one location to another. The process of cell motility, if that's a word you'd like to learn. Here is what a prokaryotic cell looks like by contrast. And I just want to give you a feeling. First of all, it looks roughly like a mitochondrion that I discussed before. But you see that there is the absence of a nuclear membrane. There's the absence of the highly complex cytoarchitecture. Cyto always refers to cells. There's the absence of the complex cytoarchitecture that one associates with eukaryotic cells. In fact, all that a bacterium has is this area in the middle. It's called the nucleoid, a term which you also will probably never hear in your lifetime. And it represents simply an aggregate of the DNA of the chromosomes of the bacterium. And, in most bacteria, the DNA consists of only a single molecule of DNA which is responsible for carrying the genetic information of the bacteria. There's no membrane around this nucleoid. And outside of this area where the DNA is kept are largely ribosomes which are important for protein synthesis. There's a membrane on the outside of this called the plasma membrane, very similar to the plasma membrane of eukaryotic cells. And outside of that is a meshwork that's called the outer membrane, it's sometimes called the cell wall of the bacterium, which is simply there to impart structural rigidity to the bacterium making sure that it doesn't explode and holding it together. And then there are other versions of eukaryotic cells. Here's what a plant cell looks like. And it's almost identical to the cells in our body, except for two major features. For one thing, it has chloroplasts in it which are also, one believes now, the vestiges of parasitic bacteria that invade into the cytoplasm of eukaryotic cells. So, in addition to mitochondria which are responsible for energy production in all eukaryotic cells, we have here the chloroplasts which are responsible for harvesting light and converting it into energy in plant cells. The rest of the cytoplasm of a plant cell looks pretty much the same. One feature that I didn't really mention when I talked about an animal cell is in the middle of the nucleus, here you can see, is a structure called a nucleolus. And a nucleolus, or the nucleolus in the eukaryotic cell is responsible for making the large number of ribosomes which are exported from the nucleus into the cytoplasm. And, as I mentioned just before, the ribosomes are responsible for protein synthesis. It turns out this is a major synthetic effort on the part of most cells. Cells, like our own, have between 5 and 10 million ribosomes in the cytoplasm. So it's an enormous amount of biomass in the cytoplasm whose sole function is to synthesize proteins. As we will learn also, proteins that are synthesized by the ribosomes don't sit around forever. Some proteins have long lives. Some proteins have lifetimes of 15 minutes before they're degraded, before they're turned over. One other distinction between our cells, that is the cells of metazoa and metaphyta, are the cell walls, analogous to the cell walls of bacteria, this green thing on the outside. As I said before, we do not have cell walls around our cells. And we will, as the semester goes on, go into more and more details about different aspects of this cytoarchitecture during the first half of the semester. Here, for example, is an artist's depiction of the endoplasmic reticulum. Why it has such a complex name, I cannot tell you, but it does. It's called the ER in the patois of the street. The ER. And the endoplasmic reticulum is a series of membranes. Keep in mind, not the only membrane in the cell is the plasma membrane. Within the cytoplasm there are literally hundreds of membranes which are folded up in different ways. Here you see them depicted. And one set of these membranes, often they're organized much like tubes, represents the membranes of the endoplasmic reticulum which either lacks ribosomes attached to it or has these ribosomes attached to it which cause this to be called the rough endoplasmic reticulum to refer to its rough structure which is created by the studding of ribosomes on the surface. As we will learn, just trying to give you a feeling for the geography of what we're going to talk about this semester, these ribosomes on the surface of the endoplasmic reticulum are dedicated to the task of making highly specialized proteins which are either going to be dispatched to the surface of the cell where they will be displayed on the cell's surface or actually secreted into the extracellular space. Many of the proteins that are destined for our body are not kept within cells but are released into the extracellular space where they serve important functions, and so we're going to focus very much on them. Here's actually what some of these things look like in the electron microscope to see whether we can either believe or fully discredit the imaginations of the artists. Here's the rough endoplasmic reticulum I showed you in schematic form before. And you can see why it's called rough. All these black dots are ribosomes attached on the outside. Here's the Golgi apparatus. You see these vesicles indicated here. And a vesicle, just to use a new word, is simply a membranous bag. And keep in mind, by the way, that we're not going to spend the semester with these highly descriptive discussions. Our intent today is to get a lot of these descriptive discussions out of the way so that we can begin to talk in a common parlance about many of the parts, the molecular parts of cells and organisms. Here is the mitochondrion which we saw depicted before. It looks similar to, but not identical to the artist's description of that. And keep in mind that the mitochondrion in our cells, as I said before, are the descendents of parasitic bacteria. Here's the endoplasmic reticulum, and the way it would look, as it does in certain parts of the cell when it doesn't have all of these ribosomes studded on the surface. The endoplasmic reticulum here is involved in making membranes. The endoplasmic reticulum here is involved in the synthesis and export of proteins to the cell's surface and for secretion, as I mentioned before. Much of what we're going to talk about over the next days is going to be focused on the nucleus of the cell, that is on the chromosomes on the cell and on the material which is called chromatin which carries the genetic material. So the term chromatin is used in biology to refer simply to the mixture of DNA and proteins, which together constitutes the chromosomes. So chromatin has within it DNA, it has protein, and it has a little bit of RNA in it. And we're going to focus mostly on the DNA in the chromatin, because if we can begin to understand the way the DNA works and functions many other aspects will flow from that. I mentioned the cell's surface, and I just want to impress on you the fact that the plasma membrane of a cell is much more complicated than was depicted in these drawings that I showed you just before. If we had a way of visualizing the plasma membrane of a cell, we would discover that it's formed from lipids. We see such lipids there, phospholipids, many of them. We'll talk about them shortly. That the outside of the cell, there are many proteins, you see them here, which thread their way through the plasma membrane, have an extracellular and intracellular part. And these transmembrane proteins, which reach from outside to inside, represent a very important way by which the cell senses its environment. This plasma membrane, as we'll return to, represents a very effective barrier to segregate what's inside the cell from what's outside of the cell to increase concentrations of certain biochemical entities. But at the same time it creates a barrier to communication. And one of the things that cells have had to solve over the last 700 to 800 million years is ways by which the exterior of the cell is able to send certain signals and transmit that information to the interior of the cell. At the same time, cells have had to use a number of different, invent a number of different proteins, some of them indicated here, which are able to transport materials from the outside of the cell into the cell, or visa versa. So the existence of the plasma membrane represents a boon to the cell in the sense that it's able to segregate what's on the inside from what's on the outside. But it represents an impediment to communication which had to be solved, as well as an impediment to transport. And many of these transmembrane proteins are dedicated to solving those particular problems. Here you see, once again an artist's depiction form, aspects of the cytoskeleton of the cell. And when we talk about the cytoskeleton we talk about this network of proteins which, as I said before, gives the cell rigidity. Keep in mind that the prefix cyto or the suffix cyt refers always to cells. Allows the cell to have shape. And here you can see this network as depicted in one way, but here it's depicted actually much more dramatically. And here you begin to see the complexity of what exists inside the cell. Here are these proteins. These are polymers of proteins called vimentin which are present in very many mesenchymal cells. Here are microtubules made from another kind of protein. Here are microfilaments, in this case made of the molecule actin. And if we looked at individual molecules of actin they would be invisible. This is end-to-end polymerization of many actin molecules. And we're looking here under the microscope from one end of the cell to the other end of the cell. And you can see how these molecules, they create stiffness, and they also enable the cell to contract and to move. Some people might think that the interior of the cell is just water with some molecules floating around them. But if you actually look at what's present in the cell, more than 50% of the volume is taken up by proteins. It's not simply an aqueous solvent where everything moves around freely. It's a very viscous slush, a mush. And it's quite difficult there for many cells to move around from one part of the cell to the other. Here you begin to get a feeling now for how the connection, which we'll reinforce shortly in great detail, between individual molecules and the cytoskeleton. And here you see these actin fibers. I showed them to you just moments ago stretching from one end of the cell to the other. And each of these little globules is a single actin monomer which polymerize end-to-end and then form multi-strand aggregates to create the actin cytoskeleton. Here's an intermediate filament and here's the microtubules that are formed, once again giving us this impression that the cell is actually highly organized and that that high degree of organization is able to give it some physical structure and shape and form. I think we're going to end today two minutes early. You probably won't object.
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https://ocw.mit.edu/courses/3-320-atomistic-computer-modeling-of-materials-sma-5107-spring-2005/3.320-spring-2005.zip
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PROFESSOR: So what we'll do today is really switching to the computational description and characterization of thermodynamic properties. As we have seen in a lot of the previous lectures that we have been focusing on finding out good energy models, that is being able to calculate, what is the energy of a system given the coordinates of its ions? And now what we want to understand, really, is, what is the evolution of the system in the presence of a finite temperature? Temperature really means that there is an average kinetic energy available for all the atoms and the molecules in your system. That means that there is really a microscopic dynamics. And we want to follow that. One of the most important things-- and this goes back to, really, the first lecture that you have seen in this class for the quantum part-- is that there is the fundamental de Broglie relation that gives us an estimate of what is the quantum wavelength of any kind of object. Remember, we have said that the wavelength times the momentum should be equal to the Planck constant. And that really means that for an object like an electron, the de Broglie wavelength that comes out from that relation is comparable to the distance between the atoms. So if the wavelength of an electron is of the order of the angstrom, 10 to the minus 10 meters, it means that the wavelike properties of the electrons needs to be studied. And the electron will behave as a wave. And that's why we use the Schrodinger equation to describe electrons. The wavelength of the nuclei is much, much smaller. That means that nuclei really didn't-- we don't really need to take into account to a good approximation the quantum nature of the nuclei. The wavelength is so small that they can be treated practically, correctly, as classical particles. And that's why, actually, nuclei, in a way, still follow the rules of classical dynamics. That is, they evolve in time following Newton's equation of motion. Again, this is an approximation that involves treating the nuclei as classical particles and not as quantum particles. But especially at high temperature, this is a very good approximation. And so a lot of approaches in computational and material science have to do with solving Newton's equation of motion for a system of interacting nuclei and evolving those nuclei in time to study the thermodynamic properties of the system. And so that's why I actually started with a reminder of Newton's equation of motion. That is really an ordinary second-order differential equation. That is, when you have a particle of mass m in a force field that is this F of r, well, we can calculate its trajectory-- that is, its position-- as a function of time and its velocity as a function of time by integrating this differential equation. And in the classical case that is always presented, say, is the evolution of a particle in a constant force field, like a particle under the gravitational law, that, to a large extent, is constant if you don't move very much. And because, in that case, the force is really a constant, the integration of this becomes trivial. It's really a second derivative that needs to be equal a constant. And integrating that gives us a parabola. But the reason why I'm focusing on this is to remind you of the initial conditions in our differential equations. That is, if we have a second-order differential equation, the general solution-- let's say in this case for this force field would be a parabola-- is uniquely identified once we specify the position at a certain instant and the velocity at a certain instant. That is, among all the possible parabolas in the world, you selected the one that your projectile is going to follow once you have the position and the velocity at a certain instant. And that's why actually dynamics and molecular dynamics and other physical properties basically depend essentially on the position of the particle in your system and the velocity of the particle in your system. That's why there is no physical observable that depends, say, on the fifth derivative with respect to time. It's really because the dynamics of that system follows Newton's equation of motion. OK, with this, what we really need to deal in general is with a system of interacting particles. And for a moment you could actually think that this i, interacting particles, could be the planets orbiting around the sun. So each of the planet i will have its own trajectory. Our r of t will have its own mass. And they all live in a force field. That's really where the many-body complexity, again, comes out. So many-body complexity, that is much easier to deal with than the many-body complexity of the electron problem because, again, now we describe each object just with a set of six variables, so its 3-space coordinate and its 3-momentum coordinate. But still, each and every object interacts with each other. When you look at the planets orbiting around the sun, you, at first approximation, could only consider the gravitational attraction to the sun. But in reality, also the planets interact with each other. And so, again, even if this is much simpler than the Schrodinger equation, as soon as you start to have more than two bodies interacting, it really becomes too complex to solve analytically. And so all the molecular dynamic techniques have, as their only object, basically, the integration in time of this equation of motion in an efficient form. There are a couple of comments that one wants to make. In general, we deal with conservative fields. That is, the force depends on position only. That also means that the work that you make in going from one point to another point, the integral of the force times the displacement is going to be independent from the trajectory. So the energy, if you want, is just a function of the position. The other thing that descends immediately from this equation of motion is that the total energy of the system is conserved. And that's very easy to prove. And I'll do it in a moment. And that's what's often called a microcanonical evolution. When you read about microcanonical ensembles, it means that you are actually under the condition in which the total energy of your system is conserved. And again, what is the total energy of your system? Well, if you have one particle, it's just, trivially, the kinetic energy 1/2 mv squared plus the potential energy V of r of that particle, and just doing everything in one dimension. And you ask, why is this going to be conserved? Well, you can just prove it by taking the time derivative of this quantity, OK? So we have taken kinetic energy plus potential energy. We look at the time derivative. And this is just simple algebra. What you get is 1/2 m times 2 times v times the derivative with respect to time, that is the acceleration, plus the derivative of the potential with respect to time. And this can just be written as mass times velocity times acceleration plus-- and we can rewrite the derivative of the potential with time as the derivative of the potential with respect to position times the derivative of the position with respect to time. And now you see where we are getting to, because this term here is equal to minus the force, while these terms here is equal to the velocity. And so it goes away with here. And so what we have is that m, the derivative of the total energy with respect to time, is equal to the mass times acceleration minus the force. That is 0 because the particle follows Newton's equation of motion. So because of Newton's equation of motion and because the particle follows exactly that evolution, we have that the sum of the kinetic energy plus the potential energy needs to be conserved during the evolution. And that's actually the fundamental sanity check that you will always have and you can always do during a molecular dynamics simulation. That is, you can check that your integration is correct by making sure that the total energy of your system is concerned. OK, so what do we do in a practical molecular dynamic simulation? Well, the idea is that, as in the motion of the planets, we want to integrate the equation of motions with this caveat that, again, each particle interacts with each other particle. And so it's really a many-body problem, even if it is a classical many-body problem. And so it's really easier to integrate than a many-body Schrodinger equation. And in general, the way particles interact with each other can be described by empirical force fields or, in the best-case scenario, can be described using quantum mechanics. That is the [? functional ?] theory [INAUDIBLE].. And so what you will have to do in that case is that for every configuration of your ionic system, you need to solve an entire total energy self-consistent problem. And that's why actually the first principle in molecular dynamics becomes a very, very expensive challenge. And we'll see in one of the later classes how this is solved. There are a lot of systems that can be described accurately, even with very simple potential. Even more than in the case of the total energy, certain thermodynamic transitions, like the transition from a solid to a liquid, can be thought of as broadly independent of the details of the interacting potentials. So even studying a system with a Lennard-Jones potential that, after all, was developed especially to do molecular dynamic simulations of rare gas atoms, or even just studying a system that is made by hard spheres that don't really interact when they travel in space, apart when they really hit each other-- so what is called a contact interaction that is 0 all the time apart from the instant in which the two spheres collide-- can sometimes give you a very accurate qualitative description of the thermodynamic transition in a problem. And molecular dynamics actually was first introduced really to study systems, like liquids, for which it's very difficult to develop analytic techniques. You see, if you want to study a solid, you can actually choose a potential. And if the potential is simple enough, you can analytically calculate the potential energy of a periodic repetition of all your atoms. But if you study liquid, disorder is really an essential part of the state of the system. And so the entropic contribution to the free energy of the system is fundamental. And really there are no reasonable analytic ways to address a lot of the questions that you have. And that's why people really moved to solving on the computer the Newton equation of motion. And this is a short history, really, of molecular dynamics, reminding you what is the history of computers, really. In 1952, one of the first computers, MANIAC, was operational in Los Alamos. And that's when Metropolis, Nicholas Metropolis, in collaboration with actually a husband and wife couple, Rosenbluth and Rosenbluth and Teller and Teller, developed what is called the Monte Carlo method. That was a very powerful method to address exactly this entropic free energy problem. And that you'll see in one of the later classes. But really, 1956 sees the first molecular dynamic simulation by Alder and Wainwright. And as I said, in looking at the dynamics of hard spheres, these are the milestone papers. Vineyard is studying radiation damage in copper, what happens when you have a copper atom that is really put in motion by a very energetic radiation kicking it and starting running around. And really this is the classic paper of Aneesur Rahman studying, with a Lennard-Jones potential, the dynamics of liquid argon. Again, a Lennard-Jones potential is actually very accurate to describe rare gases that ultimately repel each other at very close distance and, at large distances, attract each other with this weak dipole-dipole interaction that goes as 1 over r to the 6th. And then really the first ab-initio molecular dynamics and the first theory for ab-initio molecular dynamics was developed in the mid '80s by Car and Parrinello. OK, now, there is a universal way to think of these dynamics of your classical system composed of N particles. And as we have said, since they follow Newton's equation of motion, what we really need to track, the only quantities that we need to track, is their position as a function of time and their velocity as a function of time. And generally speaking, the kinetic energy is just a function of the velocities. And the potential energy is just a function of the positions. So if you have N particles, you really have to follow in time six n variables. And if we define for a moment a six-dimensional space, what we really have is that an instantaneous state of your dynamical system, that is all its position and all its velocity, can be represented by a single point in this 6N dimension of space. So if you have got just a hydrogen molecule with two nuclei, what you really need is a point in six dimensions. If you have a water molecule, you need a point in nine dimensions. And the evolution-- sorry. If you have a water molecule, you need a point in 18 dimensions. And if you have a hydrogen molecule, you need a point in 12 dimensions. And the dynamical evolution of your system is described exactly by the evolution of this point. So we often think at this phase space that we'll have, for an N particle, 6N dimension, of which-- let's try to draw a bit dimension-- 3N of them, those represented in green, will really represent the positions. And the other 3N will represent velocities. OK. And obviously this is all orthogonal dimension. And your point will be somewhere. The certain instant in time T will have a set of velocities, will have a set of positions. And the Newton equation of motion will make it evolve in this phase space. So really your molecular dynamic integration consists in following in time this evolution. And that evolution is, in principle, analytically and uniquely defined once you know the position and the velocities at every instant. And-- well, we'll see that later. OK, so what do we do once we have a computational algorithm that allows us to evolve this point in phase space? And I sort of summarized here the three main goals that could follow from a molecular dynamic simulation in the order that we'll see them. Really, one of the most common ones is to use molecular dynamic simulations to calculate thermodynamic properties, that is to calculate what we call ensemble averages. Suppose that you have a system like a solid. And you have calculated, say, its lattice parameter, like you're doing in lab 2 or lab 3. And suppose that you start heating up this solid. What happens? Well, the atoms, in the classical sense, start oscillating more and more. And you know from experience that most solids start expanding when you warm them up. So a molecular dynamic simulation could tell you how much a solid expands when you heat it up because basically you start evolving your system from 0 temperature. And then you increase its temperature. And really, temperature, remember, is just the average kinetic energy of the ions. And as you increase the temperature, the ions oscillate more and more. And if you let your unit cell expand or contract in time following what is the stress overall of all the atoms in your unit cell, you see that your system increases the dimension of the unit cell and starts oscillating around a value that's really an average, depending on temperature. And all these sort of statements can be actually made formally in a way that we'll see in a moment that is the language of thermodynamics. That is, if we have a finite temperature, we'll have a certain probability associated with every microscopic configuration. OK, so if you are at 0 temperature, really what you have is that you have a probability 1 to be in the lowest energy state possible and probability 0 of being in any other energy state. If you start increasing your temperature, states that are energetically similar to the ground state but not really the ground state start to have a certain probability of being occupied. And the more temperature you have, the more very costly configurations can be accessed by your system. And really, the average properties at a finite temperature need to be an average over all these possible states weighed with their appropriate thermodynamic weight, weighed with the probability of being in that state. And so this is what an ensemble average does really. It tries to span all the possible states in a system with a probability that is proportional to their thermodynamical weight. And then for each state, you can calculate a certain property, like what is its average potential energy, what is its lattice parameter, and so on. And out of that, you can calculate the average thermodynamic property. And we'll see, actually, an example in a moment. So we'll see a practical application of this. The other thing that you can do with molecular dynamic simulator is actually study the evolution in real time of your system. So you could actually study a chemical reaction. What happens if you have, say, an explosive molecule that is decomposing, OK? And in particular, whenever you have a chemical reaction that tends to be a bond-breaking and bond-forming reaction, you'll probably need to use a quantum mechanical approach. But in principle, you can set up your system in an initial configuration and see what happens as you evolve in time. You could look at a catalytic reaction. Say, maybe you are interested in studying fuel cells. And you want to see how a hydrogen molecule decomposes when it arrives on a platinum surface. Well, you can take your hydrogen molecule and project it delicately towards the platinum surface. And follow, with your molecular dynamic techniques, the dynamics of all these nuclei as the molecule chemisorbs and dissociates in time. And then really the last application of molecular dynamics is more of an optimization algorithm. There are problems in which it's very complex to find what is the lowest possible solution, what is the optimal solution. That is, you could try to find out what is the schedule of all your flights in an airplane company. And obviously that's a complex optimization problem because you can't move one plane at a time and figure out what is the best possible solution because there might be a completely different choice of planes and itineraries that actually gives you a best performance overall on the net. And so you can always take one optimization problem and express a cost function. That is depending on what is your interest. Minimize the amount of oil, let's say, used in running all these routes. So when you have a cost function of a very complex problem, you have really an object that depends on many variables and that has a lot of possible minima. OK, so this would be, really, the problem of optimizing the air routes of an airline. What you have is a lot of possible variables. And there are a lot of reasonable solutions. But you really want to find the one that is the best for you, something like this. And molecular dynamics can actually help to solve this problem because it's a problem that is absolutely analogous of having a system of interacting particles that has a certain potential energy surface and trying to explore that potential energy surface, moving around, until you find the lowest minimum. You could think of this as a mountain range. And you have your skiers that are really your particles moving around. And you want to sort of let the skiers ski around as much as possible until they find the minimum energy state. And so you can use a thermodynamic analogy. You heat up your system. That really means that every particle has a lot of kinetic energy. So having a lot of kinetic energy can easily overcome any kind of potential barrier. And you have a lot of these particles moving around. And then you cool them down slowly. And as you cool them down, they'll start piling up in the lowest possible energy state. And once you have reached, very slowly, 0 temperature, you look at where all your particles have ended up. And so you have at least a good sampling of the relevant minima in your potential energy surface. So actually molecular dynamics-- and this is called simulated annealing, in this context-- can actually be used as an optimization technique. And we don't have to go as far as the case of an airline or what is called the traveling salesman problem. But often what we have is that we have a system that is fairly complicated to characterize. Say we want to describe maybe an interface between a perfect silicon crystal and the SiO2 substrate because we are in the electronic industry. Well, I mean, we have a problem because how do we construct the interface between a crystalline solid and an amorphous system? I mean, we can try putting atoms in a reasonable position. But it's never going to work very well. And so what we can do is actually use molecular dynamic techniques to let the atoms evolve according to their interaction and to the Newton equation of motion and somehow find, by themselves, a slightly more favorable green minima instead of the initial configuration in which we have put them. OK, so, in principle, if you had infinite computing power or infinite mathematical prowess, you would have really solved the problem of characterizing on a computer any material that you wanted. The reason why you can't really do that in general and indiscriminately has to do with four problems that arise in the approach that I've mentioned. And I've mentioned them here, I guess, in order of importance. And the first and most dramatic problem is the one of time scales. That is, atoms move around and vibrate at time scales that are characteristic of atoms. That is, say, in a molecule, it will take tens of femtoseconds or hundreds of femtoseconds to have a periodic oscillation. So these are femtoseconds. 10 to the minus 15 seconds is the order of magnitude of the dynamics of the atoms at room temperature. Now, a lot of interesting problems take place in times of seconds, minutes, or hours. A classical problem is how a protein folds itself in its native state. And you know, this is a process that can take seconds. And so we have, really, the issue that all our calculation takes place on the timescales of 10 to the minus 15. And often we need to reach physical properties that are of the order of seconds. So we need to span a 15 order of magnitude. And that means that if it takes a time x to perform one single molecular dynamic operation, we have a 10 to the 15 cost to get to the reasonable time scale for a lot of macroscopic processes. And that's really a scale that we can't bridge. If we do quantum mechanical simulations for a small system, we can easily reach the times of tens of picoseconds or hundreds of picoseconds. If we do classical molecular dynamic simulation, we can go maybe three, four, five orders of magnitude better. And so we can start accessing nanoseconds and maybe start getting closer towards the microsecond. But there is still an enormous gap to reach. And this is really a difficult problem to overcome because we have dynamics taking place at different levels. Say, think of a protein. We have the atomic dynamics of an atom vibrating around. And then we have dynamics of subunits maybe interacting with the different amino acids along the chain. And then we have structural motifs that want to move around and then maybe the first coil-up in a certain configuration. But that's not really very good. And then after a while they sort of open up and recoil in a different configuration. And so the more you look at your system on a larger and larger length scale, the more you discover that there are dynamics of more and more complex groups that are slower and slower. And integrating from the atomic motion up to the mesoscale and the macroscopic scale of these dynamics is really an impossible problem to do by brute force. And I would say that it's still the conceptual problem for which less progress has been made. It's really the most difficult one. And you'll hear a lot of talks nowadays in science about multiscale modeling and about bridging timescales. And it's obviously a very important challenge. But I still think we are at the stage in which there are very little solutions. Somehow related to the time-scale problem, there is the length scale problem. Suppose you want to study a system close to a phase transition. Suppose that you want to see a gas evaporate. You want to study a liquid-to-vapor transition. What happens exactly at the temperature where the liquid and the vapor are in equilibrium? Suppose that you are studying water boiling. At normal condition, it boils at 373 Kelvin. And at that temperature, why there is a phase transition? Basically because the free energy of the liquid and the free energy of the gas are the same. You go a little bit above the temperature, the free energy of the gas is lower. So the system wants to be all gas. You go a little bit below the boiling temperature, the free energy of the liquid is lower. So the system wants to be a liquid. But as you get closer and closer to the temperature, these two free energies becomes comparable. That means that the cost of transforming from liquid to the gas and from gas to liquid becomes smaller and smaller. And it is 0 at the transition temperature. That means, in practice, that since it doesn't cost anything to transform from one state to the other, you have lost all possible length scale because you can nucleate a bubble of gas of any size with 0 cost. So if you think of the transformation from a liquid to a vapor as the nucleation of bubble of gas exactly at the temperature of the transition, that cost has become 0. And we can nucleate bubbles of any size. So the phase transition is a moment where we lose length scales. The fluctuation can have any size. And so any kind of finite simulation will break down because it's really only in the infinite system that we represent appropriately all these sizes popping up in our system. And so that's why it becomes very difficult to study accurately, say, a boundary at a phase transition. We say that the length scale starts diverging. And that's why one needs to be careful. This problem is not as serious as the problem of time scales. But it's still one that you want to consider very carefully and one you need to spend some time thinking, actually. That is, often we use a periodic boundary condition to describe an extended system. And so the question that you always have to ask yourself is, how large should I make my unit system that is periodically repeated before it's really describing accurately the infinite system that is the goal of my simulation? Suppose that you want to study water, OK? Or suppose that you want to study any liquid. In principle you want an infinite system. In principle you can deal only with a finite number of degrees of freedom. So the two possibilities that you have, you could study a droplet of water, OK? But then obviously you encounter the problem that, when you study a droplet, you have a very large influence coming from the surface effect. A droplet is a droplet because of the surface tension. And before reaching the thermodynamic limit of that droplet behaving as a bulk liquid, you need really to go to enormous sizes. If instead you study, say, the same number of molecules in periodic boundary conditions, you eliminate the presence of the surface. And so you can reach the thermodynamic limit of the system behaving as a bulk much faster. But still you need to always ask your question, is your simulation still large enough? And there is a size that is large enough. And that's the other very important concept. That is, whenever your simulation cell is so large that the dynamics of one water molecule doesn't affect the dynamics of a water molecule that is far away from yours as much as it is its periodic image, your system has become infinite. That is, whenever you have a molecule in a liquid, you really dynamically interact not with every other molecule in the infinite liquid but only with a sphere of neighbors. And you need to understand how large is that sphere of neighbors because those and only those are the molecules that affect yourself in the center of that sphere. And if you decide that for a water molecule at room temperature the sphere of neighbors with which there is a dynamic interaction has only a radius of 10 angstroms, whenever your unit cell is larger, is a cube, say, larger than 20 angstroms inside, you are studying, in practice, an infinite system because every molecule in that system sees all the independent neighbors that it needs to see, OK? So even a simulation with a finite number of molecules can reproduce exactly the behavior of an infinite system provided a molecule interacts with other independent molecules up to a maximum range of dynamical correlation. And we'll actually define this quantity in the lecture that follows. So again, length scales are another problem and a problem in which periodic boundary conditions tend to help a lot. But again, suppose that you want to study something like the strength of a metal. Well, you have now a problem in which a lot of length scales becomes intertwined because in the strength of a metal, you might want to consider what is the dynamics of impurities that strengthen that metal. Like, if you have steel, you have carbon atoms in between the ions. And so you need to understand what is the atomic dynamics. But then there will be dynamics of units on a larger length scale because in a metal there's going to be these location and slip and glide planes. And so there is coordinated motion of thousands of atoms that you want to consider. And then, in reality, a real metal is never a crystal. But it's always a polycrystalline material. So you have grains. And inside that grain you have a consistent ordering of the atoms. But then grains are mismatched. And there are grain boundaries. And so you have a lot of length scales to consider if you really want to predict what would be the mechanical response of a real system. OK, so this is the challenge, if you want, number 2. The challenge number 3 is the one that we have described more in detail up to now that is determining how accurate can be your prediction of the interaction between nuclei. And you have seen everything that we could tell you about empirical potential quantum mechanical simulations. And in principle, even if it's obviously very expensive to make very accurate predictions of the interaction between atoms, in principle it's the less conceptually challenging of the problems. I mean, we can just try all our electronic structure methods at will. And we can try to see how accurate we become. And the last challenge that is also not really a conceptual challenge and it can be solved but is just a very expensive one to overcome is this one in which we treat the nuclei as classical particles. And that tends to be, in general, very good. But it's sort of an approximation that breaks down the more you go towards lighter and lighter nuclei because, remember, the de Broglie relation. Your wavelength times your momentum is equal to the Planck constant. So the lighter you are, the smaller your momentum, the longer your wavelength, the more quantum you become. So something like hydrogen that has a nucleus that is just one proton, so it's fairly light-- or it's the lightest of all possible nuclei-- tend to still have significant quantum properties even at relatively high temperatures. You see, the more you increase the temperature, the more the average momentum of your particles become and the more classical they become. But the lower the temperature, the more quantum they become. And something like hydrogen is still significantly quantum at room temperature. And what does that mean? Well, it means that that particle, it's really a delocalized system that can actually tunnel through barriers. That's if you want one of the important differences between a particle that behaves according to classical mechanics and one that behaves according to quantum mechanics. If you are behaving as a classical particle, you need to have enough kinetic energy to overcome a barrier. If you are a skier and you are in a valley and you want to go in the other valley, you need to have enough velocity to be able to overcome the mountain. If you are a quantum skier, you can actually have some finite probability of tunneling through. And so there are tunneling problems that are important not only for hydrogen but sometimes even for much more massive particles or even for groups of particles, provided that the energy barrier that they need to overcome is small enough. And I've shown you in the first quantum class an example in a perovskite crystal in which we have this cubic structure with an octahedral cage of oxygen that can configure itself in a ground state that is ferroelectric, in which the oxygen cage displaces with respect to the cubic symmetry and induces a permanent electrical dipole in your system. Well, there are certain materials that have this structure. But since the octahedral cage can have roughly six possible choices on where to move offsite, if you decrease the temperature enough, the oxygen cage will start actually tunneling between all the six possibilities. Instead of finding itself at 0 temperature in a broken symmetry configuration in which the oxygen cage has chosen one of those six possibilities, in reality, because of quantum tunneling, it keeps tunneling between all the six possibilities. And on average your system looks cubic. And on average you don't have really any ferroelectricity. But the polarizability of the system is going to be different from that one in which the octahedral cage was sitting squarely in the cube. So there is a dielectric response that is different. So there are cases that are fairly exotic, I would say, for most applications, in which tunneling effects of the nuclei could become important. But there is one other important consequence that comes from quantum mechanics that tends to be important in a lot of cases. And that consequence has to do with the quantization of vibrational excitations. That is, if you look at the motion of any system-- a molecule, a solid-- in classical terms, well, that molecule or that solid can have any amount of kinetic energy. In a hydrogen molecule, the atoms can have 0 kinetic energy if we treat it classically. Or they can move just a little bit. So they have a very minute amount of kinetic energy. And we can just increase the kinetic energy that those atoms can have at will. In reality, because the system is a quantum system, the vibrational modes of a molecule or the vibrational modes of a solid are actually quantized. And you can't have arbitrary amounts of vibrational energy. Actually, in its quantum ground state, a molecule will not be at rest. But it will be in what is called the zero-point motion state. That is what is the quantum mechanical ground state. You can think of the analogy with the stability of the atoms. In its ground state, the electron around the proton is not collapsing onto the proton because of the Coulombic repulsion. If the electron was a classical particle, it would collapse on the nucleus. And the hydrogen atom would be just a proton and an electron sitting in the same place. But because the electron is quantum, it can't really collapse. And its lowest energy state is a state that is the 1s orbit for the hydrogen in which the electron is around the atom but doesn't collapse onto it. And the excitation of that electron to the next energy state is a quantized excitation. You can't go to a level of 1s that has an energy for the hydrogen of minus 1 Rydberg to another state that has energy minus 0.99 or minus by 0.98 or any kind of amount. The next state in which the electron can live is a state that has an energy of minus 1/4 and then minus 1/9 and so on and so forth. The same thing happens for the vibrational level of molecules, OK? A quantum molecule in its ground state is not in a state where the atoms do not move. But it's in a state where the atoms have a certain amount of quantum kinetic energy. And it's in a state called zero-point motion. And if you want to increase the kinetic energy of this molecule, you can't do it continuously. There is a quantized cap. So you need to provide enough energy to heat up the system to the next state and then to the next state and then to the next state. And that, if you think, affects deeply how you are going to heat up a system. That is, if you look at the specific heat of a solid, that is basically the quantity that tells you how much heat the system is going to be able to take, and you look at what happens at very low temperature, you discover that actually the system is almost unable to take any heat because all the vibrational excitation are much larger than the average amount of kinetic energy that you have if you are looking at what it takes to bring something from 1 Kelvin to 2 Kelvin. OK, but this is still relevant if you have a system like water, even at 300 Kelvin. In water, say, at 300 Kelvin, what you have is you have these water molecules. Remember what you have-- your oxygen, hydrogen, and hydrogen. And the modes of vibration of these molecules are stretching modes of the hydrogens. And then there are scissor-like models in which the two bonds do this. And then these are intermolecular modes, just sort of relevant to one molecule. And then there are all the molecules interacting. Well, at room temperature, all the internal modes of the molecule, the stretching and the scissor modes, have so much vibrational energy in their ground state that it's almost impossible to excite them to the next vibrational state with the average temperature that you have at 300 Kelvin. So even a system as fundamental as water has really a hybrid behavior in which the vibrational interaction between molecules can actually be populated arbitrarily by the temperature that is surrounding you. But it's almost impossible to excite a molecule from a zero-point stretching motion to the next one. And so, say, the specific heat of water will be slightly exotic because really part of its vibration are quantized. And so there are a number of cases in which this classical nuclear approximation is not exact. And luckily they can be studied with specific techniques that we'll see in some of the later-application classes during this course. But remember this-- if there is one thing that you want to remember about the behavior of nuclei-- that this vibrational excitation are quantized. And we actually see this quantization in physical properties like the specific heat. OK, so let's discuss the first of the possible applications of molecular dynamics that I have mentioned. And that's using molecular dynamics to really calculate average properties of a system at finite temperatures. And what I have written here is really the summary of statistical mechanics. So that is what, if you want, Boltzmann would tell you. You have a dynamical system that is described by this point moving around in phase space, OK? So with a certain possible set of positions and with a certain possible set of momenta. And suppose that we want to calculate, say, what is an average property at a certain temperature. And we call that property A. It could be the lattice parameter of the system. Or it could be the diameter, say-- it's even simpler-- of your water bubble. OK, you have this water bubble. And you want to see what is the diameter of this water bubble at a certain temperature. And this diameter is going to increase with temperature. And what you do, you just let this system evolve. And during its evolution, you keep measuring what is its diameter. And you let this evolve a lot. And you measure what is its width. Well, the way Boltzmann would set this problem is say, we really need to integrate over all possible configurations of the system. That is, we need to go through all possible configurations of the system. That is, we need to integrate over the full phase space. And for every possible point in phase space, we need to calculate the diameter of my water bubble. And then we need to take into account what is the probability of being in that configuration. And if you are in a system at a constant temperature, the probability of finding yourself at a given temperature is given by exponential of minus beta is just 1 over the Boltzmann constant times the temperature times the energy. That means that configurations that have a high internal energy are very unlikely. And configurations that have lower energy are more likely, to the point that if the temperature is 0, that is this beta is diverging to infinity, only the lowest energy configuration possible is represented. So this is the statistical mechanic points. To calculate an average quantity at a finite temperature, we need to sum over all possible configurations in space and momentum. For each possible r and p configuration, we'll have a certain value of the diameter of that bubble. We weigh that with the statistical weight of that configuration. And we actually normalize this expectation value. And this is what we obtain. So if, for a moment, you think at a phase space that is two dimensional and in this square we represent all the possible positions and velocities of our system, what Boltzmann tells us is that for every possible point, we will have a certain value of A. And the sum of all the possible values of A weighted with the probability of finding them gives us the expectation value. And so what you really need to do is an integral over all space of this. And the cost of these integrals diverges as the number of dimensions diverge because, again, every time we increase the size of the system by one particle, we have six more dimensions over which to integrate. And again, if you think that we do this just by numerical integrations by taking points on a grid and maybe we just discretize our system with 10 points along one dimension, every time you add a particle, you have 10 to the 6 points that multiply your system. So this quantity explodes. The point of view of molecular dynamics is different. What we say is that, well, let's just let this system evolve. Let this system evolve in time. And let it evolve at a constant temperature. And so now our point in phase space is going to go around, following a trajectory. And if we let it go around long enough, it will spend more time in the parts of phase space that are more favorite, the ones that are more likely, according to the Boltzmann factor. And so if our evolution, really, is distributed in the trajectory according to this thermodynamic weight, thermodynamic factor, then our expectation value, the thermodynamic average for our observable A, is just trivial, obtained by integrating over the trajectories all the values of A. That is what I was saying before. If you want to calculate the diameter of your bubble, what you do in the molecular dynamic sense, you just let the bubble evolve according to the equation of motion at a constant temperature. And you keep monitoring its diameter. And you will see that the diameter will have an average and that this is a much more efficient way then considering all possible configurations of the bubble, weighing them with how expensive that configuration is, and doing it for all the possible configurations. The molecular dynamic algorithm automatically, if you want, keeps spanning the most likely configurations. And so it does, for you, the job of going in the place where it's most likely to be. So in order to calculate a sample average, we can actually transform a problem into integration over a trajectory. And that is almost always doable under what is called the criterion of ergodicity. So the equivalence between the two formulations that I'd written before-- that is calculating a thermodynamic observable as an integral in phase space or as an integral over trajectories-- are equivalent, provided your trajectory is really able to go all over the place in phase space. So if, again, this were phase space, you could have, in principle, the problem of a known ergodic system if there is something that prevents your trajectory to reach certain so-called forbidden segments in your phase space. And I can give you an example. Suppose that you are studying something like silicon and you are in the crystalline state. And now you heat it up. And once you go above the melting temperature, your silicon will become liquid, OK? Now, suppose that you cool it down again below, now, the melting temperature. OK, in principle, the liquid silicon should become again crystalline silicon because, at that temperature in which we have brought it back to, the thermodynamic stable state is really the crystalline solid. If you do that in practice with our molecular dynamic simulation, you see that you start from the crystal. You heat it up. It becomes a liquid with no problems. And then you cool it back down. And instead of going back to a crystal, it goes back to a glassy, amorphous state. Silicon will find itself in a configuration in which locally all atoms are happy to be fourfold coordinated. But if we look at the long-range order, it's completely lost. The system is not crystalline anymore. It's glassy. What's happening there? Well, it happens that it's just, for a liquid cooling down, so much easier to trap itself into a set of configurations in which atoms are really fairly happy as they are, because what is important for silicon atoms is to be fourfold coordinated. But what is important to be in a crystal is that this fourfold coordination extends and becomes a long-range order. But for silicon, it's very inexpensive to mess up that long-range order, like for many other systems. And in principle, same thing happens for polymers. You lower it down. And they sort of get trapped in a glassy configuration. A lot of polymers might want to stay actually in a crystalline state. But really that is impossible to reach. And so if you want to study, say, what are the properties of your silicon and you have obtained or you are trying to obtain your silicon structure from a liquid that you cooled down, you'll actually-- in your molecular dynamics, you'll keep orbiting around in this glassy mass state and will never reach that region of phase space that is actually very, very small because it needs to have all atoms long range in very specific positions. And so you are really spanning a system in a nonergodic way. That is, you spend a lot of time in a region of phase space that are not really representative but out of which is very difficult to escape. OK, so there are problems in, again, brute force molecular dynamic approaches that you have to be careful about. OK, so let's see how we actually go and try to do this on our computer experiment. And I'll start with the simplest case that is what is called the microcanonical case. That is, we perform a simulation where the number of particles is constant, the volume is constant, and the total energy of the system is constant. This is what is called a thermodynamic ensemble. Again, these are all extensive variables. And we are fixing them. Sometimes it's appropriate, instead, to do a Legendre transformation and go into a different thermodynamic ensemble. Say, you could want not to be at a constant volume. But you might want to study a system at a constant pressure where the volume changes. And so we need to describe how we'll actually be able to do that. That is, how will we be able to move to a system in which different intensive thermodynamic variables are constant? But if you take a system of particles in periodic boundary conditions and you have 10 water molecules in a certain unit cell and you evolve them with the Newton equation of motion, you can obviously know that the number of molecules and the volume of the unit cell are not going to change. And if you integrate your equation of motion appropriately, the total energy of the system, that is the kinetic energy plus the potential energy, is not going to change. And so as we evolve this point in phase space, its trajectory is going to span what we call the microcanonical ensemble. And you let this point evolve long enough. And all of a sudden, you will have a meaningful trajectory, that is a meaningful assembly collection of microstates of representative configurations in space and in velocities of your system, out of which you can calculate averages of any of the observables that you want to calculate and that you want to characterize. So what do you do in these molecular dynamic simulations? Well, there are really four important steps that you need to take into account. And we keep thinking of water, liquid water, as an example. That is, you need to start from something. That is, you need to put your water molecule in a certain position. And you need to give them initial velocities. And you are almost never going to be good enough in choosing a state of the system that is somehow meaningful. You'll always, when you start building something complex, put molecules in places that they don't want to stay. And one needs to be very careful, then, in all of this. But suppose that we have, in any case, an initial configuration of position and velocity. Then what we need to do is, over and over again, integrate the equation of motion, that is, from the current position and the current velocities, predict the new position after a certain time. And once you have the new position at a certain time, you can recalculate what are the forces acting on the particle in the new position. And with these forces, you can, again, move the particles by a certain amount and so on and so forth. And that is where the molecular dynamics algorithm comes into play. We need to integrate our equation of motion. And with this, our system will start to evolve. We have put that point in phase space in motion. And it'll start moving around. And if we are integrating correctly, the total energy of the system is going to be conserved. Now what we need to do is we need to make sure that our characterization, our thermodynamic averages, do not depend on our initial condition. Again, we have started from a configuration that we have built by hand. And because of that, it's most likely a very unfavorable configuration of the system. OK, so we need to let the system evolve and somehow start figuring out really in which condition it would want to stay. Think of the previous example of silicon that is glassifying. Well, we had liquid silicon. We have to cool it down. In a way, what's happening is that we are never equilibrating it because the system goes into a glass phase. And so it never loses memory of its initial condition. It's always sort of trapped in this pseudo liquid, now glass, state. While in reality, at a temperature below the freezing or the melting point, equilibrium would be given by a crystalline state with the ions vibrating. So there are cases in which equilibration is so difficult that we actually never reach it. Silicon never goes to a crystal. That's why we need to be careful. But in most cases awfully we are lucky. And we lose memory of initial condition. Say, the opposite is actually very easy to achieve. If we start from the solid and we warm it up, it becomes a liquid. And if we wait long enough, that liquid will have lost all memory of having come from a solid. OK, so it's actually very easy in one direction and very difficult in the other direction. But once this equilibration time has passed, well, at that point we have truly everything in place. Our point in phase space is moving around. And it's really going into the regions of high probability and distributed, according to Boltzmann distribution. And so at that point, we can start an average among any of the physical observables that we want to calculate. And so there are these four phases. And I'll briefly mention how we actually choose them. Initialization. Well, again, we need position and velocities. You need to be careful, in particular, to avoid overlap between your molecules. If you put two atoms very close, they are going to repel each other with an enormous strength. And if you put them close enough, they are going to shoot in opposite directions very quickly and completely kick off and mess up everything on the side. So what is difficult is really, always in configuration space, finding the right structures. This is the difficult part of choosing initial condition. Choosing velocities is much easier. One possibility is just giving zero velocity to the system or a very small velocity and then slowly adding things. And what we usually see is that configuration in velocity space is much faster. And this is the reason. Think of that trajectory for a moment of one particle. And think of what happens when you have a collision. Suppose that you have particles, for a moment, that are hard spheres. In a collision, the velocity of the particle all of a sudden changes sign. So in the velocity part of the phase space, your point makes a sudden jump, OK, because the velocity that was plus 10 has become all of a sudden minus 10. So in velocity space, for the limit case of a system of hard spheres, your representative point hops everywhere continuously. In reality, I mean, you have interaction at all times. And you don't really hop. But you can sort of move very quickly from one region to the other region. There is not anything similar to the glass entanglement in configuration space. So it's very easy to thermalize velocity. It's very difficult to thermalize and equilibrate position. And that's where, all, you need to be careful. And if you really want to be accurate, what you can also do instead of giving zero velocity to your system, you can use your concept of statistical mechanics. That is, a system of classical particles will have a distribution of velocities that depends on the temperature at which it is. And this is 0 in Celsius. So it's 273 Kelvin. And the distribution of velocity in a classical system is actually given by the Maxwell-Boltzmann distribution written above that has basically this shape. So it means that if we are, say, at 0 Celsius, 273 Kelvin, we'll have a lot of molecules with a certain velocity. And then, in principle, we can have molecules with higher and higher velocity. But the probability of finding them decreases. And as we increase the temperature, the probability distribution of the velocities for the molecules changes. And so as usual, you can define both a median or an average in your system. That is, you can look at what is the most likely velocity, that is the top of this curve, at 0 Celsius. Or you could also define what is the average velocity. And since that this curve is asymmetric, it's going to be really shifted a little bit. And these are actually the numerical expressions for these two velocities, the most likely and the average velocity. And so I've given you the example of an oxygen molecule at room temperature. We are talking about 1,000 meters per second. So these velocities, if you want, are similar to the speed of sound in your material because, at the end, sound propagates at a certain speed because that's the velocity of the collision from one molecule to the next that are really carrying sound away. So if you want to be sophisticated, you give to your system a set of velocities that could be, say, all identical to the most likely velocity. But again, equilibration and thermalization in velocity space is very easy. So once you have this, that is, once you have a set of positions and the set of velocities, you need to use Newton's equation of motion of saying, of predicting, where your system is going to go. And so that's when the molecular dynamic algorithm comes into play. So what you need to do is, as we say, you need to use an integrator, the algorithm that tells you what your next position and what your next velocities are going to be. And there are a number of integrators. There are subtleties. But often you find the name of Verlet algorithm. This is one class of algorithms. And the other common class is the Gear predictor-corrector, just to give you the terminology. And we look, in a moment, at the Verlet algorithm. They're all fairly simple. But what is really important is that this algorithm should be robust. That is, they should give you a trajectory that is very close to the trajectory that you would get from the perfect analytical integration of the equation of motion. And in particular, what is the most important thing is that they should be accurate in, say, conserving the constant of motion. As we have seen from the Newton's equation of motion, the sum of the kinetic energy and the potential energy should be a constant. If your integration, if your numerical evaluation, of that equation is poor, you will see that the constant of motion is not any more constant. So again, this is the sanity check on the accuracy of your integration. And there are more subtle elements of the integration algorithm that can become important. But we won't go into that. And then as I said, straightforward integration of the Newton's equation of motion gives you the microcanonical ensemble in which the total energy's conserved. But sometimes you want to study systems in which the temperature is fixed. Or you could want to study systems in which the pressure is fixed. And so you need to augment your molecular dynamic integrator with some feature that allows you to control those intensive variables. So if you want, say, to control the temperature of the system, what we say usually is that we couple the system to a thermostat. That is, we make sure that the temperature in your evolution doesn't change. And you'll see examples in the next class of this. And in the case, in particular, of something, say, like the temperature, you have a whole set of possible thermostats. That is, you could use stochastic approaches. That is, you have your evolution of your system. And your system doesn't conserve, if it's microcanonical, the temperature. Sometimes it goes towards higher temperature. Sometimes it go towards lower temperature. And what you could do is either every now and then absorb some energy or give back some energy in your system-- and that would be a Langevin dynamics-- in order to control the average kinetic energy of your system. Or you could somehow forbid, in the microcanonical evolution, your system to choose a state in which the temperature changes. So if your new positions are such that your velocity gives an increase in temperature, you just shorten how much your atoms move in order to make the velocity smaller and in order to make the kinetic energy conserved. Or there are more complex dynamical ways of imposing temperature that goes under the name of extended Hamiltonian, or extended system, of which we'll see an example that is [INAUDIBLE]. But let me give you really what is the simplest and still one of the most used numerical integrators. And it tends to be very robust and very accurate. So it's still used in a lot of molecular dynamic simulations. And there is often no reason to do anything better than this. So in a way, we are in a much luckier situation than in quantum mechanics. This is good enough in almost all cases. And this is how we look at the problem. That is, what we need to do always is we need to calculate the position of each particle at a time t plus delta t once we know the position, the velocity, and the forces on that particle at time t. So suppose I'm at a certain instant in time that I call t. I know everything about my system-- position, velocity, and forces. What I really need to do is predict what are the new positions. And when I'm in the new positions, I want to calculate the forces again and evolve the system. And now, what do we do? Well, let's do a simple Taylor expansion. That is, this is r of t is a function of the variable t. So we can use a Taylor expansion. And r of t plus delta t is going to be r of t plus the derivative of r with respect to t times delta t. And that's the velocity. Plus 1/2 of the second derivative of the position with respect to time. And that's the acceleration. Times delta t squared, and so on for the third derivatives and so on for the fourth, fifth, ad infinitum. This is just the Taylor expansion as a function, OK? And so it gives us the position of t plus delta t, knowing all the derivatives-- first, second, third, fourth, nth-- at times t. And once we have the derivatives, we can not only predict what the position would be at t plus delta t, but also what the position would be at t minus delta t. And it's written here. And if you think, the terms with an odd power change sign in delta t. And the terms with an even power in delta t do not change sign. So we have just done a Taylor expansion. And now what we do is we actually sum these two quantities together, OK? And this is what we obtain, OK? We sum the two quantities together. All the odd terms disappear. This sums to 0. This sums to 0. The fifth order, seventh order, and so on sum to 0. So what we have is that we have rt plus delta t plus rt minus delta t is equal to 2 times rt plus alpha-- sorry-- acceleration times delta t squared. It's just written here. And then rearrange the terms. But you see, this is the Verlet algorithm. Now what we have when we look at this expression, we have an expression that tells us what is the position at my new time step, knowing the position at my current instant, knowing the position at the previous instant. We are keeping configurations from our trajectories. So we know where we were before. We know where we are now. And we know where we are going to be just by adding to this the term acceleration at the present time times delta t squared. We make an error because we are not introducing terms that are higher order in delta t squared. But you see, the term in delta t cubed has canceled out. So the error that we make is just of the order delta t fourth. That is, that means that the smaller we make the delta t, the smaller we make the time step, the more accurate our integration is. And so this is, if you want the most critical parameter in your molecular dynamic simulation, you need to make sure that the steps that you take are small enough, considering what is really the variation of your force as a function of time because remember that now that we use Newton's equation of motion, the acceleration at a time t is just given by the force divided by the mass. And because the force is just the gradient of the potential, it's just given by this. So at a certain instant in time t, you have the acceleration acting in your particles just by taking the force and dividing by the mass. So at every instant, you calculate force. And the force lets you evolve the system. And it keeps giving you the new positions. And so you know what is the trajectory and configuration space. And you can very simply find out what is your velocity at every instant t, if you need to calculate what is the velocity just by doing a final difference calculation of the tangent basically. The velocity is the tangent of the trajectory with respect to t. OK, this concludes class. What we'll see in the next lecture is how we actually take this Verlet integration and how we use it to do the characterization of the thermodynamic properties of a real system. So today is Thursday. Next week is spring break. So we'll see each other again on Tuesday, 29th. And I will be at a conference. And all the TAs will be at a conference next week. So it will be slightly difficult communicating via email. So try to ask a lot of questions between now and Sunday for your problem set that is due the Tuesday after spring break. And that will be problem set 2. Otherwise, enjoy spring break.
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https://ocw.mit.edu/courses/3-60-symmetry-structure-and-tensor-properties-of-materials-fall-2005/3.60-fall-2005.zip
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PROFESSOR: Let the minutes of the proceeding show that I re-entered the room at 3:00 and 12 seconds, true to my word. OK, I wanted to give you one example of a screw axis that you're probably familiar with in everyday life. That was a telephone pole that I was very familiar with when I was a little kid because we used to love to wait until it got dark and our parents couldn't see us. And then, we'd climb up the thing. And wow, you could see all over the whole neighborhood. I'm surprised nobody said, what? They had electricity when you were a little kid? Yes, they did. But now it's all underground. So kids, I don't know what they do for recreation these days. OK, let me pursue this question of screw axes. We've seen what a 2 sub 1 screw access looks like. I just erased it. But it consists of objects of the same chirality extending left and right on either side of the principal axis. The symbol for a twofold axis is this. The symbol for a twofold screw axis is the symbol for a twofold access with alternate sides extended like a propeller, a very descriptive symbol. The thing is rotating around. And you can think of these as the little ribbons that you used to have on the handle bars of your bike when you were a little kid. And as you pedaled along, these things fluttered out behind, and it was really cool. OK, let's move on to the next family of screw axes. And let me look at an operation A2 pi over 3 with a translation component, tau. And these things are going to get awfully cumbersome to draw. So let me use a device-- I wish I could claim credit for it. But actually, a student in my class one year said, hey, I've got a really cool way to draw these things for you. Let's imagine that this is a screw axis. We put a cylinder of paper around the screw. And then, let's divide up the surface of the paper in terms of end segments. So if this were a threefold screw axis, these would be 120 degrees segments. And then, let's draw horizontal lines on the sides of the cylinder. Then, if we want to see how the screw axis reproduces a pattern from a given motif, you just fill in these boxes. And then when you're all done, cut the cylinder and you can draw the pattern quite nicely on a two dimensional surface. So this is what I call the unrolled cylinder device. Somebody very imaginative invented it. His name, unfortunately, is lost to history. But I didn't want to claim credit for it myself. So what is the pattern of a threefold screw axis going to be? Let's let the difference between boxes be the translation component, tau. And so what we will do is take an initial motif, rotate it 120 degrees, slide it up by tau relative to the axis about which we're rotating. Performing the operation again would involve rotating 120 degrees, sliding up by tau. Doing it yet get again brings us back down full circle. So this would be the translational periodicity along the direction of the screw axis. And this would be the value of tau equal to 1/3 of the translation. So that is what a threefold screw access would look like. And the pattern obviously, if we would keep repeating it, would do something like this-- so perfectly interpretable, easy to draw. But things are actually more interesting and more complicated than this. We are stating two things about this operation. We're specifying a translation. But then the other thing that we're specifying is the translation component tau. And why should we be constrained to say that tau can only be one nth of the translation? If we do the operation n times, then doing the operation n times would give us a total translation of n tau. But there's no reason why that has to be one translation. Why not two? Why not three? Why not four? So the only real constraint is that n tau has to be some integer, m, times the translation that is parallel to the screw axis. And this means that the value of tau is not just equal to 1 nth of-- restricted to 1 nth-- of t. Tau can be m/n times T. And that is perfectly compatible. You do the operation n times. You're going to be directly above where you started. That's going to be a translation vector. But why not two translations, or three translations, or four translations? So there are infinitely more screw axes than just the n subscript something translations. So let's look at some of the possibilities. For a threefold screw axis, tau could be equal to 0T Tau could be equal to 1/3 of T. Tau could be equal to 2/3 of a translation. Tau could be equal to 3/3 of the translation, 4/3, 5/3, and so on, on and on and on. We could fill the whole board with possible screw axes having different values of tau. Now, let me convince you, hopefully easily, that if tau is equal to 3/3 t, tau would be an integral number of translations-- in this case, one. Down here, 6/3 t would be two translations. We're already going to have those operations in the pattern when tau was equal to 0T. So let me show you what I mean if it's not clear. Here's a trio of objects on either side of the rotation part of the operation. So what I'm saying is tau would be equal to two translations. That would, for some perverse reason, defining this as the pitch of the screw. Is it clear that that is going to be an operation that I already have when I say there's a threefold axis and a translation, T, parallel to that threefold axis? That's going to give all sorts of screw operations. But they are going to be integral multiples of the translation that's parallel to the axis. So the rule is that we can always subtract off one translation from any of these definitions of tau and reduce it to something smaller. So we can always define tau less than or equal to T without any change or artificial restrictedness on the nature of the pattern. So that says that for a threefold screw, for a threefold screw axis, there are only three that we should consider. Tau equal to 0, and that's a pure threefold rotation axis. tau could be equal to 1/3 of a translation. And now, we introduce the notation used to designate screws. If n tau was equal to mT, the symbol for the screw axis is n, the rank of the axis, with m as a subscript. So the pattern that we just drew here would be designated 3 subscript 1. And that's says automatically that the value of tau was equal to 1/3 of the translation that's parallel to the axis. So the only other ones we have to consider besides 3 and 3 sub 1 is 3 sub 2, where tau would be equal to 2/3 of T. Let's see what that looks like by quickly drawing it out again using this dandy little unrolled cylinder device. Let's let this be my first motif. And I'll take this as the value of tau, two boxes up. So I rotate, slide up by tau, rotate, slide up by tau, rotate, slide up by tau, and I'm back up here. And here is my translation. tau is equal to 2/3 of a translation. And my translation, then, should be equal to 3/2 times tau. But this is supposed to be the translation. And there's nothing in this box. OK, this introduces another very important aspect of the patterns produced by screw axes. And it's sufficiently important we must use first the basic screw operation, A alpha tau and the translation in terms of which we have defined tau. So use the spiral that you have defined by stating A alpha tau is the basic operation. But then, don't quit. You're saying another thing, that tau is a certain fraction of a translation. And you have to use that translation to generate additional objects in the pattern. So in this pattern that we've drawn here, this is supposed to be a translation. So I have to take this one and slide it up by T. I have to take this one and slide it up by T. I have to take this one and slide it down by T. So I'm using a different kind of shading to indicate the ones I produced by A alpha tau and the ones that I produced from that helix by using the translation, T. So this is the final pattern. It has a basic screw operation that's equal to 2/3 of the translation. And it has a translation that's 3/2 of tau. So everything's fine. Yeah? AUDIENCE: For the very top one where you have it colored in, [? should that stay ?] outside the box? PROFESSOR: Yeah, well, it's supposed to be another row. AUDIENCE: Shouldn't that be a box lower [INAUDIBLE]? PROFESSOR: Yes, it should indeed. Thank you. When your nose is right in the middle of the thing, sometimes you don't notice that as readily as somebody in the audience. OK, so that's the pattern of 3 sub 2. And now, if we realize that we must use both the basic operation, A alpha tau, and translation to fill things in, it's clear that if we try to have tau greater than the translation, the translations, when we apply them, would give us a screw that had the possibility of being redefined in terms of a shorter tau. OK, one thing that is apparent if you look at this pattern is that 3 sub 2 produces the same sort of helix but in an enantiomorphic sense. This one is a spiral that goes this way. And this one increases when we rotate in a clockwise fashion. So are they distinct? Yeah, they really are because you can have both the left handed spiral and the right handed spiral together in the same space group. I don't remember whether that's true for the threefold screw axes or not. I'm not sure. I have to check that. Yes? AUDIENCE: So really [INAUDIBLE] enantiomorphic because they're all left handed. So it's just minus 1, isn't it? PROFESSOR: Good for you. I was about to make that point. The sense of the spiral is enantiomorphic in the sense that this one is a right handed spiral. This one is a left handed spiral. But they're not truly enantiomorphic patterns because if we start with a right handed motif, this one also has a right handed motif. So I should qualify that statement, which I was about to do. It's the sense of the spiral which is enantiomorphic. This is not to say that one has motifs of one chirality and the other one has to have the opposite chirality. AUDIENCE: Couldn't you just [INAUDIBLE] being minus 1? Just [INAUDIBLE] minus 1? PROFESSOR: Yeah, I could do that, I could do that. And remember that we can add or subtract a translation at will from tau. And if I took a full translation and subtracted it from 3 sub 1, I would get-- I get tau equal to minus 2. Nope, doesn't work that way. AUDIENCE: [INAUDIBLE]? PROFESSOR: Yeah, it's the same tau but in the opposite sense, yeah. OK, so what comes out of this is that there are three kinds of axes that involve a 120 degree rotation. Three, which we can view as 3 sub 0, 3 sub 1, 3 sub 2. The symbols that are used for them now, we know and love the triangle that represents the locus of a threefold axis. Four 3 sub 1 and 3 sub 2, what one does is to extend the edges of the triangle. There's the streamers on the bike handle again. And if you look down on the pattern and use a right handed spiral, then you extend the streamers, the edges of the triangle, that goes down into the board this way. And 3 sub 2 would-- if it was in the same pattern-- you would indicate by extending the opposite pair of the opposite set of edges. AUDIENCE: So would they be different space groups? [INAUDIBLE]? PROFESSOR: Yes, there is a P3 sub 1. And there is a P3 sub 2. And they are distinct space groups. AUDIENCE: So how can you distinguish [INAUDIBLE]? PROFESSOR: You would find out where the atoms sit. And in one case, the spiral would occur in a right handed fashion. And in the other one, from a left handed fashion. You can determine the position of the atoms unambiguously. OK, let's do one that's more interesting where the rank of the rotation part of the operation is higher. And let's look at fourfold screw axes. So here, we would consider tau equal 0, tau equals 1/4 of a translation, tau equal to 2/4 of a translation, and tau equal to 3/4 of a translation. So let's take our cylinder that was formerly wrapped around the rotation access, straighten it out. And I won't bother to draw the pattern for a fourfold access. But for a 4 sub1 1 screw axis, this would be tau. And that would be one, two, three, one quarter of a translation that's parallel to tau. Or conversely, so T is equal to 4 tau. Or conversely, how is equal to one quarter of the translation. So start with a first motif, rotate 90 degrees, slide up by one quarter of the translation. Rotate and slide, rotate and slide. Now I've come full circle. Rotate and slide. So this, then, is the pattern for the screw axis that I'll label as 4 sub 1 by analogy to what I've done with threefold screw axes. Over here, let me immediately jump to 4 sub 3. I'm going to have to be a little more stingy with the spacing of my boxes or I'm going to run out of room. So if this is 4 sub 3, turns out that tau should be equal to 3/4 of a translation. And conversely, the translation should equal to 4/3 of tau. So that's the situation that I would obtain if one, two, three boxes gives me the length of tau and four boxes give me the length of the translation. OK, so again I'll use different sorts of shading to indicate which way I have gotten this. I would rotate 90 degrees. I would jump up one, two, three boxes. So here's the next one. Rotate, one, two, three boxes up brings me to here. Rotate, one, two, three boxes brings me up to here. Rotating again, sliding up three boxes brings me to here. OK, so there's the basic helix. And how is indeed equal to 4/3 of the translation. But I can't quit yet. I've used the basic operation, A alpha tau. Now, I've got to fill in with translation. And according to my rule, the translation should be equal to 4/3 of-- I'm sorry, this should be the translation is 4/3 of tau. So this should be the translation running up one, two, three, four boxes. So that's-- I'll use a different shading here. That would fill in one here. One, two, three, four boxes up brings me to here. One, two, three, four boxes up would bring me to here. Go down by four-- one, two, three, four-- brings me to here. And let me use still a different shading here. One, two, three, four brings me down to here. Filling in quickly, one, two, three, four. Another one would sit here. One, two, three, four, another one would sit here. One, two, three, four, another one would sit here. And you can see what is happening here, that 4 sub 3 looks exactly like 4 sub 1 but in the opposite sense. So it's the same relation as between 3 sub 1 and 3 sub 2. What do we use as symbols? This is a 4 sub 1 screw. And again, if we look down on the pattern from the top, use a right handed rule and extend every edge of the square, that would be the symbol for 4 sub 1. 4 sub 3 is the same sort of spiral, but in the opposite direction. So we will, with a right handed rule and going between neighboring closest motifs, we would extend this, extend this, and extend this. So that's the symbol for 4 sub 3. With that, we come back over to here. And that is everything except 4 sub 2. And again, I'll mark up a cylinder that's been split into quadrants, draw reference horizontal lines. This is 4 sub 2. And we define this as tau. Then, the translation, this should be equal to 1/4 of the translation parallel to the axis. And conversely, the translation should be four halves. So let's draw the pattern. This was the first one. Rotate, slide up by tau. This is the next one. Rotate and slide, rotate and slide. Rotate and slide begin, rotate and slide. So this turns out to be the nature of the helix that is produced by A pi over 2 tau. But this is not yet a pattern that has the translational periodicity that I have claimed. That's supposed to be the translation, twice tau, up to here. So I would have two slide this one up to here. I'd have to slide this one up to here. I'd have to slide this one up to here. And something's wrong here-- one, two, three, four. I think I went too far up for one of them. This one was one, two, three. This one, I went up too far. I skipped one. What did I do wrong? AUDIENCE: You did the rotation wrong. PROFESSOR: Hm? AUDIENCE: The two [INAUDIBLE]. PROFESSOR: OK, let me do this all over again. Again, I can't see standing right on top of it. It's not a very good attempt to make it look easy. Let's split this up into four quadrants. If this one comes out quite differently, I'll make sure I have it right. So this is my translation, one, two, three. This is the translation. And that should be equal to four halves of tau. So this, then, is my value of tau. It should go up two boxes to here. And that should be equal to 1/2 of T. So I want to rotate, slide up two boxes, rotate, slide up two boxes, rotate, slide up two boxes, rotate, slide up two boxes, rotate, and slide up two boxes. And now, I have to fill in with translation. So I will take this one and go up one, two, three, four translations. That's what I did wrong. I'll take this one and go down one, two, three, four translations. Take this one and go up one, two, three, four translations. Take this one and go one, two, three, four translations. Take this one and go one, two, three, four translations. And now, lo and behold, what I have is one, two, three, four. Curious sort of thing-- I have two motifs at every level. And they are 180 degrees apart. So a 4 sub 2 screw axis produces a pattern that looks like this. Pair of motifs like this 180 degrees apart, pair of motifs skewed by 90 degrees to that pair, another pair translationally equivalent to the first. But what is the chirality of this spiral? 3 sub 1 was right handed. 4 sub 1 was right handed. 4 sub 3 was left handed. 4 sub 2 was right in the middle. Is that right handed or left handed? The answer is it's both. It's both. There's one spiral that goes up this way. And there's another spiral that goes up in the opposite sense. So it's both. It's a left handed spiral and a right handed spiral together in the same pattern. And the pattern for this screw axis, the symbol for this screw axis if it occurs in a pattern, is every other edge of the square extended. Which do you extend? Well, it really doesn't matter because it's both left handed and right handed simultaneously. So you can use either one. Yes, a couple questions? AUDIENCE: Can you explain why you explain you did that shading? PROFESSOR: I just wanted to differentiate the ones that I used. They're all the same. And where they come from doesn't affect what the basic pattern looks like. But I tried to make clear which ones I got by using the operation A alpha tau. And for 4 sub 2, which I just did twice, once correctly and once incorrectly, these with shading were produced by the operation of rotating and sliding up by 1/2 of the translation. The other ones, the ones for which I used this shading, I filled in by using what the translation was, namely twice tau. So it's just a way of keeping track of what operations I used to produce everything that's in the pattern. And for 4 sub 3, I had to use three kinds of shading. Yes, sir? AUDIENCE: Should I [INAUDIBLE] the fact that this was actually two superimposed twofold axis separated by 1/2 T? PROFESSOR: Yes, you could read that into it. And you would be very observant because a 4 sub 2 screw axis contains a twofold rotation as a subgroup. AUDIENCE: [INAUDIBLE]. PROFESSOR: And since you were shrewd enough to deduce this-- we ask the question with glide planes when we're dealing with two dimensional plane groups. Is a glide plane a candidate location for a special position? In other words, if I move the atom directly onto a glide plane, is there coalescence? The answer was no. Suppose I asked the same question now of a screw axis. Is a screw axis a candidate location for a special position? The answer is sometimes yes, sometimes no. AUDIENCE: Well, only if you have [INAUDIBLE]. PROFESSOR: Yeah, so actually if I move the atom, the first atom, onto the locus of the rotation part of the operation, I get pairwise coalescence. And instead of getting all of these, I'll have just a string of single atoms separated by half a translation, half as many. So that's a special position. I don't get the full number out when I throw one into the space group. Now, there are an infinite number of screw axes. My old friend, the saguaro cactus, which has a 19 to maybe 23-fold rotational component to its symmetry, does not, if I look at it carefully, have the same thing on every one of these ribs because the tufts of spines spiral up like this. So a saguaro cactus, if I take the spines into account, doesn't have 22-fold rotational symmetry. It has some sort of 22-fold screw axis. And I've never been determined enough to risk puncturing my finger by going down around a saguaro and seeing if there's some other tuft at the same level and whether it's a 22 subscript 3 or 22 subscript 2 screw axis. But a saguaro cactus does have a screw axis as its symmetry element. OK, let us generalize on the basis of what we've done so far. For any screw axis whatsoever, crystallographic or not, we will have an n sub 1 screw, an n sub 2 screw, where tau is equal to 1 nth of a translation, tau equal to 2 nths of a translation. And we will go all the way up n sub n minus 1, n sub n minus 2. And always, regardless of n, the ones at either end tend to be spirals in an enantiomorphic sense. And n sub 2 and an n sub n minus 2 are spirals. If you end up with an odd one in the middle, this has no chirality to the spiral. There's a left handed one and a right handed one. On the other hand, for something like two and four and six, the ones at opposite ends of the series are enantiomorphs in the sense of the rotation. For any six-fold screw axis-- and I will pass out the nature of the pattern without taking time to draw it-- 6 sub 1 and 6 sub 5 are enantiomorphous. 6 sub 2 and 6 sub 4 are enantiomorphous. And 6 sub 3 stands alone. 6 sub 3 consists of a pattern of three objects on a triangle pointing in one sense, three on a triangle pointing in the opposite sense. This is the value of tau. And that is equal to 3/6 of the translation. So no left handed or right handed sense. 6 sub 2 consist of pairs of objects separated by 60 degrees. 6 sub 4, exactly the same pairs, but they go in the opposite sense. 6 sub 1 is just a single spiral going up with atoms at intervals of 1/6 of the translation. 6 sub 5, a spiral in the opposite sense. OK, with that, let me pass out a sheet that has patterns done in a decent fashion for all of the crystallographic screw axes. Bear in mind that there are fascinating, interesting, and intriguing patterns for non-crystallographic screw axes. And let me say a few words about the symbol used in the space group for glide planes. Any questions on this, by the way? I don't want to have beat it into the ground. Yes? AUDIENCE: I have a question on symbols. How do you know which way the arrows point on the triangle or the squares? PROFESSOR: Oh, here? AUDIENCE: Yeah. PROFESSOR: OK. Probably the best thing to do is to look at the patterns that I just passed around if you got one yet. If you look at 3 sub 1 and look down on it from the top and indicate the sense of rotation that gets you from the one above to the one that's directly below, your hand will curl around in a clockwise fashion. And the little tails on the symbol for the threefold axis trail out behind the way in which you're rotating. If you look at 3 sub 2 and look down on it, you would have to, using your right hand, rotate in a counterclockwise sense. And therefore, the streamers go out again in the opposite direction. Same for 4 sub 1. There, every edge of the square is enclosed. If you look at how you have to rotate to get from the one on top to the one immediately below using your right hand, again, you have to rotate in a clockwise sense. And therefore, the edges that are elongated trail out in the opposite direction. And finally, drawn in a decent fashion are 6 sub 1, 6 sub 5, which are spirals in the opposite sense. 6 sub 2, 6 sub 4, pairs of things at each level. And the level is 1/3 of the translation. 6 sub 3, triangles of objects at intervals that are equal to 1/2 of T, 3/6 of T. OK, does that answer? That is really a convention of some higher order. But nevertheless, it is a convention and it is consistent. Other questions about screw axes? Yeah? AUDIENCE: So in nature, is there a difference between the right handed screw [INAUDIBLE]? PROFESSOR: That's like asking is there something like a Coriolis effect in crystals. You know the Coriolis effect that if you're in the southern hemisphere, the water gurgles down the drain one direction. If you're in the northern hemisphere, the opposite direction? There was actually a very famous scientist, what was his name? I think it was Serret. Serret, so famous I can't think of his name. I remembers his name, Serret He developed a very famous theorem in kinetics, and that was thermal migration, migration driven not by a chemical potential gradient but by a temperature gradient. And it's very difficult to measure. And only recently have people been testing that theory and making measurements. He asked this very same question. And he said, I think it was potassium chlorate that he looked at which has no mirror plane or inversion in it. So it occurs in left handed and right handed forms. He asked, is there something like a Coriolis effect that would give you more right handed crystals in the northern hemisphere and more left handed crystals in the southern hemisphere? So what he'd do, he drew a lot of crystals. And he started counting. And he counted, and he counted, and counted. And he couldn't tell if there was a difference. And counted, counted, counted. And still, after lots and lots of counting, still could not get a result that was statistically significant. And he went nuts. Sorry to tell you this. So I think it remains to be seen whether there is indeed a Coriolis effect in crystal shapes. And that's true. It's tragic. We're all laughing at the poor guy. But he actually just couldn't stand it. He counted so many crystals, poof! That's a true, unfortunate story. No, I don't think there is any difference between left handed crystals or right handed crystals. Remember my story about sugar and its chirality and the little bugs that can only eat one chirality. Other questions of the less frivolous nature? That was not frivolous. That was a good question. OK, well let's say a little bit about glide planes and how they appear in space groups. Things were very easy in two dimensions because the glide plane was a glide line. And everything had to be confined to a two dimensional space. So we used a solid line to indicate a mirror plane, the locus of the operation, sigma. And we use a dashed line to indicate the locus of an operation, sigma tau. And tau was in the direction of the dashes. Well, I've given it all away with the little diagram that I passed around. Things are much more complicated in three dimensions. If this is the locus of a glide plane and this is the direction of tau in a space group, we could be looking at that glide plane in a total of four different ways. We could be looking at the glide plane from the side, edge on, and perpendicular to tau. We could be looking at the glide plane edge on and along the direction of tau. Or we could be between these two orientations and looking at the glide plane edge on but in between normal to tau and parallel to tau. And finally, we could look at the glide plane down from the top. If we view the glide plane from above, for example suppose we have a monoclinic crystal and we put a glide plane in the base of the cell which has tau in this direction, the way you indicate that is with a little chevron off to the side. And if this is the direction of tau, you put a little barb on the chevron that indicates the direction of tau. So the three possibilities for a monoclinic crystal is that how could be this way, in which case you use a chevron like this. Or how could be in this direction. And then, you use a chevron with a barb attached like this. Or how could be a diagonal glide, in which case the atoms would be up and opposite chirality and down and back up again. In that case, you add a little barb in between the two directions to indicate the direction of tau. We put some labels on this, this a and this b. This situation would be a tau that is equal to 1/2 of b. In this case, how would be equal to 1/2 of a. In this case, how would be equal to 1/2 of a plus b. This is designated as a b glide. And rather than having the symbol n appear in the symbol for the space group, a b would appear. This is an a glide. And the a would appear in the symbol. And this is a diagonal glide. And that, for reasons that I have never found anyone able to explain to me, is called an n glide. I know of no language in which the word for diagonal begins with an n. But that's what it's called. Now, those same glides can be viewed from directions that are parallel to the plane of the glide plane. Looking normal to tau is exactly the same situation we had in plane groups. So one uses the dash. If you're looking along the glide plane end on but in the direction of tau, you use a dotted line. It's easy to keep straight which is which. If there's a little arrow in there and you look at it from the side, you see a line segment. If you look at it from the end, you see a dot. So it's easy to remember this convention. If you're neither perpendicular to nor parallel to the glide plane, you just mix the two symbols. So you use a dashed dot line. So those are the symbols for glide. Glide plane is something like a piece of wood. It's got grain to it. The direction of the grain is the direction of tau, if you'd like. All right, in the hand out, I give you some examples of the way in which a pattern of objects that has glide in it would be interpreted in terms of a geometric symbol. At the bottom right of the page, the left hand diagram has atoms related by a glide plane in the base of the cell where tau is running left to right. And so you have a chevron with an arrow on the right hand branch. If you have atoms alternately left handed, right handed at a level and 1/2 plus that elevation, then you're looking at a glide plane edge on and down along the direction of tau. So you would indicate the locus of that glide plane by a series of dots. What other type of glide plane is possible, not for monoclinic, but for lattices that have a centered lattice point in the middle of the cell? So this takes a special type of Bravais lattice. And this would be a lattice, for example, that had lattice points at the corners of the cell and another lattice point in the middle of the cell. One face of the cell is centered. Then, you have not only translations a, b, and a plus b, which can serve as directions for glide. But you have another translation that is not present in the primitive lattice. And this is 1/2 of a plus 1/2 of b. And if that's a translation, you could have a glide plane parallel to the base of the cell and have tau equal to 1/2 of 1/2 of a plus b, namely have this little vector in here be tau. And the symbol for that kind of glide is d. And I do know what that stands for. That stands for a diamond glide because diamond, one of the very simple structures for an element or any compound and one of the early structures to be determined experimentally, does have a d glide in it. So the name of the compound, diamond, gave its name to the type of glide plane. OK, I've got still 10 more minutes. What I would like to do next is to look at some of the other monoclinic space groups and also examine the way in which this information is presented for you in the international tables. So I've got a big, fat pack of material that shows you all of the monoclinic space group, that is the space groups derived from point group two, the space groups with point group m, and the space groups with point group 2/m. And we've done a few of these. But what we haven't looked at is the Shcoenflies notation for the result or the way in which this information is presented in the international tables. To introduce you to these gradually, this is the monoclinic space group-- these are the monoclinic space groups as presented in the old international tables for x-ray crystallography. A little bit later, I will pass out to you a few higher symmetry space groups that are represented in both the old international tables and the new international tables. And as I indicated somewhat disparagingly earlier on, there is so much information in the new version that they're very, very hard to use. But OK, if you open the hand out, you find space group P2. So that's a twofold axis added to a primitive monoclinic axis. You see on the left hand page this situation, this ridiculous situation, that exists only for monoclinic crystals the first setting, where the axes are a on the left hand side running down, and b on the top running from left to right, and c coming straight up out of the page. And the second setting, where the unique axis is b, and that means if you're going to draw the space group with c coming up, a down, and b to the right, b now is the direction of the twofold axis. So the diagram looks completely different. But it's the same space group except that the first setting is tilted on its side so that a comes down, b is from left to right, and c, which is now not a direction of the twofold axis, comes out. So there's some jargon here, that geometric jargon. The atoms occur at different elevations. So in these two cases, the atoms occur only at elevations plus z or minus z. And you see next to the little circles that represent the symmetry equivalent positions for the general position a plus. And that means elevation plus z. And for the first setting, all the atoms occur at an elevation plus z for the general position. For the second setting in the right hand diagram, you're looking at the twofold axis from the side. And therefore, one atom is at plus z. The one that's related by symmetry gets rotated down to minus z. So you see a little plus next to one atom, a little minus next to the next to it. We need a symbol for a view down along the locus of a twofold axis. And that's the little pointed oval that we became familiar with with the plane groups. When you're looking at the twofold axis from the side, how are you going to indicate that? Well, the convention is to use an arrow. And that's what you see on the right hand side for the second setting of P2. If this were not a twofold axis but a 2 sub 1 screw axis, it would be a one-sided barb. So looking down at these two axes with rotations of 180 degrees, this is the symbol for two viewed from the side. And a one-sided barb would be the symbol for a 2 sub 1 screw axis. Mercifully, these are the only screw axes that need to be depicted in a view from the side. So there's no standard symbol for a sixfold access or a fourfold axis looked at from the side. In boldface, in the outer corner of these pages, you see the international symbol, which is what we used for the plane groups, the symbol for the lattice type, primitive, except now we use uppercase symbols for the lattice type to distinguish the space groups from the plane groups. Underneath it, you see a C2 superscript 1. This is the Schoenflies symbol. You'll recognize C2 as the Schoenflies symbol for a twofold axis. Schoenflies' symbol for the space group was to add a superscript, namely the first one that old [? Artur ?] got, the second one that [? Artur ?] got from this point group, the third one he got from the point group, and so on. So it's pretty much an arbitrary order, except he starts with, as you'll see, a twofold access, then replaces the twofold axis with a screw access. So if you turn to the next page, you see symbols for not twofold axes, but 2 sub 1 screw axes along the edges of the cell. You see now a different sort of symbol alongside the atoms. There's one at plus z. That's the representative one at x, y, z. If you repeat it by a screw rotation, you rotate it 180 degrees and slide it up by 1/2 of [? c. ?] So you see the symbol 1/2 plus. So plus is plus z. 1/2 plus is 1/2 plus z. Over to the second settings, trying not to sneer too vigorously, you'll see pointed barbs. That's the symbol for a twofold axis viewed from the side. And then, one of them is plus. The one that's a little long parallel to the 2 sub 1 screw axis goes to the other side and goes from plus z down to the minus z. Let me ask you a question which people usually don't think about. If you look at the first setting and ask, what are the lengths of the translations? That's a. And that's so many angstroms. And that's b. That's so many angstroms. If I ask you that for the second setting, the answer is a little trickier. What is the length of this translation? That's b. What is the length of this translation? I'm sorry, that's c. And this translation is a, right? Wrong. That is no longer a. And the reason for that is that when you have a cell with non-orthogonal angles in it, let me indicate for the most general case for a triclinic crystal. So this is a. This is b. And then the third translation normal to that plane is c. And if we have a structure with atoms in these locations and we want to project the structure along c, you do exactly that. You don't plop the atoms down onto the base of the cell when you project it because if you did so, the atom that is up in the neighboring unit cell would come down to a different location. And you would not have a pattern that was periodic based on a lattice. You get that only if, when you're projecting in an oblique lattice, you project along the translation. And then, and only then, can you end up with a pattern that looks like it has translational periodicity and really does. So if a cell is oblique, you do not project just by plopping the atoms down on the base of the cell. You project them parallel to the translation that extends up above the direction onto which you're projecting. So if our cell in the second setting is monoclinic and this is a and this b and c comes up, if we look at that cell from the side as is done in the depiction on the right hand, then we let this be the direction of b. So that is b. But the thing that is at right angles to b is going to be a times the cosine of beta. And it's not a itself. So this is a right angle. But this is not the lattice translation. It is the lattice translation times a trigonometric function. The triclinic case is a hairy beast. To calculate the length of these axes and projection, it's clear what they are. But to get these values involves alpha and beta and gamma. And it's a very, very complicated function. AUDIENCE: So the angle between a and c in general is not 90 degrees in monoclinics? PROFESSOR: The angle between depends on how you label your axes. For monoclinic, if you do the first setting, then by definition this is a, and this is b, and this is c. So these are always by definition 90 degrees. If you use the second setting, then this is the direction of b. And this is the direction of a. And this is the direction of c. That keeps the system right handed. Then, this is a right angle and this is a general angle. AUDIENCE: But all you're doing is taking that top one and just [? tilting it over ?]. PROFESSOR: No, I'm doing more than that because if there were a twofold axis, that would be the direction of the twofold axis. Now, this is the direction of the twofold axis. The other thing that happens is the symbols for the glides change. So if you look at space group number seven, which is a case where the mirror plane has been replaced by a b glide in the first setting, and that means the glide is perpendicular to the twofold axis. If the twofold axis is now the direction of b, the glide plane turns into a b glide. But let's just thumb through them quickly. And then, we're actually two minutes past my promised quitting hour. First one, P2. Then, replace the twofold access by a 2 sub 1 screw. So that's Schoenflies symbol C2 superscript 2. Then, do this with centered lattices. The international table chooses to use a side centered cell. So that's the third one you can get from a twofold axis. So Schoenflies calls it C subscript 2 superscript 3. And if you add that to a side centered lattice, you get screw axes interleaved between the twofold axes. Put it on its side, it stays C2 superscript 3 in the Schoenflies notation. But the side centered d become side centered c because it's b that comes out of the oblique net. So the international symbol tells you exactly what you have. The Schoenflies symbol is arbitrary. It tells you the point group and it tells you the order in which Schoenflies derived them. But that non-informative nature to the symbol has a blessing. And that is it doesn't change with different labellings of a, b, and c. It doesn't change with a being shortest, b being shortest. It just sits there and its dumb, uninformative fashion. And that is nice. So both of them have survived. OK, so could we get another space group out of this by replacing the twofold axis by a 2 sub 1? No, because we've already got 2 sub 1 screw axes in C2 superscript 3. So then, Schoenflies moves on and begins to work with symmetry planes. So number six puts a mirror plane in the primitive monoclinic lattice. That becomes CS. That's the Schoenflies symbol for point group m. And it's the first thing you can get. Another piece of convention for displaying the representative pattern of atoms, in the first setting, the two atoms sit directly above one another. So there are two new things here. A vertical line through the atom indicates that there are two of them that are superimposed in projection. And moreover, the little tadpole has appeared inside the frog eggs in the left hand part. So this says on the right hand side of that split circle, the atom is at plus z. On the left hand side with the comment to indicate an enantiomorph is 1 at minus z that sits directly below the first one. And you'll notice in the diagram of the symmetry elements is a chevron working off the lower right hand corner of the cell. And that's the symbol for a mirror plane that's being viewed directly from above. And then I'll just go a couple more. You could replace the mirror plane with a glide plane. And that's a b glide. Now, you can see the atoms going up at plus z, slide along by 1/2 z, and popped down to minus z, going down to an enantiomorph, so there's a comma inside the circle. The chevron in the lower right hand corner of the diagram of symmetry elements now has acquired an arrow, a barb, that indicates the direction of tau. The second setting, again, flops the thing on its side so that you're looking now down along the b glide. And so you see a dotted line that indicates a glide plane viewed edge on in the direction of tau. Then, you can put a mirror plane for number eight into a side centered b lattice. And that gives you another space group. And then, you can put a glide plane into the b lattice. And that, interestingly, gives you a different space group. Number eight has an alternating mirror plane with an axial glide. Number nine has an alternating axial glide alternating with a diagonal glide-- two glide planes. Not a mirror plane and a glide, but two different kinds of glide planes. And then, you've used up all the tricks you can pull there. So now, we take the last monoclinic point group, 2/m, drop it into a primitive lattice. Replace the twofold axis with a 2 sub 1 screw axis. That gives you P2 sub 1 over m. Then, leave the twofold axis alone and leave the plane alone and put 2/m in a side centered cell. And then, replace the mirror plane by a glide. That's number 13. And then finally, the summa cum ultra, replace the twofold axis y a 2 sub 1 screw and replace the mirror plane with a glide, number 14. And then, do the same thing in a centered lattice. So we get a total of 13 different space groups from two lousy kinds of lattices and three possible point groups. You get 13 different distinct space groups. OK, that's a quick overview. Notice that just as was the case for the plane groups, the tables go on to list the way in which atoms are related by symmetry and the number per cell and the site symmetry just as in the plane groups. And the way a three dimensional structure is going to be described to you is exactly analogous to what we did with two dimensions. Magnesium in position 3b, symmetry 6/m, oxygen in position 12d, symmetry 1. And then, the tables give you the coordinates of all the symmetry S atoms. OK, thank you for your patience. That was a long stretch in one shot. I'll say a little bit, very little, next Tuesday about the convention for orthorhombic space groups because there, no one direction is any more or less special than any other. So the symbol for the space group changes all over the place when the axes take on different lengths, relative lengths. But the symmetry is exactly the same. So we'll see you--
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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So I would like to now consider the wheel that is rolling without slipping. And what I'd like to do is consider-- let's draw the wheel rolling without slipping. And I'd like to consider the contact point between the wheel and the ground. And I'd like to understand what the velocity of that contact point is. And the result is surprising. Now, we know by our law of the addition of velocities-- let's call that the point P-- that the velocity of P is the velocity of the center of mass, and that pointed this way, plus the velocity of the point P as seen in the center of mass frame. And in that frame, every single point on the rim was just doing circular motion. So that velocity is in the opposite direction. And the vector sum is the velocity of this point P as seen in the ground frame. So we want to consider the magnitudes of these two terms. The velocity in the center of mass frame of a point on the rim has a magnitude equal to r omega. And we said that the velocity of the center of mass, we call that V cm. Now, the rolling without slipping condition was that these two magnitudes were equal. V cm equals r omega, or the velocity of the point on the rim in the center of mass frame is equal to the center of mass velocity of the wheel. And therefore, the velocity P, which is the sum of these two vectors for the contact point-- so that's when P is in contact with the ground-- is equal to 0. So what we say is that the contact point is instantaneously at rest with respect to the ground. And that's what really is the mystery of a wheel that's rolling without slipping-- that this contact point is the sum of these two vectors, and it's instantaneously at rest with the ground when the wheel is rolling without slipping. The other points are certainly not. Remember, up here, these two vectors pointed in the same direction, so that vector has twice the magnitude of either one.
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https://ocw.mit.edu/courses/7-016-introductory-biology-fall-2018/7.016-fall-2018.zip
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ADAM MARTIN: So, last semester, my grandfather passed away, and I was responsible for explaining to my two sons how a funeral works. So I'm a professor, right? I pride myself on being able to explain things clearly. So I went to tell my five-year-old son sort of what's going on during the funeral. I told him, your papa's body is up in that wooden box up there. We're going to celebrate him right now. We're going to go to the cemetery, and then we're going to bury him. And you know, I'm a professor. I thought I nailed it, OK? Except my son was showing this looking look of concern in his eyes. And he looked at me. He was like, what about his head? So my plea for this semester is please let me know when I've forgotten the part about the head, OK? You know, you guys are listening to me. I might have forgotten something that, to me, seems kind of obvious, but that, for you, it would help to know to understand the material. So today, we're going to start with cells. And so here, you can see this is one of my favorite movies. This is a neutrophil cell here that's migrating. You can see it's chasing after this smaller cell which is a bacteria cell. And around this neutrophil, what you're seeing here are these red blood cells, which aren't doing very much of anything. So the point I'm making with this video is that cells have a huge amount of diversity. So cells have diversity. As you can see from the video, there's a diversity in size. There's diversity in shape and also diversity in behavior. OK, so we're going to unpack this a little bit. And I just want to start by just pointing out that until now in the semester, we've dealt mainly with very small things, such as atoms, small molecules, lipids, and proteins. And the size of these structures, they're on the nanometer scale. Now, cells are a unit up in size. Cells span several orders of magnitude in size. So bacterial cells are on the small side. They're on the 1 micron to 10 micron size. So you can see that here. Most bacteria is about 1 micron to 10 micron. But our cells span from tens of microns to hundreds of microns, and even larger than that, because the human egg is on the order of a millimeter in size. This is not a human egg, but this is a frog egg. You can see often the egg cells are the biggest of all cells. The biggest cell is an ostrich egg. It's about 15 centimeters, I believe. So cells span a huge, wide range of sizes. I'm going to start with the simplest, which is a bacteria cell. And what I want to point out about this cell is its simplicity. So here, you can see this is an electron micrograph of bacterial cell here. And this is a cartoon just illustrating some of the key features. You have a plasma membrane and a cell wall. The cell wall here is in sort of-- the periplasmic space is in green. And this encapsulates the cytoplasm. And the only other real structure you can see, in this case, is this nucleoid structure in the middle. And what the nucleoid is, is it's just the chromosome of the bacterium. And I want to point out that later on in the course, Professor Imperiali is going to come back and tell you about antibiotics and how bacteria can develop antibiotic resistance, which is of critical importance for biology and medicine. So our cells are more complicated than this. And that's because if you look at this EM here, you can see that eukaryotic cells-- and we are eukaryotes-- have membrane-bound compartments. There's a nucleus here that houses our nuclear DNA. And also, there's a series of membrane compartments that span the cytoplasm. Now, our cells, even in a single organism, such as us, our cells have incredible diversity and specialization. So there's diversity within a single organism. There is specialization. So as we develop from a single cell, our cells acquire properties that allow them to carry out specific functions in our bodies. And an extreme example of this is shown up here. These are pictures or drawings of neurons from Ramon y Cajal, and you can see how this looks nothing like the cartoon picture of a cell I just showed you. These cells have highly dendritic sort of arrays of protrusions. And these cells have evolved such that they are very good at sort of transmitting information in the body, right? An extreme example of a nerve is a sciatic nerve, which extends from the base of your spinal cord all the way down into your foot. So it's about a meter long. That's an extreme specialization for a cell. So these cells are specialized, but what's important to note is that within a single organism, the genomic DNA of a cell is more or less the same. So genomic DNA is the same, with some exceptions, some of which we'll get to in the course. The genomic DNA is the same. What's different is the genes that are being expressed in these cells and the proteins that they encode that give these cells different functions. So what's different and what allows cells to acquire these different functionalities is this different gene expression. OK, so that's the overview. Now I want to talk about compartments. And if we go back to this cartoon that I showed you from your-- oh, I just wanted to point out that right now, at MIT and the Broad, there is a project that's ongoing to really define all the cell types that are present in humans. And this is known as the Human Cell Atlas. And I just want you to take a minute to think about if you wanted to define different cell types, what would you look at to classify these cell types? Anyone have an idea? Yes? What's your name? AUDIENCE: Rachel. ADAM MARTIN: Rachel. AUDIENCE: Cell function. ADAM MARTIN: What's that? AUDIENCE: Function. ADAM MARTIN: You could look at function. That might be a little subjective in how to interpret. What defines the function of the cell? AUDIENCE: I'm thinking like what it does and how it works with other cells. ADAM MARTIN: But is there something, I guess, within that cell that would define what its function is? Yeah? What's your name? AUDIENCE: Samantar ADAM MARTIN: Samantar? AUDIENCE: Samantar ADAM MARTIN: Samantac All right, yeah? AUDIENCE: Gene expression data. ADAM MARTIN: Gene expression, right? What genes are these cells expressing? So what they're doing is they're isolating single cells from tissues. And then they're looking at gene expression. So which type of molecule here would be the molecule you'd want to look at if you want to look at gene expression? Not Miles. Malik. AUDIENCE: You look at mRNA. ADAM MARTIN: mRNA, exactly. Malik is exactly right. So you look at the mRNA, right? Because if there's mRNA, that means that gene was expressed, and it's possibly encoding the translation of a protein. So that's what they're doing. They're isolating single cells and doing a massive single cell RNA seek project. And that's helping to identify new cell types in the human body. OK, coming back to compartments. So I showed you this picture. You can see there are lots of membrane-bound compartments in the cell. You can see the lines in this cartoon represent lipid bilayers. And so these are compartments that are completely encapsulated by a lipid bilayer. And I'm going to remind you of something that Professor Imperiali told you about already. And the reason I'm saying it again is because it is so freaking important, OK? So if we consider a lipid bilayer where the circles are the polar head groups of the lipids and the squiggly lines are the fatty acid chains-- I'm not going to draw all the fatty acid chains. So this is a lipid bilayer. And these head groups, are they hydrophilic or hydrophobic? Yes? What's your name? AUDIENCE: Stephen. ADAM MARTIN: Stephen. AUDIENCE: They're hydrophilic. ADAM MARTIN: They're hydrophilic. They like water, right? The water is on the different faces here. These are hydrophilic. How about this central region here? Yes? Name? AUDIENCE: Ory. ADAM MARTIN: Ory. AUDIENCE: Hydrophobic. ADAM MARTIN: Hydrophobic, good, right? That's excellent. So these are hydrophobic. Hydrophobic. So what that means is you have this hydrophobic layer that is surrounding each of these compartments, and that's going to serve as a barrier such that water and things dissolved in water cannot pass through this barrier unless something allows it to. So each of these compartments are membrane bound, have properties in the lumen that's in the interior of the compartment that's going to be different from that of the cytoplasm. And the cell, the interior of the cytoplasm, is going to be different from that of the extracellular space. So let me just give you some examples of how things are different inside and outside the cell and also inside and outside these compartments. So if we consider the plasma versus the cytoplasm, we can consider the concentration of various ions. And I want to get the concentrations right. Let's consider sodium, which is a monovalent cation, potassium, and calcium. I'm just going to use these as illustrations. So sodium is concentrated outside the cell, about 150 millimolar, and is less concentrated in the cytoplasm. So there's a huge difference in sodium between the outside and the inside of the cell. Potassium-- now, you might think potassium would be a lot like sodium. It's a monovalent cation. It's a similar size. But it actually shows the flip distribution. So it's four millimolar in the exoplasm and 140 millimolar in the cytoplasm, OK? So there appears to be some selectivity here, right? The cell is concentrating certain things inside the cytoplasm, and it's excluding others. So there's selectivity, even between closely related atoms here. OK, calcium is two millimolar in the exoplasm and around 10 to the negative fifth millimolar in the cytoplasm. So because there are these huge gradients in the concentration of these ions leads you to expect that this is a non-equilibrium state. Because if it were equilibrium, these ions would go in and out, and they'd equilibrate such that the same concentration would be on the inside as the outside. And so there's selectivity. And also, there's non-equilibrium, which suggests that energy is required to maintain this asymmetry. I want to point out it's not just the concentrations of various molecules that are different between the inside and the outside, but the plasma membrane can also have a voltage across it. So this is exoplasm. This is cytoplasm. And this membrane can hold a voltage. And this is going to become incredibly important when we talk about neurons, because neurons use changes in this voltage to transmit signals across their length and also to transmit signals at the synapse. And we'll talk about that later in the course. So this is known as membrane potential, this voltage difference. I also, again, want to point out this endomembrane system in here, where the gray regions here are sort of compartments that will communicate with each other. And so there's this whole internal structure of endomembrane system which is compartmentalized from the cytoplasm, OK? And so now I want to talk about how do things get in and out of this structure. So I think we're up to three, getting in and out. So cells-- and this is very important in cell communication, right? For cells to communicate, cells have to send things, like signals, to other cells. And also, cells can take in stuff and sort of receive things from other cells. So how is it that this happens? Well, if we consider the plasma membrane of a cell-- this is a lipid bilayer, PM. I abbreviate Plasma Membrane, PM. So that's a lipid bilayer. And let's say that there's some type of sort of molecule that's on the outside of the cell. So, in this case, this would be exoplasm out here. This would be inside the cytoplasm here. Let's say this cell wanted to sort of take up this pink structure into the cell. How would it do that? Well, this pink structure, because it's hydrophilic, cannot pass through the lipid bilayer. So the cell has to use another strategy to take this up into the cell. Let's see. I'll use a little blue here. So let's say the region of the membrane right here is in blue. Then, what can happen is this blue structure can invaginate, and it can take up this pink molecule. So the plasma membrane can invaginate. The plasma membrane can invaginate, and if it has this pink molecule, then if there is a cision event here, you've now gone from having your pink molecule on the outside of the cell to having the pink molecule in this vesicle or circular structure on the inside of the cell. So here now we have this. We have our vesicle in blue. And now the cell has picked up this pink molecule, such like this. So this process of sort of taking material from the outside and sort of bringing it into the inside of the cell is known as endocytosis, which is the process of taking something from the outside and bringing it in the cell, this is also a way that viruses can get in your cells, which is a more nefarious way that this system is used. All right, I need a volunteer. Yes. Ory, come on down. Don't worry. This'll be simple. I just need you to put pressure on my head right here. So you see the tassel. Make sure they can see the tassel. You see this tassel? This tassel is that pink molecule, is right now on the outside, right? But I'm going to endocytose it by going like that, right? And I also got a hand. That's great. So you see, that's basically endocytosis. I just endocytosed my tassel. All right, you can go up. We're all set, yeah. I can do the next one. OK, so the opposite of this process-- let me get another color here. You can also have vesicles that are on the inside that then fuse with the plasma membrane and release their contents to the exterior. And this is called exocytosis. So exocytosis is something that's starting out inside one of these vesicles and then goes out, OK? So this is not exactly reversible, because there are different protein machineries that mediate either endocytosis or exocytosis. So now I'm going to exocytose my tassel. There we go. So I just exocytosed it, and now my molecules is again facing the outside world. So this is a way for cells to sort of taken things and also to secrete molecules, like signaling molecules, into the extracellular space. So now, we're moving on, and now we're going to talk about compartments within compartments. So we're on four-- compartments within compartments. And you're going to see how this relates to endocytosis in just a minute. So compartments within compartments. So the example I'm going to use here is an organelle that is present in us, in animal cells, which is the mitochondria. And you all know that the mitochondria is the powerhouse of the cell. Whenever I say mitochondria is the powerhouse of the cells, I have to do 10 push-ups, because it's so cliche. I mean, my five-year-old knows that mitochondria is the powerhouse of the cell. It's a gross oversimplification, OK? Mitochondria are way more interesting than that, and I'll show you in just a minute. All right, I'll draw mitochondria, the same mitochondria that my kids draw me. So here's a mitochondria. I'm drawing first the outer membrane. So there's an outer membrane to the mitochondria. And this organelle also has an inner membrane. So this is the inner membrane. The inside here is called the matrix. I'll draw some DNA molecules there. So this is known as the mitochondrial DNA, which I'll mention in just a minute. So this is the mitochondria. Where could such an organelle come from, evolutionarily speaking? You know, one problem that eukaryotic cells have is they cannot make mitochondria de novo. The way that mitochondria is passed on is it's replicated during cell division, and it's passed on from one cell to the next. And you guys all got your mitochondrial DNA and your mitochondria, at least initially, from your mother. So this is not an organelle that can be synthesized de novo. And the fact that this is the case has led to a theory known as the endosymbiont theory, or hypothesis, which basically states that there was an ancestral eukaryotic cell. And the way that organelles such as mitochondria and plastids were derived is from engulfing bacterial cells that were either capable of oxidative phosphorylation, in the case of mitochondria, or were capable of photosynthesis, in the case of chloroplasts. And so this engulfment is much like this endocytic process that I just talked to you. It's not really endocytosis, because these bacteria are much better than an endocytic vesicle. So it's more of like a macro sort of pinocytosis or something like that, OK? So something you may have seen in the news lately-- so some of the evidence for this endosymbiont theory is that mitochondria and plastids have their own DNA, OK? So these organelles have retained DNA. The genes in the DNA encode for proteins, that function in the mitochondria. It also includes ribosomal RNAs and transfer RNAs that are required for protein synthesis within the mitochondria. But a lot of the genes that are required for a functioning organism have now been exported to our nuclear DNA, and those genes are made and proteins are produced, and then they're imported into the mitochondria. But the mitochondria has retained a number of genes, and it's encoded in the mitochondrial DNA. Another reason to think that this could be from a symbiotic relationship, this ancient, between eukaryotic and prokaryotic cells is that these organelles divide by fission similar to how bacteria divide. So they divide by fission. So I'll show you a real mitochondria and a mitochondria undergoing fission in just a minute. I just want to point out that in the news recently, there's been talk of three-parent babies. And I just want to explain to you what that is. So eggs that are-- I shouldn't say embryos, but eggs that have the nuclear DNA from two parents can then be given mitochondrial DNA from another parent. So these three-parent babies are essentially babies that have nuclear DNA from two parents but mitochondrial DNA from a third parent. So here's just an article in the NewScientist reporting that this is imminent, and now it's been done. Now, why would you might want to do that? Anyone know why someone would want to do this? Yes? AUDIENCE: There might be a defect in the mitochondrial DNA of both parents. ADAM MARTIN: Yeah. So Stephen, right? AUDIENCE: Yes. ADAM MARTIN: Stephen's exactly right, right? There are diseases that are so associated with faulty mitochondrial DNA. And if you're a mother and you have the mutations that cause this disease, this would be a way for you to have a child without passing on the disease to your child. And so this is something that is still controversial, but that's why people are exploring the opportunity for making these so-called three-parent babies. Again, I've a pet peeve with textbook pictures of mitochondria. Here's your textbook picture. Everything's labeled nicely. This is what mitochondria look like in real life. Mitochondria, like the endoplasmic reticulum, actually, form these tubular networks that essentially span the entire cell. So it's convenient for us to depict mitochondria like this in textbook, but mitochondria are way more interesting than that. They're dynamic organelles. And I also want to make the point-- we kind of talk about these organelles like they behave as these separate entities, but in fact, they interact with each other. And there's lots of interesting biology behind that. So I'm going to show you one movie that's from work done by Gia Voeltz, and this Friedman person is the first author. She just gave a talk here at MIT and showed some beautiful movies. All right, focus on this movie here. The ER is labeled in green, and the mitochondria is in red. And what you see is there is this mitochondria tubule, and it's crossed by the ER right here. Now, focus on this when the movie plays. You're going to see that the mitochondria undergoes fission right at these points, where the ER and the mitochondria intersect. So that illustrates-- and now they've mechanistically dissected what sort of makes the mitochondria undergo fission at these crossing sites. But it really illustrates the real sort of complexity and dynamics that are present in a cell, which you might not be getting from your book. Oh, I did want to mention that-- so there's a chapter in your book-- I think it's Key Concept 5.3-- where you can read about all the organelles. You should read that and know roughly what the organelles do. I mean, I could lecture about that, but it would just be so boring that I can't do it. I'd have to do a ton of push-ups. So we have to sort of choose what we lecture about. So I would just suggest you read that part in the textbook. If you have questions, come talk to me. And just be familiar with what your organelles are doing. Yes, Ory? AUDIENCE: What was the chapter? ADAM MARTIN: It's 5.3. AUDIENCE: Is it like a chapter? ADAM MARTIN: It's Key Concept 5.3. It's probably listed in the assignments, right? AUDIENCE: It's a section in the chapter. ADAM MARTIN: Yeah. AUDIENCE: It's only a little bit. It's not going to be long. ADAM MARTIN: All right. I have one last part, which is to prepare you for Friday's lecture. In Friday's lecture, we're going to start talking about genetics. And before we talk about genetics, we're going to talk about something that is-- we're going to lay the groundwork, essentially, for genetics by talking about how cells divide. So I think one of the most miraculous things that cells do is that they can undergo this trick where they replicate themselves and split it into two daughter cells. And you're seeing here, there are chromosomes in the middle here. They're going to line up along the metaphase plate. And now they're going to get pulled to separate sides. They're going to wiggle back and forth first. Then, in just a minute, they're going to get segregated. They'll go eventually. There they go. So you see-- and then the cell is going to pinch at the equator and divide in two, OK? So now I'm at this last part, the cytoskeleton, because the cytoskeleton is the answer to part of how these cells divide. And before I present the cytoskeleton, I just want to briefly mention chromosomes and what they look like. So the chromosomes are your nuclear DNA. They're linear pieces of DNA, as opposed to the mitochondria, which has circular DNA. And again, the fact that mitochondria have circular DNA is analogous with sort of bacterial chromosomes, where bacteria have circular DNA. But our chromosomes are linear pieces of DNA. And you guys have probably all seen chromosome spreads that look like this. So this would be a chromosome that's replicated. There are two copies, one and two. So this is a replicated and condensed chromosome, right? Initially, chromosomes are just like bowls of spaghetti. Everything's mixed together. But during mitosis, the chromosomes condense, and then they sort of resolve from each other, such that you can see the arms of the chromosome. And you can see where the chromosomes are coupled to each other at this structure here, which is known as the centromere. And at this centromere, a protein complex assembles. I'll just draw it like this. This protein complex is called the kinetochore. The kinetochore is a complex of hundreds of proteins that assemble into this large platform that sits on the centromere. And the kinetochore is able to attach to the cell's cytoskeleton, specifically microtubules. So this attaches to microtubules. And I will tell you what microtubules are right now. So microtubules are a component of the cell's cytoskeleton. So the cell's cytoskeleton is a network of filamentous rods that are present in the cell. And the term cytoskeleton kind of makes it sound boring, I think, because it makes it sound static. But the cytoskeleton is anything but static. It's actually a dynamic machine in the cell. And these machines that are assembled from these fibers are able to generate force. So these microtubules and the cytoskeleton, they generate force. You can think of them as motors or machines. So there are machines that generate force. And I'll just illustrate this with a couple of videos and slides. So, again, this is not a stable structure, but very dynamic. So this is going to be a movie where, in green here, microtubules are labeled, and the red label's the nucleus. And these cells in this fly embryo are going to undergo cycles of nuclear division. And so you're going to see the microtubules assemble, disassemble, assemble, disassemble. So you see how dynamic this process is. The cell is able to assemble this force-generating apparatus, which is known as the mitotic spindle, each and every cell division. So this structure, or machine, is critical to physically segregate the chromosomes to opposite poles of a cell. So now I'm going to tell you about the microtubules themselves. So microtubules, as the name implies, are sort of tube-like polymers. So these microtubules are biopolymers, which means that cells are expressing sort of genes that encode for proteins that form a subunit that can then self-assemble into a larger rod-like structure. And you can see one of these rod-like structures here. They essentially look like straws. They're about 25 nanometers in diameter. And I'm going to show you this video showing you microtubules both disassembling at first and then assembling. So that's a dissembling microtubule. But they also assemble to form these longer rod-like structures. So these biopolymers are dynamic, and they both assemble but also disassemble. And both the assembly and the disassembly can generate force. Microtubules can push if they polymerize into something. They push it, right? Just like you're kind of poking someone with your finger. Now, they also disassemble, and when they shrink and disassemble, they can actually pull. And I'll show you an example of that right here. So this is an example that's reconstituted, meaning they're just purified proteins here. There's no cell. And what this is, is a bead. And you see the dark stripe here is a microtubule. You can see it growing. There's the end right there. That microtubule's going to grow. It's going to grow out to about here. And then at the end of the movie, it's going to stop growing, and the microtubule's going to shrink back. And you're going to see that when the microtubule shrinks back, it's going to actually pull this large bead and pull it towards the left side of the slide. Here it goes. It depolymerizes and pulls. You see it? So these microtubules can generate a pulling force that's strong enough to pull this glass bead. And also, it's strong enough to pull an entire chromosome. It's actually much stronger-- it generates much more force than it requires to pull a chromosome. So it seems like there's some robustness in the system so that it's generating a pull that's much stronger than needed to actually drag a chromosome through viscus media. So during mitosis, the way this system is set up is there's what's known as a bipolar spindle. I'll just write down that. We'll come back to it in Friday's lecture. There's a bipolar spindle. And the bipolar spindle is made up of a number of microtubules. So here, you can see there are two poles, here and here. And in the middle, along the sort of equator of the cell, the chromosomes will line up. There are just two chromosomes here. And you can see how microtubules reach out from the pole and attach to both sides of that chromosome. And you can imagine that when these microtubules are told to shrink and disassemble, if they're able to remain attached to that kinetochore, which they are, when they shrink, they're going to pull the two copies of the chromosome away from each other, OK? And so this is the basic machinery that allows for chromosomes to be segregated in cells. OK, that's it for today. Any questions? All right, terrific. Good luck on your exam on Wednesday. It's in here.
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https://ocw.mit.edu/courses/8-701-introduction-to-nuclear-and-particle-physics-fall-2020/8.701-fall-2020.zip
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MARKUS KLUTE: All right. So welcome back to 8.701. So we have all ingredients now to prepare Feynman rules for QED. So that's the toolkit we need in order to make calculations to calculate scattering processes and decays. And we've already seen Feynman rules for our toy theory. Again, now the situation is a little bit more complicated, because we can consider the spin of particles in addition to their energy and momentum. The rules' sequence of things are very much the same. There is, however, a few caveats to keep in mind, and I'll point those out. OK. So the very first thing is to be very clear in our notation. So this is an arbitrary or generic QED Feynman diagram. We have only pointed out the incoming and the outgoing lines. There is internal lines which I didn't mention here. Important to note the momentum and the directions. The directions are arbitrary. We just have to be clear on them and then treat them consistently. All right? So this is not different in our previous discussion. Then, here comes the difference. Our external lines either electron, positrons, or photons. All right? You can-- fermions and photons, charged fermions and photons. So we discussed how the solutions look like, our spinors u and v. And for outgoing electrons, for outgoing particles, we have this adjunct vector here, which is given by u dagger gamma 0. And similarly for the incoming antiparticle-- v dagger gamma 0. For the photon, we have the polarization vectors for incoming and outgoing photons. All right. Then we have a vertex factor. Here, now, g e is a constant and a dimensionless property. But we do have to have a gamma mu here as part of our vertex factor. For the propagator, our internal lines, we have a difference between electrons, positrons, and photons. And that comes from the fact that electrons and positrons are massive particles. So we have vertex vectors which now have this 1 over q square behavior, or 1 over q square minus m square behavior. So here, you can already see that there's going to be a complication later when we evaluate or integrate over momentum-- simply the same discussion I had before. And we already know how to solve this problem of infinities by renormalizing-- by having a cut-off and renormalizing it. Excellent. So the next step, then, is very much the same. There's no change. We have to make sure that there's energy and momentum conservation, and we enforce this by introducing delta functions. We have to integrate over each and every internal momenta, and each internal line gets one of those integration factors. And then after we integrate, we are left with a delta function, and we have to cancel that delta function. All right. In our toy experiment, the order of things didn't matter. Everything we had in there was scalar numbers, right? Here we do have a little bit more complicated problem. So there's an importance in the order of which we execute things. So what we want to do is form fermion lines. We just follow a fermion as we go from the left to the right. And then we find things which are always of the form an adjoint spinor, a 4-times-4 matrix, and a spinor. And the result of that is going to be a number. All right? Great. There is one additional complication, is accounting for duplications and making sure that the sign is [INAUDIBLE]. I'm just mentioning this here. This will become more clear as you work through examples. So there is an antisymmetrization going on, where we have to introduce a minus sign between different diagrams that differ only by the interchange or the exchange of two incoming or two outgoing electrons or positrons and/or the incoming electron with an outgoing positron. So if you have a diagram which is exactly the same, but the two incoming electrons are interchanged, you have to add those two diagrams. You have to add all matrix elements together for recalculating amplitude. But you have to introduce a minus sign when you change those two particles. So with that, we can now just basically calculate whatever QED process we want. All the tools are already here. And what we want to do now next, in the next video, and also in the recitation and homework, is to go through a few examples to get a little practice with this. There's a number of tricks which will come in handy, and I'll explain those in a separate video. They are just mathematical tricks which allow us to quickly evaluate the multiplication of spinors and matrix elements and so on. All right. That's it for this video. Again, there is going to be another two or three videos which deal with actually evaluating or calculating matrix elements.
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https://ocw.mit.edu/courses/7-91j-foundations-of-computational-and-systems-biology-spring-2014/7.91j-spring-2014.zip
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NARRATOR: The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation, or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: All right, we should get started. So it's good to be back. We'll be discussing DNA sequence motifs. Oh yeah, we were, if you're wondering, yes, the instructors were at the awards on Sunday. It was great. The pizza was delicious. So today, we're going to be talking about DNA and protein sequence motifs, which are essentially the building blocks of regulatory information, in a sense. Before we get started, I wanted to just see if there are any questions about material that Professor Gifford covered from the past couple days? No guarantees I'll be able to answer them, but just general things related to transcriptome analysis, or PCA? Anything? Hopefully, you all got the email that he sent out about, basically, what you're expected to get. So at the level of the document that's posted, that's sort of what we're expecting. So if you haven't had linear algebra, that should still be accessible-- not necessarily all the derivations. Any questions about that? OK, so as a reminder, team projects, your aims are due soon. We'll post a slightly-- there's been a request for more detailed information on what we'd like in the aims, so we'll post something more detailed on the website this evening, and probably extend the deadline a day or two, just to give you a little bit more time on the aims. So after you submit your aims-- this is students who are taking the project component of the course-- then your team will be assigned to one of the three instructors as a mentor/advisor, and we will schedule a time to meet with you in the next week or two to discuss your aims, just to assess the feasibility of the project and so forth, before you launch into it. All right-- any questions from past lectures? All right, today we're going to talk about modeling and discovery of sequence motifs. We'll give an example of a particular algorithm that's used in motif finding called the Gibbs Sampling Algorithm. It's not the only algorithm, it's not even necessarily the best algorithm. It's pretty good. It works in many cases. It's an early algorithm. But it's interesting to talk about because it illustrates the problem in general, and also it's an example of a stochastic algorithm-- an algorithm where what it does is determined at random, to some extent. And yet still often converges to a particular answer. So it's interesting from that point of view. And we'll talk about a few other types of motif finding algorithms. And we'll do a little bit on statistical entropy and information content, which is a handy way of describing motifs. And talk a little bit about parameter estimation, as well, which is critical when you have a motif and you want to build a model of it to then discover additional instances of that motif. So some reading for today-- I posted some nature biotechnology primers on motifs and motif discovery, which are pretty easy reading. The textbook, chapter 6, also has some good information on motifs, I encourage you to look at that. And I've also posted the original paper by Bailey and Elkin on the MEME algorithm, which is kind of related to the Gibbs Sampling Algorithm, but is used as expectation maximization. And so it's a really nice paper-- take a look at that. And I'll also post the original Gibbs Sampler paper later today. And then on Tuesday, we're going to be talking about Markov and hidden Markov models. And so take a look at the primer on HMMs, as well as there is some information on HMMs in the text. It's not really a distinct section, it's kind of scattered throughout the text. So the best approach is to look in the index for HMMs, and read the relevant parts that you're interested in. And if you really want to understand the mechanics of HMMs, and how to actually implement one in depth, then I strongly recommend this Rabiner tutorial on HMMs, which is posted. So everyone please, please read that. I will use the same notation, to the extent possible, as the Rabiner paper when talking about some of the algorithms used in HMMs in lecture. So it should synergize well. So what is a sequence motifs? In general, it's a pattern that's common to a set of DNA, RNA, or protein sequences, that share a biological property. So for example, all of the binding sites of the Myc transcription factor-- there's probably a pattern that they share, and you call that the motif for Myc. Can you give some examples of where you might get DNA motifs? Or protein motifs? Anyone have another example of a type of motif that would be interesting? What about one that's defined on function? Yeah, go ahead. What's your name? AUDIENCE: Dan. [INAUDIBLE] PROFESSOR: Yeah. So each kinase typically has a certain sequence motif that determines which proteins it phosphorylate. Right. Other examples? Yeah, so in that case, you might determine it functionally. You might purify that protein, incubate it with a pool of peptides, and see what gets phosphorylated, for example. Yeah, in the back? AUDIENCE: I'm [INAUDIBLE], and promonocytes. PROFESSOR: What was the first one? AUDIENCE: Promonocytes? Oh, that one? Oh, that was my name. PROFESSOR: Yeah, OK. And as to promoter motifs, sir? Some examples? AUDIENCE: Like, [INAUDIBLE] in transcription mining site. PROFESSOR: Ah. Yeah. And so you would identify those how? AUDIENCE: By looking at sequences upstream of [INAUDIBLE], and seeing what different sequences have in common? PROFESSOR: Right. So I think there's at least three ways-- OK, four ways I can think of identifying those types of motifs. That's probably one of the most common types of motifs encountered in molecular biology. So one way, you take a bunch of genes, where you've identified the transcription start site. You just look for patterns-- short sub-sequences that they have in common. That might give you the TATA box, for example. Another way would be, what about comparative genomics? You take each individual one, look to see which parts of that promoter are conserved. That can also help you refine your motifs. Protein binding, you could do ChIP-Seq, that could give you motifs. And what about a functional readout? You clone a bunch of random sequences upstream of a luciferase reporter, see which ones actually drive expression, for example. So, that would be another. Yeah, absolutely, so there's a bunch of different ways to define them. In terms of when we talk about motifs, there are several different models of increasing resolution that people use. So people often talk about talk about the consensus sequence so you say the TATA box, which, of course, describes the actual motif-- T-A-T-A-A-A, something like that. But that's really just the consensus of a bunch of TATA box motifs. You rarely find the perfect consensus in real promoters-- the real, naturally occurring ones are usually one or two mismatches away. So that doesn't fully captured it. So sometimes you'll have a regular expression. So an example would be if you were describing mammalian 5 prime splice sites, you might describe the motif as GT, A or G, AGT, or sometimes abbreviated as GTR AGT, where R is shorthand for either appearing nucleotide-- either A or G. In some motifs you could have GT, NN, GT, or something like that. Those can be captured, often, by regular expressions in a scripting language like Python or Perl. Another very common description in motifs, there would be a weight matrix. So you'll see a matrix where the width of the matrix is the number of bases in the motif. And then there are four rows, which are the four bases-- we'll see that in a moment. Sometimes these are described as position-specific probability matrices, or position-specific score matrices. We'll come to that in a moment. And then there are more complicated models. So it's increasingly becoming clear that the simple weight matrix is too limited-- it doesn't capture all the information that's present in motifs. So we talked about where do motifs come from. These are just some examples. I think I talked about all of these, except for in vitro binding. So in addition to doing a CLIP-seq, where you're looking at the binding of the endogenous protein, you could also make recombinant protein-- incubate that with a random pool of DNA molecules, pull down, and see what binds to it, for example. So why are they important? They're important for obvious reasons-- that they can identify proteins that have a specific biological property of interest. For example, being phosphorylated by a particular kinase. Or promoters that have a particular property. That is, that they're likely to be regulated by a particular transcription factor, et cetera. And ultimately, if you're very interested in the regulation of a particular gene, knowing what motifs are upstream and how strong the evidence is for each particular transcription factor that might or might not bind there, can be very useful in understanding the regulation of that gene. And they're also going to be important for efforts to model gene expression. So, a goal of systems biology would be to predict, from a given starting point, if we introduce some perturbation-- for example, if we knock out or knock down a particular transcription factor, or over-express it, how will the system behave? So you'd really want to be able to predict how the occupancy of that transcription factor would change. You'd want to know, first, where it is at an endogenous levels, and then how its occupancy at every promoter will change when you perturb its levels. And then, what effects that will have on expression of downstream genes. So these sorts of models all require really accurate descriptions of motifs. OK, so these are some examples of protein motifs. Anyone recognize this one? What motif is that? So it says X's. X's would be degenerate oppositions, and C's would cysteines. And H's [INAUDIBLE]. What is this? What does this define? What protein has this? What can you predict about its function? AUDIENCE: Zinc finger. PROFESSOR: Zinc finger, right. So it's a motif commonly seen in genome binding transcription factors, and it coordinates to zinc. What about this one? Any guesses on what this motif is? This quite a short motif. Yeah? AUDIENCE: That's a phosphorylation. PROFESSOR: Phosphorylation site. Yeah. And how do you know that? AUDIENCE: The [INAUDIBLE] and the [INAUDIBLE] next to it means it's [INAUDIBLE]. PROFESSOR: OK, so you even know what kinase it is, yeah. Exactly. So that's sort of the view. So, serine, threonine, and tirocene are the residues that get phosphorylated. And so if you see a motif with a serine in the middle, it's a good chance it's a phosphorylation site. Here are some-- you can think of them as DNA sequence motifs, because they occur in genes, but they, of course, function at that RNA level. These are the motifs that occur at the boundaries of mammalian introns. So this first one is the prime splicing motif. So these would be the bases that occur at the last three bases of the exon. The first two of the intron here, are almost always GT. And then you have this position that I mentioned here-- it's almost always A or G position. And then some positions that are bias for A, bias for G, and then slightly biased for T. And that is what you see when you look at a whole bunch of five prime ends of mammalian introns-- they have this motif. So some will have better matches, or worse, to this particular pattern. And that's the average pattern that you see. And it turns out that in this case, the recognition of that site is not by a protein, per se, but it's by a ribonucleoprotein complex. So there's actually an RNA called U1 snRNA that base pairs with the five prime splice site. And its sequence, or part of its sequence, is perfectly complimentary to the consensus five prime splice site. So we can understand why five prime splice sites have this motif-- they're evolving to have a certain degree of complementarity to U1, and in order to get officially recognized by the splicing machinery. Then at the three prime end of introns, you see this motif here. So here's the last base of the intron, a G, and then an A before it. Almost all introns end with AG. Then you have a pyrimidine ahead of it. Then you have basically an irrelevant position here at minus four, which is not strongly conserved. And then a stretch of residues that are usually, but not always, pyrimidines-- called the pyrimidine track. And in this case, the recognition is actually by proteins rather than RNA. And there are two proteins. One called U2AF65 that binds the pyrimidine track, and one, U2AF35 that binds that last YAG motif. And then there's an upstream motif here, that's just upstream of the 3 prime splice site that is quite degenerate and hard to find, called the branch point motif. OK, so, let's take an example. So the five prime splice site is a nice example of a motif, because you can uniquely align them, right? You can sequence DNA, sequence genomes, align the CDNA to the genome, that tells you exactly where the splice junctions are. And you can take the exons that have a 5 prime splice site, and align the sequences aligned to the exon/intron boundary and get a precise motif. And then you can tally up the frequencies of the bases, and make a table like this, which we would call a position-specific probability matrix. And what you could then do to predict additional, say, five prime splice-site motifs in other genes-- for example, genes where you didn't get a good CDNA coverage, because let's say they're not expressed in the cells that you analyzed-- you could then make this odds ratio here. So here we have a candidate sequence. So the motif is nine positions, often numbered minus 3 to minus 1, would be the exonic parts of this. And then plus 1 to plus 6 would be the first six bases of the intron. That's just the convention that's used. I'm sure it's going to drive the computer scientists crazy because we're not starting at 0, but that's usually what's used in the literature. And so we have a nine-based motif. And then we're going to calculate the probability of generating that particular sequence as given plus-- meaning given our foreground, or motif model-- as the product of the probability of generating the first base in sequence, S1, using the column probability in the minus 3 position. So if the first base is AC, for example, that would be 0.4. And then the probability of generating the second base in the sequence using the next column, and so forth. If you made a vector for each position that had a 1 for the base that occurred at that position, and a 0 for the other bases, and then you just did the dot product of that with the matrix, you get this. So, we multiply probabilities. So that is assuming independence between positions. And so that's a key assumption-- weight matrices assume that each position in the motif contributes independently to the overall strength of that motif. And that may or may not be true-- they don't assume that it's homogeneous, that is you have usually in a typical case, different probabilities in different columns, so it's inhomogeneous, but assumes independence. And then you often want to use a background model. For example, if your genome composition is 25% of each of the nucleotides, you could just have a background probability that was equally likely for each of the four, and then calculate the probability, S given minus, of generating that particular [INAUDIBLE] under the background model, and take the ratio of those two. And the advantage of that is that then you can find sequences that are-- that ratio, it could be 100 times more like a 5 prime splice site than like background-- or 1,000 times. Or you have some sort of scaling on it. Whereas, if you just take the raw probability, it's going to be something that's on the order of 1/4 to a 1/9. So some very, very small number that's a little hard to work with. So when people talk about motifs, they often use language like exact, or precise, versus degenerate, strong versus weak, good versus lousy, depending on the context, who's listening. So an example of these would be a restriction enzyme. You often say restriction enzymes have very precise sequence specificity, they only cut-- echo R1 only cuts a GAA TTC. Whereas, a TATA binding protein is somewhat more degenerate. It'll bind to a range of things. So I use degenerate there, you could say it's a weaker motif. You'll often-- if you want to try to make this precise, then the language of entropy information offers additional terminology, like high information content, low entropy, et cetera. So let's take a look at this as perhaps a more natural, or more precise way of describing what we mean, here. So imagine you have a motif. We're going to do a motif of length one-- just keep the math super simple, but you'll see it easily generalizes. So you have probabilities of the four nucleotides that are Pk. And you have background probabilities, qk. And we're going to assume those are all uniform, they're all a quarter. So then the statistical, or Shannon entropy of a probability distribution-- or vector of probabilities, if you will-- is defined here. So H of q, where q is a distribution or, in this case, vector, is defined as minus the summation of qk log qk, in general. And then if you wanted to be in units of bits, you'd use log base 2. So how many people have seen this equation before? Like half, I'm going to go with. OK, good. So who can tell me why, first of all-- is this a positive quantity, negative quantity, non-negative, or what? Yeah, go ahead. AUDIENCE: Log qk is always going to be negative. And so therefore you have to take the negative of the sum of all the negatives to get a positive number. PROFESSOR: Right, so this, in general, is a non-negative quantity, because we have this minus sign here. We're taking logs of things that are between 0 and 1. So the logs are negative, right? OK. And then what would be the entropy if I say that the distribution q is this-- 0100, meaning, it's a motif that's 100% C? What is the entropy of that? What was your name? AUDIENCE: William. PROFESSOR: William. AUDIENCE: So the entropy would be 0, because the vector is determined in respect of the known certainty. PROFESSOR: Right. And we do the math-- you'll get, for the C term, you'll have a sum. You'll have three terms that are 0 log 0-- it might crash your calculator, I guess. And then you'll have one term that is 1 log 1. And so 1 log 1, that's easy. That's 0, right? This, you could say, is undefined. But using L'Hospital's rule-- by continuity, x log x, you take the limit, as x gets small, is 0. So this is defined to be 0 in information theory. And this is always, always 0. So that comes out to be 0. So it's deterministic. So entropy is a measure of uncertainty, and so that makes sense-- if you know what the base is, there's no uncertainty, entropy is 0. So what about this vector-- 1/4, 1/4, 25% of each of the bases, what is H of q? Anyone? I'm going to make you show me why, so-- Anyone want to attempt this? Levi? AUDIENCE: I think it's 2. PROFESSOR: 2, OK. Can you explain? AUDIENCE: Because the log of the 1/4's is going to be negative 2. And then you're multiplying that by 1/4, so you're getting 1/2 for each and adding it up equals 2. PROFESSOR: Right, in sum, there are going to be four terms that are 1/4 times log of a 1/4. This is minus 2, 1/4 times minus 2 is minus 1/2, 4 times minus 1/2 is minus 2, and then you change the sign, because there's this minus in front. So that equals 2. And what about this one? Anyone see that one? This is a coin flip, basically. All right? It's either A or G. [INAUDIBLE]. Anyone? Levi, want to do this one again? AUDIENCE: It's 1. PROFESSOR: OK, and why? AUDIENCE: Because you have two terms of 0 log 0, which is 0. And two terms of 1/2 times the log of 1/2, which is just negative 1. So you have 2 halves. PROFESSOR: Yeah. So two terms like that. And then there's going to be two terms that are something that turns out to be 0-- 0 log 0. And then there's a minus in front. So that will be 1. So a coin flip has one bit of information. So that's basically what we mean. If you have a fair coin and you don't know the outcome, we're going to call that one bit. And so a base that could be any of the four equally likely has twice as much uncertainty. All right, and this is related to the Boltzmann entropy that you may be familiar with from statistical mechanics, which is the log of the number of states, in that if you have N states, and they're all equally likely, then it turns out that the Shannon entropy turns out to be log of the number states. We saw that here-- four states, equally likely, comes out to be log of 4 or 2. And that's true in general. All right, so you can think of this as a generalization of Boltzmann entropy, if you want to. OK, so why did he call it entropy? So it turns out that Shannon, who was developing this in the late '40s, as developing a theory of communication, scratched his head a little bit. And he talked to his friend, John von Neumann-- none other than him, involved in inventing computers-- and he says, "My concern was what to call it. I thought of calling it information. But the word was overly used." OK, so back in 1949, information was already overused. "And and so I decided to call it uncertainty." And then he discussed it with John von Neumann, and he had a better idea. He said, "You should call it entropy. In the first place, your certainly function has already been used in statistical mechanics under that name," so it already has a name. "And the second place, and more important, nobody knows what entropy really is, so in a debate, you always have the advantage." So keep that in mind. After you've taken this class, just start throwing it around and you will win a lot of debates. So how is information related to entropy? So the way we're going to define it here, which is how it's often defined, is information is reduction in uncertainty. So, if I'm dealing with an unknown DNA sequence, the lambda phage genome, and it has 25% of each base, if you tell me, I'm going to send you two bases, I have no idea. They could be any pair of bases. My uncertainty is 2 bits per base, or 4 bits before you tell me anything. If you then tell me, it's the TA motif, which is always T followed by A, then now my uncertainty is 0, so the amount of information that you just gave me is 4 bits. You reduced my uncertainty from 4 bits to 0. So we define the information at a particular position as the entropy before-- before meaning the background, the background a sort of your null hypothesis-- minus the entropy after-- so after you've told me that this is an instance of that motif, and it has a particular model. So, in this case, you can see the entropy is going to be entropy before. This is just H of q right here, this term. And then minus this term, which is H of p. So, if it's uniform, we said H of q is 2 bits per position. And so, so the information content of the motif is just 2 minus the entropy of that motif model. In general, it turns out if the positions in the motif are independent, then the information content of the motif is 2w minus H of motif, where w is it width of the motif. So for example, the entropy of the motif of-- we said the entropy of this is 2 bits, right? Therefore, the information content is what? If this is our-- let's say this is a P. This is our routine. Are you starting to generate? What is its information content? AUDIENCE: 0? PROFESSOR: 0. Why is it 0? Yeah, back row. AUDIENCE: Because the information content of that is 0, and then the information content of the known hypothesis, so to say, is 0. Sorry, both of them are 2. So 2 minus 2 is 0. PROFESSOR: The entropy of the background is 2, and the entropy if this is also 2. So 2 minus 2 is 0. And what about this? Let's say this was our motif, it's a motif that's either A or G. We said the entropy of this is 1 bit, so what is the information content of this motif? AUDIENCE: 1. PROFESSOR: 1, and why is it 1? AUDIENCE: Background is 2, and entropy here is 1. PROFESSOR: Background is 2, entropy is 1. OK? And what about if I tell you it's the echo R1 restriction enzyme? So it's GAA TTC, a six-base motif precise-- it has to be those bases? What is the information content of that motif? In the back? AUDIENCE: It's 12. PROFESSOR: 12-- 12 what? AUDIENCE: 12 bits. PROFESSOR: 12 bits, and why is that? AUDIENCE: Because the background is 2 times 6. So 6 bases, and 2 bits for each. And you have all the bases are determined at the specific [INAUDIBLE] enzyme site. So the entropy of that is 0, since 12 minus 0 is 12. PROFESSOR: Right, the entropy of that motif is 0. You imagine 4,096 possible six-mers. One of them has probably 1. All the others have 0. You're going to have that big sum. It's going to come out to be 0, OK? Why is this useful at all, or is it? One of the reasons why it's useful-- sorry, that's on a later slide. Well, just hang with me, and it will be clear why it's useful in a few slides. But for now, we have a description of information content. So the echo R1 site has 12 bits of information, a completely random position has 0, and a short four-cutter restriction enzyme would have 2 times 4, 8 bits of information, right, and an eight-cutter. So you can see as the restriction enzyme gets longer, more information content. So let's talk about the motif finding problem, and then we'll return to the usefulness of information content. So can everyone see the motif that's present in all these sequences? If anyone can't, please let me know. You probably can't. Now, what now? These are the same sequences, but I've aligned them. Can anyone see a motif? PROFESSOR: GGG GGG. PROFESSOR: Yeah, I heard some G's. Right. so there's this motif that's over here. It's pretty weak, and pretty degenerate. There's definitely some exceptions, but you can definitely see that a lot of the sequences have at least GGC, possibly an A after that. Right, so this is the problem we're dealing with. You have a bunch of promoters, and the transcription factor that binds may be fairly degenerate, maybe because it likes to bind cooperatively with several of its buddies, and so it doesn't have to have a very strong instance of the motif present. And so, it can be quite difficult to find. So that's why there's a real bio-informatics challenge. Motif finding is not done by lining up sequences by hand, and drawing boxes-- although that's how the first motif was found, the TATA box. That's why it's called the TATA box, because someone just drew a box in a sequence alignment. But these days, you need a computer to find-- most motifs require some sort of algorithm to find. Like I said, it's essentially a local multiple alignment problem. You want multiple alignment, but it doesn't have to be global. It just can be local, it can be just over a sub-region. There are basically at least three different sort of general approaches to the problem of motif finding. One approach is the so-called enumerative, or dictionary, approach. And so in this approach, you say, well, we're looking for a motif of length 6 because this is a leucine zipper transcription factor that we're modeling, and they usually have binding sites around 6, so we're going to guess 6. And we're going to enumerate all the six-mers, there's 4,096 six-mers. We're going to count up their occurrences in a set of promoters that, for example, are turned on when you over-express this factor, and look at those frequencies divided by the frequencies of those six-mers in some background set-- either random sequences, or promoters that didn't turn on. Something like that. You have two classes, and you look for statistical enrichment. This approach, this is fine. There's nothing wrong with this approach. People use it all the time. One of the downsides, though, is that you're doing a lot of statistical tests. You're essentially testing each six-mer-- you're doing 4,096 statistical tests. So you have to adjust the statistical significance for the number of tests that you do, and that can reduce your power. So that's one main drawback. The other reason is that maybe you don't see-- maybe this protein binds a rather degenerate motif, and a precise six-mer is just too precise. None of them will occur often enough. You really have to have a degenerate motif that's C R Y G Y. That's really the motif that it binds to, and so you don't see it unless you use something more degenerate. So you can generalize this to use regular expressions, et cetera. And it's a reasonable approach. Another approach that we'll talk about in a moment is probabilistic optimization, where you wander around the possible space of possible motifs until you find one that looks strong. And we'll talk about that. And then they're deterministic versions of this, like me. We're going to focus today on this second one. Mostly because it's a little bit more mysterious and interesting as an algorithm. And it's also [INAUDIBLE]. So, if the motif landscape looked like this, where imagine all possible motifs, you've somehow come up with a 2D lattice of the possible motif sequences. And then the strength of that motif, or the degree to which that motif description corresponds to the 2 motif is represented by the height here. Then, there's basically one optimal motif, and the closer you get to that, the better fit it is. Then our problem is going to be relatively easy. But it's also possible that it looks something like this. There's a lot of sort of decoy motifs, or weaker motifs that are only slightly enriched in the sequence space. And so you can easily get tripped up, if you're wandering around randomly. We don't know a priori, and it's probably not as simple as the first example. And so that's one of the issues that motivates these stochastic algorithms. So just to sort of put this in context-- the Gibbs motif sampler that we're going to be talking about is a Monte Carlo algorithm, so that just means it's an algorithm that basically does some random sampling somewhere in it, so that the outcome that you get isn't necessarily deterministic. Your run it at different times, and you'll actually get different outputs, which can be a little bit disconcerting and annoying at times. But it turns out to be useful in some cases. There's also a special case of a Las Vegas algorithm, where it knows when it got be optimal answer. But in general, not. In general, you don't know for sure. So Gibbs motif simpler is basically a model where you have a likelihood for generating a set of sequences, S. So imagine you have 40 sequences that are bacterial promoters, each of 40 bases long, let's say. That's your S. And so what you want to do, then, is consider a model that there is a particular instance of a motif you're trying to discover, at a particular position in each one of those sequences. Not necessarily the same position, just some position in each sequence. And we're going to describe the composition of that motif by a weight matrix. OK, one of these matrices that's of width, W, and then has the four rows specifying the frequencies of the four nucleotides at that position. The setup here is that you want to calculate or think about the probability of S comma A, S is the actual sequences, and A is basically a vector that specifies the location of the motif instance in each of those 40 sequences. You want to calculate that, conditional on capital theta-- which is our weight matrix. So that's going to be, in this case, I think I made a motif of length 8, and it's shown there in red. There's going to be a weight matrix of length 8. And then there's going to be some sort of background frequency vector that might be the background composition in the genome of E.coli DNA, for example. And so then the probability of generating those sequences together with that particular locations is going to be proportional to this. Basically, use the little theta background vector for all the positions, except the specific positions that are inside the motif, starting at position AK here. And then you use the particular column of the weight matrix for those 8 positions, and then you go back to using the background probabilities. Question, yeah? AUDIENCE: Is this for finding motifs based on other known motifs? Or is this-- PROFESSOR: No, what we're doing-- I'm sorry, I should've prefaced that. We're doing de novo motif finding. We're going to tell the algorithm-- we're going to give the algorithm some sequences of a given length, or it can even be of variable lengths, and we're going to give it a guess of what the length of the motif is. So we're going to say, we think it's 8. That could come from structural reasons. Or often you really have no idea, so you just guess that you know, a lot of times it's kind of short, so we're going to go with 6 or 8, or you try different lengths. Totally de novo motif finding. OK, so how does algorithm work? You have N sequences of length, L. You guessed that the motif has width, W. You choose starting positions at random-- OK, so this is a vector, of the starting position in each sequence, we're going to choose completely random positions within the end sequences. They have to be at least W before the end-- so we'll have a whole motif, that's just an accounting thing to make it simpler. And then you choose one of the sequence at random. Say, the first sequence. You make a weight matrix model of width, W, from the instances in the other sequences. So for example-- actually, I have slides on this, so we'll just do it with the slides, you'll see what this looks like in a moment. And so you have instances here in the sequence, here in this one, here. You take all those, line them up, make a weight matrix out of those, and then you score the positions in sequence 1 for how well they match. So, let me just do this. These are your motif instances. Again, totally random at the beginning. Then you build a weight matrix from those by lining them up, and just counting frequencies. Then you pick a sequence at random-- yeah, your weight matrix doesn't include that sequence, typically. And then you take your theta matrix and you slide it along the sequence. You consider every sub-sequence of length, W-- the one that goes from 1 to W, to one that goes from 2 to W plus 1, et cetera, all the way along the sequence, until you get to the end. And you calculate the probability of that sequence, using that likelihood that I gave you before. So, it's basically the probability generating sequence where you use the background vector for all the positions, except for the particular motif instance that you're considering, and use the motif model for that. Does that make sense? So, if you happen to have a good looking occurrence of the motif at this position, here, in the sequence, then you would get a higher likelihood. So for example, if the motif was, let's say it's 3 long, and it happened to favor ACG, then if you have a sequence here that has, let's say, it's got TTT, that's going to have a low probability in this motif. It's going to be 0.1 cubed. And then if you have an occurrence of, say, ACT, that's going to have a higher occurrence. It's going to be 0.7 times 0.7 times 0.1. So, quite a bit higher. So you start, it'll be low for this triplet here-- so I'll put a low value here. TTA is also going to be low. TAC, also low. But ACT, that matches 2 out of 3 to the motif. It's going to be a lot better. And then CT is going to be low again, et cetera. So you just slide this along and calculate probabilities. And then what you do is you sample from this distribution. These probabilities don't necessarily sum to 1. But you re-normalize them so that they do sum to 1, you just add them up, divide by the sum. Now they sum to 1. And now you sample those sites in that sequence, according to that probability distribution. Like I said, in this case you might end up sampling-- that's the highest probability site, so you might sample that. But you also might sample one of these other ones. It's unlikely you would sample this one, because that's very low. But you actually sometime sample one that's not so great. So you sample a starting position in that sequence, and you basically-- wherever you would originally assign in sequence 1, now you move it to that new location. We've just changed the assignment of where we think the motif might be in that sequence. And then you choose another sequence at random from your list. Often you go through the sequences sequentially, and then you make a new weight matrix model. So how will that weight matrix model differ from the last one? Well it'll differ because the instance of the motif in sequence 1 is now at a new location, in general. I mean, you might have sampled the exact same location you started, but in general it'll move. And so now, you'll got a slightly different weight matrix. Most of the data going into it, N minus 1, is going to be the same. But one of them is going to be different. So it'll change a little bit. You make a new weight matrix, and then you pick a new sequence. You slide that weight matrix along that sequence, you get this distribution, you sample from that distribution, and you keep going. Yeah, this was described by Lorenz in 1993, and I'll post that paper. OK, so you sample a portion with that, and you update the location. So now we sampled that really high probably one, so we moved the motif over to that new orange location, there. I don't know if these animations are helping at all. And then you update your weight matrix. And then you iterate until convergence. So you typically have a set of end sequences, you go through them once. You have a weight matrix, and then you go through them again. You go through a few times. And maybe at a certain point, you end re-sampling the same sites as you did in the last iteration-- same exact sites. You've converged. Or, you keep track of the theta matrices that you get after going through the whole set of sequences, and from one iteration to the next, the theta matrix hasn't really changed much. You've converged. So let's do an example of this. Here I made up a motif, and this is a representation where the four bases have these colors assigned to them. And you can see that this motif is quite strong. It really strongly prefers A at this position here, and et cetera. And I put it at the same position in all the sequences, just to make life simple. And then a former student in the lab, [INAUDIBLE], he implemented the Gibb Sample in Matlab, actually, and made a little video of what's going on. So the upper part shows the current weight matrix. Notice it's pretty random-looking at the beginning. And the right parts show where the motif is, or the position that we're currently considering. So this shows the position that was last sampled in the last round. And this shows the probability density along each sequence of what's the probability that the motif occurs at each particular place in the sequence. And that's what happens over times. So it's obviously very fast, so I'll run it again and maybe pause it partway. We're starting from a very random-looking motif. This is what you get after not too many iterations-- probably like 100 or so. And now you can see your motif-- your weight matrix is now quite biased, and now favors A at this position, and so forth. And the locations of your motif, most of them are around this position, around 6 or 7 in the sequence-- that's where we put the motif in. But not all, some of them. And then you can see the probabilities-- white is high, black is low-- in some sequences, it's very, very confident, the motif is exactly at that position, like this first sequence here. And others, it's got some uncertainty about where the motif might be. And then we let it run a little bit more, and it eventually converges to being very confident that the motif has the sequence, A C G T A G C A, and that it occurs at that particular position in the sequence. So who can tell me why this actually works? We're choosing positions at random, updating a weight matrix, why does that actually help you find the real motif that's in these sequences? Any ideas? Or who can make an argument that it shouldn't work? Yeah? What was your name again? AUDIENCE: Dan. PROFESSOR: Dan, yeah, go ahead. AUDIENCE: So, couldn't it, sort of, in a certain situation have different sub-motifs that are also sort of rich, and because you're sampling randomly you might be stuck inside of those boundaries where you're searching your composition? PROFESSOR: Yeah, that's good. So Dan's point is that you can get stuck in sub-optimal smaller or weaker motifs. So that's certainly true. So you're saying, maybe this example is artificial? Because I had started with totally random sequences, and I put a pretty strong motif in a particular place, so there were no-- it's more like that mountain, that structure where there's just one motif to find. So it's perhaps an easy case. But still, what I want to know is how does this algorithm, how did it actually find that motif? He implemented exactly that algorithm that I described. Why does it tend to go towards [INAUDIBLE]? After a long time, remember it's a long time, it's hundreds of iterations. AUDIENCE: So you're covering a lot in the sequence, just the random searching of the sequence, when you're-- PROFESSOR: There are many iterations. You're considering many possible locations within the sequences, that's true. But why does it eventually-- why does it converge to something? AUDIENCE: I guess, because you're seeing your motif more plainly than you're seeing other random motifs. So it will hit it more frequently-- randomly. And therefore, converge [INAUDIBLE]. PROFESSOR: Yeah, that's true. Can someone give a more intuition behind this? Yeah? AUDIENCE: I just have a question. Is each iteration an independent test? For example, if you iterate over the same sequence base 100 times, and you're updating your weight matrix each time, does that mean it is the updating the weight matrix also taking into account that the previous-- that this is the same sample space? PROFESSOR: Yeah, the weight matrix, after you go through one iteration of all the sequences, you have a weight matrix. You carry that over, you don't start from scratch. You bring that weight matrix back up, and use that to score, let's say, that first sequence. Yeah, the weight matrix just keeps moving around. Moves a little bit every time you sample a sequence. AUDIENCE: So you constantly get a strong [INAUDIBLE]. PROFESSOR: Well, does it? AUDIENCE: Well, I guess-- PROFESSOR: Would it constantly get stronger? What's to make it get stronger or weaker? I mean, this is sort of-- you're on the track. AUDIENCE: If it is random, then there's some probability that you're going to find this motif again, at which point it will get stronger. But, if it's-- given enough iterations, it gets stronger as long as you hit different spots at random. PROFESSOR: Yeah, yeah. That's what I'm-- I think there was a comment. Jacob, yeah? AUDIENCE: Well, you can think about it as a random walk through the landscape. Eventually, it has a high probability of taking that motif, and updating the [INAUDIBLE] direction, just from the probability of [INAUDIBLE]. PROFESSOR: OK. AUDIENCE: And given the [INAUDIBLE]. PROFESSOR: OK, let's say I had 100 sequences of length, I don't know, 30. And the width of the motif is 6. So here's our sequences. We choose random positions for the start position, and let's say it was this example where the real motif, I put it right here, and all the sequences. That's where it starts. Does this help? So it's 30 and 6, so there's 25 possible start positions. I did that to make it a little easier. So what would happen in that first iteration? What w can you say about what the weight matrix would look like? It's going to be a width, W, you know, columns 1, 2, 3, up to 6. We're going to give it 100 positions at random. The motif is here-- let's say it's a very strong motif, that's a 12-bit motif. So it's 100%-- it's echo R1. It's that. What would that weight matrix look like, in this first iteration, when you first just sample the sites at random? What kind of probabilities would it have? AUDIENCE: [INAUDIBLE] PROFESSOR: Equal? OK-- perfectly equal? AUDIENCE: Roughly. PROFESSOR: OK. Any box? Are we likely to hit the actual motif, ever, in that first encryption? AUDIENCE: No, because you have a uniform probability, of sampling. Well, uniform at each one of the 25 positions? PROFESSOR: Right. AUDIENCE: Right now, you're not sampling proportional to the likelihood. PROFESSOR: So the chance of hitting the motif in any given sequence is what? AUDIENCE: 1/25. PROFESSOR: 1/25. We have 100 sequences. AUDIENCE: So that's four out of-- PROFESSOR: So on average, I'll hit the motif four times, right. The other 96 positions will be essentially random, right? So you initially said this was going to be uniform, right? On average, 25% of each base, plus or minus a little bit of sampling error-- could be 23, 24, 26. But now, you pointed out that it's going to be four. You're going to hit the motif four times, on average. So, can you say anything more? AUDIENCE: Could you maybe have a slightly bias towards G on the first position? Slightly biased towards A on the second and third? Slightly biased towards T on the fourth and fifth. And slightly biased towards C in the sixth? So it would be slightly biased-- PROFESSOR: Right, so remind me of your name? AUDIENCE: I'm Eric. PROFESSOR: Eric, OK, so Eric says that because four of the sequences will have a G at the first position, because those are the ones where you sampled the motif, and the other 96 will have each of the four bases equally likely, on average you have like 24%-- plus 4 for G, right? Something like 28%-- this will be 28%, plus or minus a little bit. And these other ones will be whatever that works out to be, 23 or something like that-- 23-ish, on average. Again, it may not come out exactly like-- G may not be number one, but it's more often going to be number one than any other base. And on average, it'll be more like 28% rather than 25%. And similarly for position two, A will be 28%, and three, and et cetera. And then the sixth will be-- C will have a little bit of a bias. OK, so even in that first round, when you're sampling that first sequence, the matrix is going to be slightly biased toward the motif-- depending how the sampling went. You might not have hit any instances of motif, right? But often, it'll be a little bit-- Is that enough of a bias to give you a good chance of selecting the motif in that first sequence? AUDIENCE: You mean in the first iteration? PROFESSOR: Let's say the first random sequence size sample. No. You're shaking your head. Not enough of a bias because-- it's 0.28 over 0.25 to the sixth power, right? So it's like-- AUDIENCE: The likelihood is still close 1. Like, that's [INAUDIBLE] ratio. PROFESSOR: So it's something like 1.1 to the sixth, or something like that. So it might be close to 2, might be twice as likely. But still, there's 25 positions. Does that make sense? So it's quite likely that you won't sample the motif in that first-- you'll sample something else. Which will take it away in some random direction. So who can tell me how this actually ends up working? Why does it actually converge eventually, if you get it long enough? AUDIENCE: [INAUDIBLE]. PROFESSOR: So the information content, what will happen to that? So the information content, if it was completely random-- we said that would be uniform. That would be zero information content, right? This matrix, which has around 28% at six different positions, will have an information content that's low, but non-zero. It might end up being like 1 bit, or something. And if you then sample motifs that are not the motif, they will tend to reduce the information content, tend to bring it back toward random. If you sample locations that have the motif, what will that do to the information content? Boost it. So what would you expect if we were to plot the information content over time, what would that look like? AUDIENCE: It should trend upwards, but it could fluctuate. PROFESSOR: Yeah. AUDIENCE: Over the number of iterations? PROFESSOR: I think I blocked it here. Let me see if I can-- Let's try this. I think I plotted it. OK, never mind. I wanted to keep it very mysterious, so you guys have to figure it out. The answer is that it will-- basically what happens is you start with a weight matrix like this. A lot of times, because the bias for the motif is quite weak, a lot of times you'll sample-- even for a sequence, what matters is-- like, if you had a sequence where the location, initially, was not the motif, and then you sample another location that's not the motif, that's not really going to change anything. It'll change things a little bit, but not in any particular direction. What really matters is when you get to a sequence where you already had the motif, if you now sample one that's not the motif, your information content will get weaker. It will become more uniform. But if you have a sequence where it wasn't the motif, but now you happen to sample the motif, then it'll get stronger. And when it gets stronger, it will then be more likely to pick the motif in the next sequence, and so on. So basically what happens to the information content is that over many iterations-- it starts near 0. And can occasionally go up a little bit. And then once it exceeds the threshold, it goes like that. So what happens is it stumbles onto a few instances of the motif that bias the weight matrix. And if they don't bias it enough, it'll just fall off that. It's like trying to climb the mountain-- but it's walking in a random direction. So sometimes it will turn around and go back down. But then when it gets high enough, it'll be obvious. Once you have a, say, 20 times greater likelihood of picking that motif than any other sequence, most of the time you will pick it. And very soon, it'll be stronger. And the next round, when it's stronger, you'll have a greater bias for picking the motif, and so forth. Question? AUDIENCE: For this specific example, M is much greater than L minus W. How true is that for practical examples? PROFESSOR: That's a very good question. There is sometimes-- depends on how commonly your motif occurs in the genome, and how good your data is, really, and what the source of your data is. So sometimes it can be very limited, sometimes-- If you do ChIP-Seq you might have 10,000 peaks that you're analyzing, or something. So you could have a huge number. But on the other hand, if you did some functional assay that's quite laborious for a motif that drives luciferase, or something, and you can only test a few, you might only have 10. So it varies all over the map. So that's a good question. We'll come back to that in a little bit. Simona? AUDIENCE: If you have a short motif, does it make sense, then, to reduce the number of sequences you have? Because maybe it won't converge? PROFESSOR: Reduce the number of sequences? What do you people think about that? Is that a good idea or a bad idea? It's true that it might converge faster with a smaller number of sequences, but you also might not find it all. So generally you're losing information, so you want to have more sequences up to a certain point. Let's just do a couple more examples, and I'll come back. Those are both good questions. OK, so here's this weak motif. So this is the one where you guys couldn't see it when I just put the sequences up. You can only see it when it's aligned-- it's this thing with GGC, here. And here is, again, the Gibbs Sampler. And what happened? Who can summarize what happened here? Yeah, David? AUDIENCE: It didn't converge. PROFESSOR: Yeah, it didn't quite converge. The motif is usually on the right side, and it found something that's like the motif. But it's not quite right-- it's got that A, it's G A G C, it should be G G C. And so it sampled some other things, and it got off track a little bit, because probably by chance, there were some things that looked a little bit like the motif, and it was finding some instances of that, and some instances of the real motif. And yeah, it didn't quite converge. And you can see this probability vectors here, they have multiple white dots in many of the rows. So it doesn't know, it's uncertain. So it keeps bouncing around. So it didn't really converge, it was too weak, it was too challenging for the algorithm. This is just a summary of the Gibb Sampler, how it works. It's not guaranteed to converge to the same motif every time. So what you generally will want to do is run it several times, and nine out of 10 times, you get the same motif. You should trust that. Go ahead. AUDIENCE: Over here, are we optimizing for convergence of the value of the information content? PROFESSOR: No, the information content is just describing-- it's just a handy single number description of how biased the weight matrix is. So it's not actually directly being optimized. But it turns out that this way of sampling tends to increase information content. It's sort of a self-reinforcing kind of a thing. But it's not directly doing that. However MEME, more or less, directly does that. The problem with that is that, where do you start? Imagine an algorithm like this, but where you deterministically-- instead of sampling from the positions in the sequence, where it might have a motif in proportion to probabilities, you just chose the one that had the highest probability. That's more or less what MEME does. And so what are the pros and cons of that approach, versus this one? Any ideas? OK, one of the disadvantages is that the initial choice of-- how you're initially seeding your matrix, matters a lot. That slight bias-- it might be that you had a slight bias, and it didn't come out being G was number one. It was actually-- T was number one, just because of the quirks of the sampling. So what would this be, 31 or something? Anyway, it's higher than these other guys. And so then you're always picking the highest. It'll become a self-fulfilling prophecy. So that's the problem with MEME. So the way that MEME gets around that, is it uses multiple different seeding, multiple different starting points, and goes to the end with all of them. And then it evaluates, how good a model did we get at the end? And whichever was the best one, it takes that. So it actually takes longer, but you only need to run it once because it's deterministic. You use a deterministic set of starting points, you run a deterministic algorithm, and then you evaluate. The Gibbs, it can go off on a tangent, but because it's sampling so randomly, it often will fall off, then, and come back to something that's more uniform. And when it's a uniform matrix, it's really sampling completely randomly, exploring the space in an unbiased way. Tim? AUDIENCE: For genomes that have inherent biases that you know going in, do you precalculate-- do you just recalculate the weight matrix before, to [? affect those classes? ?] For example, if you had 80% AT content, then you're not looking for-- you know, immediately, that you're going to hit an A or a T off the first iteration. So how do you deal with that? PROFESSOR: Good question. So these are some features that affect motif finding. I think that we've now hit at least a few of these-- number of sequences, length of sequences, information content, and motif, and basically whether the background is biased or not. So, in general, higher information content motifs, or lower information content, are easier to find-- who thinks higher? Who thinks lower? Someone, can you explain? AUDIENCE: I don't know. I just guessed. PROFESSOR: Just a guess? OK, in back, can you explain? Lower? AUDIENCE: Low information content is basically very uniform. PROFESSOR: Low information means nearly uniform-- right, those are very hard to find. That's like that GGC one. The high information content motif, those are the very strong ones, like that first one. Those are much easier to find. Because when you stumble on to them, it biases the matrix more, and you rapidly converge to that. OK, high information is easy to find. So if I have one motif per sequence, what about the length of the sequence? Is longer or shorter better? Is long better? Who thinks shorter is better? Shorter-- can you explain why short? AUDIENCE: Shouldn't it be the smaller the search space, the fewer the problems? PROFESSOR: Exactly, the shorter the search space, and your motif, there's less place for it to hide. You're more likely to sample it. Shorter is better. If you think about-- if you have a motif like TATA, which is typically 30 bases from the TSS, if you happen to know that, and you give it plus 1 to minus 50, you're giving it a small region, you can easily find the TATA box. If you give it plus 1 to minus 2,000 or something, you may not find it. It's diluted, essentially. Number of sequences-- the more the better. This is a little more subtle, as Simona was saying. It affects convergence time, and so forth. But in general, the more the better. And if you guessed the wrong length of your matrix, that makes it worse than if you guess the right length in either direction. For example, it's six-base motif, you guess three. The information content, even if it's a 12-bit motif, there's only six bits that you could hope to find, because you can only find three of those positions. So clearly, effectively it's a smaller information content, and much harder to find. And vice versa. Another thing that occurs in practice is what's called shifted motifs. Your motif is G A A T T C. Imagine in your first iteration you happen to hit several of these sequences, starting here. You hit the motif, but off by two at several different places. That'll bias first position to be A, and the second position to be T, and so forth. And then you tend to find other shifted versions of that motif. You may well converge to this-- A T C C N N, or something like that-- which is not quite right. It's close, you're very close, but not quite right. And it's not as information rich as the real motif. Because it's got those two N's at the end, instead of G A. So one thing that's done in practice is a lot of times, every so often, the algorithm will say, what would happen if we shifted all of our positions over to the left by one or two? Or to the right by one or two? Would the information content go up? If so, let's do that. So basically, shifted versions of the motif become local, near-optimal solutions. So you have to avoid them. And biased background composition is very difficult to deal with. So I will just give you one or two more examples of that in a moment, and continue. So in practice, I would say the Gibbs Sampler is sometimes used, or AlignACE, which is a version of Gibbs Sampler. But probably more often, people use an algorithm called MEME, which is this EM algorithm, which, like I said, is deterministic, so you always get the same answer, which makes you feel good. May or may not always be right, but you can try it out here at this website. And actually, the Fraenkel Lab has a very nice website called WebMotifs that runs several different motif finders including, like I said, a MEME and AlignACE, which is similar to Gibbs, as well as some others. And it integrates the output, so that's often a handy thing to use. You can read about them there. And then I just wanted to say a couple words-- this is related to Tim's comment about the biased background. How do you actually deal with that? And this related to this notion of a mean bit score of a motif. So if I were to give you a motif model, P, and a background model, q, then the natural scoring system, if you wanted additives scores, instead of multiplicative, you would just take the log. So log P over q, I would argue, is natural additive scores. And that's often what you'll see in a weight matrix-- you'll see log probabilities, or logs of ratios of probabilities. And so then you just add them up, and it makes life a bit simpler. And so then, if you were to calculate what's the mean bit score-- if I had a bunch of instances of a motif, it will be given by this formula that's here in the upper right. So that's your score. And this is the mean, where you're sampling over the probability in using the motif model, probabilities. So it turns out, then, that if qk, your background, is uniform, motif of width w-- so its probability of any w-mer, is 1/4 to the w, then it's true that the mean bit-score is 2w minus the entropy of the motif, which is the same as the information content of the motif, using our previous definition. So that's just a handy relationship. And you can do a little algebra to show that, if you want. So basically summation Pk log Pk over qk-- this log, you turn that into a difference-- so that summation Pk log Pk minus Pk log qk. And then you can do some rearrangement, and sum them up, and you'll get this formula. I'll leave that as an exercise, and any questions on it, we can do it next time. So what I wanted to get to is sort of this big question that I posed earlier-- what's the use of knowing the information content of a motif? And the answer is that one use is that it's true, in general, that the motif with n bits of information will occur about once every 2 to the n bases of random sequence. So we said a six-cutter restriction enzyme, echo R1, has an information content of 12 bits. So by this rule, it should occur about once every to 2 to the 12th bases of sequence. And if you know your powers of 2, which you should all commit to memory, that's about 4,000. 2 to the 12th is 4 to the sixth, is 4,096. So it'll occur about once every 4 [? kb, ?] which if you've ever cut E. coli DNA, you know is about right-- your fragments come out to be about 4 [? kb. ?] So this turns out to be strictly true for any motif that you can represent by a regular expression, like a precise motif, or something where you have a degenerate R or Y or N in it, still true. And if you have a more general motif that's described by weight matrix, then you have to define a threshold, and it's roughly true, but not exactly. All right, so what do you do when the background composition is biased, like Tim was saying? What if it's 80%, A plus T? So then, it turns out that this mean bit-score is a good way to go. So like I said, the mean bit-score equals the information content in this special case, where the background is uniform. But if the background is not uniform, then you can still calculate this mean bit-score, and it'll still be meaningful. But now it's called something else-- it's called relative entropy. Actually it has several names, relative entropy, Kullback-Leibler distance is another, and information for discrimination, depending whether you're reading the Double E literature, or statistics, or whatever. And so it turns out that if you have a very biased composition-- so here's one that's 75% A T, probability of A and T are 3/8, C and G are 1/8. If your motif is just C 100% of the time, your information content by the original formula that I gave you, would be 2 bits. However, the relative entropy will be 3 bits, if you just plug in these numbers into this formula, it will turn out to be 3 bits. My question is, which one better describes the frequency of C in the background sequence? Frequency of this motif-- the motif is just a C. You can see that the relative entropy says that actually, that's stronger than it appears. Because it's a C, and that's a rare nucleotide, it's actually stronger than it appears. And so 2 to the 3rd is a better estimate of its frequency than 2 squared. So relative entropy. So what you can do when you run a motif finder in a sequence of biased composition, you can say, what's the relative entropy of this motif at the end? And look at the ones that are strong. We'll come back to this a little more next time. Next time, we'll talk about hidden Markov models, and please take a look at the readings. And please, those who are doing projects, look for more detailed instructions to be posted tonight. Thanks.
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https://ocw.mit.edu/courses/8-03sc-physics-iii-vibrations-and-waves-fall-2016/8.03sc-fall-2016.zip
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YEN-JIE LEE: Before people take 8.03 class, especially my version, most of the students may think uncertainty principle is actually completely related to quantum mechanics. So actually it is actually not the case. Actually, uncertainty principle almost have nothing to do with quantum mechanics. It has to do with just the wave description we employ when we describe the system. So that's actually one of the big thing which you will learn from 8.03. The other thing which people will learn a lot more is actually that before they come into the class, they may think it's almost impossible to understand infinitely number of coupled oscillators. And it turns out that there is a very systematical way you can actually make use of simple symmetry argument, and that you can actually greatly simplify complicated system, like infinity long system. So it is actually not an impossible task, and then everything can be actually understand, using the concept of symmetry, which we introduce in 8.03.
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https://ocw.mit.edu/courses/8-962-general-relativity-spring-2020/8.962-spring-2020.zip
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SCOTT HUGHES: All right. Good morning, 8.962. This is a very weird experience. I am standing in here talking to an empty classroom. I have some experience talking to myself, because like many of us, I am probably a little weirder than the average. But that does not change the fact that this is awkward and a little strange, and we already miss having you around here. So I hope we all get through MIT's current weirdness in a healthy and quick fashion so we can get back to doing this work we love with the people we love to have here. All right, so all that being said, it's time for us to get back to the business of 8.962, which is learning about general relativity. And today, the lecture that I am recording is one in which we will take all the tools that have been developed and we will turn this into a theory of gravity. Let me go over a quick recap of some of the things that we talked about in our previous lecture, and I want to emphasize this because this quantity that we derived about two lectures ago, the Riemann curvature tensor, is going to play an extremely important role in things that we do moving forward. So I'll just quickly remind you that in our previous lecture, we counted up the symmetries that this tensor has. And so the four most-- the four that are important for understanding its properties, its four main symmetries are first of all, if you exchange indices 3 and 4, it comes in with a minus sign, so it's anti-symmetric under exchange of indices 3 and 4. If you lower that first and next so that they are all in the downstairs position and you exchange indices 1 and 2, you likewise pick up a minus sign. Again, keeping everything in the downstairs position, if you just wholesale swap indices 1 and 2 for indices 3 and 4 like so, that's symmetric, and so you get the whole thing back with a plus sign. And finally, one that's a little bit non-obvious but can be seen if you're sort of pigheaded enough to sort of stare at this thing and kind of pound on the algebra a little bit, if you take the Riemann curvature tensor-- and this can be at the end of-- the first index can be either upstairs or downstairs, but if you cyclically permute indices 2, 3, and 4 and add them up, they sum to 0, OK? So that tells you that it's a constraint on this thing when I look at the behavior, this thing with respect to indices 2, 3, and 4. We introduced a variant of the Riemann curvature tensor called the Ricci curvature. So the way I do this is if I take the trace on the Riemann curvature tensor on indices 1 and 3, which is equivalent to taking it with the indices on the downstairs position and hitting it with the metric like so, I get this quantity, which I forgot to write down, is symmetric. One of the major important physical applications of the Riemann curvature tensor is that it allows us to describe the way in which two neighboring geodesics-- if I have two geodesics that are separated by a four vector c, and I look at how that separation evolves as they move forward along their geodesic paths, this differential equation describes how it behaves. And the key thing is that what we see is that the rate of separation is proportional to the Riemann curvature. It ends up playing the role-- when we think about-- what this tells us is this ends up, remember, geodesics describe free fall. And so what this is telling me is a way in which two nearby but somewhat separated-- separated by a distance c-- nearby but slightly separated geodesics-- both are in free fall, but their free fall trajectories are diverging from one other and perhaps being focused towards one another depending upon how R is actually behaving, this is the behavior of tides. Free fall is gravity, and this is saying that the free fall trajectory's change of separation is governed by the Riemann tensor, and that's telling me about the action of gravitational tides. The last thing that we did in Tuesday's-- excuse me-- in Thursday's lecture was I went through and I developed this proof of what is known as the Bianchi identity, which is an identity on the-- it's an identity on the covariant derivative of the Riemann tensor. And so notice what's going on here. I'm leaving indices 3 and 4 on the Riemann tensor-- oh shoot. I'm actually changing my notation halfway through, let me fix that. My apologies. OK. Apologies for that. Are you leave indices 3 and 4 unchanged, and what you do is you cyclically permute the index against-- in the direction which you're taking a derivative with indices 1 and 2. So my first term, it goes alpha, beta, gamma; then beta, gamma, alpha; gamma, alpha, beta. OK, notice the way they are cyclically permuting like that. Sum them up and you get 0. So let's take it from here. We're going to start with this Bianchi identity. What I want to do now is contract the Bianchi identity in the following way. So let's take this form that I've written out here-- and let me just make sure I've left it now in a form that comports with my notes. I did, good. So what I'm going to do is multiply the entire thing. Using the metric, I'm going to contract it on indices beta and mu. So remember, the metric commutes with the covariant derivative. So unless the derivative itself is with respect to either the beta or the mu index, that g just sort of waltzes right in top of there. So when I do this, it's going to beta-- g beta mu nu upstairs is going to walk right through this, it's going to raise the beta index, and what I wind up with here is this first term becomes the covariant derivative of the Ricci tensor, OK? I have contracted on indices beta and mu. When it hits this one, is just going to raise the index on that covariant derivative. So I've got a term here now that looks something like the divergence of the Riemann tensor, divergence with respect to index 3. When I hit this term, it walks through the covariant derivative again, and you see what I'm doing is a trace on indexes 2 and 3. Now I can take advantage of the anti-symmetry here-- let's reverse this. And so it's like doing a-- throwing in a minus sign and then doing a trace on indices 1 and 2-- excuse me, doing a trace on-- rewind, back up for just a second. I'm going to take advantage that anti-symmetry, I'll exchange indices 1 and 2, and then I am doing a trace on indices 1 and 3, which is going to give me the Ricci tensor. But because I have used that-- or that anti-symmetry, I will do so with a minus sign. So what I get here is this. I'm going to probably bobble more than once in this lecture, because again, doing this in an empty room is a little weird. All right. So I really want to get a relationship that simplifies the Riemann tensor, OK? A Riemann tensor's got four indices on it. I'm not scared, but I don't like it, OK? So we're going to do one more contraction operation to try to simplify this. Let's now contract once more using the metric on indices g and nu, OK? So when I do it with the first one, it walks right through-- right through that covariant derivative, and I get the trace of the Riemann tensor, the Ricci-- excuse me-- I get the trace of the Ricci tensor, the Ricci scalar. When I do it on the second term, OK? I am now tracing on indices 1 and 4. I will invoke anti-symmetry to change that into a trace on indices 1 in 3, and I get the Ricci tenser with a minus sign. And then the next one, I just trace on this and I wind up with something that looks like this. OK. Now these two terms are both divergences on the-- they are both divergences on the second index. The second index is a dummy index, so I can put these two together. So this is equivalent to-- or dividing out a factor of minus 2, the way this is more often written. You can also factor out that derivative. Let's write it like this. OK? So what I'm doing here is I divided by a minus 2 so that I can put this guy in front and I get a minus 1/2 in front of my Ricci term. And because I want to factor out my covariant derivative, I need to throw in a factor of the metric there so that the indices line up right. So what we do at this point is we do what-- whenever you reach a certain point in your calculation where you've got something good, you do what every mathematician or physicist would do, you give this guy a name. So switching my indices a tiny bit, we define g mu nu to be the Ricci tensor minus 1/2 metric Ricci scalar, and this is an entity known as the Einstein tensor. This is a course on Einstein's gravity, so the name alone should tell you, this guy is going to matter. One quick side note. So suppose I take the trace of the Einstein tensor. When we took the trace of the Ricci tensor, I didn't write it down, but if I take a trace of this guy, I just get the Ricci scalar R, which I used over here. So when I do this here-- oops. Suppose I just want to call this the Einstein scalar g. Well, applying that to its definition, this is going to be equal to the Ricci scalar minus 1/2 trace of the metric times R. And it's a general rule in any theory of spacetime that the trace of the metric is equal to the number of dimensions in your spacetime, OK? You can easily work it out in special relativity, you're basically just raising one index, and as we'll see, it holds completely generally. In fact, it follows directly from the fact that the upstairs metric is the matrix inverse of the downstairs metric. So this is equal to 4, so this whole thing is just the negative of the Ricci scalar. What this means is that the Einstein tensor is the trace-reversed Ricci tensor, OK? I just want to plant that for now. This is a fact that we're going to take advantage of a little bit later, but for now, it's just a mathematical fact that I want to point out, I want to set aside. We'll come back to it when it matters. OK. We now have everything that we need to take all of the framework that we have been developing all term and turn it into a theory of gravity. I just had a nightmare. Am I being recorded?-- yes, OK. Sorry, just suddenly thought I might have forgotten to turn my microphone on! So let's turn this into a theory of gravity. Ingredient 1 is something that we have discussed quite a bit before. I want to restate it and I want to sort of remind us. Several times over the past couple of lectures I have implicitly used this rule already, but I want to make it a little bit more explicit now. We're going to use the principle of equivalence. In particular, we're going to use what is known as the minimal coupling principle. So here's the way-- what this basically means. We're going to try to take laws of physics that are well-understood from laboratory experiments, from special relativity, everything that we have known and loved and tested for-- since we started studying physics, and we're going to try to see how that can be carried over to working in the curved spacetime that will describe gravity with as little additional sort of coupling to spacetime entities as is possible. So here's what we're going to do. Take a law of physics that is valid in inertial coordinates and flat spacetime, or equivalently, the local Lorentz frame, which corresponds to the local region of a freely-falling frame or a freely-falling observer, take that law of physics that is good in that form and rewrite it in a coordinate-invariant tensorial form. This is one the reasons why throughout this term, we have been brutally didactic about insisting on getting all of our laws of physics expressed using tensors, quantities which have exactly the transformation laws that we demand in order for them to be true tensors that live in the curve manifold that we use to describe spacetime. The last time I actually did something like this was when I derived-- I've erased it now, but when I derived the equation of geodesic deviation, OK? I first did it using very, very simple language, and then I sort of looked at it and said, well this is fine according to that local Lorentz frame, according to that freely-falling observer. But this is not tensorial, it's actually as written only good in that frame. And so what we did was we took another couple minutes and said, let's see how I can change this acceleration operator that describes my equation of geodesic deviation, put in the extra structure necessary so that the acceleration operator is tensorial, and when we did that, we saw that the result was actually exactly what the Riemann tensor looks like in the local Lorentz frame. We said, if it holds a local Lorentz frame, I'm going to assert it holds in all other frames. And that indeed is the final step in this procedure. We assert that the resulting law holds in curved spacetime. OK? So this is the procedure by which general relativity takes the laws of physics, good and flat spacetime, and rejiggers them so that they work in curved spacetime. Ultimately, this is physics, and so ultimately the test for these things are experiments. And I will simply say at this point that this procedure has passed all experimental tests that we have thrown at it so far, and so we're happy with it. So let me just describe one example of where we did this-- actually, I'm going to do two examples. So if I consider the force-free motion of an object in the freely-falling frame-- so recall, in the freely-falling frame, everything is being acted upon by gravity in an equal way. If I am in the local Lorentz frame, I can simply say that my object feels-- my freely-falling object, freely-flying observer-- feels no acceleration. That is a perfectly rigorous expression of the idea that this observer or object is undergoing force-free motion in this frame. This is not tensorial, though. And so we look at this and say, well, if I want to make this tensorial, what I'm going to do is note that the tensor operator that describes-- now let me keep my indices consistent here. My tensor operator that describes this equation is given by taking the covariant derivative of the four velocity and contracting it with the four velocity itself. These say the exact same thing in the local Lorentz frame. This one is tensorial, though, that one is not. And so we then say, OK, well this is the version that is tensorial, I'm going to assert that it holds in general. Another example in flat spacetime, local conservation of energy and momentum was expressed by the idea that my stress energy tensor had no divergence in the local Lorentz frame. Well, if I want to make this tensorial, all I do is I promote that partial derivative I used to define the divergence to a covariant derivative. This is how we are going to define conservation-- we're going to define local conservation of energy and momentum in a general spacetime theory. So that's step 1. We need-- or sorry, ingredient 1. Ingredient 2 is-- well, let's just step back for a second. We have done a lot of work to describe the behavior of curved spacetimes, OK? Spacetimes that are not just the spacetimes of special relativity, spacetimes when my basis objects are functional, where the Riemann curvature tensor is non-zero. We've done a lot to do that, but I haven't said anything about where that curved spacetime actually comes from. So the next thing which I need is a field equation which connects my spacetime to sources of matter and energy. That's a tall order. The way we're going to do this, we're actually going to do it two different ways. So in this current lecture, I'm going to do it using a method that parallels how Einstein originally did it when he derived-- what resulted out of this is what we call the field equation of general relativity or the Einstein field equation, and in this first presentation of this material, I'm going to do it the way Einstein did it. So what we are going to do is we will require that whatever emerges from this procedure, it must recover Newtonian gravity in an appropriate limit. This is a philosophical point about physics. When you come up with a new theory, you may conceptually overturn what came before. You may have an entirely new way of thinking about it. You may go from saying that there is a potential that is sourced by mass that fills all of space and that objects react to to saying something like, we now decide that the manifold of events has a curvature that is determined by the distribution of matter and energy in spacetime. It's very different philosophical and ultimately mathematical ways of formulating this, but they have to give consistent predictions, because at the end of the day, Newtonian gravity works pretty damn well, OK? We can't just throw that away. So what we're going to do is demand that in an appropriate limit, both the field equation for Newtonian gravity-- so this is the Laplace operator now, which I'm going to write in a semi-coordinate-invariant form, as the chronic or delta contracted-- basically it's the trace on a matrix of partial derivatives acting on a potential. This equals 4 pi rho, and I call this semi-coordinate invariant because part of what goes into this is this Newtonian limit only works if everything is sufficiently slowly varying in time, that things having to do with time derivatives can be neglected, OK? It's never really been-- prior to some of the more modern experiments that we've had to do, time-varying sources of gravity are very hard to work with. And so Newton was never really tested in that way. Nonetheless, whatever emerges from Einstein had best agree with this. And we are also going to require that the equation of motion in this framework agree with Newtonian gravity. We actually went through this-- this was a little bit of a preview of this lecture, we did this in our-- we concluded our discussion of geodesics. Let me just recap the result that came out of this. So our equation of motion was that-- you can write it as the acceleration of an observer is related to the gradient of the potential. All right. So let's follow in the footsteps of Einstein and do this. So what we're going to do-- let's do the equation of motion first. I've already gone through this briefly, but I want to go over it again and I want to update the notation slightly. So let's do the equation of motion by beginning with the geodesic equation. We will start with the acceleration coupled to the four velocity by the Christoffel symbols. All tests of Newtonian gravity, especially those that Einstein had available at the time that he was formulating this, were slow motion ones. We were considering objects moving at best in our solar system. And so things there on a human scale certainly move quickly, but they're slow compared to the speed of light. And so let's impose the slow motion limit, which tells us that the 0th component of the four velocity is much larger than the spatial components of the four velocity, OK? Remember working in units where the speed of light is equal to 1. And so if this is being measured in human units, kilometers per second, and things like that, this is on the order of the speed of light. So when we throw this in, we see that we expand this out, that the contributions from the dt d tau terms here are going to be vastly larger than any others. And so we can simplify our equation to a form that looks like this. OK? In the spirit of being uber complete, let's write out that Christoffel symbol. So dig back into your previous lectures' notes, remind yourself what the formula for the Christoffel symbol is. OK. Notice, two of the terms here are time derivatives. The Newtonian limit-- all the tests that were available when Einstein was formulating this, the limit that we care about here, the gravitational field, the gravitational potentials that he was studying, what the Newtonian limit emerges from, they are static. So we're going to do is neglect time derivatives to recover this limit. And when we do this, what we find is that the component-- the Christoffel component that we care about looks like one derivative of the 0 0 piece of the spacetime metric. It's not too hard to convince yourself that this, in fact, reduces-- oops, pardon me. I skipped a step. Pardon me just one moment. Just one moment, my apologies. I'm going to write the spacetime metric-- I'm going to work in a coordinate system such that spacetime looks like the flat space time of special relativity plus a little bit else, OK? This is consistent with the idea that every system we have studied in Newtonian gravity is one where the predictions of special relativity actually work really, really well, OK? Gravity is new, it's special, it's why we have a whole other course describing it. But clearly it can't be too far different from special relativity or we wouldn't have been able to formulate special relativity in the first place. So my apologies, I sort of jumped ahead here for a second. We're going to treat the g mu nu that goes into this as the metric of flat spacetime plus something else where I'm going to imagine that all the different components of this-- so a typical component of this h mu nu has an absolute value that is much smaller than 1. It's not too hard to prove that when you invert this, what you wind up with is a form that looks like so where this h mu nu with the indices in the upstairs position is given by raising h's indices using the metric of flat spacetime. We're going to talk about this in a little bit more detail in a future lecture, it doesn't really matter too much right now. I just want to point out that the inverse g, which we need to use here, also has this form that looks like flat spacetime metric and this h coupling into it. Now the reason I'm going through all this is that in order to work out this Christoffel symbol, I need to take a derivative. The derivative of eta is 0, OK? So the only thing that gets differentiated is h. So when you work out this Christoffel symbol, what you get is this. If you're being-- keeping score, there are corrections of order h squared, and pardon me, I should have actually noted, there are corrections of order h squared that go into this inverse. Let me move this over so I can fit that in a little bit better. But in keeping with the idea that-- in keeping with the idea that for the Newtonian limit, the h squared could-- h is small, we're going to treat the h squared corrections as negligible and we will drop them. OK. So let's look at what motion in this limit turns into, then. We now have enough pieces to compute all the bits of the equation of motion. So in keeping with the idea that I am going to neglect all time derivatives, this tells me that the gamma 0, 00 term is equal to 0. And from this, we find that there is a simple equation describing time. In our equation describing space, OK? So what I've done there is just taken this geodesic equation, plugged in that result for the Christoffel symbol, and expanded this guy out. So what results, I can divide both sides now by two powers of dt d tau. All right. If you take a look at what we've got here, this prediction of the-- no even a prediction, this result from the geodesic equation is identical to our Newtonian equation of motion provided we make the following identification. h00 must be minus 2 phi where phi was Newtonian gravitational potential. Or equivalently, g00 is the negative of 1 plus 2 phi. All right. So that's step 1. We have made for ourselves a correspondence between what the metric should be and the equation of motion. We still have to do the field equation, so let's talk about that. So very helpfully I've already got the Newtonian field equation right above me here. Let me rewrite it because I'm going to want to tweak my notation a tiny bit. I don't want to think about what this is telling me. So eta ij is the same thing as delta ij, I just want to put it in this form so that it looks like a piece of a spacetime tensor. This is manifestly not a tensorial equation. I have a bunch of derivatives on my potential being set equal to-- OK, there's a couple constants, but this, OK? When we learned about quantities like this in undergraduate physics, usually we're told that this is a-- excuse me, this is a scalar. But we now know, rho is not a scalar, it is the mass density, which up to a factor of c squared, is the same thing as the energy density. And when we examined how this behaves as we change between inertial reference frames, we found this transforms like a particular component of a tensor. And as I sort of emphasize, not that long ago we have been in something of a didactic fury insisting that everything be formulated in terms of tensors. Pulling out a particular component of a tensor is bad math and bad physics. So we want to promote this to something tensorial. So s on the right-hand side, we've got one component of the stress energy tensor. We would like whatever is going to be on the right-hand side of this equation to be the stress energy tensor, OK? We can sort of imagine that what's going on in Newton's gravity is that there is one particular-- maybe there's one component of this equation that in all the analyses that were done that led to our formation of Newtonian gravity, there may be one component that was dominant, which is how it was that Newton and everyone since then was able to sort of pick out a particular component of this equation as being important. Over here on the left-hand side, we saw earlier that the equation of motion we're going to look for corresponds to the Newtonian limit if the metric plays the same role-- up to factors of 2 and offsets by 1 and things like that-- the metric must play the same role as the Newtonian gravitational potential. So if I look at-- if I look at the Newtonian field equation, I see two derivatives acting on the potential. So I want my metric to stand in for the potential, we expect there to be two derivatives of metric entering this relationship. So now two derivatives of the metric is going to give me something that smells like a curvature. So we want to put a curvature tensor on the left-hand side of this equation. We have several to choose from, OK? It clearly can't be the Riemann tensor. There's too many indices, it just doesn't fit. It could be the Ricci curvature, OK? The Ricci curvature has two indices. That has two indices, that's a candidate. But it's worth stopping and reminding ourselves, wait a minute, this guy has some properties that I already know about. t mu nu tells me about the properties of energy and momentum in my spacetime, and as such, conservation-- local conservation of energy and momentum requires that it be divergence-free. So whatever this curvature tensor is here on the left-hand side, we need it to be a divergence-free 2-index mathematical object. At the beginning of today's lecture, I showed how by contracting on the Bianchi identity, you can, in fact, deduce that there exists exactly such a mathematical object. So let us suppose that our equation that relates the properties of the spacetime to the sources of energy and momentum of my spacetime is essentially that that Einstein tensor, g mu nu, be equal to the stress energy tensor. Now in fact, they don't have the same dimensions as each other, so let's throw in a kappa, some kind of a constant to make sure that we get the right units, the right dimensions, and that we recover the Newtonian limit. The way we're going to deduce how well this works is see whether an equation of this form gives me something that looks like the Newtonian limit when I go to what I'm going to call the weak gravity limit, and I'm going to then use, assuming it does work-- not to give away the plot, it does-- we'll use that to figure out what this constant kappa must be. So if we do, in fact, have a field equation of the form g mu nu is some constant t mu nu, it's not too hard to figure out that an equivalent form of this is to say that the Ricci tensor is k times t mu nu minus 1/2 g mu nu t where this t is just the trace of the stress energy tensor. Remember, I spent a few moments after we derived the Einstein tensor pointing out that it's essentially the same thing as Ricci but with the trace reversed. This is just a trace-reversed equivalent to that equation. This step that I'm introducing here, basically it just makes the algebra for the next calculation I'm going to do a little bit easier, OK? So I just want to emphasize that this and that are exactly the same content. All right. So to make some headway, we need to choose a form for a stress energy tensor. Our goal is to recover the Newtonian limit, and so what we want to do is make the stress energy tensor of a body that corresponds to the sort of sources of gravity that are used in studies of Newtonian gravity. So let's do something very simple for us. Let's pick a static-- in other words, no time variation, a static perfect fluid as our source of gravity. So I'm going to choose for my t mu nu-- dial yourself back to lectures where we talk about this, OK? So this is the perfect fluid stress energy tensor. We're working in the Newtonian limit, and we are working in units where the speed of light is equal to 1. If you put speed of light back into these things, you explicitly include it, this is actually a rho c squared that appears here. And so what this tells me is that if I'm studying sort of Newtonian limit problems, rho is much, much, much greater than P in the limit that we care about. Furthermore, I am treating this fluid as being static. So that means that my four velocity only has one component, OK? The fluid is not flowing. You might be tempted to say, oh, OK, I can just put a 1 in for this. Not so fast, OK? Let's be a little bit more careful about that. One of the key governing properties of a four velocity is that it is properly normalized. So this equals g mu nu u mu u nu is minus 1. We know that the only components of this that matter, so to speak, are the mu and nu equal 0. So this becomes g00 mu 0 squared equals minus 1. But g00 is-- well, let's write it this way-- Negative 1 plus h00. Go through this algebra, and what it tells you is u0 equals 1 plus 1/2 h00. Again, I'm doing my algebra at leading order in h here. We raise and lower indices. So in my calculation, I'm going to want to know the downstairs version of this. And if I, again, treat this thing-- treat this thing consistently, OK. What I'll find is I just pick up a minus sign there. OK. OK. Let's now put all the pieces together. The only component of my stress energy tensor that's going to now really matter is rho u0 u0, which, putting all these ingredients back together, is rho 1 plus h00. The trace of this guy, putting all these pieces together, is just equal to negative rho. Since I only have one component that's going to end up mattering, let's just focus on one component of my proposed field equation. OK? So this is the guy that I want to solve. I'll let you digest that and set up the calculation. We've got T00 minus 1/2 T00 T. This is going to be, plugging in these bits that I worked out on the other board, here's my T00. Just make sure I did that right earlier-- I did. OK. So this is my right-hand side of my field equation. It will actually be sufficient for our purposes to neglect this term, OK? We'll see why in just a moment. So plugging that in, I need to work out the 00 component of my Ricci. So I go back to its foundational definition. This is what I get when I take the trace on indices 1 and 3 of the Riemann tensor. I can simplify that to just doing the trace over the spatial indices, because the term I'm leaving out is the one that is of the form 00 here, which by the anti-symmetry, on exchange of those indices, must vanish. Plugging in my definition, what I find is it is going to look like this here. So I'm just going to neglect the order of gamma squared term because I'm working in a limit where I assume that all these h's are small. This is going to vanish because of my assumption of everything being static in this limit. So this, I then go and plug in my definitions. OK. Again, I'm going to lose these two derivatives by the assumption of things being static. And pardon me just a second-- yeah, so I'm going to lose these two because of the assumption of things being static. The only derivative-- the only term that's going to matter, the derivative here is h. And so when I hit it with the inverse metric, this becomes simply the derivative of the h00 piece, OK? I can go from g straight to eta because the correction to this is of order h squared, which as I've repeatedly emphasized, we're going to neglect. All right, we're almost there. Let me put this board up, I want to keep this. OK, where was I? So I've got it down to here, let me just simplify this one step more. Eta is-- if mu is not spatial, then this is just 0. So I can neatly change my mu derivative into a j. I can just focus on the spatial piece of it. So this tells me R00 is minus 1/2 Kronecker delta delta i delta j acting on h00. This operator is nothing more than-- it's a Laplace operator. So this is minus 1/2, our old-fashioned, happily, well-known from undergrad studies Laplace operator on h00. So putting all this together, my field equation, which I wrote in this form, reduces down to del squared h00 equals minus kappa rho. The Newtonian limit that we did for the equation of motion, the fact that we showed that geodesics correspond to this, that already led me to deduce that h00 was equal to minus 2 phi. My Newtonian field equation requires me to have the Laplace operator acting on the potential phi, giving me 4 pi g rho. Put all these pieces together, and what we see is this proposed field equation works perfectly provided we choose for that constant. Kappa equals 8 pi j. And so we finally get g mu nu equals 8 pi g t mu nu. This is known as the Einstein field equation. So before I do a few more things with it, let us pause and just sort of take stock of what went into this calculation. We have a ton of mathematical tools that we have developed that allow us to just to describe the behavior of curved manifolds and the motion of bodies in a moving curved manifolds. We didn't yet have a tool telling us how the spacetime metric can be specified, OK? We didn't have the equivalent of the Newtonian fuel equation that told me how gravity arises from a source. So what we did was we looked at the geodesic equation, we went into a limit where things deviated just a little bit from flat spacetime, and we required objects be moving non-relativistically so that their spatial four velocity components were all small. That told us that we were able to reproduce the Newtonian equation of motion if h00, the little deviation of spacetime from flat spacetime in the 00 piece, was equal to negative 2 times the Newtonian potential. We then said, well, the Newtonian field equation is sort of sick from a relativistic perspective because it is working with a particular component of a tensor rather than with a tensor. So let's just ask ourselves, how can we promote this to a properly constructed tensorial equation? So we insisted the right-hand side be t mu nu. And then we looked for something that looks like two derivatives of the potential, or, more properly, two derivatives of the metric which is going to give me a curvature tensor, and say, OK, I want a two-index curvature tensor on the left-hand side. Since stress energy tensor is divergence-free, I am forced to choose a character tensor that is divergence-free, and that's what leads me to this object, and there's the Einstein curvature. And then insisting that that procedure reproduce the Newtonian limit when things sort of deviate very slightly from flat spacetime, that insisted the constant proportionality between the two sides be 8 pi j. This, in a nutshell, is how Einstein derived this equation originally when it was published in 1915. When I first went through this exercise and really appreciated this, I was struck by what a clever guy he was. And it is worth noting that the mathematics for doing this was very foreign to Einstein at that time. There's a reason there's a 10-year gap between his papers on special relativity and his presentation of the field equations of general relativity. Special relativity was 1905, field equation was 1915. He was spending most of those intervening 10 years learning all the math that we have been studying for the past six or seven weeks, OK? So we kind of have the luxury of knowing what path to take. And so we were able to sort of pick out the most important bits so that we could sort of-- we knew where we wanted to go. He had to learn all this stuff from scratch, and he worked with quite a few mathematicians to learn all these pieces. Having said that, though, it did strike me this is a somewhat ad hoc kind of a derivation. When you look at this, you might sort of think, well, could we not-- might there not be other things I could put on either the left-hand side or the right-hand side that would still respect the Newtonian limit? And indeed, we can add any divergence-free tensor onto-- depending how you count it-- either the left-hand side or the right-hand side-- let's say the left-hand side-- and we would still have a good field equation. Einstein himself was the first one to note this. Here's an example of such a divergence-free tensor. The metric itself, OK? The metric is compatible with the covariant derivative. Any covariant derivative of the metric is 0. And so I can just put the metric over here, that's perfectly fine. Now the dimensions are a little bit off, so we have to insert a constant of proportionality to make everything come out right. This lambda is known as the cosmological constant. Now what's kind of interesting is that one can write down the Einstein field equations in this way, but you could just as easily take that lambda g mu nu and move it onto the right-hand side and think of this additional term as a particularly special source of stress energy. Let's do that. So let's define t mu nu lambda equal negative lambda over 8 pi g times g mu nu. If we do that, we then just have g mu nu equals 8 pi gt mu nu with a particular contribution to our t mu nu being this cosmological constant term. When we do this, what you see is that t mu nu is nothing more than a perfect fluid with rho equals 8 pi g in the freely-falling frame, pressure of negative lambda over 8 pi j. Such a stress energy tensor actually arises in quantum field theories. This represents a form of zero-point energy in the vacuum. You basically need to look for something that is a stress energy tensor that is isotopic and invariant to Lorentz transformations and the local Lorentz frame, and that uniquely picks out a stress energy tensor that is proportional to the metric in the freely-falling frame. So this is an argument that was originally noted by Yakov Zeldovich. Whoops. And much of this stuff was considered to be kind of a curiosity for years until cosmological observations-- we haven't done cosmology yet. We will do this in a couple of weeks-- a couple of lectures, I should say. And it turns out that the large-scale structure of our universe seems to support the existence potentially of there being a cosmological constant. So the behavior of all these things is a lot more relevant, it's been a lot more relevant over the past, say, 15 or 20 years than it was when I originally learned the subject in 1993. So I want to just conclude this lecture with a couple of remarks about things that are commonly set equal to 1 when we are doing calculations of this point. So one often sets G equal to 1 as well as c equal to 1. Carroll's textbook does not-- several other modern textbooks do not-- I personally like for pedagogical purposes leaving the G in there, because it is very useful for calling out-- helping to understand the way in which different terms sort of couple in. It can-- if nothing else, it serves as a very useful order counting parameter, something that we'll see in some of the future calculations that we do. But there's a reason why one often works with G equal to 1 in many relativity analyses. Fundamentally, this is because gravity is a very weak force. G is the most poorly known of all of the fundamental constants of nature. I think it's only known-- I forget the number right now, but it's known to about five or six digits. Contrast this with things like the intrinsic magnetic moment of the electron, which is known to something like 13 digits. What this sort of means is that because G is so poorly known-- well, let me just write that out in words first. So G is itself poorly known. And so when we measure the properties of various large objects using gravity, we typically find that something like G times an object's mass is measured much better than M alone. Basically, the observable that one is probing is G times M. To get M out of that, you take G times M and you divide by the value of G that you have determined independently. If you only know this guy to five or six digits, you're only going to know this guy to five or six digits. Whereas, for instance, for our sun, GM is known to about nine digits, maybe even 10 digits now. When you set both G and c to 1, what you find is that mass, time, and length all come out having the same dimension. And what that means is that certain factors of G and c can be combined to become convergent factors. So in particular, the combination over c squared, it converts a normal mass-- let's say an SI mass-- into a length. A very useful one is G times the mass of the sun over c squared is 1.47 kilometers. GM over c cubed, OK? You're going to divide by another factor of velocity. This takes mass to time. So a similar one, G mass sun over c cubed, this is 4.92 times 10 to minus 6 seconds. One more before I conclude this lecture. (I can't erase this equation, it's too beautiful!) If I do g over c to the fourth, this converts energy to length. I'm not going to give the numeric value of this, but I'm going to make a comment about this. So bear in mind that your typical component of T mu nu has the dimensional form energy per unit volume-- i.e., energy over length cubed. So if I take G over c to the fourth times T mu nu, that is going to give me a length over a length cubed-- in other words, 1 over length squared, which is exactly what you get for curvature. So when one writes the Einstein field equations, if you leave your G's and your c's in there, the correct coupling factor between your Einstein tensor and your stress energy tensor is actually 8 pi G over c to the fourth. And I just want to leave you with the observation that G is a pretty small constant, c to the fourth is a rather large constant, and so we are getting a tiny amount of curvature from a tremendous amount of stress energy. Spacetime is hard to bend.
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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DEEPTO CHAKRABARTY: Now that we've seen how to describe a rotating vector, we can use that to analyze the motion of our gyroscope. So again, I'll draw a side view of my pivot, my rod. Here's the wheel. I'll call this point S. That's a distance d. And we'll assume that the angular velocity is in that direction, so that the spin angle velocity vector is pointing outwards in the plus r hat direction. That's the k hat direction and theta hat is into the screen. Now, again, the weight is acting downwards at the center of mass of the wheel. There's a normal force acting upwards. So the spin angular momentum with respect to point S is just equal to the moment of inertia of the disk about its center of mass times the angular speed of the spin, and that's directed in the plus r hat direction. And this is the angular momentum with respect to point S. Now, the torque with respect to point S, again, it's R cross F relative to point S. That's going to be Mgd in the plus theta hat direction. And for a fast omega, if omega is a large angular speed, and therefore if the angular momentum vector is a large vector, then the torque, which acts perpendicular to the angular momentum, will cause the angular momentum vector to rotate. And so as a rotating vector, we can write that the magnitude of the time derivative of that rotating vector is equal to the angular velocity of the rotation times the length of that vector. And this is Ls. And so the length of that vector is just I times little omega, and then we multiply that by capital omega. Now, this quantity here, dL dT, is the torque. So that's the magnitude of the torque. But the torque we said is equal to Mgd in the theta hat direction. And so the torque, Mgd, is equal to I little omega times big omega. So I can solve for big omega. And big omega, which is the angular speed of rotation of the angular momentum vector, is Mgd divided by moment of inertia, I, times little omega. Now recall, little omega is the angular velocity, the angular speed of the spin of the disk or the wheel. Capital omega is the angular speed of the rotation of the angular momentum vector. It basically tells us the speed at which the center of mass of the wheel orbits around the vertical axis through the pivot. We call capital omega the precessional angular velocity. So the precessional angular velocity is capital omega. And notice that the faster little omega is, the bigger little omega is, the slower the precession angular velocity is. Now, this expression tells us what the magnitude of capital omega is, what the magnitude of that precessional angular velocity is, but it doesn't tell us which way the system is processing, whether say, viewed from the top, the motion is clockwise or counterclockwise, or equivalently, which way the vector, capital omega, is pointing. We know it must point along the vertical axis, but does it point upwards, in the plus K hat direction or downwards in the minus K hat direction. To see that, we need to look at which way the angular momentum vector is rotating. And there are two possibilities. So let's again go to a top view of our gyroscope. So suppose here is the pivot point, and here is my gyroscope. And the way I'm drawing things is that r hat is this way, is the top view, so theta hat is that way, and k hat is out of the screen. So in the example that I did earlier, the angular momentum was pointing in the plus r hat direction. So that's L sub s pointing that way. If the torque is pointing in the theta hat direction, then that's going to act to rotate the angular momentum vector this way. And so the sense of rotation will be like that. And so in that case, looking down on the system, we would see a counterclockwise rotation. And that's equivalent to omega vector pointing in the plus k hat direction. Alternatively, suppose the wheel were spinning in the other direction. OK, so the sense of rotation of this wheel around the axle, we're in the opposite direction. In that case, even when the wheel was on this side of the pivot --again, this is a top view-- the angular momentum vector would be pointing in the opposite direction. So now this is my angular momentum vector. It's pointing in the opposite direction that the axle is pointing. Still along a line, but in the minus R hat direction. But the torque would still be in the theta hat direction. And so my new angular momentum vector would rotate this way, which would be equivalent to the axle rotating in this direction. And that's equivalent to a rotation in the opposite sense, clockwise as viewed downward from the top, or in other words, with capital omega hat vector pointing in the minus K hat direction. So the direction of precession depends upon which way the wheel is spinning, which way the spin angular momentum vector is actually pointing. Now, I'd just like to point out that there's an approximation that we've been making here. I've alluded to it, but I want to make it very specific right now. We have been assuming that the spin angular momentum vector L is large enough that the torque vector, which is perpendicular, provides only a small angular impulse that rotates L without changing it in length. Now, as L rotates, the direction of r hat and the direction of theta rotate with it, and the torque is always in the theta hat direction. So after any small delta T, when the angular momentum vector rotates, the instantaneous torque at that next instant is still perpendicular to the rotating momentum vector. And so we're making an approximation that L is large enough that we can consider the instantaneous angular impulse as a small perpendicular perturbation. That's equivalent, it turns out, to saying that, so this is the approximation that we're making, that we've been making so far, in this vector, is that little omega, the spin angular velocity, is much, much larger than capital omega, the precession angular velocity. This is called the gyroscopic approximation. We'll see a more precise statement of it later on. But it's basically equivalent to saying that the spin angular momentum vector is so large that we can consider the angular impulse due to the torque as causing a pure rotation of the vector without any change in its length. It's important to keep in mind that a vector can include both a rotating component and a constant component. And in that case, it's important to identify what the rotating component is in order to use the analysis that we presented earlier. So for example, let's consider the case of a tilted gyroscope, instead of one that's horizontal and parallel to the ground. So here is my pivot point. That's the vertical. And instead of a horizontal gyroscope, now I'll draw my gyroscope at some angle like this. Here's my wheel. I'll call that angle with the vertical phi. This distance is still d. And I'll use the usual coordinate system, that r hat is pointing out this way, k hat is pointing vertically, and theta hat is pointing into the screen. Now, in this case, the gyroscope will still process around the vertical axis to the pivot point, and the angle phi will remain constant. Now, the angular momentum vector due to spin points now not in the r hat direction, but again, outward along the axis of rotation. And that angular momentum vector can be decomposed into two components, an r hat component and a k hat component. So in other words, my angular momentum vector can be written as the sum of a vector pointing along Z axis or the k hat direction plus a vector pointing in the radial direction. I can also write that as L sub z times k hat plus L sub r times r hat. Now, notice as this gyroscope precesses around, the z hat component is constant, but the r hat component rotates around. OK, so the r hat component, the vector along the r hat direction, is a purely rotating vector, whereas the z component is a constant vector. So L sub z is constant and L sub r is rotating. And so now, the magnitude of dL dT, which is equal to the magnitude of dLz dT plus dLr dT, well dLz dT, since that's a constant vector, is just 0. So the time derivative just involves the r component. And so that's equal to the angular speed of rotation times the length of the rotating vector, which is just the r component, not the z component. So this is multiplied by L sub r. Now, in this particular case, the r component of the angular momentum vector is L, the angular momentum vector, times the sine of phi. So in this particular case, this would be omega times the full L times the sine of phi. So in the case that we have here, the angular momentum vector consists of a constant part and a rotating part. And the magnitude of its time derivative is equal to the angular velocity rotation times the length of the rotating part of the vector.
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https://ocw.mit.edu/courses/7-012-introduction-to-biology-fall-2004/7.012-fall-2004.zip
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I actually would like to take this occasion to praise and thank the TAs. We've been teaching this course for about a dozen years and we've had some really excellent groups of TAs, but this year's crop is really off scale, really outstanding. And we're all very grateful. [APPLAUSE] You know their names. I won't go through them all. But the fact of the matter is these TAs teach this course because they're told to teach the course, and at the same time they're doing all their thesis research, so they end spending about 168 hours a week on various kinds of work. So it's not a natural thing for them to spend an enormous amount of time, as they have been this year, really just off scale extraordinarily good. Here they are, Winston, Susan, Michelle, Sara, Divia, Jim, Sydney, Yasmine and Cha. So thank you all. At the same time, I'd also like to thank Claudette Gardel who runs this thing. This is a large undertaking, believe it or not, with almost 350 students enrolled, but it's moved seamlessly and without any problems this year for which many of us are very grateful. So, thank you Claudette. I'm going to spend today trying to broadcast into the future, forecast into the future I should say, about where all of what we've talked about this semester is taking us. Where is it going to get it us in terms of where we're going to be ten or twenty years from now and what you're going to be in the middle of ten or twenty years from now as you begin to move, as it will happen surely, as day follows night, into midlife. Imagine that. When you're 35 or 40 this will happen. And in order to do so I just want to go back 125 years or so to give you a feeling for what the history of biology has been like, some of it, since the end of the 19th century. Just to give you a little flavor for what can happen to biology if things aren't done right. Charles Darwin had a cousin named Francis Galton. He was knighted by the Queen and called Sir Francis Galton. And he was an early pioneer in statistics. And he coined the term eugenics. And, as you may note, eugenics is simply the science of trying to use genetics to bread better livestock, better plants, and ultimately maybe to breed better human beings. And we humans have been doing eugenics on plants and livestock for at least 10,000 years. That is to say we have continually been selecting out the best of the breed as the progenitors of the next generation of the breed. And in that way corn, which was originally this large when it was grown 5,000 years ago in Mexico, the cobs have now become this large and quite tasty. And all of that is through selective breeding. But in the last half of the 19th century, inspired by Darwin and subsequently by Mendel's work on Mendelian genetics, a whole science of eugenics grew up in this country which included not only the improvement in the quality of livestock and plants but also improvements in the gene pool of humanity. There was a strong conviction that genes were directly responsible for all kinds of physical traits, as well as mental and psychological traits. There was a strong belief that some races were superior and other races were inferior because of genetic gifts or genetic deficits. And this included as well within races, however one defined them, different ethnic groups. There was a firm belief that science could ultimately solve a lot of social problems including urban violence, labor unrest, manic depression, schizophrenia and even mental retardation. And the eugenicists, as they came to be known, came to believe that the problems of the world, alcoholism, poverty, prostitution, criminality, feeblemindedness, chess playing ability, tendency to commit industrial sabotage, that was big in the beginning of the 20th century when the unions were coming into power, they were all associated with one or another rather penetrant Mendelian allele. A well known geneticist named Davenport, who subsequently was associated with an unnamed university up Mass Ave. studied various ethnic groups and races and concluded that on the basis of genetics the Germans ranked highest in quality such as leadership, humor, generosity, sympathy and loyalty. Italians and Irish ranked lowest in most of these traits, he was lucky he survived in this town since together Italians and Irish, I think encompass 70% of the population. British were lowest in two of the traits. Irish were highest in suspiciousness. Jews were highest in obtrusiveness, whatever that is. And all of these things were said to be genetically templated. And so at the beginning of World War I IQ tests were first instituted in this country during the draft in order to determine who was genetically fit to serve and who was below standard. And using IQ tests, which were implemented in great numbers and throughout the society in the 1920s, a well known geneticist named Goddard discovered that 80% of Jewish, Hungarian and Polish immigrants, as well as Italian and Russian immigrants were mentally defective or feebleminded, and that these traits, these mental defects in 80% of these groups were transmitted as regularly and as surely as the color of hair or eyes. I'm not making up fairy tales now. I'm telling you what's happened in our history. In the 1920s there was a Eugenics Record Office in this country which existed for the next twenty years, an American Eugenics Society which has 1200 people, and J.H. Kellogg of Battle Creek, Michigan, you know how he made his money, don't you? Cornflakes. He founded a Race Betterment Association whose intent was to better the gene pool of the American population through selective breeding. By 1928 there were 376 college courses taught across this country on the subject of eugenics, i.e., how to improve the human race by beginning to breed more fit individuals. And so when the Nazis came to power, as they did in 1933 in Germany, they had much to draw from. In fact, most of their scientific rationale, to the extent they had any, didn't come from Germany, it came from eugenicists in the United States. And so they began to look amongst their society for people who were useless eaters, i.e., they consumed food but didn't produce, they lived lives not worth living, the elderly, the chronic poor, the crippled and the misfits. And they began involuntary sterilization. So, people who were regarded as genetically less fit were sterilized. By the time the Nazis finished their 12 years in power, 400,000 people had been sterilized in Germany because for one or another reason they were regarded as somehow defective. And as the Second World War made resources more tight, they just did something much simpler, they just euthanized people, they just killed them if they were regarded as in one way or another genetically defective. In fact, my father had a first cousin who around 1936 or so was gassed because he had a bad stuttering defect. So, this is all things that really happened to people. As a consequence of all of this eugenics, by 1924 Immigration Act was passed which severely circumscribed the amount of people immigrating into this country because the immigrants were widely viewed as diluting and contaminating the American gene pool. And this probably, well, this undoubtedly had a devastating affect on this country which we'll never really be able to know because the truth of the matter is that to the extent we have economic and scientific robustness in this country, it has come, for the last century, year after year, generation after generation from the immigrants who come to this country, not people who were here three, four, five generations. It's the immigrants who brought in the new ideas, the energy, the power, and I venture to say that if I were to ask what fraction of you are first generation Americans the number would be pretty high, right? But in 1924 that was for a while stopped simply because people coming into the country were viewed as genetically less than acceptable. By 1940 thirty states had compulsory sterilization laws in this country, i.e., people who were deemed to be genetically less gifted were sterilized against their will. 60,000 of those sterilizations were performed in this country. And the eugenics moment gained more and more adherence. What shut it off ultimately was what happened in World War II where six million Jews were killed, along with probably five or six million Slavs and other races who were deemed, and gypsies, there were probably half a million gypsies killed by the Nazis, different groups of people who were deemed to be genetically less deserving of living and genetically less likely to be productive and useful human beings. And were it not for World War II, it's quite plausible that the eugenics movement would have continued to grow and that today, when we talk about genetics, much of it would be referred to a belief that somehow we can determine people's phenotype and genotype and that we can predict how useful or useless they're going to be on the basis of our insights into genetics. And this ideology of genetic determinism, I say it had a great decline, this is the phrase we use, genetic determinism, i.e., to say that an individual's life course is strongly dictated by his or her genome. These are her alleles. You heard a lot about the alleles last time from Eric. But genetic determinism is once again coming to the forefront. Why? Because now, for the first time, we actually have a science of human genetics. When all of this other stuff was going on 50 and 100 years ago it was all pseudo-science, it was all made up. No one had the vaguest idea what genes were present in people's DNA. They didn't even know about DNA. They didn't really know about most Mendelian traits being passed in human populations. And they had no way of knowing, in the vast majority of cases, whether a certain person's phenotype was or was not dictated by genotype. So is this notion of a strong genotype-phenotype connection totally nonsense? Well, I'll give you an example of where you might begin to think it isn't. And it comes from studies of identical twins who were separated at birth and brought up in different families. So these identical twins obviously have an identical genotype. So here's a famous story that I like to refer to. There was a chance meeting in 1979 between a steelworker named Jim Lewis and a clerical worker named Jim Springer. They both lived in Ohio. They were separated five weeks after birth and they were raised 80 miles apart in different towns in Ohio. And at the age of 39 they discovered themselves through some change meeting. They discovered each other. Well, they both had dark hair, they both stool six feet tall and they both weighed 180 pounds. That's not so surprising. They both spoke with the same inflections, which they clearly had not yet learned to speak with when they were five weeks old. They walked with the same gait. They made the same gestures. They both loved stockcar racing. They both hated baseball. They both married women named Linda. They were both divorced and in their second marriages both of them married women named Betty. They both drove Chevrolets. They drank Miller Lite. They both chain smoked Salems. They vacationed on the same half-mile of beach in Florida. They both had elevated blood pressure, severe migraines, both had undergone vasectomies, they both bit their nails, and their heart rates, their brainwaves and their IQs were so similar that you couldn't tell whether it was the same person or two separate people being studied. Now, what do you begin to think of all that? Well, that's an extreme case. The fact is most identical twins raised apart do have a bit of divergence in their phenotype, in the way they grow up, but it begins to plant in your mind the notion that maybe many aspects of the way we think and act actually have a strong genetic template in them. And one can begin to study identical twins and ask things about, especially those who are separated at birth, and not use such extreme anecdotes like the one I just used. And one begins to find that there's an impressive list of attributes that can only be explained by their being a strong genetic determinant in them. And these traits include being alienated by people around one, extroverted, being a traditionalist, looking backwards in terms of one's customs, leadership, career choice, risk aversion, attention deficit disorder, religious conviction and vulnerability to stress. Heritability it turns out, if you study identical twins, is about, I'm sorry, happiness, if you study identical twins, is about 80% heritable it turns out and depends little on one's wealth, achievement or marital status. But 80% of it, if you study identical twins, seems to have a genetic template. And you'll say, well, that's all very satisfying, but it begins to be a little unsettling because it begins to cause each of us to ask are we really who we think we are or are we just kind of cassette recorders who are playing out the program that was stuffed into us when the sperm hit the egg that lead to each of our appearing on the face of the planet? To what extent are we individuals or to what extent are we simply manifestations of genotype? And to what extent do we have freewill? That's kind of an interesting question. Now, people like Eric, I'm not pointing an accusing finger, people like Eric have begun to refine the science of genetics so it really is a science. And so, restriction fragment polymorphisms, SNPs, haplotype analysis are now uncovering a staggering array of human traits. I believe that the number of human traits that have now been localized, specific genes, most of these are diseased genes, exceeds 2,000 is my recollection. And there are only 21,000, 22,000 genes in the human genome. And the pace with which genes and genotype and phenotype will be linked to one another is going to increase if nothing else. Many of the traits that one thinks about in terms of human beings are obviously polygenic. They're not single strong Mendelian alleles with strong penitence. They represent the confluence, the collaboration of multiple alleles that are conspiring to create one or another phenotype. And these polygenic traits or even polygenic diseases have traditionally resisted analysis because mathematically they are so complex to dissect out, to dissect out the contributing genes which together as a cohort create a genotype. But, as Eric told you last time, people like you who are great software developers will one day begin to figure out how one can take extraordinarily complex datasets and begin to associate specific chromosomal regions, and ultimately genes, with specific genetic sequences that contribute to a polygenic trait. I think at one time Eric spent, about three or four years ago, he worked with people at Cornell studying the polygenic trade of ripening in tomatoes. It's a polygenic trait like probably chess playing ability in human beings. And was able to localize ripening rate of tomatoes to five or six distinct genetic regions in the chromosomes of the tomato plant. But that's only a harbinger of what could come. So let's imagine now, again, I'm not blaming Eric for this, I'm just telling you he's the one, he more than anyone else almost on the planet is the person who is leading the charge to refine and strengthen these extraordinarily powerful tools that enable us to discern how our genome creates us the way we are. But he's not going to be the one who applies these tools. They'll be applied all over the planet. There are geneticists everywhere who are interested in looking at how different aspects of human phenotype, including disease phenotype are governed by the alleles, by the SNPs, by the polymorphisms that we carry, and obviously by the genes and proteins that we make. So let's begin to imagine, let's put ourselves fast-forward ten years and begin to imagine where this is going to take us. We already know about a very substantial number of genes that determine the risk of different kinds of cancer, i.e., there's at least 15 different cancer syndromes that people have which have been associated with specific genetic loci. I talked briefly about retinal blastoma, which is a rare one, but even commonly occurring cancers will soon be connected with specific alleles in the genome. And the risk of getting them in one's lifetime will be relatively accurately predictable. It might take another decade but it will happen. Manic depressiveness, some people have great swings in mood. 2% or 3% of the population doesn't wake up happy every morning. And this is also, I believe, going to yield two specific analyses and association with certain genes. There's already a suggestion that the D4 dopamine receptor, which is involved in receiving one of the neurotransmitters in the brain, may have a polymorphism that's connected with manic depressiveness. There will be probably alleles which are connected with, in some way, novelty or adventure seeking. There are going to be alleles that are associated with anxiety, probably maybe connected with the serotonin transporter in the brain. Cardiac disease susceptibility is already mapped out in a number of traits in the most extreme cases, but cardiac disease is very frequent in this population. And there undoubtedly will be alleles that are discovered that determine whether one has a high risk or low risk of getting heart disease, of getting arthrosclerosis, and whether or not one can go to McDonald's every day and Big Macs with impunity. Can one do that or not? Some people probably can. Some people can eat as much salt as they want and it doesn't give them high blood pressure. Other people cannot. We still don't really understand that. Schizophrenia is probably also very strongly genetically templated, not totally but very strongly. Susceptibility to rheumatoid arthritis probably also has a strong genetic component. Difficulty or ease with which you solve math problems probably also will one day be associated with a certain number of genetic loci. How many difficulties in learning languages? There's already a trait that was discovered in a family in the Netherlands, I believe, and they had a very specific grammatical defect in the way that they assembled the syntax of sentences associated with a certain allele of a certain gene. Difficulty in just adding rows of numbers may also be associated with certain combinations of alleles. Now, you will say, well, it's impossible, it's inconceivable that these different aspects of cognitive function can be associated with a small number of genes. But let me tell you something else. We talked a week ago about the evolution of humanity over the last couple hundred thousand years. And the pace with which the human brain has evolved over the last half million years, and more recently the last 200, 00 years, has been so frighteningly rapid that the evolution of cognitive function and perception in different ways can only have happened through the actions of a small number of genes. If one needed to have dozens of genes change in concert in order to acquire the penetrating minds that we now have in which our ancestors 500,000 years didn't have, the evolution could not have occurred so quickly. And, for that reason alone, one begins to suspect that the genetic differences between people who lived 500, 00 years ago visa vie their cognitive function and ours are not so large. And, therefore, a rather small number of genes may have been responsible for conferring on us the powerful minds which we now, which most of us, I didn't say anything, which most of us now possess. So where is this going to take us? What are the consequences of this? Let's imagine ten or twenty years down the road when we can do some kind of SNP analysis on one of these chips that have been developed in California and here and in various places. And we can begin to imagine the allelic diversity in a newborn child's DNA or even prenatally if you want. So what are you going to do if you begin to find on a chip of a child's DNA that this kid is likely to be very good in language, probably is going to have poor math skills, will be a rather anxious and obsessive person, will have difficulty associating with his or her peers, and is likely to come down with heart disease at the age of 45? How is that going to affect your relationship to that person, that child? And will you give that child a different kind of education than a newborn who has SNPs which indicate that without doubt they're going to get 1600s on their morning boards and their shoe-ins for admission to MIT? Are you going to treat those kids the same or are you going to treat them differently? Do you give them the same kind of education and nurturing? And how do you treat them throughout their elementary and high school? Are you going to segregate them into different groups or is everybody going to be given an equal chance? Well, you might say it's our tradition in this country to give everybody equal footing, in part because of a reaction to what happened in World War II in no small part. But what if the time comes when people say we need to be more efficient economically in this country and we need to devote our resources, need to maximize the investment, the benefit we get from various investments, and so it's much more efficient to put kids in a certain genetic class in one school and kids who have another level of genetic giftedness in another school? Of course much of this will be foolhardy because all of these genetic tests, although they will give you probabilities of certain phenotypes, they'll never, at least for the foreseeable generation or two give you certainties. No one will be able to predict with absolute total certainty about the potentials of one or another young person on the basis of DNA tests, at least not in the near future. Right now one can predict with total certainty that somebody who has a certain allele will come down with Huntington's disease at the age of 30 or 40 or 50. There the predictability, the penitence is 100%. But what if somebody has an allele that says with 60% likelihood they're not going to very good at math? Is that already going to be enough to justify their segregation amongst a group of the mathematically less gifted? Let's say that they've gone through elementary and secondary education and high school and they've made it through college and they start looking for employment. Actually, there are jobs out here to be had in this economy, you wouldn't know it, but there actually are people who can find jobs. And let's say one has now an employer who is evaluating a certain job candidate for employability. Maybe they'd like to have a good medical checkup before they employ this person ostensively to see whether this person is healthy enough to last for ten or twenty or thirty years of employment. And maybe they'd like to include among that medical exam that person's DNA just in case. And what if the DNA tells the employer that this individual is likely to get colon cancer in 18 years and has a slight susceptibility to mood instability and perhaps even manic depressiveness, that this person is not one of those who can go to McDonald's and eat Big Macs with impunity but has a tendency to arthrosclerosis? You can think of whatever possibilities you will. Will that be, therefore, a ground to reject that person as an employee? Well, you'll say they really have no right to do that. But keep in mind that with increasing frequency in this society medical benefits, medical insurance is paid by the employer. So does the employer want to have a whole workforce of people who are in various stages of terminal disease or would this employer like to be able to pay the lowest possible health benefits because the employer has taken care to employ only people who have a really terrific genotype, whatever that is defined as being arbitrarily admittedly? And what about marriagability? As my sister always says to me, if you want to marry a man the first thing you should do is look into his genes. That's a double entendre. Anyhow. See, somebody finally got it. [LAUGHTER] The fact is that maybe certain people will be deemed to be less desirable genetically. Well, the fact is we've been doing that for the last million years. If you're attracted to someone and you end up marrying them then they have phenotypes which you think are in one way or another valuable. Consciously or unconsciously, you are practicing a form of eugenics. But obviously there could be a much more subtle form of eugenics where part of the marriage contract states that you want a sample of that person's buccal swab or some lymphocytes to check out what kind of DNA he or she has. Now you say, well, that could never happen. But it happens today regularly. There are villages in Greece where there are a substantial percentage of people who carry the trait sickle cell anemia which, as you may know, is not so serious phenotypically in heterozygous form, but in the homozygous form is actually devastating. And the reason they have sickle cell anemia is that those areas of Greece historically had high rates of malaria. And, as you may know, sickle cell anemia actually protects, in the heterozygous state actually protects one from the ravages of the malarial parasite. So about 20 years ago it became possible to do a simple genetic test to determine whether an individual was heterozygous for sickle cell anemia. And what happened is that somehow what was supposed to be confidential medical genetic tests got out. They became public. And young individuals in the population became known as carrier, as heterozygotes for sickle cell trait, even though phenotypically they were reasonably normal because the heterozygote condition is not so devastating. So those individuals were soon ostracized, to use an old Greek word. They were soon put to the side. They were placed in the pool of the unmarriable because nobody wanted to marry them. And so they then, as a consequence, began to marry amongst themselves. Remember what I told you about homozygosis with the sickle cell trait. But that's only one example of that. Among Orthodox Jews, among Ashkenazi Jews between 2% and 3% of the population carries an allele for Tay-Sachs disease which is phenotypically silent in the heterozygous state but in the homozygous state is a devastating condition which leads to death in the first years of life. So now among the Orthodox Jews in New York before two young people will get married they will do a test to see whether they are heterozygous for the Ta Sacks allele. And, in fact, it's not limited any longer to Orthodox Jews. Because if they're both heterozygotes their marriage to one another, in spite of anything else they consider, is strongly discouraged. Among those who don't live in such a closely structured society such people might nonetheless decide to get married and then face the devastating possibility of one of their four offspring on average coming out as a homozygote and having an incurable genetic disease which is going to lead to their early death. But what about other traits? And what time will this genetic discrimination, where will it begin and where will it end? What if you find an individual who has a trait of manic depressiveness among relatives? And when will these genetic tests become public? When will they be private knowledge? You say, well, they can all be kept private. But ultimately there are already insurance companies which are demanding to determine whether and individual can be insured by looking at whether they have genes for certain kinds of disease-causing alleles. After all, why should they insure somebody, give somebody life insurance if they are likely to come down with Huntington's disease at the age between 35 and 40, which will surely and inevitably lead them to an early grave? You'll say, well, we cannot have genetic information like that become public or even become accessible. Maybe that's a solution. The problem is we've been talking about these issues for 10 to 15 years in this society, and we've not yet converged any kind of solution. And the solutions to these problems should not be left in the hands of molecular biologists, because molecular biologists or biologically cognizant people by now, like you, are no more gifted and no more insightful to deal with these issues than anyone else is. They're intuitively obvious these issues. You don't need to know about SNPs to begin to understand the potentially devastating impacts that the misuse of genetics can have on our society. And what happens if one of these days people discover alleles for certain aspects of cognitive function? Chess playing ability. The ability to learn five different languages. The ability to remember strings of numbers. The ability to speak extemporaneously in front of a class, for what it's worth, for 50 minutes several times a week. Whatever ability you want, valued or not so valued, what if those alleles begin to come out? And here's the worse part. What if somebody begins to look for the frequency of those alleles in different ethnic groups scattered across this planet? Now, you will say to me, well, God has made all his children equal. But the fact is if you look at the details of human evolution, some of which I discussed with you a week ago, last week, you'll come to realize that most populations in humanity are the modern descendents of very small founder groups. Remember about the story of the Fins. 70% of Finish men carry the same Y chromosome. All modern Fins, most modern Fins, all of them are likely to be the descendents of a small founder group that came into Finland 2,000 or 3, 00 years ago and carry with them the peculiar set of polymorphisms that founder group happens to have had. And arguments like that begin to persuade you that there'll be different allele frequencies in different populations of humanity. What if somebody begins to discover that Macedonians have an enormously high rate of the ability to play chess because of a certain allele? And here I'm talking very speculatively. I'm not literally meaning that. And Tibetans have a very poor ability to construct software programs because of a genetic allele they carry? I hope nobody's Tibetan here. I tried to choose two. Are there any Macedonians? All right. I succeeded. All right. Anyhow. So the fact is it's inescapable that different alleles are going to be present with different frequencies in different inbreeding populations of humanity or populations of humanity that traditionally have been genetically isolated from one another. It's not as if all the genes that we carry have been mixed with everybody else's genes freely over the last 100,000 years. Different groups have breed separately and have, for reasons that I've told you, founder affects and genetic drift acquired different sets and different constellations of alleles. So what's going to happen then, I ask you without wishing to hear an answer because nobody really knows? Then for the first time there could be a racism which is based not on some kind of virulent ideology, not based on some kind of kooky versions of genetics, because the eugenicists in the beginning of the 20th century, as well as the Nazis hadn't had any idea about genetics, they were just using the word, even though they knew nothing about the science of genetics as we understand it today. But what happens if now for the first time we, i.e., you who begin to understand genetics, begin to perceive that there are, in fact, different populations of humanity that are endowed with different constellation of alleles that we imagine are more or less desirable? What's going to happen then? I don't know. But some scientists say, well, the truth must come out and that everything that can be learned should be learned, and we will learn how to digest it and we will learn how to live with that. But I'm not so sure that's the right thing. And you all have to wrestle with that as well. And even more insidious is the following notion. Remember the story about the two Jims, the two guys from Ohio who met one another at the age of 39 after they'd been separated at five weeks of birth? That story begins to persuade you of something I said before, and that is that a lot of what you think you are isn't what you made of yourself, isn't what your parents made of yourself, isn't what your environment made of you and your experiences. Maybe it's all just in your genes. And if that's so then maybe you can't take credit for any of the good things you've done. And conversely maybe you're not responsible for all the bad things you've done. Maybe three years from now somebody will begin to plead that even though they were not criminally insane when they committed a string of serial murders, in fact it really wasn't their fault because they happen to have this particular genotype which is known to be correlated with a strong tendency to violent. And, by the way, there is an allele which has a correlation, I forget which one it is, has a correlation with violent behavior. So what if one begins to write off everything we do as not a reflection of our own freewill, our own volition, but instead a consequence of the genes which our parents hoisted on us? Of course, we can blame it on them. As a father of children, I can tell you that it's amazing how many different things can be blamed on the parents. [LAUGHTER] Of course, the parents have their own out. The parents can blame it on their parents. So now it goes back to the grandparents, back to the beginning of time. We laugh about these things and they are amusing, but they are taking us on a collision course with some very difficult problems. And you guys have to wrestle with them and you guys have to explain to the people who haven't taken 7. 12 where the world of biology is taking us. And on that note, I want to tell you that Eric and I have enormously enjoyed being with you this semester. We wish you much luck and success in your future lives. We hope some of you have become interested in biology and that you found this course a little different from what you took in high school. And have a wonderful winter vacation. See you. [APPLAUSE]
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https://ocw.mit.edu/courses/5-61-physical-chemistry-fall-2017/5.61-fall-2017.zip
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FEMALE SPEAKER: The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So we're now starting to get into a really familiar territory, which is the hydrogen atom. And it's a short step from a hydrogen atom to molecules. And we're chemists. We make molecules. But one question that would be a legitimate question to ask is, what does a hydrogen atom have to do with molecules, because it's an atom and it's the simplest atom. And what I hope to show is that what we learn from looking at the hydrogen atom has all sorts of good non textbook stuff that prepares you for understanding stuff in molecules. And I am going to stress that because you're not going to see it in McQuarrie and you've probably never seen it elsewhere except in freshman chemistry when they expect that you understand the periodic table based on some simple concepts. And they come from the hydrogen atom. And so I'm going to attempt to make those connections. My last lecture was I told you something you're not going to be tested on. But it's about spectroscopy. That's how we learned almost everything we know about small molecules and a lot of stuff about big molecules too. The crucial approximations in that were the dipole approximation where the radiation field is such that the molecule field's a uniform but oscillating electromagnetic field. In developing the theory that I presented, we assume that only one of the initial eigenstates is populated at t equal zero. And only one final eigenstate gets tickled so that it gets populated. And so even though there is an infinity of states, the theory specifies we start with one definite one and only one final state gets selected because of resonance. And this is what we do all the time. We have an infinite dimension problem, and we discover that we only really need to worry about a very small number of states. And then we deal with that. There are a couple other assumptions that were essential for this first step into a time-dependent Hamiltonian. And that is the field is weak and so you get linear response. That means that the increase in the mixing coefficient for the final state is proportional to the coupling matrix element times time. Now mixing coefficients can't get larger than 1, and so clearly we can't use a simple theory without saying, OK, we don't care about the mixing coefficients. We care about the rate of increase. And so by going from amplitude to rates, we are able to have a theory that's generally applicable. Now what I did was to talk about a CW radiation field, continuous. Many experiments use pulsed fields. Many experiments use an extremely strong pulsed field. That I hope to revisit later in the course. But the CW probably is the best way to begin. So now let's talk about the hydrogen atom. We did the rigid rotor, and that led to some angular momenta. Well, let's just picture. So we have this rigid rotor, and it's rotating about the center of mass. And the Schrodinger equation tells us the probability amplitudes of the rotor axis relative to the laboratory frame. And we got angular momenta, which we can denote by a little l, big L, J, and many other things. And this is important because, if it's an angular momentum, you know it. You know everything about it. And you don't know yet that, if you have two angular momenta, you could know everything about the interactions between them. That's called the Wigner-Eckart theorem. That's not in this course. But the important thing is, if you've got angular momenta, you're on solid ground and you can do stuff that doesn't ever need to be repeated. And so one of the things that we learned about the angular momentum was that we have-- even though it's going to be L as the angular momentum mostly in the hydrogen atom, I'm going to switch to J, which is my favorite notation. So J squared operating on a state-- now I can denote it like this or I can denote it like this. It's the same general idea. These are not quite equivalent things. And you get H bar squared J A plus one JM. And we have JZ operating on JM. And you get H bar M JM. And we can have J plus minus operating on JM. And that's H bar times this more complicated looking thing, J plus 1 minus M, M plus or minus 1. So that's the only bit you have to remember. And you might say, well, is it plus or minus 1 or minus or plus 1? And so if you choose-- M is equal to J. Then you know that if you have M is equal to J, this is going to be 0 if we're doing a raising operator. And so there's nothing to remember if you are willing to go to an extreme situation and decide whether it's plus minus or minus or plus. Now the rigid rotor is a universal problem, which is solved. It deals with all central force problems. Everything that's round, the rigid rotor is a fantastic starting point. So one-electron atoms for sure-- many-electron atoms, maybe. We'll see. But anything that's round, the rigid rotor is a good zero point for dealing with the angular part of the problem. So for the hydrogen atom, we have a potential. It looks like this. And so the radial potential is new. For that rigid rotor, the radial potential was-- it's just 0 for the R is equal to R zero. And not zero-- then the potential is infinite. Now here we have potential, which doesn't go to infinity here and it does something strange here because you can't get to negative R in spherical polar coordinates. So is it a boundary condition or is it just an accident of the way we use coordinates? So we're going to apply what we know about angular momenta to the hydrogen atom. So our potential for the hydrogen atom is going to be expressed in terms of the distance of the electron from the nucleus and then the theta phi coordinates, which you already know. And the theta phi part is universal, and the R part is special to each problem that has physical symmetry. And there are different kinds of approximations you use to be able to deal with them. The nice thing about it is it's one dimensional, and we're very good at thinking about one dimensional problems. And even if the problem isn't exactly one dimensional or it has some hidden stuff to it, we can extend what we know from one dimensional problems and get a great deal of insight. And if you have a one dimensional problem, it's very easy to describe the potential in one dimension and the eigen functions in one dimension. And so we can begin to really understand everything. So we're going to have a wave function, which is going to have quantum numbers. And we're going to be able to write it as a product of two parts. Well, this is the same thing we had before for the rigid rotor. So we're going to be able to take this rotary equation and separate the wave function into two parts, a radial part and an angular part. And this is old and this is new. Because we've got this, we can use all of that stuff without taking a breath, except maybe getting the right letter L, J, S, whatever. So really the hydrogen atom is just one thing with some curve balls thrown at you in the latter stages. One of the things you also want to be able to do-- because nodes are so important in determining both the names of the states and how they behave in various situations, including external fields and excitation by electromagnetic radiation-- you want to be able to understand the nodal surfaces. And you already know this one. So how many nodal surfaces are there if you have a particular value of L? Yes. AUDIENCE: L. PROFESSOR: Right, and if you have a particular value of M, how many nodes are there in the xy plane? You're hot-- AUDIENCE: I mean, I know if M is 0, the entire angular momentum is tipped into the xy plane. So that means that the axis of the rotor has to be orthogonal to the angular momentum. So that means that the probability density is oriented along Z, I think. PROFESSOR: Now you're saying things that I have to stop and think about because you're not telling me what I expected to know. So if M is equal to 0, L is perpendicular to the quantization axis, and there are no nodes. AUDIENCE: Well, it depends on if you're talking about the nodes of the probability density of the rotor, where the axis is. PROFESSOR: That's true, but where the axis is determined by theta. And when theta is equal to pi over 2, you're in the xy plane. And the number of nodes in the xy plane is M, or absolute value of M. The phi part of the rigid rotor is simple. It's a differential equation that everybody can solve. We already know that one, and we know what the wave functions look like. We can write the hydrogen atom Schrodinger equation, and it's very quick to show separation of variables. And I'll do that in a minute. And then we have the pictures of the separated parts-- RNL of R and YLM of theta phi. And essential in these pictures is the number of nodes. In the spacing between nodes, remember the semiclassical approximation. We know that Mr. DeBroglie really hit it out of the park by saying that the wavelength is H over P. For every one dimensional problem, we know what to do with that. And we're going to discover that, for the radial problem, we have a very simple way of determining what the classical momentum is. And so we know everything about the nodes and the node spacing and the amplitudes between nodes and how to evaluate every integral of some power of R or Z. And so there's just an enormous amount gotten from the semiclassical picture once you are familiar with this. And so we can say that we have the classical wavelength, or the semiclassical wavelength. It has an index R. So we have a momentum, a linear momentum with a quantum number on it for L. See, that's a little strange. But that tells us what the potential is going to be and that tells us what to use in order to determine the wavelength. And this is really the core of how we can go way beyond textbooks. We can do-- in our heads or on a simple piece of paper, we can do this. And we can draw pictures, and we can evaluate matrix elements. Without any complicated integral tables, you can make estimates that are incredibly important. And from that you can get expectation values and also off-diagonal matrix elements of integer powers of the coordinate. That's an enormous amount of stuff that you can do. Now you can't do it yet. But after Wednesday's lecture, you will. And so this will be Wednesday. And then we'll have evidence of electron spin. And I will get to that today. So this is the menu. Now let's start delivering some of this stuff. I have to write some big equations. So the Hamiltonian is kinetic energy plus potential energy, and kinetic energy was-- I'm sorry. For the rigid rotor, V was zero. And everything was in the kinetic energy. Well, it's not quite true. And so, for hydrogen-- so we know this is P squared over 2 mu, and mu for the hydrogen atom, reduced mass. And we know this is just Coulomb's Law, minus e squared over 4 pi epsilon zero R, H bar squared. I'm sorry-- it was supposed to be e squared. So this is the classical-- So these are the parts. But the since it's spherical, we're not going to work in Cartesian coordinates. And in fact, that's a very strong statement. If you have a spherical problem, don't start in Cartesian coordinates because it's a horrible mess transforming to spherical polar coordinates. Just remember the spherical polar coordinates. So you know how spherical polar coordinates work. I'm not going to draw it. So the kinetic energy term is-- and this is the Laplacian, and that's a terrible thing. And Del squared-- it looks like it's going to be a real nightmare partial with respect to R, R squared partial with respect to R. And we have 1 over R squared sine squared theta partial with respect to theta, sine theta partial with respect to theta. And then a third term-- 1 over R squared sine squared. I got a square here. That's wrong. And this is sine squared theta. I heard a mumble over there. Yeah, but I'm just doing Del squared. AUDIENCE: With your first Del, it's H plus squared. PROFESSOR: My what? AUDIENCE: Your first time you used the Del operator, it's H plus squared. PROFESSOR: Oh, yes, yes. AUDIENCE: And then the second one, if you could clarify that that's a Del-- PROFESSOR: I'm sorry. AUDIENCE: The second Del squared, if you could clarify that's a Del squared. It looks like a-- PROFESSOR: It looks like a terrible thing. And the last part is a second derivative with respect to phi. This looks like a terrible thing to build on. But with a little bit of trickery-- and that is, suppose we multiply this equation by R squared-- then we have killed the R squared terms here and here. And we're going to be able to separate it. And so we are able to write an equation which has the separability built in-- and so partial with respect to R, R squared partial with respect to R plus L squared plus 2 mu H R squared V of R minus E psi. That's a Schrodinger equation. So we have an R dependent term and another R dependent term and a theta phi dependent term all in this one, nice operator that we've understood. So now there's one more trick. In order for the shorter equation to be separated into a theta phi part and an R part is we need a commutator. And so that commutator is this. What is a commutator between L squared and any function of R? Yes. AUDIENCE: Their operators depend on different variables? PROFESSOR: Absolutely, that's really important. We often encounter operators that depend on different variables. And when they do, they commute with each other, which is an incredibly convenient thing because that means we can set up the problem as a product of the eigenfunctions of the different operators. So this means that we can write a theta phi term, which we completely know, and an R term, which we don't know and contains all of the interesting stuff. So let's forget about theta and phi. Let's just look at the R part. Well, the way we separated variables, we had the Schrodinger equation and we divided by the wave function, which would be of R of R, Y L M of theta phi. And on one side of the equality, we have only the angle part. And so when we do that, when we divide by R, we kill the R part on this side. And on the other side, it's the opposite. And so we get two pieces-- one is only dependent on theta and phi and one is only dependent on R. So they both have to be a constant. So we get two separate differential equations. And we've already dealt with one of them. So what you end up getting is 1 over R of R times stuff times R of R is equal to 1 over YLM L squared YLM-- sorry. So this is the separation. This is all R stuff. This is all theta stuff and separation constant. Well, this one is inviting a separation concept because L squared operating on YLM is H bar squared, LL plus 1. That's the separation constant. So now let's look for the only time at the R part of the differential equation. And so this writing out all the pieces honestly-- we have-- there's one more. So that's the Schrodinger equation for the radial part. It looks a little bit annoying. The important trick is that we've taken the separation constant, and it has an R dependence but when we divide through by 2 mu HR squared. And we say, oh, well, let's call these two things together-- VL of R. This is the effect of potential, and it depends on the value of L. So this is just like an ordinary one dimensional problem except now, for every value of L, we have a different potential. And this potential is-- OK, when L is not equal to zero, this potential goes to infinity at R equals 0, which is bad, except it's good because it keeps the particle ever from getting close to the nucleus. And that's what Mr. Schrodinger-- I'm sorry what Mr. Bohr was thinking about, that we have only circular orbits or we have only orbits that are away from the place where the Coulomb interaction would be infinite. And I mean, there are several things that's wrong with Bohr's picture. One is that we have orbits. And the other is that we don't include L equals 0, but this is telling you a lot of really important stuff because the value of L determines the importance of this thing that goes to infinity at R equals 0. And that has very significant consequences as far as which orbital angular momentum states we're dealing with. So we're going to get the usual LML quantum numbers, and we're going to get another one from the radial part. And since this is a 1D equation, there's only one quantum number. And we're going to call it L. And now there's this word principal, and there's really two words-- principle with an LE and principal with an AL. Which one do you think is appropriate? I'm sorry. I can't hear. AUDIENCE: AL. PROFESSOR: AL is the appropriate one. Principle has to do with something fundamental. Principal has to do with something that's important. And I can't tell you how many times people who should know better use the wrong principal. You'll never do that now because it was like, what's nu? C over lambda. So this is the right principal. If we do something clever and we have the radial part and we say, let us replace it by 1 over R times this new function chi L of R, well, when we do that, this equation becomes really simple. So what we get when we make that substitution is H bar squared over 2 mu H second derivative with respect to R plus VL of R minus E chi L of R is equal to 0. That looks like a differential equation we've seen before. It's a simple one dimensional differential equation-- kinetic energy, potential energy. But it's not. There's some kinetic energy hidden in the potential energy. But this is simple, and we can deal with this a lot. I'm not going to. But if you're going to actually do stuff, you're going to be wanting to look at this differential equation. But I'm going to forego that pleasure. One of the problems with the radial equation is the fact that R is a special kind of coordinate. It can't go negative. And so treating the boundary condition for R is equal to 0 is a little subtle. And it turns out that when L is equal to 0, then R of 0 is not 0. But for all other values of L, R of 0 is 0. And all of NMR depends on L being 0 because the electron feels the nucleus. It doesn't experience an infinite singularity. It feels the nucleus. And when L is not equal to 0, then it's sensing the nucleus at a distance. And that leads to some very small splittings called hyperfine. And so there's different kinds of hyperfine structure. But anyway, this is a really subtle and important point. How much time left? Yes? No. So now pictures of orbitals-- So the problem with pictures of orbitals is now we have a function of three variables and it's equal to some complex number. So we need two degrees of freedom to present a complex number and we have three variables. And so representing that on a two dimensional sheet of paper is horrible. But we have this wonderful factorization where we have RNL of R. And we could draw that easily. We don't need any special skill. And we have YLM of theta phi. And we've got lots of practice with that, although what we've practiced with may not be completely understood yet. So we have ways of representing these. And so we go through the understanding of the hydrogen atom by looking at these two things separately. Now for the radial part, the energy levels-- where this is the Rydberg constant. It's a combination or fundamental constants. And for hydrogen, the Rydberg constant is equal to 109,737.319-- there's actually more digits, wave numbers-- times mu H over mu infinite. Well, this is actually the Rydberg constant for an infinite mass. And so mu H is equal to the mass of the electron times the mass of the proton over the mass of the electron plus the mass of the proton. Now the mass of the proton is much bigger than the mass of the electron. And so you can use this as a trick. You can say, well, what is it? Well, we know what this is. It's easy to calculate. But there are two limits. What is the smallest possible reduced mass? And that would be for-- if the mass of the proton is infinite, then we just get the mass of the electron. That's the biggest reduced mass. And the smallest is, if we have positronium, where we have an electron bound to a proton-- and then when we do that, we get half. I'm sorry. It's not a proton. It's a positively charged particle, which we call a positron. And so each of the terms here is the mass of the electron. And so the range is from 1/2 me to me depending on what particles you're dealing with. And so that's a useful thing. And so the Rydberg constant for hydrogen is smaller than the reduced mass of the infinite. I'm sorry. It's smaller than that for the infinitely mass nucleus. And it is the value that made Mr. Bohr very happy-- 679. So the important thing is this mass scaled, or reduced mass scaled, Rydberg constant explains to a part in 10 to the 10th all the energy levels of one-electron systems-- hydrogen, helium plus, lithium 2 plus. That's it. And that's fantastic. Now we have to talk to become really familiar with this Rnl of our function. And one of the questions is, how many radial nodes. And you know that for 1S there aren't any nodes. And you know for 2P there aren't any radial nodes. And so what we need is something that goes like n minus L minus 1. The number of radial nodes, which is all you need to know about the radial A function, is how many nodes are and how far are they apart and what's the amplitude of each loop between nodes. Semi classical theory gives you all of that. And often when you're calculating an integral, all you care about is the amplitude in the first loop. And so instead of having to evaluate an integral, you just figure out what is the envelope function based on the classical momentum function. Now the thing that everybody's been waiting for. Spin. So remember, we know an angular momentum is R cross P, and we know that there isn't any internal structure. Or at least we don't know about internal structure of the electron or a proton. And so we can't somehow say, well, it's R cross P. So we have the Zeeman effect. So we can look at the Zeeman effect for an atom in a magnetic field. And we have several things that we know. So we have the magnetic moment of the electron is equal to minus the charge on the electron times 2ME times L. Well, that's an angular momentum. So if we had circulating charge, well, that circulating charge will produce a magnetic moment, which has a magnitude which is related to the velocity. And what is L divided by mass? It's a velocity. And we have the charge here. So this is a perfectly reasonable thing. And that's good because we know that, if we have electrons with some kind of internal structure, we know what this is going to do. Now the magnetic potential is equal to minus the magnetic moment times the external magnetic field. And so what we have is E, BC, LZ over 2 NE. Well, this is another very good thing because, not only do we know what this is, we know that it has only diagonal elements. And so we can do first order perturbation theory. If this is so easy, if this is the Hamiltonian, we can just tack that on and it adds an extra splitting to our energy levels. The only tricky thing is, we don't know that it's LZ. We just know that it has a magnetic moment or it could have a magnetic moment. And so this wonderful experiment-- suppose we start with a 1S state, and you go to a 2P state. Now this implies that we know something about the selection rules for electromagnetic transitions. And so for an electric dipole transition, we go from 1S to 2P. Now this is not how it was done initially because the frequency of this transition is well into the vacuum ultraviolet. And in the days when quantum mechanics was being developed, that was a hard experiment. So one used an S to P transition on some other atom, like mercury, but let's pretend it was hydrogen. So this is an angular momentum of 0. So we would expect it would be ML equals 0. And here we have it could split into three components. And this is an L equals 1, 0 minus 1, because that's minus-- there with a minus sign somewhere. That's what we expect. And so being naive, you might expect transitions like this. Well, the transitions where you do not change the projection quantum number are done with Z polarized radiation. And so if it's Z polarized, you get only delta ML equals 0. And if you have x and y, you have delta ML equals plus and minus 1. So depending on how the experiment was done, you would expect to see one, two, or three Zeeman components. And they did the experiment, and they saw more than three components. They saw five components. And that's possibly for a number of reasons. But they saw five components. So they knew that there was something else going on. And so you have to try to collect enough information to have a simple minded picture that will explain it all. So one thing you might do is say, OK, suppose there is another quantum number, another thing. And we're going to call it spin. And we don't know whether spin is integer or half integer. We know from our exercise with computation rules that both half integer and integer angular momentum are possible. And so we can say we have a spin. And it could be 1/2. It could be 1. It could be anything. And then we start looking at the details of what we observe. And what we find is, all we need is spin 1/2 to account for almost everything. However, if it were spin 1/2, then you get two here. And you'd get two here, two, here and two here. And that's six. That's bigger than five. So you need to know something else. And one thing is quite reasonable-- you can say, well, the spin thing is mysterious, and electromagnetic radiation acts on the spatial coordinates. And there aren't any spatial coordinates of the internal structure of the electron. And so we have a selection rule delta MS equals 0. That still doesn't do it. You still need something else, and that is that the proportionality concept between the angular momentum and the energy, which is called the G factor. For orbital angular matter, the G factor is 1, or the proportionality constant is 1. So the G factor is L. And the G factor for the electron turns out to be 2-- not exactly 2, just a little more than 2-- Nobel Prize for that. And so with those extra little things, then every detail of the Zeeman effect is understood. And we say, oh, well, the electron has a spin of 1/2, and it like it acts like an angular momentum. It obeys the angular momentum computation rules. It also obeys the computation rules. Well, I won't say that. So I should stop-- there are other things that provide us information that there is something else. And one, I'm called by some people the spin orbit kid because I've made a whole lot of mileage on using spin orbits splittings. And the spin orbit Hamiltonian has the form L.S And this gives rise to splittings of a doublet P state, for example, as what you would see in the excited state of hydrogen into two components at zero field. And so there's all sorts of really wonderful stuff. And I'm going to cheat you out of almost all of it because we're going to go over to the semiclassical picture next time. And we'll understand everything about Rydberg states and about how do we estimate everything having to do with the radial part of the wave function. And there are some astonishing things.
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https://ocw.mit.edu/courses/7-016-introductory-biology-fall-2018/7.016-fall-2018.zip
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ADAM MARTIN: And so I wanted to start today's lecture by continuing what we were talking about in the last lecture. So I'm just going to hide this real quick. And so we're talking about the fruit fly and the white gene and the white mutant, which results in white-eyed flies. And we talked about how if you take females that have red eyes and cross them to males, the white-eyed male, then 100% of the progeny has red eyes in the F1 generation. And so I asked you guys, would you get the same results if you did the reciprocal cross? So what if we took white-eyed females and mated them to red-eyed males? So what about this? Actually, I'm going to move this over to over here so that maybe it's more visible. So what if we have white-eyed females and crossed this to red-eyed males? So let's unpack this sort of a little bit at a time. So what's the genotype of these white-eyed females here? Miles? AUDIENCE: So if you designate the eye gene as the letter A, a female would be X lowercase a, X lowercase a. ADAM MARTIN: Yes. So Miles is exactly right. So the dominant phenotype is red eyes, because the gene encodes for an enzyme that's important for the production of the red pigment. And so X lowercase a here would be a recessive mutant that lacks the pigment. And because it's a recessive allele-- because you need only one copy of this gene to produce the pigment. So the recessive allele results in the white phenotype. Therefore, this has to be homozygous recessive. How about this red-eyed male? Yeah, Ory? AUDIENCE: Wouldn't you have a Y and then an X capital A? ADAM MARTIN: Yes. So this would be this phenotype, right, where capital A is the gene that produces-- is a normal functioning gene that produces the pigment. So then in your F1 here, are you going to see something similar to this or something different? AUDIENCE: Something different. ADAM MARTIN: Different, great. Who said different? Javier? Do you want to propose what you might see? AUDIENCE: Yeah. For the males, they're going to inherit the Y gene from the father and the [INAUDIBLE].. ADAM MARTIN: Exactly. So the males are going to get the Y from the father, and they're going to get one X from their mother. So all the males are going to be of this genotype here, which means they're going to have what color eye? Javier is exactly right. That means they're going to have white eyes. So all the males will have white eyes. And what about the females? AUDIENCE: [INAUDIBLE] ADAM MARTIN: What's that? AUDIENCE: Red eyes. ADAM MARTIN: Yeah. So Ory is saying the males are going to get red eyes, right, because they're-- or the females are going to have red eyes, because they're going to get the X chromosome from their father, which has the dominant gene that produces the red pigment. So all the females are going to be heterozygous, but have a functional copy of this gene. So all of the females will have red eyes. OK, does everyone see how-- now, how would this compare with Mendel's crosses and pea color? Would there be a difference in Mendel's crosses if you switch the male versus the female if these were autosomal traits? Ory is shaking his head no, and he's right, right? In that case, it doesn't matter. You can do the reciprocal crosses, you get the same result. But because this is sex linked, which one is the male and which is the female is relevant. And this actually relates to something that we just saw on the MIT news. I just got this email this morning, but it came out I think yesterday, which is that biology-related research in the mechanical engineering department-- specifically, the CAM lab-- they've been able to design a 3D sort of model for ALS disease, which is also known as Lou Gehrig's disease. And so what they've done in the CAM lab is to take cells from either patients that have ALS or from normal individuals, and they coax these cells to become neurons. Here, you're seeing a neuron in blue and green here. And you see the neurites extend from this neuron. And they have a model where this neuron can then synapse with a muscle. And so they're using this 3D sort of tissue model to model ALS and to look for drugs that might affect ALS, potentially curing ALS. And so last night, I started reading about ALS and was pleased to find that there's actually a very rare X-linked, dominant form of the disease that can be passed on from generation to generation. And the inheritance pattern of this X-linked, dominant version of ALS would have an inheritance pattern that's similar to what we observe for the white mutant in the fruit fly, right? Whereas, if you have an affected father, and this is a dominant mutant on the X chromosome, then all of his daughters will get that X chromosome and be affected, whereas the sons will all be unaffected. However, if you have the reciprocal situation, where you have an affected mother and an unaffected father, then the sons and daughters get the disease randomly. So this is a sort of form of inheritance, which is relevant if you're considering human disease and some forms of it. Most versions of ALS, well, are sporadic, but inherited forms are usually autosomal dominant. So this is a rare case here. But I thought it was interesting in that it's relevant to what we've been talking about. So now, just to recap-- here, I'll throw this up so everything's up. So in the last lecture, we talked about Mendelian inheritance. And we talked about when you take two parents that differ in two traits and you perform a cross, you get a hybrid individual that is heterozygous for both genes. And now this is the F1 individual. Let's say we want to know what types of gametes this F1 individual produces. We can perform a type of a cross known as a test cross, where we cross this individual to another individual that is homozygous recessive for both these genes, which means that you know exactly which alleles are coming from this parent. And they're both recessive, so you can see whether or not the gamete produced by this individual has either the dominant or the recessive allele. Let me see. I'll boost this up. So now we can consider the different types of progeny that result from this test cross. And some will have the two dominant alleles from this parent and will, therefore, be heterozygous for the A and B gene. And it would exhibit the dominant A and B phenotype. I think that is what I'm showing here. So if the chromosomes, during meiosis I, align like this, then the two dominant alleles segregate together, and you get AB gametes. And you also, reciprocally, get these lowercase a and b gametes, as well. So that's the other class here. So you can get these two classes of progeny. And the phenotypes of these two classes will resemble the parents, right? So these are known as parental gametes. So these are the parentals. But you know because Mendel showed that if you have genes and their alleles on separate chromosomes, they can assert independently of each other. So an alternatively likely scenario is that the chromosomes align like this, where now the dominant allele of B is on the other side of the spindle. And therefore, these chromosomes are going to segregate like this during the first meiotic division. And that gives rise to gametes that have a different combination of alleles than the parents. So you have some that look like this. So each of these would be different classes of progeny. And you have one last class that would look like this. And so neither of these look like the original parents, and so they're known as non-parental. And so if these two genes are behaving according to Mendel's second law, where there's independent assortment-- if you have independent assortment, what's going to be the ratio of parental to non-parental? Rachel? AUDIENCE: One to one. ADAM MARTIN: Yeah, Rachel says one to one, and I think a number of others also said one to one. So you have 50% parental, 50% non-parental, right? Because it's equally likely to get either of those alignments of the homologous chromosomes during meiosis I. So now I'm going to basically break the rules I just explained to you in the last lecture and tell you about an exception, which is known as linkage. Gesundheit. And in the abstract sense, linkage is simply when you have two traits that tend to be inherited together. So just considering probability. So you have traits inherited together. They're exhibiting what is known as linkage. But that's an abstract way to think about it. It's just based on probability, right? So a physical model for what linkage is, is that you have chromosomes. The genes are on the chromosomes. And for two genes to be linked, those genes are physically near each other on the chromosome. So the physical model is that two genes are near each other on the chromosome. OK, so let's consider again these generic genes, A and B. If A and B resulting from this cross-- if these two happen to be on the same chromosome, now they're physically coupled to each other. Then they're going to tend to be inherited together. No matter how these align, they're always going to go together during the first meiotic division. And that's just going to only give you the parental gametes. So if there's linkage, you're going to have-- let's consider the case where you have complete linkage. If you have complete linkage, 100% of the gametes are going to be parentals, and you're going to have 0% non-parental. That's if the genes are really, really, really close to each other, and maybe you don't count so many progeny. You won't see any mixing between the two. But there is a phenomenon that can separate these genes, and it's known as crossing over. And another term to describe it is recombination. So the alleles are getting recombined between the chromosomes. Recombination. And what crossing over or recombination is, is it's a mixing of the chromosomes, if you will. Or it's an exchange of DNA. So there's a physical exchange of DNA from one of the homologous chromosomes to the other, OK? So you can think of this as an exchange of DNA between the homologous chromosomes, OK? And that's important. It's not an exchange between the sister chromatids, but between the homologous chromosomes that have the different alleles. And what's shown here is a micrograph showing you a picture of the process of crossing over. You can see the centromeres are the dark structures there. And you can see how the homologous chromosomes intertwine. And there are regions where it looks like there's a cross. Those are the homologous chromosomes crossing over and exchanging DNA such that one part of that chromosome gets attached to the other centromere. So I'll just show you in sort of my silly cartoon form how this is, just to make it clear. So let's say, again, you have these A and B genes, and they're physically linked on the chromosome. During crossing over, you can get an exchange of these alleles, such as a bit of one chromosome goes to the other homologous chromosome and vice versa, OK? So now you have the dominant A allele with the recessive b allele and vice versa. So now during meiosis I-- after meiosis II, this will give rise to two types of gametes, one of which is non-parental. And the same for the one down here. You get two types of gametes. One is non-parental-- the lowercase a, uppercase B gamete. So this happens if there's incomplete linkage. That means there can be a recombination event that separates the two genes. And I'm going to give you an example of a case where data was collected with what fraction of each class there is. So now we're considering an example where you have a linkage. So A and B are on the same chromosome. And so we'll consider a case where, in this class, there are 165 members. For this one, there is 191. So I'm kind of-- line down like this. And then for the first recombinant class, 23 individuals. And for the last, there are 21 individuals. So you can see there are many more of the parental class than the recombinant class, but we can calculate a frequency, or recombination frequency, between these two genes. And in this case, the recombination frequency is 44/400, which is equal to 11%, OK? So 11% of the progeny from this cross had some sort of crossing over between the A and B alleles. It would've been up here. Now, this frequency is interesting, because it is proportional to the distance that separates these two genes. So this recombination frequency is proportional to the linear distance along the chromosome between the genes. Now, it also depends on the recombination frequency in a given organism or in a given part of the chromosome. So when you're comparing recombination frequencies between different organisms, there's actually differences in the real different-- they're not equivalent. You can't compare them, basically. And also, there are regions of the chromosome where recombination happens less frequently than others. And so, again, you can't compare distances along those. But overall, you can use this as a distance in order to map genes along the linear axis of a chromosome. And maps are useful, because you can see where stuff is, right? So in this example here, I'll highlight a couple of places. Here's Rivendell. Here's Lonely Mountain. Here's Beorn's house. So let's say we are able to determine the distance between Rivendell and Lonely Mountain, and the distance between Lonely Mountain and Beorn's house, and the distance between Rivendell and Beorn's house. You'd be able to get a relative picture of where all of these places are in relation to each other. So this is a two-dimensional map I'm showing here. It's not one dimensional, but chromosomes are one dimensional, so it's a bit more accurate, OK? So this idea that recombination frequency can be used to measure distances between genes and that this could be used to generate a map is an idea that an undergraduate had while working in Thomas Hunt Morgan's lab back in 1911. And what I find fascinating about the story is this guy basically blew off his homework to produce the first genetic map in any organism. So the person who did it was Alfred Sturtevant, and he was an undergraduate at Columbia working for Thomas Hunt Morgan. And I'll just paraphrase this quote here. In 1911, he was talking with his advisor, Morgan, and he realized that the variations in the strength of linkage attributed by Morgan to differences in the separation of genes-- so Morgan had already made this connection, that the recombination frequency reflects the distance between the genes. But then Sturtevant realized that this offered the possibility of determining sequences in the linear dimension of the chromosome between the genes, OK? So then-- this is my favorite part-- "I went home and spent most of the night, to the neglect of my undergraduate homework, in producing the first chromosome map." And this is it. So the first chromosome map was of the Drosophila X chromosome, which we've been talking about. There's the white gene, which we've been talking about in the context of eye color. There's a yellow body gene here. There's vermilion, miniature, rudimentary, right? These are all visible phenotypes that you can see in the fly. And you can measure recombination between various alleles of these different genes. All right, so now I want to go through with you an example of how you can make one of these genetic maps. And it's essentially the same conceptually as to what Sturtevant did. And it involves what is known as a three-point cross. So a three-point cross. So there are going to be three genes, all of which are going to be hybrid, and I'll start with the parental generation that is little a, capital B, capital D. And we'll cross this fly or organism to an organism that is capital A, lower case b, lowercase d. Yes, Carmen? AUDIENCE: So when you write the gametes up there, does that imply that they were analogous parents? ADAM MARTIN: So what I'm writing here is the phenotype, basically. And so these are homozygous for each of these, yes. I could also write this as-- but I'm not going to draw the chromosomes, because it kind of gets more confusing. I'll draw the chromosomes here in F1, because we have now, basically, a tri-hybrid with one chromosome that looks like this, right? They got that chromosome from this individual here. And another chromosome will look like this. See? So this F1 fly is heterozygous for these three genes, and it has these two parental chromosomes. So now we can look at the gametes that result from this fly by doing a test cross, just like we did before. And so we want to cross this to a fly that's homozygous recessive for each of these genes. And now we can look at the progeny. And just by looking at the phenotype, we're going to know the genotype, because we know all of the flies from this cross have a chromosome from this individual that has recessive alleles for each gene. So we can consider now this first one here. That's one potential class of progeny. Another class would be this one. And these two, you can see, resemble the parents, right? So these are the parental classes of the progeny. So this is parental. All right, now you can consider all other combinations of alleles. And so I'll quickly write them down. You could have something-- progeny that look like this and this. These are just kind of reciprocal from each other. You could have progeny that look like this and this. And the last class would be this and this. So all of these progeny that I drew down here are recombinant, because they don't resemble the parents, right? Because there are three genes, now there's many more ways to get recombinant progeny, as opposed to having just two genes, right? So you can have many different combinations of these different alleles. And so now I'm going to give you data from a cross with three such genes. So you might get 580 individuals that look like this, 592 like this, 45 and 40, 89, 94, 3, and 5. So this is data that is, I believe, from fly genes. I've just ignored the fly nomenclature, because it's confusing, and just given them lettered names, OK? But this reflects data from some cross somewhere. So now we want to know-- let's go back to our map. We want to make a map, OK? And so to make our map, we're going to want to consider all pairwise distances between different genes. So we'll start with the A and B gene. I'll write over here. So let's consider the A/B distance. And remember, to get a distance, we're looking at the number of the frequency which there is recombination between these two genes, OK? So we now have to look through all of these recombinant class of progeny and figure out the ones that have had a recombination between A and B, right? So on the parent chromosome, you see little lowercase a started out with capital B and vice versa. So any case where we don't have lowercase a paired with capital B, there's been some type of exchange. So here, lowercase a's with lowercase b. So that's a recombinant. Here, uppercase A's with uppercase B. That's a recombinant, too. So we have to add all these up. So 45 plus 40. How about here? Recombination here or no? Yes. I'm hearing yes. That's correct. Here, recombination, yes or no? Carmen's shaking her head no. She's exactly right. So we just have to-- these are all the recombinants between A and B. So it's 45 plus 40 plus 89 plus 94, which equals 268 over a total progeny of 1,448. And that gives you a map distance of 18.5%. Because this method was developed in Morgan's lab, this measurement is also known as a centimorgan. It was named in honor of Morgan. So that's what I refer to when I have lowercase c capital M. That's a centimorgan. So you can also use centimorgan here. All right, so that's A and B, but now we have to consider other distances. So how about the A/D distance? And again, we have to go through and figure out where the alleles for A and D have been recombined, OK? So little a is with capital D, and upper case A is with lowercase d. So we have to find all the cases where that's not the case. Here, this is lowercase a with capital D and capital A with lowercase d. This is parental from the respect of just the A and D genes, but all the rest of these guys are recombinants, OK? So this is 89 plus 94 plus 3 plus 5, which comes out to be 191 over 1,448. And this is 13.2 centimorgans. So that's the distance between A and D. Distance between A-B, distance between A-D. So the last combination, then, is just B and D. So if we consider the B/D distance, again, we have to look for all cases in which lowercase b and d become separated and uppercase B and D become separated. Here, they're separated. Here, they're separated. Wait, no, not here, sorry. Here, they're not separated. Here, they are separated. Everyone see how I'm doing this? Are there any questions about it? You can just shout it out if you have a question? So this distance is 6.4 centimorgans. Everyone see how I'm considering every pairwise combination of genes and then just ignoring the other one and looking for where there's been a recombination in the progeny? So now that we have our distances, we can make our map, right? So the two genes that are farthest apart are A and B, OK? So that's kind of like here, Rivendell and Lonely Mountain. Those are the two genes that are at the extremities. So I'll draw this out. It doesn't matter which way you put it. We're just mapping these genes relative to each other. But B and A are the farthest apart from each other. Now if we consider the distance between B and D, that's 6.4 centimorgans. So it appears that D is closer to B than it is to A, because it's 13 centimorgans away from A, OK? So D is kind of like Beorn's house here. It's closer to Rivendell than it is from Lonely Mountain. So we'll put that in there. This distance is 6.4 centimorgans. And then the distance here is 13.2 centimorgans. As far as I know, no one of the field of genetic mapping has been assaulted by large spiders, but the field is still young. So one thing that should be maybe bothering you right now is if you add up the distance between B and D and D and A, you don't, in fact, get 18.4 centimorgans. Instead, you get 19.6 centimorgans. This is 19.6 centimorgans. So it seems like, somehow, we are underestimating this distance here, OK? So we seem to be underestimating this. So why is it that we are underestimating this distance? Well, to consider that, you have to sort of look at how all of these classes were generated. So now I'm going to go through each class, and we'll look at how it was generated. I'm also going to-- well, I'll just draw new chromosomes. So we have to draw this order now. We have B, D, a. So the first chromosome is B, D, a. That's right, B, D, a. The other chromosome is b, d, A. So now we look and see how this recombinant class was generated. So this is lowercase a. We'll start with b-- lowercase b, lowercase a, but capital D. So it's capital D, lowercase a. So that recombinant results from this chromosome here, where there's crossing over, and little b gets hooked up to big D and little a. See how I did that? And then if we consider this class, this is uppercase B, lowercase d, capital A. So these two classes of progeny result from a single crossover between B and D, OK? So this is a single crossover between B and D genes. And now we can go through and look at how this is generated. So to get all recessive alleles on the same chromosome, there would be a crossover here. And so this is a single crossover between D and A. So a single crossover between D and A. Now, these last couple classes of progeny are interesting in that they're the least frequent class. And so when we consider how they're generated, we'll start with uppercase B. Let's see if I can get rid of this. So uppercase B, lowercase d, lowercase a. And so what this last class is, is actually a double crossover. So this is a double crossover. And it's least frequent because there's a lower probability of getting two crossovers in this region. But now you see that even though it doesn't look like there was recombination between A and D, in fact, there was. There were two crossovers, and it just looks like there was no recombination, if you didn't see the behavior of gene D. So if we take into account that there are actually double crossovers between B and A, then if we add that into our calculation here, where you add in 3 plus 5 multiplied by 2 because these are both double crossover events, then you get the 19.6 centimorgans that you would expect by adding up the other recombinations, OK? How is that? Is that clear to everybody? You're going to have to make maps like this on the problem set and possibly the test. So make sure you can given-- yeah, Ory? AUDIENCE: I realized that you immediately [INAUDIBLE] overestimated the difference between A and B and not overestimated B to D or D to A? ADAM MARTIN: It's because when you have two genes that are very far apart, you can have multiple crossovers. And when you have sort of crossovers that are in pairs of two, then it's going to go from one strand back to the other, and so you're not going to see a recombination between the two alleles. So it's an underestimate, because if you have multiples of two in terms of crossovers, you're going to miss the recombination events. You see what I mean? You understand that you can miss the double crossover events? AUDIENCE: Yeah, I get that. ADAM MARTIN: Yeah, right? So then you're going to underestimate the number of crossovers that actually happened in that genetic region. All right, now I want to end with an experiment that, again, makes the point that genes are these entities that are on chromosomes. So just like you can have linkage between two genes on the same chromosome, you can also have linkage between genes and physical structures on chromosomes, like the centromere. So you could have genes like A and B here that are present on the chromosomes and present very near the centromere of those chromosomes, OK? So they could be right on top of the centromere, OK? And to show you how this manifests itself, I have to tell you about another organism, which is a unicellular organism called yeast. And yeast is special and that it can exist in both a haploid and a diploid form. So it has a lifecycle that involves it going both as a haploid and as a diploid. And so you can take yeast-- and we'll take two haploid yeast cells. And much like gametes, these can fuse to form a zygote. So, in this case, I'm taking-- again, we'll consider two generic genes, A and B. And we'll make a diploid yeast cell that is heterozygous, or hybrid, for A and B. And what's great and special about yeast and why I'm telling you about this is because as opposed to flies and us and other organisms, the product of a single meiosis is packaged in this single package, if you will. So the yeast can undergo meiosis, and the product of a single meiosis is present in this case, where each of these would represent a haploid cell that then can divide and make many cells. But this is the product of a single meiosis in a package, OK? So you can actually see the direct result of a meiotic division, a single meiotic division. So this is the product of a single meiotic division. And that's special because when we make gametes, we have individual cells. All the products of meiosis are split up, and then just one randomly finds an egg and fertilizes it. So you don't know which of the gametes are from the product of a single meiosis. And so being able to see the product of a single meiosis allows us to see things like genes being linked to physical structures on the chromosome like the centromere. So if we consider this case, these two genes are both linked to the centromere. And during metaphor phase of meiosis I, they could align like this, in which case you would get spores that are parental for both dominant alleles or parental for both recessive alleles. So each of these cells is known as a spore. So I'll label spore numbers here. So this is spore number. And in this case, you get two spores that are dominant for both alleles and two spores that are recessive for both alleles. Because there are two types, it's known as a ditype. And this is a parental ditype, because you have two types of spores . And they are both parental An alternative scenario is that these chromosomes would align differently, right? So you get parental spores there. However, alternatively, you could have this configuration, where this is now flipped. And during meiosis I here, these chromosomes move together. And they, again, produce two types of spores, so it's a ditype. But in this case, all the spores are non-parental. So another scenario is you get this. And because there are two types and they're non-parental, this is known as non-parental ditype. That's a non-parental ditype. And if these genes are linked to the centromere completely, then you can only get these two classes of packages, OK? So if these genes are unlinked-- so the two genes are unlinked, but both linked to the centromere, then you get parental ditype-- 50% parental ditype, type 50% non-parental ditype. So I'm abbreviating parental ditype PD and non-parental ditype NPD. So what has to happen to get another type of spore? And another type of spore would be-- you could have spores that are all different genotypes from each other and that you have A cap dominant A/B; dominant A, recessive b; recessive a, dominant B; and lowercase a and b. And this is known as tetratype, because there are four types. So how do you get this tetratype? Anyone have an idea? Yeah, Jeremy? AUDIENCE: You're crossing over. So one of A and B would switch in one of the [INAUDIBLE].. ADAM MARTIN: Where would the crossing over happen? AUDIENCE: Between the two [INAUDIBLE] one [INAUDIBLE].. ADAM MARTIN: Between the allele and what? AUDIENCE: Sorry? ADAM MARTIN: The crossing over would occur between the gene-- AUDIENCE: Oh, and the centromere. ADAM MARTIN: And the centromere, exactly. Jeremy is exactly right. So Jeremy said that in order to get a tetratype, you have to have a recombination event, but this time, not between two genes, but between a gene and the centromere. So at least one of the genes has to be unlinked to the centromere. And in that case, now you get a meiotic event that gives rise to four spores. And there are four different ways to get this. So if you have two genes unlinked and at least one is unlinked to the centromere, then you get a pattern where you have a 1 to 1 to 4 ratio between all these different events. So you have a 1 to 1 to 4 ratio between parental ditype, non-parental ditype, and tetratype. And we can see this in yeast. If you have two genes that are linked to the centromere, you only get parental ditypes and non-parental ditypes, where virtually everything else gives rise to tetratypes, except if they're linked. What happens if the two genes are linked to each other, irregardless of the centromere? If you have two genes that are linked, what's going to be-- what are your progeny going to look like? AUDIENCE: Parentals. ADAM MARTIN: You're only going to get the parentals, or you're going to get a lot of parentals. Javier is exactly right, right? If the two genes are linked, the parental ditypes are going to be much greater than any of the other classes. Now, this might seem esoteric, but I like the idea that you can have linkage between a gene and something that's just the place on the chromosome that's getting physically pulled. It all makes it much more physical, which I think is nice to think about. All right, we're almost done. I just have-- yes, Natalie? AUDIENCE: Can you go over what the PD [INAUDIBLE]?? ADAM MARTIN: Yes. So PD is parental ditype. So this is the parental ditype. It's a class of product here where you get four spores that are each of these genotypes, OK? So each of these 1, 2, 3, and 4 would represent one of these cells from a single meiotic event. Does that make sense, Natalie? Does everyone see what I did there? So these 1 2 and 3 are the spores of the meiotic event right here.
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https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/8.06-spring-2018.zip
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PROFESSOR: Today we have to continue with our discussion of the hydrogen atom. We had derived or explained how you would derive the corrections to the original Hamiltonian, the Hamiltonian you've studied already in a couple of courses, this h0 Hamiltonian for the hydrogen atom that has a kinetic term and a potential term. This Hamilton we've studied for a while. And what we've said was that the Dirac equation provided a way to get systematically all the corrections, the first order of that Hamiltonian, and that's what we have here. That was the end product of that discussion in which the better Hamiltonian or the perturbed Hamiltonian including more effects was the original one plus a relativistic correction because this is not the relativistic kinetic energy. Then there was a spin orbit coupling in which you couple the spin of the electron to the orbital angular momentum of the vector. And finally, there is a Darwin term. It's a surprising term. This is Charles Gaston Darwin, a British physicist, that discovered this term. And all these terms were suppressed with respect to h0, in the sense that h0 had energies. The energies associated to h0 went like alpha squared mc squared. The fine structure constant, that's for h0 veera. But for h1 or for delta h, the energies go like alpha to the fourth mc squared, where m is the mass of the electron, and that's about 20,000, 19,000 times smaller. So this is our perturbation, and that's what you have to understand. So we'll begin with the Darwin term. Then we'll turn to the relativistic term, then to the spin orbit term, and, by the end of today's lecture, we'll put it all together. So that's our plan. So let's start with the Darwin term. What is this term? So Darwin term, I will try to evaluate it. So it depends on the potential energy. It has the Laplacian of the potential energy. So the potential energy is minus e squared over r. So how much is the Laplacian of the potential energy? Well, the Laplacian of v would be minus e squared times the Leplacian of 1 over r. And that's minus e squared times minus 4 pi delta function of r. That's something you study in in E and M. The Laplacian of 1 over r is related to a delta function. It's the charge density of a point particle that produces that potential. So our result here is that this Laplacian is e square 4 pi e squared delta of r. It's a delta function contribution. So let's write it here. The delta h Darwin is pi over 2 e squared h squared over m squared c squared delta of r. All right, so that's our correction. And we want to do perturbation theory with it, how it affects the energy levels of the various states of hydrogen. So how are they changed? What does it do to them? Now, there is a simplifying fact here, the delta function. So you remember, first order corrections are always obtained by taking the interactive Hamiltonian and putting states in it. And the first thing we notice is that unless the wave function of the states does not vanish at 0, the correction vanishes. So this will pick up the value of the wave functions. When you have two states, arbitrary state 1 and state 2 and delta h Darwin here-- there's two wave functions that you're going to put here if you're trying to compute matrix elements of the perturbation. And if either of these wave functions vanishes at the origin, that's not possible. But all wave functions vanish at the origin, unless the orbital angular momentum l is equal to 0. Remember, all wave functions go like r to the l near the origin. So for l equals 0, that goes like 1, a constant, and you get a possibility. So this only affects l equals 0 states. That's a great simplicity. Not only does that, but that's another simplification. Because at any energy level, l equals 0 states is just one of them. If you consider spin, there are two of them. Remember in our table of hydrogen atoms states go like that. So here are the l equals 0 states. And those are the only ones that matter. So the problem is really very simple. We don't have to consider the fact that there are other degenerate states here. We just need to focus on l equals 0 states because both states have to be l equals 0 and l equals 0. So our correction that we can call e1 Darwin for n 0 0-- because l is equal to 0, and therefore m is equal to 0-- would be equal to psi n zero zero delta h Darwin psi n 0 0. So what is this? We have this expression for delta h Darwin. So let's put it here, pi over 2 e squared h squared over m squared c squared. And this will pick up the values of the wave functions. This overlap means integral over all of space of this wave function complex conjugated, this wave function and delta h Darwin. So at the end of the day, due the delta function, it gives us just psi n 0 0 at the origin squared. That's all it is. Simple enough. Now, finding this number is not that easy. Because while the wave function here for l equals 0 is simple for the ground state, it already involves more and more complicated polynomials over here. And the value of the wave function at the origin requires that you normalize the wave function correctly. So if you have the wave function and you have not normalized it properly, how are you going to get the value of the wave function at the origin? This whole perturbation theory, we always assume we have an orthonormal basis. And indeed, if you change the normalization-- if the normalization was irrelevant here, this number would change with the normalization, and the correction would change. So you really have to be normalized for this to make sense. And finding this normalization is complicated. You could for a few problems maybe look look up tables and see the normalized wave function, what it is. But in general here, there is a method that can be used to find this normalization analytically. And it's something you will explore in the homework. So in the homework there is a way to do this, a very clever way, in which it actually turns out. So in the homework, you will see that the wave function at the origin for l equals 0 problems is proportional to the expectation value of the derivative of the potential with respect to r. So it's something you will do. This potential, of course, is the 1 over r potential in our case. And the derivative means that you need to evaluate the expectation value of 1 over r squared in this state. But the expectation values of 1 over r squared in the hydrogen atom is something you've already done in the previous homework. It's not that difficult. So the end result is that this term is calculable. And psi of n 0 0 at the origin squared is actually 1 over pi n cube a0 cube. With that in, one can substitute this value and get the Darwin correction. You can see there is no big calculation to be done. But you can rewrite it in terms of the fine structure constant Darwin. And it's equal to alpha to the fourth mc squared 1 over 2n cubed. And it's valid for l equals 0 states. So this number goes in here, and then write things in terms of the fine structure constant, and then out comes this resolved.
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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This week, we will continue our discussion of energy. We will introduce the concept of potential energy-- which is related to the work done by a conservative force, and also the concept of total mechanical energy-- which is defined as the sum of the kinetic and potential energies. We will demonstrate that the total mechanical energy of a system remains constant if only conservative forces are acting. This is known as the Principle of Conservation of Mechanical Energy. We will also see how the work done by any non-conservative forces causes the total mechanical energy of the system to change, usually causing mechanical energy to be lost from the system. Finally, we will see how graphs of a system's potential energy, as a function of position, provide an elegant way of understanding the behavior of the system for a given value of the total mechanical energy.
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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So suppose we have a point mass particle of mass m moving with a velocity vector, v. We can introduce a quantity we call the momentum of that particle. I'll label it with the symbol p. And it's equal to the product of the mass times the velocity. This is something you've undoubtedly seen before. Now, let's think about the dimensions of momentum for a moment. So dimensionally, momentum has units of mass times the velocity. And so that's in SI units, the units of mass are kilograms and then the units of velocity are meters per second. You can also express momentum dimensionally as a product of a force times time. And so, again in SI units, units force are the Newton and the units of time is the second. So these are the SI units for momentum, two different ways of writing the same dimensions. Now, we've seen that for a single particle, we can write Newton's second law as the force is equal to the mass times the acceleration. Or equivalently, we can write that as the mass times the time derivative of the acceleration, dv/dt. Now, if m is a constant, then I can rewrite this as F is equal to the time derivative of the mass times the velocity, or equivalently as the time derivative of the momentum, since mv is just equal to p. So I actually want to stress this much-- I'm going to put it in a box because it's very important that I can write Newton's second law, instead of F equals ma, as F equals the time derivative of the momentum. And this is absolutely where we'll see the momentum becomes very useful. Because it turns out that this form of Newton's second law is actually the most general form of the equation because it's applicable not just to a single point mass, but also to a more complicated system. A system consisting of many masses or a system where the masses changing or the masses flowing, as in a fluid. In all of those cases, this form of Newton's second law is correct. F equals ma, which is probably more familiar to you, is actually a special case of this law for the case of a single point mass. So this is where we'll see that momentum is quite a useful concept, especially as we start considering more complicated systems, as we'll get to a little later in the course. What I want to do now though, is to take a closer look at this equation, force is equal to the time derivative of the momentum. Whenever we have a relation involving a derivative like this, we can always also rewrite it in an equivalent integral form, which can be very useful and give us a different way of looking at the same information. So let's take a look at that. So if I take this equation and integrate both sides with respect to time, then I can write that as the integral of F with respect to time is equal to the integral of the right-hand side, dp/dt with respect to time. Now, let's make this a definite integral. I'll go from time t1 to time t2 on both sides here. Now, this right-hand side is just-- so the integral of dp/dt with respect to time is just p at time2 minus p at time1. And that is just the change in the momentum vector going from time1 to time2. Now, this integral on the left-hand side, we give a special name. We call this the impulse. This name, impulse, calls to mind a short, sharp, shock of some sort. But it can also refer to a weak force acting over a long interval. And notice here the function F, the force F, is in general a function of time. So this doesn't necessarily mean a constant f. This could mean a force that's varying in time. And what this equation tells us is that the change in the momentum of the system doesn't depend on the detailed time dependence of F, but rather just on the integral of F. And so suppose I were to graph the force as a function of time going from time0 to a time delta t. And suppose I had some complicated function that looked like that. The impulse is just the area under this curve. It's the integral of this function. That's the impulse. And the change in the momentum depends only on the area under this curve and not on the detailed shape of the curve. So what that means is that I can define an average force by choosing a constant force that has the same area as this example on the left. So suppose I calculated that. And there is some constant force here. I'll call this F average. Going over the same time interval. The average force is that constant force which has the same area as the area under my F of t. So in other words, F average times delta t, which is the area on the right-hand side here, is equal to the integral of F of t dt integrated from 0 to delta t, which is the area under this right-hand curve. And so my average force is just that integral, F of t dt, divided by delta t. And this is integrated from 0 to delta t. So that's my average force.
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https://ocw.mit.edu/courses/5-08j-biological-chemistry-ii-spring-2016/5.08j-spring-2016.zip
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The following content is provided under a Creative Commons license. Your support will help MIT Open Courseware continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT Open Courseware at ocw.mit.edu. ELIZABETH NOLAN: We're going to continue where we left off last time. So briefly I'll make a few points about initiation of translation in prokaryotes. And then where we're going to spend the bulk of the time today is with a review of tRNAs and then discussing the aminoacyl-tRNA synthetases, which are the enzymes responsible for loading amino acids onto the three prime end of the tRNA. And these points are important because these process has to happen in order for the amino acids to be delivered to the ribosome, which is where we'll go on Wednesday. So the first questions are, how does initiation happen? So how does this ribosome, 70S ribosome, get assembled with the mRNA and initiator tRNA bound? And then we're going to ask, how do we get an aminoacyl-tRNA, such that the amino acids can be delivered to the ribosome? So first, for initiation in prokaryotes, there's a few steps to this process. We'll just look at these at a basically superficial level of detail. But recall that there are translation factors. And during initiation, there are three initiation factors-- so IF 1, 2, and 3-- that are required to help assemble the 70S ribosome here. So first in terms of initiation, what happens is that the mRNA needs to bind to the 16S RNA of the 30S subunit. And so I point this out because at this stage in the process, the 70S ribosome isn't assembled yet. So we have the mRNA binding to the small subunit. And this process requires initiation factor 3. And effectively what happens is that the mRNA has a region called the Shine-Dalgarno sequence in prokaryotes, which is the site of ribosome binding. And then upstream of that is a start codon that signals for the start of translation. So if we think about the mRNA of the five prime end, and somewhere there's a sequence that signals for ribosome binding. OK, and then we have our start codon that signals the start of translation. OK. And so this gets translated here. OK, so this start codon pairs with initiator tRNA. And this initiator tRNA is special. One reason why it's special is because the amino acid attached is an N-Formylmethionine OK. So sometimes the initiator tRNA is called f-met tRNA f-met as an abbreviation there. So just as some overview here, what we're seeing in this alignment is a number of the ribosome binding sites, or Shine-Dalgarno sequences in prokaryotes. We have the start codon on that pairs with the initiator tRNA. And here's a schematic depiction of what I've indicated here on the board. OK, so the mRNA binds to the 16S of the 30S subunit. So the 70S is not assembled at this stage. And IF3 is involved, as I said. The Shine-Dalgarno sequence determines the start site. And we determine the reading frame, as well. So here is just an indicating translation of a polypeptide. What happens after that? So after that, it's necessary to assemble the 70S ribosome, have the initiator tRNA in the P site, and have the cell ready to go for translation. And here's just one cartoon overview that we'll use as a description of this process. OK. So what do we see? We've talked about this step so far. We see there's a role for initiation factor 1. And in this cartoon, if we imagine the E site, the P site, and the A site, what we see is that IF1 is binding to the site of the ribosome. And one way we can think about this is that the initiator tRNA has to get to the P site. And so that region is blocked to facilitate the initiator tRNA getting to the P site. OK, we see that initiator tRNA binding to the P site. And this happens via formation of a ternary complex with IF2 and GTP. So initiation factor 2 hydrolyzes GTP. There's an event that results in joining of the two subunits. And there has to be dissociation of these initiation factors for the ribosome to be ready to accept its first aminoacyl-tRNA in the A site. OK, so the outcome of this process here is that we have an assembled 70S ribosome with the initiator tRNA in the P site. The A site is empty, so it can accommodate an incoming aminoacyl-tRNA. And the E site or exit site is also empty. So that's the main take home for initiation. And that's the extent to which we're going to discuss it within this class. So in order to get to the elongation cycle, we need to get the aminoacyl-tRNA into the A site. And that's going to require the help of EF-Tu, so elongation factor Tu. Before we discuss how elongation factor Tu is going to help deliver that aminoacyl-tRNA, we need to talk about how we get the aminoacyl-tRNA in the first place. So what is the tRNA structure, just as a review to get everyone up to speed. How are amino acid monomers attached to the tRNA? And how is the correct amino acid attached? So this is an aspect of fidelity, which came up as a concept last week in lecture. And so we'll look at the mechanism of aminoacyl-tRNA synthetases to see how is the correct amino acid attached, and then what happens if the wrong amino acid is selected. Are there mechanisms to correct that? And if it's not corrected, what are the consequences here? So moving forward with that, we're going to focus on the tRNAs and addressing those questions. So just as a review, so we can think about tRNA secondary structure, which is often described as cloverleaf. So we have a five prime end. The tRNA has several arms. OK. So we have a D arm. This arm here has the anticodon that pairs with the codon of the mRNA. We have a variable arm, this arm here. And we have this three prime end here, where the amino acids get attached. So this, in terms of base numbering, we have C74, C75, A76 here. OH. This is often called the CCA acceptor stem. And the amino acids are attached here. I'm going to abbreviate amino acid as AA via an ester linkage. And these ester linkages are important for the chemistry that happens in the ribosome. OK. So we can imagine just if we have abbreviating the tRNA structure like this and if we think about the sugar of A76-- bless you. OK. We have one prime, two prime, three prime here. This type of connectivity here. And this is abbreviated throughout as amino acid tRNA, aa in general terms, or the three-letter abbreviations, like what we saw for f-met tRNA f-met with the initiator tRNA here. So here's a schematic of a tRNA secondary structure with a bit more detail than what I show you on the board. And something we need to keep in mind is even though we often draw the tRNA in this cloverleaf type depiction, it has tertiary structure. And so it's very important to think about this structure as we think about how the tRNAs enter the various sites of the ribosome. OK. So this structure is L-shaped. And I like this depiction here because regions of the secondary structure are color coded with the corresponding regions of this tertiary structure here. OK, so we see the shell shape of an L, rather upside down here, where we have the CCA acceptor stem over here and the anticodon arm and anticodon region down here. So what is a consequence of this structure? The tRNA is quite narrow. So we're thinking about 20 to 25 Angstroms in width. And if we think about this in the context of the ribosome and the peptidyl transferase center, 3 tRNAs need to fit into that catalytic center during the elongation cycle. So it makes sense that they're relatively narrow. This allows three to fit there. So as we think about the translation process and also think about some of the translation factors, we want to keep this type of structure in mind here. Here's just another view of that, with some additional descriptions of the overall structure. And this includes the numbering of the tRNA bases within that structure here. Just a point to make, this won't be a major focal point in the course, but do keep in mind that tRNA contains many post-transcriptionally modified bases, so you'll see an example of that in problem set one. Up to 25% of the bases can be modified. Typically, we see about 5% to 20% of them modified here. OK, you're not responsible for these structures, these modified structures, in the context of this class. So the key question for today is how are amino acids attached to the tRNA, as shown here? And in order for that to happen, there's a family of enzymes called aminoacyl-tRNA synthetases, or abbreviated aaRS. OK, so this name tells you right away, synthetase, that these enzymes use ATP. And these enzymes catalyze the attachment of amino acids to the three prime OH, or sometimes two prime OH, of the tRNA here for that. And so we're going to consider this overall reaction. And then we're going to think about the reaction mechanism and experiments that were done to give support to the mechanism that we see. So all aminoacyl-tRNA synthetases require ATP and hydrolyze ATP to AMP and PPI. And so they catalyze this overall reaction where we have an amino acid monomer. We have the tRNA that encodes this-- that is for this amino acid. ATP to give us the aminoacyl-tRNA AMP and PPI. So if the ATP is being hydrolyzed to AMP and PPI, what phosphate is being attacked? So we saw on Friday there's the alpha, beta, and gamma phosphates of ATP. OK, pardon? AUDIENCE: Beta. ELIZABETH NOLAN: Beta. Any takers? AUDIENCE: Alpha? ELIZABETH NOLAN: Any takers? Gamma? Yeah, so it's alpha. If you're getting AMP, it's attack at alpha. If you're getting ADP, it's attack at gamma here. OK, so P alpha is next door to the ribose of the nuc-- there. Yeah. OK. So if we consider this overall reaction, how does it work? Just before that, one other observation I just want to point out, if we're thinking about these enzymes and asking what is it that they recognize of the tRNA, so we have the anticodon. And that goes in hand-in-hand with the identity of the amino acid. Just keep in mind that it's not just the anticodon. So here we're seeing an example of an aminoacyl-tRNA synthetase with its tRNA bound. And we see that there's many contacts between the tRNA and this enzyme here. OK, so here we have the amino acid end, the anti-codon end, and all throughout here. So what is the mechanism to get us where we need to go? We have our overall reaction that I'll put up on the board, just to keep it straight as we move forward. So amino acid plus ATP plus the tRNA for that amino acid. Aminoacyl-tRNA synthetase to give us the aminoacyl-tRNA plus AMP plus PPI. So let's consider a mechanism. This is going to be a two-step mechanism. And so in the first step of this mechanism, we have the amino acid plus ATP. And we have formation of an OAMP intermediate. Plus PPI here. So this intermediate is called an amino adenylate. Adenlyate because adenosine here. And we need to think about why this intermediate might form. Why would we propose this in a mechanism? And then in step two-- we'll come back to that in a minute-- we can take our amino adenlyate, have our tRNA, this is the three prime end here. We can have attack with release of AMP. OK, so here we have the ester linkage at the three prime end, like what we see on that board here, to give us our aminoacyl-tRNA. OK, so we see in step one, there's formation of this amino adenylade intermediate. And in step two, there's transfer of the amino acid monomer to the three prime end of the tRNA here. So why might these enzymes go through that OAMP intermediate? What needs to happen for this chemistry to occur? AUDIENCE: You need a more activated reading group to have that acyl substitution form an ester from a carboxylate. ELIZABETH NOLAN: Right. We need to activate the CO2H group there. So this affords that. So what might be another possible mechanism, right? Imagine you're the experimentalist and you've combined your eighth amino acid ATP tRNA and this enzyme you've isolated in a test tube. And you see you've got this as a product. And this as a product. And you're wondering how did we get from reactants to products? This is one possibility. Maybe there's also a possibility of a concerted mechanism where there's no intermediate like the one I'm showing you here. These are just things to keep in mind when thinking about reactions. This two-step mechanism is the accepted mechanism for the amino aceyl tRNA synthetases. And so what we're going to think about are what are the experiments that were done to support this mechanism here. So what are the things we need to think about? And so we're going to think about this by examining one aminoacyl-tRNA synthetase as a paradigm. And this is the one for a isoleucine here. OK, so what are the experiments that need to be done to characterize this reaction and determine mechanism? OK. So one thing we need to confirm is reaction stoichiometry. So there's a stoichiometry up in what I've written above. But experimentally, that needs to be determined. So one, reaction stoichiometry. And so how can we think about this? We can think about the equivalence of the amino acid. So in this case, isoleucine. How many equivalents of isoleucine? And presumably, this isoleucine binds to the enzymes. We can think about it of equivalence of isoleucine bound. And we also see that ATP is consumed, right? That's hydrolyzed to AMP and PPI. So how many equivalents of ATP are consumed in this reaction? What else do we want to know? We need to know something about kinetics. So what are rates of formation? What is the rate of formation of the product, the aminoacyl-tRNA, and since I've told you this intermediate forms, what is the rate of formation of the intermediate? And since this is an intermediate, it's something transient. So we need to think about how are we as experimentalists going to detect this intermediate over the course of this reaction. It forms and decays in order to get product here. So rates of formation. And so we have formation of our product, which in this case-- and then formation of the intermediate, which I'll just abbreviate Ile-AMP. And what else would we like to know? We can figure out how, in addition to rate of formation of the product and the intermediate, we can think about the rate of transfer of Ile from the intermediate to the tRNA. So what this tells us is that we need a way to look for or detect the intermediate. Here. So imagine let's just have a hypothetical situation. If we find the intermediate, that tells us something about the reaction. If we don't find the intermediate, what can we conclude? Pardon? AUDIENCE: That there was no intermediate? ELIZABETH NOLAN: So that's one possibility. Are there other possibilities if our method doesn't let us detect the intermediate? AUDIENCE: Second step is to test. ELIZABETH NOLAN: Can it be hard to detect an intermediate? It can be very hard, right? So they don't always-- there aren't around all the time very much or in very abundant quantities. So if it's not detected, could it be there? Yeah, it might be there. And the method just didn't allow for it to be seen. So you always need to keep that possibility in mind. This will be a case where there is a robust method that allows us to detect this type of intermediate. But always keep that in mind. OK, so first thinking about reaction stoichiometry. We're not going to go over the experiments that were done to define this. I'll just tell you some facts that result from some experimental studies. So this isoleucine aminoacyl-tRNA synthetase binds 1 equivalent of isoleucine as indicated in the overall reaction. And it consumes one equivalent of ATP, also as shown in this overall reaction, to make one equivalent of the aminoacyl-tRNA. OK and these stoichiometries were determined experimentally. So now we need to think about points two and three to characterize the reaction kinetics. So what experiments were done? So there are several different sets of experiments, some of which we're familiar with from 7.05 or 5.07 and others that will be new and presented in more detail in recitation this week and next week. So we can imagine doing steady state kinetic experiments, as well as pre-steady state kinetic experiments. And the general aims here are, one, to determine the rate of aminoacyl-tRNA formation, to determine the rate of amino adenylate formation, so this intermediate-- and again, we need a method to detect the intermediate. And at the end of the day, we'd like to know what is the rate determining step. So a method that is commonly employed for these types of studies involves the use of radioactivity. And we'll just go over a few points about radioactivity now to help with understanding these experiments. And you'll hear more about this method in recitation this week. So the experiments I'm going to tell you about are going to involve the use of radio isotopes like C14, P32. And the question is, why do we like to use radio isotopes in biochemical experiments? And they're really excellent probes. It's the bottom line. And one reason for that is that if you can use a radio isotope like C14 or P32, it's introducing minimal perturbation into your system. So you're not needing to attach a fluorophore whether it be a small molecule or a protein. You're not modifying the structure of a component of your system. So the overall size and the chemical properties are maintained when you use different isotopes of the same element. And some of the ones we'll see today are, for instance, C14 labeled isoleucine, P32 labeled ATP. They have the same chemical properties as the unlabeled forms, and same size. The other point to make is that we can detect very small amounts of radioactivity in a sample. And you'll see some of those calculations and how to do them in recitation this week. So we can detect small amounts, and that's good for looking for something like an intermediate. And there's readily available techniques for quantifying radioactivity in a sample. So if you see nomenclature like this, the NX nomenclature indicates the radioisotope in this sample. And I'll just say in passing here, we all know the isotopes are atoms bearing the same number of protons but different numbers of neutrons. And radioactive isotopes have an unstable nucleus, which means there's a radioactive decay. And typically-- well, we often use beta emitters in biochemical studies. And that's what you'll see today. So what are some of the experiments? We're first going to consider looking at the steady state kinetics to ask what do we learn in the steady state. So from our steady state experiments, we're able to get our Kcat and our Km and the catalytic efficiency, which is the Kcat over Km. We're going to compare our Kcat values or turnover today. So experiment one is to monitor formation of product. So how is this done? This reaction is done by taking C14 labeled isoleucine and unlabeled tRNA and watching for transfer of that radio label to the tRNA. And so what comes from these studies is a Kcat on the order of 1.4 per second. And now we have a way to detect this amino adenylate intermediate. And we'll talk about that assay in a minute, after we get through this comparison. We do a steady state experiment to monitor formation of this amino adenylate intermediate. And this assay also uses radioactivity. And it's called ATP PPI exchange assay. And we'll go over how this works in a minute. So the results of these experiments give a Kcat on the order of 80 per second. So what does this comparison tell you? These values are quite different, correct? So we're seeing that this ATP PPI exchange assay is telling us that ATP PPI exchange, which is a measure of formation of this intermediate, is about 60-fold faster than formation of product here. That's an important observation to have. So how are we going to figure this out? How are we going to see this intermediate? That's the question we need to ask next. And so we need to go over this ATP PPI exchange assay. And this is an assay that will come up again in module 4 when we talk about the biosynthesis of non-ribosomal peptides. So we'll return to this type of assay and data many times. So the question is, if we have this reaction, OK, how do we detect this? OK, it's not so easy. And we need an assay. And this is some of the background towards the development of this assay. So we need to suppose that our amino acid and ATP react with the aminoacyl-tRNA synthetase in the absence of tRNA. And that's indicated by step one, more or less. But that doesn't show it experimentally. So in the absence of tRNA, this amino acid and ATP react with the enzyme and they form the aminoacyl AMP intermediate and PPI. And they do this reversibly. OK, so the reversibility of this reaction is key for ATP PPI exchange to work. So if this occurs and they do this reversibly, therefore we can deduce formation of the aminoacyl AMP. If we add radio labeled PPI, the amino acid, and ATP to the enzyme and we see that radio labeled phosphorus from the radio labeled PPI incorporate into ATP. That's only going to happen if this chemistry is reversible. And bear in mind, we can detect very small quantities with radioactivity. So it's not that it has to be reversible to some large degree. We're relying on the detection of this radio label. So how does this work chemically? Let's take a look. OK, so imagine here we have our ATP. We have our amino acid. And we have our enzyme. And step one, we have binding. So there's some ATP binding site to the enzyme and some site for the amino acid to bind. And I'm leaving magnesium out of this depiction, but remember that magnesium and ATP come together. Now what? Step two, OK, we're going to have a chemical step where we have formation of the amino adenylate and PPI. And they currently are bound to the enzyme. We have step three. So imagine in this step our PPI is released. And this is another key aspect of this assay. So what does this mean? We now need to think about going backwards. If the PPI is released and we spike this reaction with radio labeled PPI and work our way backwards, will the radio label end up here in the ATP? OK. So this is going to be going backwards. We've left off with this enzyme with the amino adenylate bound. We have the PPI that was released. And then we spike this reaction with our radio labeled or hot PPI. So then what happens? Step four, working backwards. Imagine that some of the radio labeled PPI binds. Then what? Working backwards another step. 32 P ATP and the amino acid. And then we have release here. OK, so then the question is, can you detect this? And so if you can detect some incorporation of this radio label into the ATP, that indicates that this enzyme worked through that type of intermediate. AUDIENCE: So are PPI not also sometimes [INAUDIBLE] and then if you had some competing hypothesis where it made ATP and ADP, then your PPI would maybe sometimes turn into just a single radio label phosphate that could then have the same reverse reactions as the [INAUDIBLE]? ELIZABETH NOLAN: Yeah. So whether you initially end up with PPI or PI is going to depend on how the ATP is hydrolyzed. And so you could imagine maybe there could be some background ATP hydrolysis that gives ADP and PI in this type of assay. That's something you always need to look out for. For the purpose of this, let's assume that we're not having some background problem in terms of the ATP source, and also that the enzyme is specific in terms of what it's doing to the ATP. But yeah, certainly background ATP hydrolysis can be a problem. So how will this be detected? And how will you know the radio label is associated with ATP and not something else in your mixture? AUDIENCE: [INAUDIBLE] ELIZABETH NOLAN: Pardon? AUDIENCE: [INAUDIBLE] ELIZABETH NOLAN: No. So we're going to look at the radioactivity. So this will come up more in recitation this week. But we need to be able to measure radioactivity by, say, scintillation counting here. But what's also needed is a separation because you need to know where that signal's coming from. You need to know it's coming from ATP and, say, not a background from however much of the PPI you introduced. Or if you have no idea what's going on with your chemistry, maybe the data are going to tell you it's not this mechanism. So you need to have a separation. So how might you separate ATP from all of these other components? AUDIENCE: Based on affinity column. ELIZABETH NOLAN: Some affinity column. So I like the column. But we're not going to have some sort of tag on the ATP. That might be a problem for that enzyme. But your notion is correct in the sense that we'll use some sort of chromatography in order to separate. OK, so maybe HPLC, how many of you have used an HPLC or at least know what one is? AUDIENCE: [INAUDIBLE]. ELIZABETH NOLAN: Right. So typically looking at UV vis. But you can imagine hooking up an HPLC to a detector that allows you to do scintillation counting and some sort of column that will allow you to look for ATP. Is all of the ATP going to be radioactive in this assay? No. So again, we can detect small quantities. And as long as there's a little bit of reversibility, we can see this here. OK, so what's critical in this assay is the reversibility of steps 3 and 4. What would happen in this assay if the PPI is not released? AUDIENCE: [INAUDIBLE]. ELIZABETH NOLAN: Right. Under the conditions, or if for some reason the PPI is not released, we're not going to see this exchange reaction. We're going to have a readout that doesn't give us this. Does that mean this didn't form? No. OK, so there's many caveats and details that you need to think through when thinking about a reaction and then the experiment is done to test this. So in the case of these aminoacyl-tRNA synthetases, these ATP PPI exchange assays work well. And these assays can be used to get steady state kinetic parameters, to get Kcat, Km, Kcat over Km, which is where this type of value comes from, in this case here. So back to these analyses up here, what they're telling us is that formation of this amino adenylate intermediate is about 60-fold faster than formation of the product. OK. And what we all want to recall when thinking about steady state experiments is that they're set up with a great excess of substrate and with the enzyme concentration. The reaction is zero order in respect to substrate. And you'll have some additional notes about that in your recitation materials this week for review. So something else biochemists like to do when looking at reactions and understanding reaction mechanisms is to look in the pre-steady state. And this came up briefly in lecture 1 as a method. And again, you'll hear more about it in recitation over the next two weeks. In these experiments, the goal is to look at the very first, early moments of a reaction. And they're set up quite differently. So limiting substrate is used. There's no turnover, so huge contrast to what we know about steady state experiments. And one of the goals is to look at the formation and consumption of intermediates here. So this type of chemistry often happens on a fast timescale. You can imagine millisecond timescale here, which means that we need a special apparatus that has fast mixing capabilities, because there's no way for one of us to do this on our own with our pipette. And so the type of experiment or apparatus used is called a stop flow. And I just show one depiction of a stop flow apparatus here. You'll get some other variations on this theme in the recitation notes. OK, but effectively what happens is that you have two drive syringes, a and b, and each of these syringes will contain certain components of your reaction. And this stop flow has a drive motor and a stop syringe. And it effectively allows you to rapidly mix the components of these syringes in a mixer, shown here. And then you either have some way to detect product-- so maybe if you can use optical absorption, you have a UV vis detector or a fluorescence detector. Or in other cases what you'll do is you'll punch the reaction at a certain time point. So you need a third syringe not shown here with a quencher. So you can imagine if you're working with an enzyme, maybe you quench by addition of acid or base, something that will denature and precipitate that enzyme. And then you can take that sample and analyze it in some way that fits in terms of what you need to detect there. So this type of methodology was used in order to monitor transfer of isoleucine to its tRNA. And so where we'll pick up in lecture on Wednesday is the design of that experiment in terms of what will we put in each syringe, and then what are the results of those experiments? And ultimately, what does that tell us about rates of transfer here? That's where we'll continue.
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https://ocw.mit.edu/courses/3-091sc-introduction-to-solid-state-chemistry-fall-2010/3.091sc-fall-2010.zip
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The following content is provided under a creative commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. SAL: Hi. I'm Sal. Today we're going to be solving problem one of exam three of fall 2009. Now before you attempt the problem, there's a couple of things that you should know or have background knowledge of these materials before starting. First thing is understanding the properties of crystal defects and for this problem particularly, Frenkel defects, which we'll talk about. The formula, which is an Arrhenius relationship, as a function of temperature and energy enthalpy of the fraction of vacancies that are formed in a crystal and the conversion again between one electron volt and to the joule. This will help you solve the problem. So the problem reads as follows. Silver bromide, which is AgBr, has rock salt crystal structure. So it's an FCC Bravais lattice with the ion pair Ag plus and the Br as the basis-- Br minus the basis. The dominant defect in AgBr or silver bromide is the Frenkel disorder. So it's telling you what the dominant defective is. So the question asks, for part A, so I'll put A here. Part A for the question asks, does the Frenkel disorder in silver bromide create vacancies of silver plus, vacancies or Br minus or both? Explain. Dionic radii are 0.67 Angstroms for silver plus and 1.96 Angstroms for bromine minus. So that's data that's given to us. I'm going to go ahead and write that down. So I know from reading about point defects and crystals that a Frenkel defect is pretty much formed when you have an ion pair of dissimilar sizes. So if one of the radii-- like, say, your anion is a lot bigger than the radius of your cation, then one should expect that you would have a Frenkel defect to form. And what a Frankel defect is-- which you should know from reading the material-- is that it's when one of the ions in your crystal leaves and leaves back a vacancy-- so just migrates-- just hops out of its lattice site and leaves back empty spaces, which is called a vacancy. So we're given two things. If I draw a little picture-- I can call this my silver and I'll call this my bromine. And I'm given the fact that the radius of Ag plus is equal to 0.67 Angstroms-- and an Angstrom is just 10 to the minus 10 meters-- very small number. And I'm also given the radius of bromine, the anion, is 1.96 Angstroms. Now I would argue that these two have very dissimilar radii. Obviously, the bromine looks to be about three times bigger than your silver. So by understanding the definition of a Frenkel defect, I can claim that I would expect the smaller ion of the two to be the one that leaves the vacancy behind, hence forms the Frenkel defect. So for part A, do I expect it to create silver plus vacancies or Br minus vacancies or both? I would expect just to be silver plus, given the size of your cation. So if you were to write on this problem, just expect only Ag plus, which is your silver cation, or the smaller one. Expect Ag plus smaller ion to create a vacancy and hence, this leads to the Frenkel defect. So yes, I should expect it to form a defect now. If I was given a cation that had a similar radius that I-- I wouldn't know if I could argue the fact that this will form a Frenkel defect or not because the conditions are that you have to have the similar radius between your cation and your atom-- and it makes sense because the fact that your silver is small-- it has more freedom to hop around. So it requires less energy for this smaller atom to then start hopping around your lattice site without penalty or without major penalty compared to your bromine ion. So therefore, this is the one that should be expected to form that. OK. So that's part A. And part B reads-- calculate the temperature at which the fraction of Frenkel defects in a crystal with silver bromide exceeds one part per billion. The enthalpy of Frankel defects and formation, delta h sub f, has a value of 1.16 electron volts per defect. And the entropic pre-factor A has a value of 3.091. So it's giving you data. Now the way I would solve this problem is that the first thing I would do is that I would write down what my data is. So I'll call this data and the first thing is that our fraction of vacancies that are formed in silver bromide is one part per billion-- so 10 to the -9. That's a fraction so no units. The second thing that we're given in the problem is that our energy of formation-- our enthalpy energy-- is 1.16 electron volts per defect. This is the energy penalty for every time one of your silver cations jumps out of sight to create that. Nothing is free. Everything requires energy. We're also given the fact that A, the entropic pre-factor, is 3.091-- with no units-- and given our equation that I showed you on the bullet point. That you should know-- you can go ahead and see what's missing. So if I write down my equation-- this is all-- I'll box this off as my data because I'm going to refer to this to solve the problem. And the problem talks about temperature. So I know that f sub v is going to equal to the pre-factor times the exponent of negative delta h, of formation over kb t. Now you notice that kb wasn't given to you and kb is both in its constant. Now a lot of student forget that they have a table of contents in front of them and that value is in there. So if you don't know it by heart, then you want to make sure that you reference to that table because a lot of information will be given to you, because you're expected to look at the table of contents or your periodic table. But if I include this as my data-- I'm going to go ahead and write it over here-- that kb-- half the value of 1.38 times 10 to the -23 and this has units of joules for degree Kelvin. So this is something that you should pay particular attention to because now kb is in joules for Kelvin, but our energy was given in electron volts. Now this is the number one thing that will take off points. You'll get points taken off if you don't notice this-- that you need to go ahead and do the conversion from electron volts to joules or joules to electron volts. Either which way, you're going to get the answer. But the problem asks, calculate the temperature. The very first sentence. So what does that mean? Well, I'm going to go ahead and look at my equation and I'm going to look at the data that I have and the constants and see if I'm missing anything else because obviously you can't solve an equation that has two unknowns and just one equation. Solve the equation, you've got to only have one unknown for the equation. So f sub v-- no. Check-- no. That's given. Is A given? It's right here. That's given as well. What about delta A sub F? Well, that's the energy-- the energy penalty to create a defect and kb, which we got from our table of contents, and t. So this concludes that the only thing we're not given is temperature because that's what we're asked to solve. So I need to do some math on here to go ahead and solve for temperature. And the first thing I want to do is take the natural log of both sides. So by taking the natural log of both sides-- so natural log of f sub v-- this equals to natural log of A plus the natural log of the exponent part and we know from math that the natural log of an exponent cancels each other out because they're inverses. So the natural log of exponent, of negative delta h sub f, over kb t-- this cancels that and we just get the natural log of A plus negative because you have a negative up here-- delta h of formation divided by kb t. So the only thing we don't know here is temperature. So if I can rearrange this equation and solve for temperature, I can get an answer. So if I do that-- what's the best way of doing that? Well. I can add-- I can move this to the other side and move this to the other side and that gives me delta h sub f divided by kb t equals natural log of A minus natural log of your vacancy fraction and I can then multiply both sides by t so it cancels this one and it arrives over here and then divide both sides by ln of this. So I'll go ahead and do that. So I'll multiply this by t. Multiply that by t. So that cancel that and I end up having delta hf over kb t-- the t got canceled-- this equals to the natural log of A minus the natural log of your vacancy fraction times t. So now if I divide both sides by this factor, I solve for temperature. So this gives me an isolation that t ends up being delta hf over kb times the natural log of A minus the natural log of the vacancy fraction. Now all we have to do is plug in the numbers, but again, if you look at the units of Boltzmann's constant, which are joules per Kelvin and the value that we're given up here is electron volts for defect, we want to go ahead and convert the electron volt to the joule because it'll just be easier to do the math that way. So I can go ahead and write it out and I do that by multiplying the top by that conversion factor and if I do the math, t ends up being-- we have 1.16 of this-- has units of electron volts for defect and then I want to get joules because that's what Boltzmann's constant has. So in order to get joules, if I multiply this by the conversion factor which is 1.6 times 10 to the -19-- this has units of joules per electron volt. Now the electron volts cancel and I get-- the numerator has units of joules per defect. That's good because that's going to cancel with the units in Boltzmann's constant. And then I put in 1.38 times 10 to the -23-- this is joules per Kelvin-- and this factor is multiplied by the natural log of 3.091 minus the natural log of 10 to the -9. So unitless-- so if you do the math out, you end up getting a value that at the end of the day, your temperature-- this t right here-- will be equal to-- let's go ahead and write it over here. So now that we go ahead and do the math-- that way it's nice and clean. We know that the t, the temperature that we got from plugging all that math-- ends up being just 615 Kelvin. And where did the Kelvin come from? Well, it came from your joules per Kelvin, from your Boltzmann's constant, because that's what the dimensional analysis does. So this is the answer. This is good. The problem doesn't say show the answer in degrees Celsius, but since everybody knows degrees Celsius-- everybody in science relies on degrees Celsius-- you can just convert this over to degrees Celsius, which happens to be 342 degrees Celsius. So this right here is your answer to part B. And again, I want to express again that the crucial part in getting this is knowing the fact that you're given an energy in electron volts, but the constant that you use has units of joules. So you need to convert to get the right unit out or else you will get not Kelvin-- you would get something different here, which wouldn't make any sense given the question that was asked. So with that, again I advise that-- always read the problem in detail. Look at the units that you're working with and make the appropriate conversions because everything should be on your table of contents.
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https://ocw.mit.edu/courses/5-60-thermodynamics-kinetics-spring-2008/5.60-spring-2008.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation, or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So last time, there was a question that was asked about the order of magnitude rates. You asked the question about the concentration of the catalysts. So the rates that I gave you, which were the rates of reactions are basically order of magnitude. They're estimates and odd numbers that, they obviously depend, as you pointed out, depend on the concentration of the catalyst. Depends on the concentration of the reactants. Depends on where you are on those concentrations. As we saw when we did enzyme catalysis, is that depending on KM, the Michaelis constant, and the substrate concentration, if your substrate's concentration is much higher than KM, you are in the maximum velocity limit, where the rate depends on k2 times the initial concentration of enzyme. In that case the rate of the reaction depends just on the concentration of enzyme, and not on the concentration of substrate. But if you are at the small concentration of substrate relative to KM, then your rate depends linearly with the concentration of substrate. And the same thing happens with the non-enzymatic catalyst, that there will be some sort of mechanism that goes along with it. And most likely it's going to be second order in the catalyst. And the reactant if it's just the two body process. But it could be more complicated. So definitely those concentrations will go in there. But a priori, it's not clear how they will go in there. But the orders of magnitude are roughly right. And the basic idea is that you change your rate of reactions by 15, 18 orders of magnitude with the right catalyst, especially if it's a biological catalyst. One concept that I didn't go through last time, which is also quite useful, it's called the turnover number. Turnover number. And this is useful for any kind of catalyst. But for enzymatic, or enzyme catalysts, the turnover number is k cat. The number per second. It's a rate per second. How many times does a reactant get, the number of times that a reactant turns into a product per second, per catalyst. So let me write it out. So it's the number of product formed per second, per second, per enzyme, per molecule of enzyme. OK, so if one enzyme is going to be there, there's going to be some substrate going in. Coming in, the pocket coming out. Coming in, the pocket coming out, so the number of cycles of this process per second. is the turnover number. And so if you are where the concentration of substrate is very much larger than the Michaelis constant, where are you are at the saturation limit, then this turnover number doesn't depend on the substrate concentration. It's just k cat. So that's why k cat is important. As this turnover number in the maximum velocity limit. Because in the maximum velocity limit, then the process that's dominant is the second process, ES goes to enzyme plus product. The first process here is not the rate limiting step. This is the rate limiting step. Because the substrate concentration is very high. So this first part goes very high, goes very fast. And the number of product formed per second, per molecule of enzyme, the number of molecules formed for second, that's the rate of product formation. Rate of product formation per second. Divided by the concentration of the enzyme. If you normalize by the number of enzyme. So this becomes rate of product formation in moles per second, divided by moles of enzyme. It's the same thing as what we're saying here in words. And so the rate of product formation is dP/dt, and you divide that by the concentration of enzyme, which you can write as E0. dp/dt, and the maximum rate limit is k max. It's a constant divided by E0. And k max is k cat times E0 divided by E0. The E0's come out. And this is k cat. Which is basically k2. So that's the turnover number. Any questions on this turnover number? You see that a lot in catalyst literature and especially the enzymes. I have to get out these lecture notes. Send these back. OK, so the last topic I want to talk about is oscillating reactions. It's sort of like an interesting topic. And it's applicable not just to just chemistry, but it's basically playing with the differential equations. And you'll see this sort of stuff later if you keep doing science that involves coupled differential equations. It's just all over the place. So normally we have an equilibrium process. A goes to B, B goes back to A. And if you start out with something which is out of equilibrium, then you will eventually reach equilibrium at a rate which is dependent on those two rates. k1 plus k minus one. So B will to come up to some B equilibrium. And A will, if you start out with a lot of A, and a little bit of B, and concentration of A, will come down smoothly, monotonically, to the equilibrium state. But sometimes, if you're out of equilibrium, and this doesn't have to be chemistry. It could be anything that's out of equilibrium, your emotion's out of equilibrium. Stock market is out of equilibrium. Population dynamics are out of equilibrium. The weather. Whatever. So, you're out of equilibrium, let's say this is equilibrium for A, this is equilibrium for B, let me try to keep them somewhat the same as here, I messed up. You're out of equilibrium, you start down here for A. And you, instead of going linearly down or monotonically down, exponentially down to the equilibrium, instead you overshoot. And you keep going back and forth. Like a spring. And then B does the same thing. Overshoots, comes down. Overshoots, comes down. Like a pendulum. A lot of things work like that. The heart is something that works like that, right? It beats. It's not in equilibrium. Good thing it's not in equilibrium. So the heart is one example of something that oscillates, back and forth, back and forth. And so there are actually many examples of complicated processes that involve chemistry, that aren't at equilibrium, that instead go back and forth, in and out, of on one side or the other of equilibrium. Now, there's an important feature of, if you want to build a chemical process that looks like this. It's often very useful to have a particular step in the mechanism that's called an autocatalysis step. And that provides feedback. It's like, if my microphone here, the speaker which is here, I were to stand right in front of the speaker and my microphone would pick up the volume from the speaker, it would get amplified, and I get feedback. And that would be a bad thing. So this is a form of feedback here. You need some sort of feedback that the reaction knows that you're building up, that this is going too fast. And then there's a feedback process that brings it back up. And a feedback process that goes back and forth. And the way you get this feedback is by having a step in the mechanism that has both a molecule as part of the reactants and as part of the product. It's not the same as a chain reaction, where we talked about an intermediate being recycled and building up. This is an actual reactant and an actual product that's part of this step here. So this is an autocatalytic step. And let's just see what that step looks like. Suppose we just have that step. So we want to find out, as a function of time, if you start out with some A and B, what happens to the concentration of B as a function of time. The time dependence of that. Clearly it's going to build up. But how is it going to build up? So we need to solve for B as a function of time. So let's go ahead and write down our rates. The rate of the reaction is d[A]/dt, is, let's put a rate constant here. k, k times [A] times [B]. Now, [B] and [A] are related to each other through stoichiometry. The concentration of B is the concentration of B that he started out with, [B]0, then every time you destroy an A, you create 2B. So it's plus two times the amount of A that you've destroyed, [A]0 minus [A], [A]0 is what you started out with. This is what's left over. So the difference is what you've destroyed. But every time you destroy an A to form 2B, you also destroy a b. So you have to subtract away the B that you've destroyed. Which is [A]0 minus [A]. This is the 2B that you created by destroying A, and this is the B that you destroy by having to destroy A, for the reaction. You've got to bean-count correctly. Keep track of everything. So [B], then, is just, I'm going to drop those brackets. B0 plus A0 minus A. So you can plug that in here. And is equal to k times A, times B0 plus B0 minus A. And now you have a differential equation that only contains A and time. B0 is a constant. And the way you solve that is by partial fractions. Use partial fractions, solve for this. And I'm not going to go through the math of solving for it. It's not that interesting. Let me give you the answer. You end up with [B] as a function of time, is this function A0 plus B0 over one plus A0 over B0 times e to the minus k times A0 plus B0 plus time. And that's where the time component comes in. Whoops, I'm sorry. Usually I don't have this on, let's turn this off. That's where the time dependence comes in. And clearly, at t is equal to infinity, t equals infinity just goes to zero. So you have B as A0, plus B0, everything goes through to the product. t equals to zero, this is equal to one. And B is equal to just B0. Alright, and so the curve looks something like this. This is what it turns out to look like. We're starting with B0 here. There's an induction period where very little happens. It's like the dormant stage. So, locusts that are sitting in the ground for seventeen years waiting for something to happen. Dormant stage, the induction period. And suddenly something starts to happen. There's an inflection point. And you get to A0 plus B0. The locusts wake up. And go to maximum concentration. So there's an s-like shape that has an inflection point and an induction period. That's pretty typical of an auto-catalytic reaction. OK so let's apply it to a process now. And again, this is going to be coupled differential equations. Broadly, broadly applicable, the example that I'm going to give you. This mechanism has been used for all sorts of things. So what do we have here? We're going to have an island. In the ocean. Island here. And it's going to be sunny, so let's get some yellow chalk here. There's the sun. And every now and then it's going to rain. We need some white chalk for the rain. There's a cloud that comes in. That's going to be one of the reactants. And then some seeds blow in from the continent, and grass starts to grow on the island. Grass. And then there's a shipwreck and a couple of rabbits get on the island. And now we have rabbits. Rabbits. A little tail on the back. I hate rabbits. I have a little vegetable garden, a little tiny vegetable garden. And there are rabbits. And I have to fence in my vegetable garden and make it impermeable impenetrable to rabbits. And rabbits are very clever. And my vegetable garden looks like a fortress. It's got a green fence. Heavy duty things. It's terrible. Rabbits are awful. They're very cute but they're awful. Alright, so now what happens. Well, given the rain and the grass, it turns out that this island can only support 25 rabbits. If there are more rabbits than that, then they all eat too much. If there are less rabbits then that's fine. But 25 is about the maximum that can support on this island. The rabbits, unfortunately, are not that smart. They don't know that. They don't know that only 25 rabbits can live there. So, what happens is that the Year one, Year one there are ten rabbits. And the food condition is great. Lots of food, number of rabbits. Lots of food. So the bunnies, the rabbits make bunnies. Lots of rabbits. Year two, they kind of went overboard. Now there are 50 rabbits. The food is lousy. Terrible. Well, there's a feedback here. Feedback and rabbits start to die off. Don't reproduce as much. Year three, we're back to ten. Great food. Rabbits don't notice. They multiply. Food is terrible. And they just don't learn. Humans are a little bit like that. Except our cycle isn't one year, it's usually on the order of 30 years. It's like the memory of a generation disappearing. And we have to relearn the mistakes of the previous generation. Science is like that too. Science goes in cycles. 20, 30-year cycles. Topics that were out of fashion 20 years ago, 25 years ago, come back. And suddenly everybody's excited about them. And all the ideas, of course we make progress. We go a little bit further than we did 20 years ago. But all the ideas that were out of fashion 20 years ago come back. People get really excited, and they relearn all the mistakes that people made at the beginning. Technology has improved, we know more, and therefore we go a little bit further this cycle than we did the previous cycle. But it's really interesting to look at history of science and see this sort of cycle. So let's write a mechanism. Let's write a mechanism here. So we have rain plus grass makes more grass. At the rate k1. That's autocatalytic. It provides some feedback. Then we have grass plus rabbits make more rabbits. OK, let's do more because I don't know how much, more, more. Also autocatalytic. Then we have rabbits. Eventually, unfortunately, rabbits become dead rabbits. Rate k2 k3, and that's the termination of the process. OK, so now let's put some chemistry on this. Let's assume that instead of having objects, live objects here, we have chemical molecules. So A plus B goes to 2B. That's the first step. Then we have B plus C, C is the rabbits. And then C goes to some sort of products. Soil. This is a very famous mechanism. It's called the Lotka-Volterra mechanism. It's also called the predator-prey mechanism. In my case here, I made it sort of warm and fuzzy. I didn't have a predator in there. But you could change this. Instead of having rabbits and grass, we could have rabbits and foxes. Where A is the grass, if you take A to be the grass. B to be the rabbits. And C you be the foxes. Works the same. You have grass plus rabbits makes more rabbits. Rabbit plus foxes make more foxes. Eventually the foxes die off. Same idea. So now, let's solve this. Let's assume that A, the rain here, it's constant. There's a steady supply of rain. It doesn't go away. That makes sense in my example here. So the concentration of A is kept constant. In order to get oscillations to keep going, that turns out to be important. To have one of the reactants just keeping being reintroduced in the system. Because it gets used up. And when all the reactant gets used up, then you're done. So in the case of the heart, we have to keep feeding ourselves to produce energy. Otherwise the heart would stop. OK, now we want to know what is the concentration of B and C as a function of time. So we write down our kinetic equations. dB/dt gets produced through the first step, k1 A times B, gets destroyed in the second step. k2 B times C. dC/dt, C gets produced in the second step, k2 B times C. Gets destroyed in the third step. k3 times C, got to do your bean-counting correctly. And the strategy here that we're going to use to solve the problem is to assume that we're at steady state. We're going to find the solution at steady state. Then we're going to perturb away from steady state a little bit. And see whether this creates an oscillation. First we need to find out what the steady state solution is. So first, let's find steady state. Then perturb away from steady state. We're looking at a harmonic response, if you've heard that term. When it's going to look like a spring, close to its equilibrium state. That means that we're going to set it these two things equal to zero, for this steady state. Things are not changing. Concentration of B and C are not changing. A is not changing on purpose here. And so all these guys here are then steady state concentrations. If we set it equal to zero. We solve for A in this. For C in this process here. The B's cancel out here. We divide by B. And you solve for C as a function of A. So this one here gives you C steady state is equal to k1 over k2 times the concentration of A. And this one here, now you put in the C and you solve for B, B steady state, is equal to k3 over k2. So the next step is to perturb away from equilibrium. Or from the steady state, rather. This is not equilibrium, it's not equilibrium because we keep adding A. We have to put some input into the system. So here we are. We've got some concentration of B here at the steady state. We have some concentration of C at the steady state. And now we're going to add some delta B to the system, or delta C. Or both. Delta C, we're going to ask the question, given our system here, how is it going to respond? It has a number of choices. It could respond by going back to the steady state in a monotonic fashion. Without overshooting. This is kind of like an overdamped system. It's like having a shock absorber on your car. You go over a bump, the car doesn't oscillate up and down. It's sort of damped. It's a damped oscillator. It goes back to the steady state. Or, you could, there we go, there's B, C. Or you could perturb, and it could just stay where it is. That could happen too. That's an inelastic response. Kind of like the supply and demand with oil. The price of oil goes from $10 a barrel to $120 a barrel. The use of oil doesn't seem to be affected very much by the increasing price. You're still using just as much oil. There's an inelastic response. What we're looking for is something that has an elastic response, or a harmonic response. We perturb it, it tries to get back to steady state. But because of feedback, and not being over-damped, if it's an oscillator, it goes up and down. Like a scale that is not well, well, like a car that has bad shock absorbers, all it has is the springs. Up and down. So what we want is, we want to solve for dB/dt as a function of time. We want, well, I'm going to do everything in green. Since I lost my white chalk. We want delta B, delta C as a function of time. So, that means that we start with B is B steady state plus delta B, C is C steady state plus delta C. And we put that in our equations. For dB/dt and dC/dt. So now dB/dt is the same thing as dB steady state plus delta B dt, which is d delta B / dt. Because this is a constant here. And that's equal to k1 times A, times B steady state plus delta B. Minus k2 times B steady state plus delta B times C steady state plus delta C. And then you have the same thing for dC/dt, it's going to look exactly identical. d delta C / dt. Thank you much, this is magic. k2 times B steady state plus delta B, times C steady state plus delta C, minus k3 times C steady state plus delta C. So these are two differential equations. And there are coupled differential equations. Because delta B is occurring here, here, and here. So the time derivative of delta C depends on delta B. And the time derivative of delta B depends on delta C. It's a coupled system. And if you actually expand this out and take away all the terms that cancel out, you end up with something that looks much simpler but is just as hard. d delta B / dt is equal to minus k3 delta C and d delta C / dt is equal to k1 A delta B. And then you really see that it's coupled, because the time derivative of delta B depends on delta C. And the time derivative of delta C depends on delta B. And there's the feedback sitting right here. And that's what you do. You learn how to solve in 18.03. So I'm not going to solve it here, I'm just going to give you the solutions. Because that's what we're interested in. After all, you're not expected to learn how to solve this equation. Although you should be able to put the solutions in and make sure that they do work out. And so the solutions are harmonic solutions. Delta B is a function of time is equal to delta B0. The amount of the perturbation times the cosine of omega t, where omega is the frequency of the oscillation. Minus some pre-factor here, k3 over k1 A to the 1/2 power. Times delta C0 sine of omega t. And then delta C looks the same. The same frequency. Delta C0, cosine of omega t. Plus k1 over A times k3 to the 1/2 power. Delta B0 sine of omega t, where omega, the frequency of the oscillation, is these rates. k1, k3 and A. So obviously, keeping the concentration of a constant is important because we don't want a time dependence here. The rain is constant. And then if you do that, then you have this oscillation. Up and down and up and down with this frequency here. You get exactly what we're trying to get. Which is this process right here. So let me give you an example of this. And the best way to do is to actually go and see a movie of a reaction that looks like this. This is the reaction that was optimized and created for demonstration purposes. So it's a pretty complicated reaction. There are ten steps to the mechanism. And if we gave you that mechanism to solve for the final exam, you would still be here probably two weeks form now. It's pretty complicated. You'd need a computer for sure. So, ten steps, well, that's what's known. There are probably more than that. It's probably more complicated than what we really know in terms of fishing out intermediates. And the basic reaction is, you start out with an oxide of iodine, which is clear. You create I2, which is going to look gold. In solution, and then it creates Ii minus, which is going to look deep blue. And then it'll feed back and start again. So we're going to have this cycle of products. And amongst those ten steps, or more than ten steps, there are a few autocatalytic steps, which provide the feedback. So now I have to start it. Let's go ahead and read. Alright. So here we go. So, there are two solutions, A and solution C. Actually, we mix three solutions together. The potassium iodide is what is going to give rise to the different colors. And now we're in the gold phase of the reaction. It's important to stir, otherwise it wouldn't, and then it turns deep blue. And then there's an induction period, where you wait for a while. And then eventually it should go clear. So it goes clear. And then it turns gold again. And then deep blue, and there's the oscillation. And then a long induction period, and then it goes through that cycle over and over again. And the recipe can be found pretty easily. And if you ever are in a lab, have access to doing this, it's a pretty easy reaction to put together. The only problem is that unless you use fresh hydrogen peroxide, it doesn't work. Because hydrogen peroxide tends to go bad over time. If you're going to do this, buy your hydrogen peroxide immediately before you try to do the experiment. And then it'll work. So what happens is that as long as there is a steady supply of hydrogen peroxide in here, it'll keep oscillating. But as you've used up the hydrogen peroxide, this is going to start to die. And eventually it'll have this pale blue solution. And it's like the heart stopping. If you want to start it up again, add more hydrogen peroxide and it'll start up again. OK, any questions? There's like an infinite number of these oscillating reactions that people have discovered. And optimized, that you can find in different places. Questions? Questions about the final. I didn't get to do a review, but you'll get that done with the TAs. Yes. STUDENT: [INAUDIBLE] PROFESSOR: Because there's enough of the equivalent of A, of the rain here. So it'll oscillate and there'll be some damping. So what will happen in this reaction is that the induction period will get longer, and longer, and longer. Because the w, this term right here, the frequency which depends on the concentration of A, that well get slower and slower. And eventually it'll just die. OK?
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https://ocw.mit.edu/courses/8-286-the-early-universe-fall-2013/8.286-fall-2013.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT at [email protected]. PROFESSOR: OK, in that case, let's get started. As usual, I like to begin by giving a review of what we talked about last time. This time on slides instead of on the blackboard. We're talking mainly about relativistic energy, or relativistic energy and momentum, and pressure, sometimes. The key equation is probably the most famous equation in physics, Einstein's e equals m c squared. And I gave some numerical examples. I actually looked up some more numbers since then, when I was revising the lecture notes. So these are slightly more up to date. But it's still true that if you could burn matter at the rate of one kilogram per hour, you would have about one and a half times the total power output of the world. And that's apparently still valid in 2011. I only had 2008 figures, actually, at the lecture last time. And if you imagine the 15-gallon tank of gasoline, and you could figure out how much that-- what it's mass is, and convert that to energy, it turns out that a 15-gallon tank of gasoline could power the world for about two and half days, if you could convert all of it into energy. The catch, of course, is that we can't convert matter into energy. We can't get around the problem that, at least at the energies that we deal with, baryon number. And that number of protons and neutrons is conserved, so we can't make protons and neutrons disappear. And that means that we're limited in what we can do. In particular, one of the most efficient things we can do is fission uranium 235. But when uranium 235 undergoes fission, less than 1/10 of 1% of the mass is actually converted into energy, which is why we can't actually avail ourselves of these fantastic numbers that would apply, if we could literally just convert matter into energy. We went on to talk about the relativistic definitions of energy and momentum, and how they come together to form a Lorentz four vector, and the underlying theme here is that we consider ourselves users of special relativity. Most of you I know have taken special relativity courses, and for those of you who have, this is a review. For those you who have not, and there are some of those also, no need to panic. I intend to tell you every fact that you need to know about special relativity. I won't tell you how to derive them all, but I'll tell you all you'll need to know for this class. So in particular, it's useful for this class to recognize that energy and momentum can be put together into a four vector, where the zeroth component is the energy divided by the speed of light. And the three spatial components are just the three components of the spatial momentum, although they have to be defined relativistically. The relativistic definition of momentum, at least how it relates to velocity, is that it's equal to gamma times the rest mass times the velocity, where gamma is the famous factor that we've been seeing all along when we've talked about relativity. The Lorentz contraction factor, one over the square root of 1 minus v squared over c squared. The energy of a particle, relativistically, is the same gamma, times m 0 times c squared, and it can also be written as the square root of m 0 c squared squared, plus the momentum squared, times c squared. Since the momentum forms a four vector, its Lorentz and variant square should be Lorentz and variant, and that means that the momentum squared minus p 0 squared should be the same in all inertial reference frames. And that's just equal when you put in what p 0 means, the momentum squared minus the energy squared, divided by c 0 squared. And to know what value it's equal to in all frames, it's efficient to know what it's equal to in one frame. And the one frame where we do know what it's equal to is the rest frame of a particle. So in the rest frame, the momentum vanishes, and the energy is just m 0 c squared. So in the rest frame, we can evaluate this, and we get minus m 0 c squared squared. And that means that has to be the value in every frame. And this in fact is the easy way to derive the relationship between energy and momentum. If we go back, the equation we had relating energy and momentum is really exactly that equation, rearranged. Just to give an example of how this works, when we actually have energy exchanges, I pointed out that we could talk about the energy of a hydrogen atom. And because energy and mass are equivalent, the hydrogen atom clearly has a little bit less energy than an isolated proton, plus an isolated electron. Because when you bring them together there's a binding energy, and that binding energy is called delta e, and has a value of 13.6 electron volts for the ground state of hydrogen. So that tells us the mass of an hydrogen atom is not the sum of the two masses, but rather has this correction factor, because we've taken out a little bit of energy for the binding. And that means we've taken out a little bit of mass. OK, then we talked about the mass density of radiation and how-- building up to how that will affect the universe. And we said that the mass density of radiation is just the energy density divided by c squared. And that can be taken, really, as a definition of what we call relativistic mass, and hence relativistic mass density. But the important point is that this mass density actually does gravitate the same as any other mass density of the same value. It really does create gravity in the same way. Now I mentioned that things are much more complicated if you want to talk about the gravitational field produced by a single moving particle. That's asymmetric, the velocity of the particle shows up in the equations that describe the metric surrounding a single moving particle. But if you have a gas of particles moving at high velocities, where the velocities nonetheless average to zero, which tends to happen, in a gas at least in the rest frame of the gas. Then that gas will produce gravitational fields, just like a static mass density. Where the mass density is this relativistic energy divided by c squared, the relativistic definition of the mass density. It's also useful to know that the photon-- if we want to describe it as a particle, is a particle of zero rest mass. Which means that it can never be at rest, it always moves at the speed of light. And it also means that its energy can be arbitrarily small, because the energy is proportional of momentum, and the momentum of a photon can be as small as you like. For giving frequency, of course, the energy of a photon is fixed. It's h times nu but if you're allowed to vary the frequency, which you can do if you just look at it in different frames, you can make the energy as small as you like. And the famous equation then, p squared minus e squared c squared, which would have on the right, minus m 0 squared c to the 4th m 0 squared c squared, excuse me-- has zero on the right hand side, because m 0 is 0 And that means that for photons, the energy is just the speed of light times the momentum. And that's a famous relationship that photons obey. Now, thinking about how this gas of photons will behave in the universe, we realized immediately that it does not behave the same way as a mass density of ordinary non-relativistic particles. Which is what we have been dealing with to date. The important difference is that in both cases, the number density falls off like 1 over a cubed, as the universe expands, these particles are not created and destroyed in significant numbers, they just persevere. So the number density of either non-relativistic particles, or photons, just falls off like one over the volume, as the volume increases and the number density dilutes. But what makes photons different from non-relativistic particles, is that a non-relativistic particle will maintain the energy of that particle as the universe expands, but photons will redshift as the universe expands. So each photon will itself lose energy. And it loses energy proportional to one over the scale factor. And that's just because the frequency shift proportionally to the scale factor. And that means that the energy per photon shifts, because quantum mechanically we know that the energy of a photon is proportional to its frequency. So if the frequency redshifts so must the energy. In exactly the same way, 1 over a f t. Yes, question? AUDIENCE: You said previously that neutrinos behave like radiation in the sense that theta energy falls is 1 over a. What is it about them that makes this happen? Because there are also particles with standard kinetic energy, right? PROFESSOR: OK, the question, in case you didn't hear it, is why-- how did neutrinos fit in here? I've made the claim that neutrinos act like radiation in the early universe, but neutrinos have a non-zero mass, so they should obey the standard formulas for particles with nonzero masses. The answer to that is-- there is, I think, a simple answer, which is that as long as the energy is large compared to the mass, particles with masses will still act like massless particles. It doesn't really matter if the mass is zero or not, the key thing, really, is this equation. So if the term on the right hand side is small compared to either of the two on the left, it's not much different from being zero. And that's what happens for neutrinos in the early universe. And we'll see soon that if you go to early enough times, it's true even for electron-positron pairs. They will also act like radiation. Any particle will act like radiation as long as the energy is large compared to the rest energy. So getting back to the discussion of the early universe, if the energy of each photon falls off like one over the scale factor, and the number density falls off like one over the scale factor cubed, it means that the energy density, and hence, the mass density of radiation will fall off like one over a to the fourth, in contrast to the one over a cubed, that we found when we were talking about non-relativistic matter. And that, of course, is going to make a difference. Because those issues play a key role in our discussions about how the universe evolved. An important feature, which we see immediately, is that if we extrapolate backwards in time, since the radiation mass density is falling off like one over a to the fourth, and the matter density is falling off like one over a cubed, it means that as you go back in time, the radiation becomes more and more important relative to the matter, by factor of the scale factor. So if we go back far enough, we will even find a time when the mass density in radiation equaled the mass density in non-relativistic matter. And we calculated about when that would be. We said that the energy density in radiation today is given by this number, 7 times 10 to the minus 14, joules per meter cubed. And I just gave you this number, I didn't derive it yet. We will derive it, probably later today. But for now, we're just accepting it. And that implies we can calculate from that the ratio of the mass densities in radiation and ordinary matter. Here, we use the fact that ordinary matter can be described by having an omega ratio to the critical density of about 0.3. And we know how to calculate the critical density, and that allowed us to calculate the density of ordinary matter. And then this ratio turned out to be 3.1 times 10 to the minus four. So radiation in today's universe is almost negligible in its contribution to the overall energy balance, compared to non-relativistic matter. But if you extrapolate backwards, we know that this ratio will vary as one over the scale factor. And we could figure out what constants to put this equation by putting in the right constant, so that this equation gives us the right value today. Where the right value today is 3.1 times 10 to the minus four. And notice that this works, if we let t be equal to t sub zero, this factors one and we get 3.1 times 10 to the minus four. So these two factors together they have t zero and the 3.1 times 10 to the minus four, are just the right factors to put in to give us the right constant of proportionality in that equation. Having this equation, we can then ask, how far back do we have to go, how much we have to change t, for the ratio to be one? And that's a straightforward calculation. And the ratio of the a is then just one over 3.1 times 10 to the minus four, or 3,200. So if we talk about it in terms of a redshift, which is how astronomers always talk about distances, or times, we're talking about going back to a redshift of 3,200. We can figure out what time, then, corresponds to also, if we know how a f t depends on time. And we do, approximately. For this calculation, I assume for now, we could do better later, and we will-- but I assume for now that we could just treat the period between matter radiation equality, so-called t x and now as being entirely described by a matter-dominated universe. That's only a crude approximation, but it will get us the right order of magnitude. And we'll learn how to do better later in the course. So if we assume that, then t x is just this number to the 3/2 power, cancelling the 2/3, times the age of universe, t naught. And that turned out to be about 75,000 years. So somewhere in the range of 100,000 years, 50,000 years, is the time in the history of the universe when radiation ceased to be more important than matter. And for earlier times than that, the radiation dominated. And that's what we refer to as the radiation dominated era. Any questions? OK, now I think we get on to what is really the important subject that we want to understand, and most of this you did yourself on the homework. But I'll summarize the argument here. We want to understand what this tells us about the Friedmann Equations. And first, we'd like to understand what it says about pressure. Because it turns out that pressure is the crucial issue in determining how fast row falls off as a expands. So if row is proportional to one over a cubed, we can just differentiate that, putting in a constant proportionality temporarily, just to keep track of what we're doing. Since we know how to differentiate qualities and we're less familiar with how to differentiate proportionalities. But what we find immediately, is that row dot is then minus 3, where that 3 is that 3, times a dot over a times row. On the other hand, if row of t falls like one over a to the fourth, row dot is minus 4 times a dot over a times row, just by differentiation. So we get different expressions from row dot, between radiation and matter. And we want to explore the consequences of that difference. It's related to the pressure of the gas, because we can relate the pressure to row dot. Because we know that as a gas expands, it loses energy, which is just equal in amount to pdV. And we illustrated this by a piston thought experiment, but it's true in general. So we can apply this famous formula, dU equals minus pdV, to a patch of the expanding universe. And by a patch I mean some fixed region and coordinate space. So the total energy in that region of coordinate space will be the physical volume, which will be a cubed times the coordinate volume. Which is going to cancel out of this equation on both sides. So it's a cubed times the coordinate volume times the energy density, which is row c squared. The rate of change of that is Du. And then on the right hand side, we have minus p minus b times dV, which is the rate of change of a cubed, again times the coordinate volume that we're talking about. But that will cancel out on the two sides of the equation. So this is really just a description for the universe of the dU equals minus pdV equation. And this can just be rearranged, expanding the time derivatives, to give us row dot, and we get minus 3 a dot over a, times row plus p over c squared. So this tells us how to relate row dot to the pressure. And it tells us that the formula that we started with a long time ago, which just said that row fell off like 1 over a cubed, was synonymous with saying the pressure is zero, for a gas of non-relativistic particles, the pressure is negligible. But for radiation, clearly if we're going to get a four instead of a three, the pressure will be non-negligible. And in fact, it implies that the pressure is exactly equal to one third of the energy density for a gas of radiation. OK, knowing that, we can now look back at the Friedmann equations and ask, how do they stand up? Are they still consistent, or do we have to modify something? And this is really the crucial point. What we know are these three equations, which are the two Friedmann equations and the equation for row dot. And we could see immediately that those equations are not independent of each other. The easiest thing to see is that if we start with the top equation, we could differentiate it. And since the top equation has a dot in it, when we differentiate, we'll get an equation for a double dot. But the equation will also involve row dot, if we take the time derivative of that top equation. But if we know what row dot is, we could put that in, and in the end we'll get an equation for a double dot by itself. And it will in fact agree with the equation on the middle line. Again, things would be inconsistent. Things are consistent, we didn't make any mistakes. If we derive the equation for a double dot by using the first and third of those equations, then we'll get the second of those equations. But the catch is that now we want to modify the third of those equations, the equations for row dot. And then the Friedmann equations as we've written them will not be consistent anymore. Because we'll have a different equation for row dot, we'll get a different equation for a double dot. So we have to decide what gives. What can we change to make everything consistent? And here, the rigorous way of proceeding is to look at general relativity and see what it says, and the answer we're going to write down is exactly what general relativity says. But we can motivate the answer in, I think, a pretty sensible way, by noticing that as the universe expands, we'd expect the energy density to vary continuously, because energies are conserved. And we also expect a dot to vary continuously, because basically, the mechanics of the universe are like ton's laws. And velocities don't change discontinuously. You can apply a force, and that causes velocities to have a rate of change. But velocities don't change instantaneously, unless you somehow apply an infinite force. And the same thing will be true with the universe. On the other hand, accelerations can change instantaneously. You could change the force acting on a particle, in principle, as fast as you want, and the acceleration of the particle will change at that same rate. So if we look at these equations, we would expect that the first equation and the third equation would not be allowed to involve the pressure. Because the pressure basically is a measure of a force. Pressures can change instantaneously. So what you need to do, if we're going to make these equations consistent in the presence of pressure-- which changes the row dot equation, the only equation we can change is the second one. And then we can ask ourselves, what do we have to change it to make the three equations consistent? And this is what you looked at on your homework. And the answer is that the a double dot equation has to be modified to give the equation at the bottom of the screen here. And this is the correct form of the a double dot equation in cosmology. And this is what we'll be using for the rest of the term, this is exactly what you would get from general relativity. As long as we're talking about homogeneous and isotropic universes, this formula as exact as far as we know. OK, any questions about that? Yes. AUDIENCE: Why when we derive that equation do we use-- PROFESSOR: This equation? AUDIENCE: Or the one above that. PROFESSOR: Yeah. AUDIENCE: We use dU equals minus pdV? I mean, I agree with that, but couldn't we use a more complete version? Like, the complete version of the first law of thermodynamics, that dU equals TDS minus pdV. PROFESSOR: OK, yeah. The question was when we wrote down dU equals minus pdV, why did we not include a plus TDS term here, which could also be relevant. The answer is that for the applications we're interested in-- you're quite right, it could be important, but for the applications that we're interested in, which is the expanding gas in the universe, the expanding gas in the universe will be making use of this fact. It really does expand adiabatically, that is, there's nothing putting heat in or out, and everything is remaining very close to thermal equilibrium, which means that entropy does not spontaneously change. So the TDS term we will be assuming is very, very small, and that's accurate. And you're right, if that were not the case, there would be further complications in terms of figuring out what row dot is. Let me point out here that this equation actually does contain a somewhat startling perhaps fact about gravity it says that in the context of general relativity. And that's really the context that we're in, even though we haven't learned a lot of general relativity. But it says that in the context of general relativity, pressures, as well as mass densities, contribute to the gravitational field. A double dot is basically a measure of how fast gravity is slowing down the universe. And this says that there's a pressure. It can also help to slow down the universe. Meaning that pressure itself can create a gravitational field. In the early universe, where we go back to this radiation dominated period, we know that the pressure is one third of the energy density. That says that this pressure term is the same size exactly as the mass density term. So in the radiation dominated phase, the pressure is just as important in effect for slowing down the universe as is the mass density. In today's universe it's negligible. Well, we'll come back that. The dark energy has a non-trivial pressure, but the pressure of ordinary matter in today's universe is negligible. The other important fact about this equation is that energy densities, so far as we know, are always positive. We don't know for sure what the ultimate laws of physics are, but for all the laws of physics that we know, energy densities are positive. On the other hand, pressures actually can be negative for some kinds of material. And we'll talk a little bit more about how to get a negative pressure later. But this formula tells us that positive pressures act the same way as positive mass densities, creating an attractive gravitational field which slows down the expansion of the universe. But if there could be a material with a negative pressure, this same equation, which would presumably still hold, and believe it does, would tell us that that negative pressure would actually cause the universe to accelerate, because of its gravitational effects. Now, we're not talking about the mechanical effects of the pressure. Mechanical effects of pressure only show up when there are pressure gradients, when the pressure is uneven. So the very large air pressure in this room, which really is quite large, we don't feel all, because it's acting equally in all directions. Uniform pressures do not produce forces. So the mechanical effects of the pressure in the early universe, which we're assuming is completely homogeneous, is zilch. There is no mechanical effect. But what we're seeing in this equation is a gravitational effect caused by the pressure. And it's obviously gravitational, business the effect is proportional to Newton's capital G, a constant determining the strength of gravity. So the equation is telling us that a positive pressure creates a gravitational attraction, which would cause the universe to slow down in its expansion. But a negative pressure would produce a gravitational repulsion, which would cause universe to speed up. And we now know that, today-- in fact, for the last five billion years or so-- our universe itself is actually accelerating under the influence of something. And the only explanation we have is that the something that's causing the universe to accelerate is the repulsive gravity caused by some kind of a negative pressure material. And that negative pressure material is what we call the dark energy. And we'll talk a little more later about what it is. It's very likely just vacuum energy. But we'll come back to that later in the course. Yes? AUDIENCE: When we looked at the toy example of the piston in the cavity, the pressure of the gas was pushing outwards against the wall of the container. But we can't have that view of the universe, really, because there's no exteriors in the universe. PROFESSOR: There's no walls, right. AUDIENCE: So how should we view pressure in the sense that-- PROFESSOR: OK. Yeah. OK, the question is, in our toy problem involving the piston, we had walls for the pressure to push against. And that was where the energy went. It went into pushing the walls. When we're talking about the universe, there are no walls. How does that analogy work? What plays the role of the walls? And the answer, I think, is that the role of the walls, when we're talking about the universe, first of all, you can ignore it if you just took in a small region. You could still just say, the small region is pushing out on the regions around it. And I think that's enough to make the logic clear. But it still leaves open the question of, ultimately, where does this energy go? So saying it goes from here to there doesn't help you unless you know where it goes after it goes from and there and there. So you might want to ask the question more generally, where does the energy ultimately end up? And then I think the answer is that it ends up in gravitational potential energy. You could certainly build a toy model, where you just have a gas in a finite region, self-contained under gravity. And then you'd have to make up some kind of a mechanism to cause it to expand. But when you cause it to expand, you'll be pulling particles apart, which are attracting each other gravitationally. And that means you'll be increasing the gravitational potential energy as you pull the gas apart. So I think, ultimately, the answer is the energy imbalance that we seem to be seeing here is taken up by the gravitational field so that, all in all, energy still conserved. Any other questions? OK, in that case, let us continue on the blackboard. OK, first thing I want to look at it is just the behavior of a radiation-dominated flat universe. So a flat universe is going to obey H squared equals 8 pi over 3 G rho, and then the potential minus kc squared over a squared. Hard to write dotted lines on the blackboard. But this potential term is not there, because k equals 0. We're talking about the flat case. So for a flat case, we could just express H in terms of rho. And we know how rho behaves for radiation. Rho falls off as 1/a to the fourth. So H squared is proportional to 1/a to the fourth. That means that H itself is proportional to 1/a squared. So we can do that. a-dot over a is equal to some constant over a squared, a-dot over a being H. And now we can multiply both sides by a, of course. And we get a-dot is equal to a constant over a. And this we can integrate. The way to integrate is to put all the a's on one side and all of the dt's on the other side. So we get ada, writing this as da/dt. So ada is equal to the constant times dt. And then, as we've done before, when we're talking about matter, it's the same calculation there. I just did different power of a appears so we'd know how to do it. Integrating, we get 1/2a squared is equal to the constant times t, and then plus a new constant of integration, constant prime. Now we make the same argument as we've made in the past. We have not yet said anything that determines how our clocks are going to be set. So we can choose to set our clocks in the standard way, which is to set our clocks so that t equals 0 corresponds to the moment where a is equal to 0. And if a is going to be equal to 0 when t is equal to 0, it means constant prime is going to be equal to 0. So by choosing the value of constant prime, we really are just determining how we're going to set our clocks, how we're going to choose the 0 of time. And we'll do that by setting constant prime equal to 0. And then we get the famous formula for a radiation-dominated universe, a of t is just proportional to the square root of t, or t to the 1/2 power. And this is for a radiation-dominated flat universe, replacing the t to the 2/3 that we have for the matter-dominated flat universe. Once we know that a is proportional to the square root of t-- and for the flat universe, the constant proportionality mean nothing, by the way. It's not that we haven't been smart enough to figure out what it is. It really has no meaning whatever. You could set it equal to whatever you want, and it just determines your definition of the notch, your definition of how you're going to measure the comoving coordinate system. Once we know this, we should know pretty much everything. So in particular, we can calculate h, which a-dot over a. And the constant proportionality drops when we compute a-dot over a. As we expect, it has no meaning. So it should not appear in the equation for anything that does have physical meaning. So H is just 1/2t, the 1/2 here coming from differentiating the t the 1/2 power. We can also compute the horizon distance. So the physical horizon distance, l sub p horizon, where p stands for physical, is equal to the scale factor times the coordinate horizon distance. And the coordinate horizon distance is just the total coordinate distance that light could travel from the beginning of the universe. And we know the coordinate velocity of light is c divided by a. So we just integrate that to get the total coordinate distance. So it's the integral from 0 to t of c over a of t prime dt prime. And since a of t is just t to the 1/2, this is a trivial integral to do. And the answer is 2 times c times t. So in a radiation-dominated universe, the horizon's distance is twice c times t. For a static universe, the horizon distance would just be c times t. It would just be the distance light can travel in time t, but more complicated in an expanding universe. For the matter-dominated case, we discovered that the horizon distance was 3ct, if you remember. For the radiation-dominated case, it's 2ct. And finally, an important equation is that, going back to here, where we started, this equation relates H to rho. We found out that, merely by knowing the universe is radiation-dominated, without even caring about what kind of radiation it is, how much neutrinos, how much photons, whatever-- doesn't matter-- merely by knowing the universe is radiation-dominated, we were able to tell that H is 1/2t. And if we know what H is, that formula tells us we also know what rho is. So without even knowing what kind of radiation is contributing, we know that, for a radiation-dominated universe, rho is just equal to 3 over 32 pi Newton's constant G times time, little t, squared. It's rather amazing that we can write down that formula without even knowing what kind of radiation is contributing. But as long as that radiation falls off as 1 over the scale factor to the fourth, and as long as we know the universe is flat, then we know what that energy density has to be. This is crucial here, by the way. The energy could be anything if we did not assume that the universe was flat. OK, any questions about this? Yes? AUDIENCE: If we assumed that it was almost flat, would we be able to have any bounds on it? PROFESSOR: OK, question is, if we assumed that it was almost flat, would we be able to have any bounds on it? The answer is, yeah, if you were quantitative about what you meant by "almost flat," you could know how almost true that formula would have to be. OK, if there are no other questions, I want to switch gears slightly now and go back to talk about some of the basic underlying physics that we are going to need, and in particular, the physics of black-body radiation. So this is really just a little chapter of a stat mech course that we're inserting here, because we need it. And because it comes from another course, we're not going try to do it in complete detail. But I'll try to write down formulas that make sense. And that will give us what we need to know to proceed. So that will be the goal. So what is black-body radiation? The physical phenomenon is that, if one imagines a box with a cavity in it-- that's supposed to be a box with a cavity in it, in case you can't recognize the picture-- if the box is held at some uniform temperature t-- t is temperature-- then is claimed and verified experimentally that the cavity will fill up with radiation-- in this case, we're really just talking about electromagnetic radiation-- the calving will fill up with electromagnetic radiation whose characteristics would be determined solely by that temperature t and will therefore be totally independent of the material that makes up the box. Roughly speaking, I think the way to think about it is to say that the box will fill up with radiation at temperature t. And saying that the radiation has temperature t is enough to completely describe the radiation. It doesn't matter what kind of a box that radiation is sitting in. So the box will fill with radiation at temperature t. And that radiation is called black-body radiation. Like many things in physics, it has a variety of names, just to confuse us all. So it's also called cavity radiation, which makes a lot of sense, given the description we just gave. And it's also sometimes called just thermal equilibrium radiation. This is radiation at temperature t. I haven't really justified the word "black-body" radiation yet, so let me try to do that quickly. The reason why it can be called black-body radiation-- and this will be important for some things in cosmology; I'm not sure if it will be important to us or not, but certainly important to know-- the reason why it's called black-body radiation is because we imagine inserting into this cavity a black body, in the literal sense. What is the literal sense of a black body? It's a body which is black in the sense that all radiation that hits it is absorbed. Now, this black body is still going to glow. If you heat a piece of iron or something to very high temperatures, you see it glow. That glow is not reflection. That glow is emission by the hot atoms in the piece of iron or whatever. Emission is different from reflection. When we say it absorbs everything, we mean it does not reflect anything. But it will still admit by thermal de-excitation. The crucial distinction between reflection and thermal emission is that reflection is instantaneous. When a light beam comes in, if it's reflected, it just goes back out instantaneously. Emission, thermal emission, is a slower process. Atoms get excited, and eventually, they de-excite and emit radiation. So it takes time. And that's the distinction. We're going to assume that this body is black in the sense that there's no reflection. OK, now we're going to make use of the fact that we know that thermal equilibrium works. That is, if you let any isolated system sit long enough, it will approach a unique state of thermal equilibrium determined by its constituents, which you've put in to begin with, but otherwise independent of how exactly you arrange those constituents. So if we put in, for example, a cold black body, it will start to get harder, warming up to the same temperature as everything else. If we put in an extra hot black body, it would emit energy and start to cool down to the temperature of everything else. But eventually, this black body will be at the same temperature as everything else. And we're going to be assuming here that the box itself is being held at some fixed temperature t. So wherever energy exchange occurs because of this black body, it will be absorbed by whatever is holding the outer box at the fixed temperature. So in the end, if we wait long enough, this black body is going to acquire the same temperature as everything else and hold that temperature. Now, if it's holding that temperature, it means that the energy input to the box, to the black body, will have to be the same as the energy output of the black body. Now, the black body is going to be absorbing radiation, because we have radiation here, and we said that any radiation that hits it is absorbed. That was the definition of "black." So it's clearly absorbing energy. If it's not going to be heating up-- and we know that it's not, because it's in thermal equilibrium; the temperature will remain fixed-- in order for it to not heat up, it has to radiate energy, as well. And the energy it radiates has to be exactly the same as the energy it's absorbing once it reaches thermal equilibrium. So in equilibrium, the black body, BB, radiates at same rate that it absorbs energy. This radiation process is this slow process of thermal emission. There are atoms inside this black body that are excited. Those atoms will de-excite over time, releasing photons that will go off. And the important thing about that slow mechanism is that, if we imagine taking this black body out of its cavity, but not waiting long enough for its temperature to change-- so we'll assume its temperature is still the same, t. So this is a picture of the same black body at temperature t, but now outside the cavity. Its radiation rate is not going to change when we take it outside the cavity, because the radiation was caused by things happening inside the black body, which are not changed when we put it in or out of the cavity. So it will continue to radiate at exactly the same rate that it was radiating when it was in the cavity. And that means it's going to radiate at exactly the same rate as the energy that it would have absorbed if it were bathed by this black-body radiation. So essentially, it means it will emit black-body radiation with exactly the intensity that the black-body radiation would have on the outside if the black body were still inside the cavity. So it radiates with exactly the same intensity as the energy that it would receive if it were inside the cavity. And furthermore, you could even elaborate a bit on his argument to show that the radiation that it radiates has exactly the same spectrum, exactly the same decomposition into wavelengths, as the black-body radiation inside the cavity. And the way to see that is to imagine surrounding this black body by absorption filters that only let through certain frequencies. And the point is that, no matter what frequencies you limit going through this filter, you have to stay in equilibrium. It will never get hotter or colder. So that means that each frequency by itself has to balance, has to have exactly the same emission as it would have abortion if the black body were just exposed to black-body radiation surrounding it. So it radiates black-body radiation. And the intensity and spectrum must match what we call black-body or cavity radiation. So the cavity radiation has to exactly mimic the radiation emitted by this black body. And that's the motivation for calling it black-body radiation. Now, if this black body absorbed some radiation and reflected some, then it would emit different radiation. So it is important that this body be black, in the sense that it doesn't reflect anything. All radiation hitting it is absorbed. And only under that assumption do we know exactly what it's going to emit. Yes? AUDIENCE: So is this only true for right after you take it out? PROFESSOR: Well, it will start to cool after you take it out. And as it cools, its temperature will change. But if you account for the changing temperature, it will be true at anytime, actually. But the temperature will change. AUDIENCE: Because, in the black cavity, it has things exciting it. And when you take it out, there's no photons, no constant radiation to excite it. So it can radiate-- PROFESSOR: Yes. Once you take it out, it's no longer being excited. And I think, technically, you're right. Once you take it out, it will not only cool, but it will cease to be at a uniform temperature. And that's basically what we're saying if we're saying that the atoms that are excited won't necessarily be in the right thermal distribution as they would be if it was on the inside. That would be a statement that is not any longer in thermal equilibrium. But as long as the radiation is slow, you could just account for the changing temperature. You would know how it radiates. And I think that's a very good approximation. Although in principle, it will cease to be in thermal equilibrium, as soon as you take it out, the edges will be cool, and the center will be hot. And you'd have to take into account all of those things to be able to understand how it radiates. Any other questions? OK, next, I want to talk a little bit about what this black-body radiation is. And one can begin by trying to understand it purely classically, which, of course, is what happened historically. In the 1800s, people tried to understand cavity radiation or black-body radiation using classical physics, Maxwell's equations, to describe the radiation. And then, in a nutshell-- we're just trying to establish basic ideas here-- one can try to treat a field statistical-mechanically by imagining not fields in empty space, but fields in some kind of a box. In this case, it doesn't necessarily have to be the cavity that we're talking about. It could be a big box that just enclosed the system somehow to make it easier to talk about. And in the end, you could imagine taking that box to infinity, this theoretical box that you use to simplify the problem. But once you put the system in a box, then a field, like the electromagnetic field, can always be broken up into normal modes, standing wave patterns that have an integer or a half integer number of wavelengths inside the box. And no matter how complicated the field is inside the box, you could always describe it as a superposition of some set of standing waves. In general, it takes an infinite number of standing wave components to describe an arbitrary field-- that is, with shorter and shorter wavelengths. But you can always-- and this is Fourier's theorem-- you can always describe an arbitrary field in terms of the standing waves. And that's good for the point of view of thinking about statistical mechanics, because you could think about each standing wave almost as if it were a particle. It really is a harmonic oscillator. So if you think you know the statistical mechanics of harmonic oscillators, each standing wave in the box is just a harmonic oscillator, so simple. We now try to ask what is the thermodynamics of this system of harmonic oscillators. And the rule for harmonic oscillator is simple. Stat mech tells you that you have 1/2 kT per degree of freedom in thermal equilibrium. The energy of a system should just be 1/2 kT per degree of freedom. Having said that, all the complicate questions come about by asking ourselves what is meant by degree of freedom. But for the harmonic oscillator, that has a simple answer. A harmonic oscillator has two degrees of freedom-- the kinetic energy and the potential energy. So the energy of a harmonic oscillator should just be kT per degree of freedom. And we could apply that to our gas in the box and we could, ask how much energy should the gas absorb at a given temperature? What should be the energy density of the gas at a given temperature-- this gas of photons. But it was noticed in the 1900-- the 1800s that this doesn't work because there's no limit to how short the wavelengths can be. And therefore, there's not just some finite set of harmonic oscillators. There's an infinite set of harmonic oscillators where you have more and more harmonic oscillators at shorter and shorter wavelengths ad infinitum, no limit. And that came to be known as Jean's Paradox. So what it suggests is that if this classical stat mech worked-- which obviously it's not working. But if it did work, it would mean that as you tried to put a gas in just an empty box in contact with something at a fixed temperature, the box would absorb more and more energy without limit. And ultimately, it would presumably cause the temperature of the whatever is trying to maintain the temperature to go to 0 as energy gets siphoned off to shorter and shorter wavelengths of excitations. That obviously isn't the way the world behaves. We'd all freeze to death if it did. So something has to happen to save it. And it wasn't at all obvious for many years what it was that saves it. But this Jean's Paradox turns out to be saved by quantum mechanics. And the important implication of quantum mechanics is that the energy of a harmonic oscillator is no longer allowed to have any possible value, but is now quantized as some integer times h times nu, the frequency-- h being Planck's constant, nu being frequency, n being some integer where this integer might be called the excitation level of the harmonic oscillator. Depending how you choose your 0, you might have an n plus 1/2 there. But that's not important for us right now. It will be important later actually. But for now, we'll just allow ourselves to readjust the 0 and just think of it as n times h nu, or h bar omega. Now, this makes all the difference, statistical mechanically. One can apply statistical mechanics using basically the same principles to the quantum mechanical system. And the key thing now is that for the very short wavelengths, which are the ones that were giving us trouble-- the infinities came to short wavelengths. For the short wavelengths where nu is high, h nu is high. And it means that there's a minimum ante that you could put in to excite those short wavelength harmonic oscillators. And it's a large number. You either put in a large amount of energy or none at all. Quantum mechanics doesn't let you do anything in between. Now remember, the classical mechanics answer was that you have kT in each harmonic oscillator. And kT would be small compared to h nu, if we're talking about a very short wavelength. So the classic answer is just not allowed by quantum mechanics. You either have to put in nothing or an amount much, much larger than the classical answer. And when you do the statistical mechanics quantum mechanically, which is not a big deal really, you find that when you're confronted with that choice, the most likely answer is to put in no energy at all. So quantum mechanics freezes out these short wavelength modes. And then the n produces a finite energy density for a gas of photons. Yes? AUDIENCE: Seems like if you were to sum over like all the possible wave numbers, that-- well, so the energy is inversely related to wavelength, right? So even if you quantize it, like for large wavelengths, isn't the sum still like a sum of one over lambda wavelength, with its derivative? PROFESSOR: You're saying, isn't there also a divergence at the large wavelength n? AUDIENCE: Because that sum doesn't seem like it would work. PROFESSOR: Right. No, that's important. The reason that's not a problem is that if you're talking about the energy in a box, the wavelength can't be bigger than the box. The largest possible wavelength is twice the box so that half a wavelength fits in the box. If you're talking about the energy in the whole infinite universe, then we expect the answer to infinite. And it is. There's no problem with having infinite total energy if you want to have a finite energy density throughout an infinite universe. So the size of the box cuts off the large wavelengths. And quantum mechanics cuts off the small wavelengths. So in the end, one does get a finite answer for the energy density of black-body radiation. And that's crucial for our survival, crucial for the existence of the universe as we know it, and also crucial for the calculations that we're about to do. OK, so when one does these calculations initially for photons only, what we'd find is that the energy density is equal to a fudge factor, which I'm going to call g. And you'll see later why I'm introducing a fudge factor. For now, g is just 2. But later, we'll generalize the application of this formula, and g will have different values. But for now, we're dealing with photons. There's a factor of 2 there, but I'm going to write 2 as g, writing g equals 2 underneath. And then the pi squared over 30-- you can really calculate this-- times kT to the fourth power divided by h bar c cubed, h bar being Plank's constant divided by 2 pi and little k being Boltzmann constant. So this is calculated just by thinking of the gas in a box as a lot of harmonic oscillators and applying standard stat mech to each harmonic oscillator, but you apply the quantum mechanical version of the stat mech to each harmonic oscillator. And you can also find, by doing the same kind of analysis, that the pressure is 1/3 the energy density, which we also derived earlier by different means. And it's all consistent so you get the same answer every time, even if you think about it differently. So here, I have mine deriving it directly from the stat mech. You can also, from the stat mech, calculate the number density of photons in thermal equilibrium. And that will be equal to -- again, there's a factor of 2. But this time, I'll call the factor of 2 g star, where g star also equals 2 for photons. But when we generalize these formulas, g will not necessarily equal g star, which is why I'm giving it two names. And this g star multiplies zeta of 3, where zeta refers to the Riemann zeta function, which I'll define in a second, divided by pi squared times kT cubed divided by h bar c cubed. OK. OK, so I need to define this zeta of 3. It's 1 over 1 cubed plus 1 over 2 cubed plus 1 over 3 cubed plus dot dot dot. It's an infinite series. And if you sum up that infinite series, at least to three decimal places, it's 1.202. OK, then there's one other formula that will be of interest to us. And that'll be a formula for the entropy density. Now, if you've had a stat mech course, you have some idea of what entropy density means. If you have not, suffice it to say for this class that it is some measure of the disorder in the sense of the total number of different quantum states that contribute to a given macroscopic description. The more different microstates there are that contribute to a macroscopic description, the higher the entropy. And the other important thing about entropy to us besides that vague definition-- which will be enough-- but the important thing for us is that under most circumstances, entropy will be conserved. It's conserved as long as things stay at or near thermal equilibrium. And in the early universe as the universe expands, that's the case. So for us, the entropy of our gas will simply be a conserved quantity that we can make use of. And we will make use of it in some important ways. And we could write down a formula for the entropy density of photons. And it's g, where this g in fact the same g as over there. It is related to the energy. So it's the same g that appears in two cases, 2 in both cases for protons by themselves. And then there are factors that you can calculate-- 2 pi squared over 45 times k to the fourth T cubed over h bar c cubed. OK, this time, the number of k's and T's do not match. That's mainly due to the conventions about how entropy is defined. It's not really anything deep. I might mention at this point that the 2's that I've been writing for everything-- g equals 2, g star equals 2-- the reason those 2's are written explicitly rather than just absorbing the factor of 2 into the other factors is that photons are characterized by the fact that there are two polarizations of photons. So if I have a beam of photons, they could be right-handed or left-handed. And anything else could be considered a superposition of those two. So there are two independent polarizations. And it's useful to keep track of these formulas as the amount of energy density per polarization. Thus, different kinds of particles will have different numbers of polarizations. So if we know the amount per polarization, we'll be able to more easily apply it to other particles. Yes? AUDIENCE: Sorry, just to kind of bring up the same question, if we-- I read that in the early universe, the temperature is constantly changing or it's cooling. PROFESSOR: Right. AUDIENCE: So then if the temperature is changing, then how can we say that the entropy is constant? PROFESSOR: OK, important question-- we'll be getting to it very soon. But since you asked the question, I'll ask it now. The question was if the universe is expanding, and the entropy density is going down because it thins, how can that happen-- I guess it was asked the other way around. If the temperature is falling, how can entropy be conserved if this is the formula for entropy density? And the answer-- when I tell you, you'll see it's obvious. We don't expect the entropy density to be conserved if the entropy is conserved. The entropy thins out as the universe expands. So if we just had a gas with nothing else changing, we expect the entropy density to go down like 1 over the scale factor cubed, just like the number density of particles. So if s is going to go down like 1 over the scale factor cubed, that would be consistent with this formula if the temperature also fell as 1 over the scale factor. So that to cubing it made things match. And that's what we'll find. The temperature falls off, like 1 over the scale factor. And that's consistent with everything that we said about energy densities and so on. OK. Next thing I want to talk about is neutrinos, which I told you earlier contributes in a significant way to the radiation energy density in the universe today. Neutrinos are particles which for a long time, were thought to be massless. Until around 2000 or so, neutrinos were thought to be massless. Now we know that in fact, they have a very small mass, which complicates the description here. It turns out that cosmologically, neutrinos still act as if they were massless for almost all purposes and for really all purposes that we'll be dealing with in this class, although if we were interested in the effects of neutrinos on structure formation, we'd be interested in whether or not the neutrinos have a small mass or whether it's smaller than that. We know it's non-zero. We don't know what the neutrino mass is, by the way. What we actually know from observations is that there are three types of neutrinos. And those are called flavors. And I'll use the letter nu for the word neutrino. And those three types of neutrinos are called nu sub e, called the electron neutrino; nu subbed mu, called the muon neutrino, and nu sub tau, called the tau neutrino. And these letters e, mu, and tau link to the names of particles. This is the electron neutrino. This is the muon neutrino connected to a particle called the muon, which is like the electron but heavier and different. And this is linked to a particle called the tau, which is also like the electron but much more heavier but otherwise similar in its properties. And the neutrinos are linked in the sense that when a neutrino is produced, depending on how you start, it is very typically produced in conjunction with one of these other particles. So an electron neutrino is typically produced in conjunction with an electron. And similarly, a muon neutrino is typically produced in conjunction with a muon. And a tau neutrino is typically produced in conjunction with a tau. Now, what does this have to do with neutrino masses? We've never actually measured the mass of neutrino. So we only know that they have mass indirectly. What we have seen is one flavor of neutrino turn into another flavor. And it turns out, in the context of quantum field theory and I think this does make a certain amount of sense just by intuition, if a particle is massless, it can never change into anything. The process by which one changes into another is pretty quantum mechanical and a little hard to understand anyway. But if the particles were really massless, they would move at the speed of light. And if the particles were moving at the speed of light, if the particle had any kind of a clock on the particle, that clock would literally stop with the particle moving at the speed of light. So if particles are massless, any internal workings that that particle might have to be frozen. That is, if it's a clock, it has to be a clock that's stopped completely. And for reasons that are essentially that, although they can be made more formal and more rigorous, a truly massless particle could never undergo any kind of change whatever. It would have to stay exactly like it looks like to start with. Because it just has no time. So the fact that these neutrinos turn into each other implies that they must have a nonzero mass. It must not really be moving at quite the speed of light. And that's the way the formalism works. And we could set limits on the masses based on what we know about the transitions between one kind of neutrino and another. Yes? AUDIENCE: How do we explain photon decaying to an electron-positron pair then? PROFESSOR: A photon decaying to an electron-positron pair? AUDIENCE: Yeah. PROFESSOR: The answer is a free photon never does decay to an electron-positron pair. Photons can collide with something and produce an electron-positron pair. But that collision, that's a more complicated process. What I'm saying is-- I'm sorry. This process of conversion-- maybe I should have clarified-- happens just as the neutrinos travel. It's not due to collisions. Due to collsions, complicated things can happen whether the particle is massless or not. But a massless particle simply in transit cannot undergo any kind of transition. And these neutrinos are seen to undergo transitions simply being in transit without any collisions. In terms of my clock analogy and a stopped clock, I think the reason the photon can convert into electron-positron pairs if it collides with something is that when it collides, it essentially breaks the clock. You don't have a photon that's just moving along without time anymore. OK, so these neutrinos have masses. And maybe I should, at this point, write down some bounds on these masses. m squared 21 times c to the fourth is equal to 7.50 plus or minus 0.2 times 10 to the minus 5 electron volts squared. These numbers come from the latest particle data tables, which I gave references for in the notes. So this is the difference of the mass squared. And delta m 23 squared times c to the fourth to turn it into the square of an energy is 2.32 plus 0.12 minus 0.08 times 10 to the minus third electron volts squared. So one thing you notice immediately is that these are incredibly small masses. Remember, the proton weighs 938 MeV, three million electron volts. And these are fractions of one electron volt. So by the standards of particle physics, these are unbelievably small energies, unbelievably small mass differences. But they're there. They have to be there for the physics we know to make sense. The other thing that you may notice about this notation-- which I don't want to elaborate on but I'll just mention-- this is called 21. This is called 23. There's no 1, 2, or 3 there. There's an e and a mu and a tau. The complication here is something very quantum mechanical. The e, mu and tau labels are labels which basically label the neutrino according to how the neutrino is created. It turns out that their mass eigenstates-- states which actually have a definite mass-- are not the e, the mu or the tau. In fact, if the e had a definite mass, that would be saying that an e would just propagate as a particle with a certain mass. It would not convert into anything else. The fact that an e converts into other particles-- a nu sub e converts into other particles is really the statement that nu sub e is not a state with a definite mass. But there are states with definite masses which could be expressed quantum mechanically as superpositions of these flavor eigenstates. So nu sub 1, nu sub 2 and nu sub 3 are states of neutrinos that have definite masses. And each one of them is a superposition of nu sub e, nu sub mu and nu sub tau. Yes? AUDIENCE: How come we don't have a delta n from 31? PROFESSOR: Good question. I think it's just the lack of knowledge. I don't think there's any reason it's not defined. I'm sure it is defined. I think it's just lack of knowledge. And if I knew more details about how these things were measured, I could give you a better story about that. But I don't, frankly. So in the end, it's a rather complicated quantum mechanical system which we're not going to go into any details about. What more should I say today? OK, let me just mention for today and we'll continue next time after the quiz, for our purposes, we're going to treat these neutrinos as if they're massless. And it turns out that that's actually extraordinarily accurate from the point of view of cosmology, at least for the kind of cosmology that we're doing where we're just interested in the effect of these neutrinos on the expansion rate of the universe. And treating them as massless particles, I will shortly give you the formulas for how they contribute to the black-body radiation. But it I think there's no point in my writing them now. I'll just write them again in the beginning of next period. But they do contribute to the black-body radiation and in a way that we actually know how to calculate. And they have a noticeable effect on the evolution of our universe. OK, that's all for today. Good luck on the quiz on Thursday. I'll be here to help proctor. I think Tim [INAUDIBLE] will be here too. And I'll see you more intimately either at my office hour tomorrow or at lecture a week from now.
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https://ocw.mit.edu/courses/5-08j-biological-chemistry-ii-spring-2016/5.08j-spring-2016.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JOANNE STUBBE: So the key question is, do these-- and I think this is a general question you can ask, metabolically, inside any cell is do these enzymes that are on different polypeptides cluster. And is there an advantage, kinetically or whatever, is there some kind of an advantage to have clustering inside the cell. And where have you seen something like this before? Do you remember? Do you remember the section. Where have you see multi enzyme complexes and clustering before? Yeah? So or the classic one, PKS has been around. But it's completely analogous to fatty acid synthesis, right? And so in bacteria, they're all single polypeptides. In humans, they're all activities that are on single chains. OK, so that's sort of what's going on here with the purine pathway. We'll see there are ten. This just sort of helps us focus if we get to the data at the end, which I think we will from what we did the last time. Is that you have six enzymes for 10 activities. So that just means you have more than one enzyme per polypeptide, OK? And so I guess the key thing that I wanted to focus on is do you think it's important to cluster? Here's a pathway. These are the names. We're not going to go through the names. The names really aren't important for what we're doing. There'll be two names that we'll be looking at over and over again. These are the papers that you guys did, in fact, read. One is the original paper, which got a lot of press. And I just want to show you that there have been there's actually been four papers published in the last six months on this topic. And one of which was published. Yeah. One of which is published with Science, where they are now claiming that this complex is localized to the mitochondria. OK, so you take pictures. And it's that this is looking at super resolution fluorescence methods. And you can clearly see clumps of blobs focused on the mitochondria. Why would you want it to be at the mitochondria? So then you have to ask your question. You might need purines, because that's where you make, through a proto mode of force in respiration. Remember, when you convert oxygen to water, you get a huge amount of energy released. You make ATP. But it's going to be made from something. So maybe you would want. That's the way they rationalize it. And they do. And then they connect it to the other latest hot topic, which is EM torque, which is the major signaling switch for fatty acids and for amino acids. And now in the last two years, purines and pyrimidines, I decided-- I've done a lot of reading about it, decided and believe. I mean, I believe it. But I don't believe the connections yet. So again, this is what you're going to see in the next decade is connecting signaling to primary metabolic pathways, like the purine pathway. That's going to be a big thing and how do you connect them is going to be the key question. So anybody that wants to do some more reading, this is an updated version. I kept updating this three or four times. And so I think these are the key questions we want to focus on. And so what I'm going to do. Well, define the questions a little bit and whether the things we need to think about just to determine whether this is really important, biologically. Then we'll define fluorescence and what you can do with fluorescence. And then we'll come back and look at the data. In the paper, we're also going to look. We probably won't get all the way through all of the data. But we will look at some of that data again in either the next lecture or Wednesday's lecture. So you will see it again if we don't get through the data. So they claim they have a multi enzyme complex. Did you believe that from the data? I mean, they didn't look at all 10 enzymes simultaneously, right? Or six enzymes. AUDIENCE: Whenever they show an image of the cells, and then they fluorescent, trying to show local sections, it's always so hard for me to figure out-- JOANNE STUBBE: What you see. AUDIENCE: Yeah. JOANNE STUBBE: OK, so that was said. We'll look at some of those pictures. But I completely agree with that, that you can't see anything from fluorescence pictures. So everybody, all chemists or chemical biologists, now have huge numbers of these pictures in their papers. And with Alice's group, I'm always on their case that I can't tell a damn thing. This is on thesis. I can't see anything. And Alice says she can't see anything either. So it's very hard to see things in these pictures. The contrast isn't very good. And what her lab now does is it goes to EM, where you can see things much more clearly. The fluorescence things are tough. So you're not the only one. And if somebody says it's obvious that this. And you don't see it. Raise your hand and say, I don't see it. Show me what I should be looking at, OK? So that's a good take home message, because everybody and his brother is doing this. And this goes back to knowing how to do it correctly. We're not going to talk about any of that stuff. I mean every one of the methods I'll sort of show you that's out there. You have to really study it to make sure you're handling it correctly. So I mean I think, to me, this has been a problem that I've been interested in. And I started working on this a long time ago in the purine pathway is not are things sticking together important. Actually, I don't think those are important. You immunoprecipitate all these things. OK, so you say obviously these are talking to each other. But the key thing is the kinetic competence. And lots of times when you mess around. You get it in a state. And you post-translationally modify it. So it's sitting in this state there probably isn't on the pathway. You need to then show it's on the pathway. So I think a lot of protein, protein interactions, especially now that we know that proteins move around. And they're in this complex. And they're in that complex. And they're in that complex. The key, I think, is to transient interactions. So why? So this is just my personal take on this. I'm letting you think about this. But is it easy to look at transient interactions? No. OK, so anyhow, I think people need to start doing a lot more thinking about how to look at that. And one way you could look at transient interactions is if you can fluorescent label something. And they come together on a certain timescale and then move apart. And can you do that inside the cell with the right spatial and time resolution. You might be able to start looking at that. So that the methods that are being developed and continue to be developed are incredibly powerful. And I think will allow us to ask this question happens inside the cell, which you've all seen pictures in your introductory courses of, man, how complicated the inside of the cell is. That's part of the issue. So the issue is that you might have a purinosome somewhere in the cell, depending on the growth conditions. But those enzymes might be involved in other things. And so you have only a tiny amount of it, as opposed to trying to make the cell by growth conditions, putting it into all one state. So you can see it. So the question is, how do you see it? And so that's the key issue. And if you perturb it enough. And you do see it. Then you have to ask the question. And this is a question that you might want to think about in terms of these two papers you were reading. That's what. If you do this, that's what the Marcotte paper said, that the cells were incredibly sick when you take out all the purines. And in fact, Alice is-- because of this mitochondria connection between the purinosome and Alice's interest in the mitochondria, she's had people trying to repeat this. And Vicki Hung worked on this and couldn't repeat it. So she didn't spend that much time on it. But all I'm saying is it's not a slam dunk to be able to do this. But that being said, I think this has been an issue that people have been thinking about for decades. And it's just really hard to test experimentally inside the cell. This is where we need chemical biologists to figure out new ways of being able to look at this, so that you can actually make a measurement that's interesting. So I guess the question I want to start with, before we researched looking at fluorescence, is why do you think it would be important to do this. Or do you think it would be important to have a complex. What's the advantage of doing that? Yeah. AUDIENCE: You were saying in lecture that you want to increase the effective molarity. And so by having all these things right next to each other, there's-- obviously you're going to have more interactions per second. JOANNE STUBBE: Well, you may or may-- you may not. It depends. So I think this is the key question. Is diffusion fast inside the cell? AUDIENCE: Yeah. JOANNE STUBBE: Yeah. It's still very fast. For small molecules, it's incredibly fast. Even for proteins, it's incredibly fast. So even if this guy is over here. If you're turning over here at a much slower rate, and you have enough of them so you can interact at diffusion control, do you need this organization? There are a lot of smart people who think you don't need that. There are a lot of smart people who think you do need that. But this is the question I want to raise. However, so catalytic efficiency is absolutely it. But where might you really need catalytic efficiency. And so that goes back. There are places where you really need this. So if you look at the first intermediate in the pathway, this guy, what do you think about that guy? Do you think he's stable? So if you look at the first intermediate in the pathway, which we'll talk about next time. So this is amino phosphoribosine-- phospho-- I'm drawing a complete-- I think I'm tired. Anyhow it's the amino analogue of ribose 5-phosphate. Phosphoribosylamine, that's what it's called, PRA. OK, do you think that's stable, as chemists? So what do you think that could do? AUDIENCE: Could you release the amine? JOANNE STUBBE: Yeah, so how would you do that? AUDIENCE: So if it's proteinated, and then the ring opens till-- JOANNE STUBBE: OK, so that would be one way. You want to release it that way. OK, so it would have to be under conditions. We could do that under neutral conditions. What else can happen to this ring. That doesn't happen. There are lots of ways this molecule can break down. OK, it depends on the details of the environment. How else could this molecule ring open? You wouldn't need to ring open here. You just go through an oxocarbenium ion and have water attack. So what if it opens that way. So that's the way you form aldehydes. All sugars are in equilibrium with aldehydes. These things are in equilibrium. So you have a ring open species. But then what happens if a ring closes? It can ring close from the top face or the bottom face. You have an imine. What can happen to the imine? It can hydrolyze. This molecule, and this is a molecule my lab worked on decades ago, has a half life in solution of 10 seconds. So is 10 seconds short or long, biologically? What do you think? AUDIENCE: I'm going to say short. But I don't know JOANNE STUBBE: Yeah. I think it's amazingly long inside the cell. So I think as a chemist, nobody could ever. Nobody ever saw this intermediate. My lab was the first one that figured out how to look at it. And I won't go through that. But the fact is that 10 seconds is a long time inside the cell, if you think about how small the cell is and how fast diffusion is. OK, so one place, though, where you might want to have organization is if you have something chemically really unstable. OK, because then when you generate it, it could potentially be passed off, or as you say in the immediate vicinity, it's a competition. But if it's right there, you're effective molarity, that would get into that first question, the effective molarity. It would be high enough to get passed on. It would get high enough to get passed on to the next guy in the pathway. So that would be one thing is instability. And in the purine pathway. We will go through this a little bit. But really, that's one of the things that's most amazing about Buchanan's elucidations of the pathway is only intermediates are unstable. Nobody, still, if you're looking at omics, looking at nucleotides, nobody knows how to deal with these molecules. They're all chemically unstable. And they don't get that they're chemically unstable. They don't ever see them. The reason they don't see them is because they don't know how to handle them to keep them alive during the analysis part of the project OK, so you have this instability problem. And in the purine pathway, the instability problem is a real problem for not just this guy. This guy is obvious. But for other guys. OK, so then the next question is where else. And if you're thinking about metabolism in general, where else might you want to have organization of your enzymes? You might want to have it if you generate an intermediate in the pathway. And then there. It's a branch point for other metabolic pathways. OK, so there's an intermediate in this pathway that can go to thiamine biosynthesis to histidine to tryptophan in biosynthesis. I'm not going to go through that. But that would be another place that I think it's obvious that you could sequester, under a different set of conditions, and prevent the other pathways from happening. So if you have an intermediate branch point, you can prevent other pathways. So those two things I think are important. One of the questions is, do you increase the flux through the pathway? OK, so there's been a lot of engineering people. People really care about this in terms of engineering. If you want to engineer a metabolic pathway, should you be linking all your proteins together? And there have been a lot of papers published. If you look at bioengineering papers, where they link all the pathways, all of the enzymes together in a way, because they want them to cluster, because they think they're increasing the flux through the pathway. And so there are some people that do calculations that show you increase the flux. Other people do calculations so you don't increase the flux. So I think, again, this is an area that I think is very active. And it's pertinent, because everybody and his brother is trying to make biofuels. You'd need to do a lot of engineering from a lot of enzymes from different places, putting them together. How do you make them efficient? OK, so we asked the question about flux. And I think, mathematically, people are looking at that. You need to know a lot about the kinetics of your system. These systems, there's a lot known about the kinetics. So and then this goes to the question of how, what is unstable. And you need to think about diffusion. I think this is not so easy to think about this. But we do need to think about flux through the pathway. And then the other thing that's interesting in terms of regulation is it turns out in eukaryotes, where things are much more regulated than in prokaryotes, because of the increased complexity of everything. Almost all of these pathways are organized on multiple activities on one polypeptide. That's telling us something, I think, since we see this over and over and over again. So there must be some reason to do that. So for all of these reasons in terms of the purine pathway this has been sort of a target for people for a long time. That's one of the reasons I decided to talk about it, because this was one of the first papers where people were excited that they thought they had evidence for this kind of organization in the cell. Not in the animal. But in the cell. OK. Let's see what I want to say next. I'm trying to keep this on some kind of a schedule. OK, so this is the hypothesis. The hypothesis is that these things are organized in some way. And this was taken out of-- probably it was a review paper. It wasn't taken out of a paper you had to read. Here's the cell. That's the nucleus of the cell. And what do you see. I think you can see this right. You see these little dots which they call punctate staining. So what else do you need to know that they don't have in this picture that's really sort of key to thinking about this model. So here they've just have a bunch of enzymes stuck together and all in a little ball. OK, so if you read the paper there was a couple of things. AUDIENCE: How you're getting the fluorescence? JOANNE STUBBE: How you're getting them? AUDIENCE: If it's by effusion or fluorescence. JOANNE STUBBE: Yeah, so how you're getting the fluorescence becomes key. OK, so we're going to talk about that. How did what was a major way they got the data. We'll talk about this in a minute in more detail but. But whenever you're going to use fluorescence, you have to figure out how to get a probe onto your protein. So that's like a major focus. And this again is where chemical biology needs to play a role. We still need better ways to be able to do this. You've seen over the course of the semester. I think in a lot of ways you could potentially do this. We'll come back to that in a minute. But if you look at this, what's missing? And this is something that drove me crazy when I reviewed the original paper. AUDIENCE: I just noticed, so you're getting that. But they didn't stain the membranes really. There's not a good-- I mean, you can kind of see the shape of the cell. But it would be nice to have a clear sort of-- JOANNE STUBBE: OK, so they might have done that. Did you look at the supplementary material? They might have stained the membrane. OK, so I think everybody would believe you see little blobs. OK, so what do you need to think about in terms of the little blob. AUDIENCE: The size. JOANNE STUBBE: The size. Right. Yeah, so that's one thing. They don't ever they don't ever talk about this. They might in some of the very later papers. But if we know this, we have structures of all the enzymes in the pathway. So you could make a guesstimate about how big these blobs should be, if you had one of each of these. And these things are huge. So this would tell you that you would have many, many of these. This is one thing that I think they need to do some more thinking about that they could have many, many of these things. And then the question is, why would you want many, many of these things. And how were they organize? Are they just sort of randomly organized or are they really organized in something like that with this big huge protein in the middle. That's one of the ones they look at. FGAM synthase, that has a molecular weight of 150,000, which is huge for an enzyme. And so for a long time-- and the catalytic activity-- my lab has studied that-- is way over here. And so you have a lot. Could it be a scaffold. OK, so that's where that idea actually came from. So but the hypothesis is that these guys are organized. And they're under certain growth conditions. That's the key. And we'll look at those pictures that come together if they do this when you need to make purines. And then they can go apart. OK, so the key thing, I think, is. And I wanted to just remind you why we're spending this time looking at fluorescence. And we probably should have spent two or three recitations on fluorescence methods. But we didn't. Is that we've seen this many times before. We've seen stopped-flow fluorescence in the Rodnina paper, where we were looking at the kinetics of fidelity of EF-Tu. And somehow they put a fluorescent probe onto the piece of tRNA. That was not trivial. How you got the probe there. And that probe could-- and we'll talk about this in a minute. But it could change. It changes when it's in different environments. And so you can use it as a way to monitor changes. So reactive oxygen species, we just looked at this. And I decided to put this up, since we didn't have the structures up last time. Fluorescein is one of the dyes that. This is fluorescein that people use. This is a version of fluorescein. But we talked about how do you know that epidermal growth factor is generating hydrogen peroxide? OK, so what we need is a sensor of hydrogen peroxide. So we talked about that last time. And this is the sensor that people use. Why did they use it. We talked about it. But we didn't have the structure. So they use the dye acetate of this molecule. This one they use, the triacetate. The one that they use in paper was the diacetate. Anyhow, you need to get the fluorescent probe into the cell. So that's something you're going to have to deal with. And so if you acetylate it, you don't have phenols or phenolates which might not get through the membrane, which apparently they don't. So then when they get in the cell, what do they do? They hydrolyze, OK. So what happens is when they hydrolyze, they are now-- you have these hydroxylated compounds that are able to be oxidized by an oxidant. And one of the oxidants that can do this. And there are others that can do it as well, is hydrogen peroxide. So people use this as an indicator of hydrogen peroxide. But it's not specific. Yeah? AUDIENCE: So are they also trapped after that esterase, like from diffusing that out to the-- JOANNE STUBBE: No. I mean, I don't think they diffuse back out, because I think they're the phenolates. So I think the diffusion out, like with many of these things, like if you use-- lots of times you esterify phosphates to get them into the cells. Once they hydrolyze, they charge. They don't get back out. So I don't really know. But that's what I would guess. So I guess the key thing and the basis for some comments that I made in class was that we don't really have. We don't know that this is specific for one reactive oxygen species. And so there are lots of people in the chemistry, biology interface trying to make specific sensors. OK, that's not easy to do. The hydrogen peroxide, they're getting better. In fact, Ting's APEX, which is a peroxidase, sort of similar to what we talked about with peroxireductions in the myeloperoxidase can actually function as a hydrogen peroxide sensor. So anyhow, what happens is that when it gets oxidized, it becomes fluorescent. So it's not fluorescent. It becomes fluorescent. So it just gets turned on. And you can see something, OK? So that's something we talked about. In Liz's part of the course, we talked about the fact that we can watch protein unfolding in the e. Coli proteasome. OK. And what did you look at in the proteasome, clip X clip P? You looked at titin. That had a little tryptophan on it. And tryptophan can absorb. It's not a very good thing, because it absorbs in the UV. But tryptophan fluorescence is used a lot. There are lots of tryptophans, so it's also really hard to use. But titin was this little tiny protein. And it was the only tryptophan. And they also did experiments with green fluorescent protein, which is what we're using in this paper. We remember. They pull on it. And you pull and you pull when you pull and then all of a sudden it unfolds. And you lose your chromophore. So you go from the on state to the off state. So all of these things. Binding measurements. You talked about. You had one problem set. I don't know whether you guys did that problem set or not. But there was a-- what was the calcium sensor? Does anybody remember? Anyhow, there was the calcium sensor, where you were asked in the problem set for a something or other that you asked to measure the KD for. And you can do binding assays. So fluorescence is an incredibly powerful tool as is the take home message. And we've seen it throughout the course. We just haven't talked about it. So now the key thing. And we're going to talk a little bit about fluorescence at probably a freshman level. Many of you guys, who were the undergraduates. You guys, have you done fluorescence experiments? You haven't done in the lab? I thought we had two Eureka labs that were fluorescence oriented. AUDIENCE: [INAUDIBLE]? JOANNE STUBBE: No? AUDIENCE: Yeah, yeah. so we-- [INTERPOSING VOICES] JOANNE STUBBE: So doesn't Tim's? He does sensors to sniff. I don't know what to sniff, but to sniff something, TNT or-- AUDIENCE: Right. but we didn't use fluorescence with that. JOANNE STUBBE: You didn't use fluorescence for that. OK, or the Tokmakoff lab? AUDIENCE: The one experiment we did in lab is we labeled a protein, the green absorbing dye. And it used laser anisotropy to measure KD rotations. And so the-- JOANNE STUBBE: OK, so you guys are experts, then, on fluorescence. Well, hopefully you-- anyhow, so one of the questions is we need to ultimately the key thing for any of this is we're going to have to have a fluorophore. So that's it. So we need the key starting point is a fluorophore And what are fluorophores. So you want something that's usually aromatic and large. It could be-- it could have a lot of nitrogens in it. Oh, I knew I forgot something. So there's a book called, Molecular Probes. OK so I gave you a handout on fluorescence. I forgot to bring the book, if anybody wants to see it. This book is worth its weight in gold if you're a chemical biologist. Because this has everything in the world you need to know about fluorescence. It's described in a thoughtful way. They sell all the probes. If you want to do something to tweak something, they'll help you do all of that. So this book, this is molecular probes book. I think it's online now. I have a copy that's five years old. I use it a lot. It's a really important book. And I got this out of the book. And it just shows you in the book, they have all these pictures of these fluorophores. So they're just big, huge, greasy molecules. You have to worry about solubility a lot of the time. So you have to stick sulfates, or something that ends up making it soluble. So that's going to be a key thing. So we need to have a fluorophore. And we have many options that we can buy these things. OK, so what's this? OK, so what we want to think about is this. So in your, the latest version of your handouts, I've written down what I'm going to say. Butt it's pretty simple for-- I'm talking about this in a pretty simplified viewpoint. But what we're going to see is these fluorophores are going to allow us to. They allow us to do assays. I'm going to show you a quick example of that. That is you can have something that is. You can have a molecule that is quenched, so you have a quencher on one side. I'll show you. And I'll show you the way the quenching comes from, something fluorescent on the other side. You can't see anything. You cut it in half. It could be a protease. It could be a nuclease. The quencher goes away. And you see fluorescence. You could have a sensor for metal binding, which Liz talked about. So you have two fluorophores. OK, you've got to figure out what the right fluorophores are. Something binds. They change confirmation. And they change confirmation in some way that you can actually detect a shift in the wavelength. And then you're looking. In our case, we're just sticking something on the end to see something. You were making a protein fluorescent. That's all we're doing. So you can use it for assays. You can use it for FRET. And in the current-- so you can measure distances. We're not going to go into that. But any of you that are interested in the current version of the handout, I have sort of short tutorial on what FRET is and where you should go to look this up. And then we just basically have a fluorescent tag. OK, and we'll come back and talk about the tag. We already talked about the fact that we have green fluorescent protein, red fluorescent protein tags. But we'll come back and talk about the other tags. So we have a fluorophore And so what does that mean in terms of what's going on. So you have your molecule. And your molecule has a ground state, which we'll call this S0. This is the ground state. And you have many vibrational modes. And you have this big huge fluorophore that can absorb your electron. And your fluorophore can't absorb a photon. And so what happens is. So we're going to have excitation with a photon in a certain way, in a wavelength that can be absorbed by the electron in your molecule to the excited state, which they call S1. And so you can have your electron going to an excited state. And we have a wavelength of light when that happens. And that depends on the structure of your molecule. So you don't want to be in the UV region. You want to be out in the region where you have less interference. And so that's the key game you have to play to get into that region in the visible. You really have to put a lot of stuff on here. You just can't make a small little molecule that absorbs at 600 nanometers. So that's part of the problem. So you're making big things of necessity, so you can actually see something happen. And so then what happens under those conditions. So we're going to have the excitation wavelength of light at a certain lambda max. You absorb. It's just like absorption. You have a certain wavelength that it absorbs more frequently. Then what happens in the excited state on a very fast timescale, you lose energy. OK, so under these conditions, you're doing a relaxation. And then we'll see in a minute. I'll talk about what are the mechanisms of relaxation. But that can tell you. You can use those relaxation mechanisms in a different way to design your fluorescent experiments. So what you see in this cartoon is that you're relaxing on a very fast timescale. And physical chemistry has told us that to see fluorescence, it needs to go down. So these are the vibrational modes. So you're exciting your electron electronically and vibrationally. And then you need to go down in vibrations. You're losing energy somehow. What happens to that energy? OK, we can talk about what can happen to that energy. And when it gets to the lowest level of the excited state, you have fluorescence. OK, and so that also happens on a pretty fast timescale. So the key thing here. So when you get to the lowest. So this is the lowest level, it fluoresces. And so this is where the photon emits. OK, so the photon wavelength for emission or h nu emission. And the key thing that you've probably heard about, again, when you were introduced to fluorescence is because you're losing energy here, what happens to the energy? You're going to a longer wavelength. OK, so the excitation and the emission wavelengths are distinct. And that's called the stoke shift. So it's the wavelength of excitation vs. the wavelength of emissions. So you have a stokes shift, which is the wavelength of excitation minus the wavelength of emission. And so you need to look at molecules. People have spent a lot of time. You saw those 25 lists of things where people have designed things that actually work quite effectively. OK, and so then the question is you losing energy. You are always going to be at longer wavelengths. OK, so that's good, that makes it easier to see, because there aren't that many things inside the cell that give you a background, which is what you need to worry about in all of the experiments you're doing inside the cell. The brightness, we'll come back to that in a minute. So what kinds of models can give you. What kinds of mechanisms are there for relaxation of the excited state. And so there are a number of mechanisms that can be involved. And one is, again, non-radiative relaxation. And how does that happen? So you're changing vibrational modes. And when you're in the excited state, if you're in solution, you have interactions with solvent or other molecules, all of which can affect this kind of transition. If you're in the active site, there can be other things. So the key here is the environment. And again, it could be solvent. It could be protein. And the only way you can tell is by actually looking at the fluorophore on your molecule to end up seeing what you end up seeing. OK, so a second way that you can see. And you probably saw this in your introductory. Yeah? AUDIENCE: So what would be an example? Like, if a unit of the energy being released is a photon in one case for non-radiative, what's the unit of energy? JOANNE STUBBE: What is the unit of energy? So energy, heat is one way that you lose all of this. So it's vibrational energy. I would say, it's mostly heat. So you're changing excitation levels somehow. And the beauty of fluorescence. And this is the key to the sensitivity is you're not doing anything to your molecule. So your electrons got excited. They give off a little heat or whatever. They somehow change a little bit. And then they go back down to the ground state again. So what can you do? You can excite them again. So this can happen over and over and over again, unless the molecule in the excited state becomes destroyed. So that's called photo bleaching. So the key thing here. And this is, I think, this ability to recycle is the key to sensitivity. But again, I haven't used fluorescence inside the cell. I've never done this myself, experimentally. So I don't really know. But you hear about photo bleaching all the time. So I think this is not a trivial thing that you can just blow off. It would be nice. But what you're doing is you're using the same excitation. And then loss and excitation and loss over and over and over again. And so it provides a much more sensitive assay than what you normally see for something like absorption. OK, so let's see. There was one other thing. Oh, so we talked about this mechanism, non radiative relaxation. How else could you relax? You can go from a singlet state to a triplet state. OK, I'm not going to talk. But intersystem crossing, yeah. So you can go from the singlet excited state to the triplet state. I'm not going to talk about this. But the triplet state then can phosphoresce. We're not going to be discussing that at all. But that's one possibility. We just talked about the fact that you can have something in there that quenches the fluorescence. It interacts with something in a distance dependent fashion. And that, again, affects the intensity of your fluorescence. So you also have reaction with the second molecule. And that can become. And it could be good or bad. If it reacts with oxygen, what happens is oxygen, the energy is immediately transferred to the oxygen. So that's why in many fluorescence experiments, you remove oxygen from all of your samples. It acts as a quencher. So you have. And it could be oxygen, which acts as a quencher. Or it could be another fluorophore. In which case. and if everything is set up correctly, you can get the energy to shift the energy of emission can get shifted to longer wavelengths. So that's what FRET is all about. OK, so it could not. A second molecule could be another fluorophore. OK, so those are sort of ways that you can relax. And then you can set up different kinds of experiments, depending upon what the objective is of using fluorescence. So I've written this out in more detail. And for those of you who want to look at FRET, I've defined FRET. I've given you the equations. And people use this quite a bit inside the cell. You need to study this. There are a lot of issues associated with it that you need to think about. And I'll come back. You need to think about. It's not. There are a lot of constants that determine the rate constant for your FRET, OK. And so you just you need to think about all these constants to be able to interpret the data in a thoughtful way. And I've given you a tutorial that I felt was pretty good that I get off the web that just shows what FRET is. And that we have many, many dyes that we can measure distances from 10 to 100 Angstroms using FRET. That's not in this paper. So I didn't. And this just sort of is a cartoon of what I was just telling you. So here, you might have an interaction. But if you cut it, the interaction could be gone. Here, you might have no interaction. But when some small molecule binds, you see an interaction. And you can pick this up using fluorescence changes. OK, so people do these kinds of experiments all the time. And this kind of an assay is extremely-- there are two kinds of assays that one does. So if you work in a pharmaceutical company, people do this all the time. They want a very sensitive assay. Everybody uses fluorescence. They might use an assay like this, where you go from nothing to something. OK, so you have high sensitivity. And the other thing they use is, which I gave you in your handout, is fluorescence polarization, which I'm not going to be talking about. But those are the two major methods that people develop assays around in the pharmaceutical industry. So fluorescence is here to stay. We still need better tools. It can be quantitative. You can measure a quantum efficiency of the electron, light, that's involved in the excitation and the photon that's involved in the emission. If it's 100 percent efficient, then you're quantum efficiency is 1 anyhow. So you have a whole range of quantum efficiencies. OK, so now what I want to do is we're late. But we'll at least get to the other sources telling you what I just told you. OK, so I want to just introduce to you some of the issues that we're going to be facing. And we are going to talk about this in class, probably Monday or on Wednesday morning, OK. So I'll extend this in class. But they've attached green fluorescent protein to all of these things. So this is issue number one. What should they have done in these papers that they didn't do? If you read the paper carefully. I mean, it's hard to read a science paper, because all the key pieces of data are in supplementary information. So they made a few. In all of these, I can't remember what they made. But they made fusion proteins, right? So here, you have a purine enzyme. And here we have some kind of fluorescent protein. So that's the probe they're using. OK, so what's wrong with that, with the way they did their experiments? Can anybody look at the details of what's going on? So what, if you made this fusion, what would be the first thing you would do with a fusion protein? AUDIENCE: My first thought would be, if it changes the activity of the original protein. GFP's a very large [INAUDIBLE]. JOANNE STUBBE: Right. Exactly. GFP I'm going to show you in a second. I think I can show you this in a second. These are just the ways they were looking. But you have all these probes. GFP, it's over here. These are the organic dyes. Here's an antibody. We'll come back to that. So GFP is big. So does it change activity? They didn't assay that. To me, that's mind boggling. OK, because I've dealt with these. I know these proteins, that one protein there. So two of them they're dealing with. One of them is a trifuctional protein. The other one's 150 kilodaltons. These proteins are not easy to deal with, OK. So to me, this is a key thing. So this goes back to the Marcotte paper, where he's saying, well, I mean, maybe these things don't express very well. And they aggregate. They don't fold. We saw how complicated the folding process is. What is the second thing? How did they get the proteins into the cell? How did they get? They don't get proteins into the cell. How did they get? Yeah, how did they get GFP constructs into the cell? AUDIENCE: Transient transfection? JOANNE STUBBE: Yeah, transient transfection, what is the issue there? Without going into details, but what's the issue? AUDIENCE: Like, when the cell's normal mechanism, like the cell's own enzymes maybe-- JOANNE STUBBE: So you do have a normal-- you do have the normal enzyme. They didn't make any effort to knock out the purine enzymes. OK, but I think the key thing with transient transfection is the levels. First of all, a lot of cells don't have anything. But then you don't care about that, because you don't look at them, because they're not fluorescent. OK, but do you think the levels are important. I think the levels are incredibly important. So the question is 100-fold, 1,000 fold over the endogenous levels. And so to me, the first experiments I would have done before I did any of these other experiments is I would have looked at. You might have chosen the trifunctional protein, which they did, because it has activities 2, 3, and 5. And this other big huge protein, which-- so these are the proteins they focus on is activity 4. So 4 is huge. You might think it could function as a scaffolding protein to interact with activities 2, 3, 5. All of that's totally reasonable. OK, but they didn't deal with those issues. So you need to figure out how to attach something that's fluorescent. So one way is genetically. OK, and we've seen this. So we're just fusing GFP onto the protein of interest. Another way in this paper, also, and you mentioned that, is they were using endogenous antibodies. OK, so antibodies can't get into cells. So how do you assay this? So these are also tough experiments. So somehow you fix the cells. So they aren't falling apart when you're trying to perturb the cell to allow the antibodies to get in. And then you permeabilize the cells. Have any of you ever done that? I've done it in yeast. In yeast, it's brutal. I mean, it works. But it's the conditions are like it's a witch's brew. Anyhow, so then you get the antibody in. And that's what you're looking at. And if you look in the-- I have. We're not going to get that far. But I have pictures of-- So when they compared the transient transfection with the endogenous levels, that might give them some feeling for what levels, the levels of expression actually are. And of course, the way that people really want to attach things is using small things, whatever these lists of dyes are that we have. And what are the methods that you guys have learned about to attach these fluorophores. So instead of using a genetic fusion, which is probably. That's a really good way, except the protein, the green fluorescent protein is big. Green fluorescent protein is also a dimer. So people have spent a lot of time engineering green fluorescent protein to be a monomer. So the ones you buy commercially now are all monomers. That would add complexity to everything on top of this. How would you attach some of these things? So we know what the structures of these things are. AUDIENCE: You can do like a halo tag. JOANNE STUBBE: So you could do a halo tag. Have you talked-- we haven't talked about that. So give me another method. Give me a method we've talked about. AUDIENCE: A handle, [INAUDIBLE] handle to attach? JOANNE STUBBE: Yeah, but how would you do that? How do you attach these handles? You want to attach a fluorophore. OK, so it turns out that all of these things here, which you can't see. But these little aromatic things have been synthesized. So click it on. If they could have a settling there. But then it needs to be clicked to something. AUDIENCE: [INAUDIBLE]. JOANNE STUBBE: So you can't just. So how do you click it? AUDIENCE: [INAUDIBLE]. JOANNE STUBBE: So but is that easy to do inside the cell? No. And in mammalian cells, it's impossible. OK, so you can't use unnatural amino acids inside the cell. The technology is not there at this stage. So the question of how you attach this. You could make your. If you could make your protein outside the cell. You might be able to do that. But then you have the problem of getting your protein inside the cell. So getting a probe that's fluorescently, you're labeling the protein of interest is not easy. And Alice Ting's lab, again, has spent a lot of time, not that successfully. But using ligases that you can then incorporate into the cells that can then react with things you put onto your protein to attach fluorophores. But this is an area that's really important, because in my opinion, looking at regulation inside the cell, we don't really want to perturb. We don't want to be at very high levels. And we want to be able to see something to understand regulation. So I think. So anyhow, the issue is that we want to be as small as possible. We don't want to be Brad's lab. What is Brad's lab? Does he use these nanobodies that are antibodies? AUDIENCE: Like an [INAUDIBLE]? JOANNE STUBBE: No. They have all these. They have things called the nanobodies now. So I think they are like the little guys you make on your solid phase peptide synthesizer. But they are specific. They specifically bind to proteins. So there are only five examples that I've seen in the literature. So they act like antibodies. But they're-- huh? AUDIENCE: Like [INAUDIBLE],, like little-- JOANNE STUBBE: They're little tiny proteins that are maybe. I don't know. 50 amino acids that somehow, some guy at the University of Chicago-- not Kent-- developed these things. And they specifically. They act like an antibody. They can specifically interact with a protein of interest. And then you attach a green fluorescent protein onto it. So again, what you have something smaller. So because with these antibodies. What you see is the non-specific, right? I mean, we've seen that. And with fluorescence, that means you have fluorescence background in everything you do. So anyhow, I think we're not that. So that's just you're using fluorescence microscopy. This tells you why you're interested in fluorescence microscopy. And we'll just close here. And we're going to come back and talk about this in class. But this is sort of the example of the data that you need to think about. So what we hear is in the presence of purines, you don't see any of these little dots. You remove the purines. OK, so this is not so easy either, because the way we grow cells, we don't have defined media, right? I mean we're using. I don't know what you guys use now, but fetal calf serum or something. It's got all this stuff in it that we don't really know what it is. We don't use defined media. And apparently, when they-- the Marcotte paper-- when they were describing this, said it was not so easy to remove the purines. And the method they used to remove the purines also removed other stuff, OK. So you're stressing the cell. That was the take home message. So under those conditions, you see something different. OK, and so then they did another experiment, because they were worried about levels. Here, they are they have an antibody to the trifuctional protein. And so this is what they see under low purine conditions. Does this look like this? I don't know. So you can't tell by looking at one picture. OK, so you've got to do statistical analysis of all these things. So I think this sort of-- we'll come back and talk about this in class. But I think this is the first example where people are trying to look at this. The data is interesting. But we've already raised issues of what some of the problems are. And hopefully, you can think about more of the problems.
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https://ocw.mit.edu/courses/7-01sc-fundamentals-of-biology-fall-2011/7.01sc-fall-2011.zip
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PROFESSOR: Now, onward. Transcription. So, we've got DNA, we'll do two, three, probably three by now, transcription. So we have DNA goes to DNA. DNA makes RNA, RNA makes protein. This, by the way, gets the name the central dogma of molecular biology. Due to Francis Crick, and as an aside, Francis actually never said DNA goes to RNA goes to protein. What he said was nucleic acids go to protein. The information flows from nucleic acids to proteins. He never actually said DNA goes to RNA goes to protein, and that's an important point. And we'll come to it at some point, probably next time. So, transcription. Here's my genome. Here's my double helix. I'm going to stop wrapping around itself, just because it's tedious. And here's a chunk of DNA that encodes a gene. Maybe it's a gene that makes our enzyme for arginine biosynthesis. Remember, we had our arginine genes and all that. But what happens is it has a starting point, it has a stopping point. Five prime to three prime. What happens is there is a signal in the DNA that the cell knows how to read called a promoter. And under certain circumstances, this promoter invites an enzyme to sit down, and the enzyme starts copying. Which direction does this enzyme go? Five prime to three prime. They all go five prime to three prime. But it makes RNA. Okay? And then it gets to certain point, and it stops copying. This process of copying is called transcription, because it's just a direct transcribing. So what's the difference between DNA and RNA? Two differences. One, this is two prime deoxyribose. This is ribose. It's not two prime deoxy. It's truly ribose. The other difference, where DNA has T, RNA, has U, uracil. The difference, what's the difference between T and U? The difference is a methyl group. It's a methyl group in an unimportant position for the base paring. It doesn't matter. It has an extra methyl group. You can look it up in your book. So for all practical purposes, you, in thinking about polymerization and five prime to three prime, and everything, could imagine the DNA and RNA are the same basic structure, because it's got one little methyl group that distinguishes T and U. And it's got a hydroxyl on the two prime position in the carbon, in the sugar, which actually isn't anything we use. We never use the two prime. So none of what I've told you is affected by t versus u, or deoxyribose versus ribose. Now, it turns out it does make a difference in the overall structure. It's harder to base pair with the ribose there as opposed to the deoxyribose because they stack differently, et cetera. It makes a difference to the cell. RNA is less stable, all sorts of things. But for your practical purposes, apart from having to know that it's T versus U, and deoxyribose versus ribose, you won't see in this course actually see any real strong reasons why it matters, but it does matter to the cell. All right. So this enzyme comes along, and it copies a segment of the DNA, starting at a promoter. It knows which strand it's on. Remember, this is stranded. It's not like it's going back this way. It has a directionality to it. And in reaches what's called a transcriptional stop signal. Transcriptional stop, which is a certain sequence in the DNA, and it comes to an end. And it makes an RNA transcript, which then floats away. Which we'll talk next time, gets translated into a protein. And how do you think this works? It takes nucleotides, RNA nucleotides here, with their triphosphates, and sticks them on, just like we saw with DNA. And it makes a polymer of RNA. And the enzyme is called RNA polymerase, right? This is all pretty logical stuff. RNA polymerase comes along and does that. So we get RNA polymerase. Now, when I am a cell, and this is my genome, I have a gene that goes this way. Here's its promoter, here's its transcriptional stop. I could also have a gene that goes this way. Here's its promoter, here's its transcriptional stop. Directionality could go in either direction. RNA polymerase comes along, and with the help of friends, knows where to start. Those friends could be other proteins that are sitting down there that RNA polymerase likes to associate with. And which strand is being transcribed? The bottom strand, or the top strand? Matters. You get a different single stranded RNA. So RNA, when it floats off, is single stranded. This is going to make a single stranded RNA that, five prime to three prime RNA that is complimentary to, matching to, the bottom. This guy, however, will make an RNA that is complimentary to the top. Much of the business of running your cell is figuring out which genes you should be transcribing into RNA. It turns out your liver is making different transcripts. It's transcribing different segments of your genome than say, a muscle cell. Than say, a brain cell. All of that machinery of figuring out how one genome gets read out in different ways by RNA polymerase is the problem of gene regulation. And we will talk about that in a little while. What we're going to talk about next time is how that RNA transcript gets translated into a protein. And you guys probably, again, all know this. But nonetheless, we'll talk a little bit about it. And then we'll talk about some variations on that theme. Until next time.
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https://ocw.mit.edu/courses/8-591j-systems-biology-fall-2014/8.591j-fall-2014.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Today, what we're going to do is, first, introduce this idea of oscillations. It might be useful. A fair amount of the day will be spent discussing this paper by Michael Elowitz and Stan Leibler that you read over the last few days, which was the first, kind of, experimental demonstration that you could take these random components, put them together, and generate oscillatory gene networks. And finally, it's likely we're going to run out of time around here. But if we have time, we'll talk about other oscillator designs. In particular, these relaxation oscillators that are both robust and tunable. It's likely we're going to discuss this on Tuesday. All right, so I want to start by just thinking about other oscillator designs. But before we get into that, it's worth just asking a question. Why is it that we might want to design an oscillator? What do we like about oscillations? Does anybody like oscillations? And if so, why? Yes. AUDIENCE: You can make clocks. And clocks are really-- PROFESSOR: Perfect. Yes, all right. So two part answer. You can make clocks. And clocks are useful. All right. OK, so this is a fine statement. So oscillators are, kind of, the basis for time keeping. And indeed, classic ideas of clocks, like a pendulum clock. The idea is that you have this thing. It's going back and forth. And each time that it goes, it let allows some winding mechanism to move. And that's what the clock is based. And even modern clocks are based on some sort of oscillatory dynamic. It might be a very high frequency. But in any case, the basic idea of oscillations as a mechanism for time keeping is why we really care about it. Of course, just from a dynamical systems perspective, we also like oscillations because they're interesting from a dynamical standpoint. And therefore, we'd like to know how we might be able to make them. Can anybody offer an example of an oscillator in a G network in real life? Yes. AUDIENCE: Circadian. PROFESSOR: The circadian oscillator. That's right. So the idea there is that there's a G network within many organizations that actually keeps track of the daily cycle and, indeed, is entrained by the daily cycle. So of course, the day, night cycle. That's an oscillator. It's on its own. And it goes without us, as well. But it's often useful for organisms to be able to keep track of where in the course the day it might be. And the amount of light that the organism is getting at this particular moment might not be a faithful indicator of how much light there will be available in an hour because it could just be that there's a cloud crossing in front of the sun. And you don't want-- as an organism-- to think that it's night. And then, you shut down all that machinery because, after that cloud passes, you want to be able to get going again. So it's often useful for an organism to know where in the morning, night, evening cycle one is. And we will not be talking too much about the circadian oscillators in this class. Although, I would say to the degree of your interest in oscillations, I strongly encourage you to look up that literature because it's really beautiful. In particular, in some of these oscillators, it's been demonstrating you can get the oscillations in vitro. I.e, outside of the cell. Even in the absence of any gene expression, in some cases, you can still get oscillations of just those protein components in a test tube. This was quite a shocking discovery when it was first published. But we want to start out with some simpler ones. In particular, I want to start by thinking about auto repression. So if you have an auto regulatory loop where some gene is repressing itself, the question is does this thing oscillate. And indeed, it's reasonable that it might because we can construct a verbal argument. Starts out high. Then, it should repress itself so you get less new x being made. So the concentration falls. So maybe I'll give you a plot to add to it. Concentration of x is a function of time. You can imagine just starting somewhere high. That means it's a repressing expression. So it's going to fall. But then, once it falls too much, then all of a sudden, OK, well we're not repressing ourselves anymore. So maybe then we get more expression. More of this x is being made. So it should come back up. And then, now we're back where we started. So this is a totally reasonable statement. Yes? AUDIENCE: [INAUDIBLE]? PROFESSOR: Well I don't know. I mean, I didn't introduce any damping in here. The amplitude is the same everywhere. AUDIENCE: So you're saying that you could actually have something-- PROFESSOR: Well I guess what I'm really trying to say is that just because you can construct a verbal argument that something happens does not mean that a particular equation is going to do that. Part of the value of equations is that they force you to be explicit about all the assumptions that you're making. And then what you're going to do is you're going to ask, well, a given equation is a mathematical manifestation of the assumptions you're making. And then, you're going to ask does that oscillate. Yes/no? And then you're going to say, OK, well what would we need to change in order to introduce oscillations? And I'll just-- OK. So this is definitely an oscillation. The question is, should you find this argument I just gave you convincing? And what I'm, I guess, about to say is that you shouldn't. But then, we need to be clear about what's going on and why. And just because you can make a verbal argument for something doesn't mean that it actually exist. I mean, that's a guide to how you might want to formalize your thinking. And in particular, the simplest way to think about oscillations that might be induced in this situation would be to just say, all right, well the simplest model we have for an auto regulatory loop that's negative is we say, OK, well there's some alpha 1 plus protein and minus p. So this is, kind of, the simplest equation you can write that captures this idea that this protein p is negatively regulating itself in a cooperative fashion maybe. Now it's already in a non-dimensionalize version. Right? And what you can see is that, within this realm, there are only two things that can possibly be changing. There's how cooperative that repression is-- n-- and then, the strength of the expression in the absence of repression. And as we discussed on Tuesday, alpha is capturing all these dynamics of the actual strength of expression together with the lifetime of the protein together with the binding. You know, the binding affinity k. So all those things get wrapped up in this a or alpha rather. All right, so this is, indeed, the simplest model you can write down to describe such a negative auto regulatory loop. Now the question is now that we've done this, we want to know does this thing oscillate. And even without analyzing this equation, there's something that's very strong, which you can say. So in theory we're going to ask is it possible for this thing to oscillate. All right. Possible. Your oscillations, we'll say oscillations possible. And this time, referring to mathematically possible. So maybe this thing does oscillate. Maybe it doesn't. But in particular, without analyzing it, is there anything that you can say without analyzing it? We're just going to say is it possible. Yes or no? If you say no, you have to be prepared to give an argument for why this thing is not allowed to oscillate. I'm talking about this equation. Do you don't you understand the question that I'm trying to ask? And we haven't analyzed this thing yet. But the question is, even before analyzing it, can we say anything about whether it's mathematically allowed to oscillate? I'll give you 10 seconds to think about it. And if you say no, you get to tell me why. All right, ready? Three, two, one. All right, so we got a smattering of things. So I think this is not, obviously, a priori. But it turns out that it's not actually. It's just mathematically impossible for this hing to oscillate. And can somebody say why that might be? AUDIENCE: Because it might be you could only have one value of p dot? PROFESSOR: Perfect OK. So for a given value of p, there's only some value of p dot that you can have. And in a particular-- so p here is like a concentration of x. So I'm going to pick some value, randomly, here of p. And what you're pointing out is this is a differential equation in which if you give me or I give you the p, you can give me p dot. And there's a single value p dot for each p. And in this oscillatory scheme, is that statement true? No. What you can see is that, over here, this is x slash p concentration of x. We're using p here because we're about to start talking about mRNA So I want to keep the notation consistent. What you see is that the derivative here is negative. The derivative here is positive. Negative, positive. So any oscillation that you're going to be able to imagine is going to have multiple values for the derivative as a function of that value just because you have to come back and forth. You have to cross that point multiple times. So what this is saying is that since this is a differential equation-- and it's actually important that it's a differential equation rather than a difference equation where you have discrete values. But given that this is a differential equation where time is taking little, little, little steps and you have a single variable, it just can't oscillate. So for example, if you're talking about the oscillations the harmonic oscillator the important thing there is a you have both the position in the velocity see these two dynamical variables that are interacting in some way because you have momentum, in that case, that allows for the oscillations in the case of a mass on a spring, for example. Question. AUDIENCE: I'm still not understanding. So the value of p can not oscillate? PROFESSOR: Right. So we're saying is that, right, p simply cannot oscillate in this situation where we have a differential equation describing p with-- if we just have p dot as a function of p and we don't have a second order. A p double dot, for example. So if we just have a single derivative with respect to time and some function of p over here, what that means is that, if p is specified, then p dot is specified. And that's inconsistent with any sort of oscillation because any oscillation's going to require that, at this is given value of p-- this concentration of p-- in this case, the concentration's going down. Here, it's going up. So here, this is-- from that standpoint-- a multi valued function. OK? And other questions about this statement? Even if I just written down some other function of p over here, this statement would still be true. And it's valuable to be able to have some intuition about what are the essential ingredients to get this sort of oscillation. And for simple harmonic motion, right there we have the second derivative, first derivative, and that's what allows oscillations there. OK, so we can, maybe, write down a more complicated model of a negative auto regulation. And then, try to ask the same thing. Might this new model oscillate? And this looks a little bit more complicated. But we just have to be a little bit careful. All right, so this is, again, negative auto-regulation. What we're going to do is we're going to explicitly think about the concentration of the mRNA. OK. And that's just because when a gene is initially transcribed, it first makes mRNA. And then, the mRNA is translated into protein. Right? So what we can do is we can write down something that looks like this. M dot derivative of m with respect to time. It's going to be-- all right, so this is the concentration of mRNA. And p is the concentration of protein. OK? All right, and what you can see is that the protein is now repressing expression of the mRNA. mRNA is being degraded. But then, down here, this is a little bit funny. But what you can see is that, if you have more mRNA, then that's going to lead to the production of protein. Yet, we also have a degradation term for the protein. Yes? AUDIENCE: Why are we multiplying the degradation rate of the protein times some beta, as well? PROFESSOR: That's a good question. OK, you're wondering why we've pulled out this beta. In particular-- right. OK, perfect. OK, yeah. This is very important. And actually, this gets in-- once again-- to this question of these non-dimensional versions of equations. Mathematically, simple. Biologically, very complicated. Well, first of all, what is that we've used as our unit of time in these equations? AUDIENCE: The life of mRNA. PROFESSOR: Right. So it's based on the lifetime of the mRNA because we can see that there's nothing sitting in front of this m. And if we want to, then, allow for a difference in the lifetime mRNA and protein, then we have to introduce some other thing, which we're calling beta. So beta is the ratio of-- well which one's more stable? mRNA or protein, often, typically? AUDIENCE: Protein. PROFESSOR: Proteins are, typically, more stable. So does that mean that beta should be larger or smaller than 1? OK, I'm going to let you guys think about this just make sure we're all-- OK, so the question is beta, A, greater than 1? Typically, much greater. Or is it, B, much less than 1, given what we just said? All right, you think about it for 10 seconds. All right. Are you ready? Three, two, one. All right, so most people are saying B. So indeed, beta should be much less than 1. And that's because beta is the ratio of the lifetime. So you can see, if beta gets larger, that increases the degradation rate of the protein. What do I want to say? So beta is the ratio of the lifetime in the mRNA through the lifetime of the protein. Yes? AUDIENCE: So I get why we have to-- PROFESSOR: Yeah. No, I understand. No, I understand you. I'm getting to your question. First, we have to make since of this because the next thing is actually even weirder. But I just want to be clear that beta is defined as the lifetime of mRNA over the lifetime of the protein. What's interesting is, actually, there's a typo or mistake in the elements paper, actually. So if you look at figure 1B or so-- yeah, so figure 1B, actually. It says that beta is the protein lifetime divided by the mRNA lifetime. So you can correct that, if you like. So beta's is the mRNA divided by the lifetime of the protein. OK, so I think that we understand why that term is there. But the weird thing is that we're doing p minus m over here. Right? And it feels, somehow, that that can't be possible. You know, that it shouldn't be beta times m over here because it feels like it's under determined. Right? OK. So it's possible I just screwed up. But does anybody want to defend my equation here? How might it be possible that this makes any sense that you can just have the one beta here that you pull out, and it's just p minus m over here? AUDIENCE: I think it's an assumption of the model where they choose the lifetime of the protein and the mRNA to be similar. PROFESSOR: Well no because, actually, we have this term beta, which is the lifetime of mRNA divided by lifetime of protein. So we haven't assumed anything about this beta. It could be, in principle, larger than one. Smaller, actually. So it's true that given typical facts about life in the cell, it's true that you expect beta be much less than 1. But we haven't made any assumption. Beta is just there. It could be anything. Right? So yeah, it's possible we've made some other assumption. But what is going on. Yes? AUDIENCE: Is it the concentration is scaled by the amount of necessary-- PROFESSOR: Yes, that's right because, remember, you can only choose one unit for time. And we've already chosen that to get this to be just minus m here. But you get to choose what's the unit of concentration for, both, mRNA and for protein. Can somebody remind us what the unit of concentration is for protein? AUDIENCE: The dissociation constant of the protein to the-- PROFESSOR: That's right. So it's this dissociation constant. And more generally, it's the protein concentration, which you get half maximal repression. And depending on the detailed models, it could be more complicated. But in this phenomenological realm, if p is equal to 1, you get half repression. And that's our definition for what p equal to 1 means. So we've rescaled out that k. So what we've really done is that there's some unit for the concentration of mRNA that we were free to choose. And it was chosen so that you could just say p minus m. But what that means is that it requires a genius to figure out what m equal to 1 means, right? It doesn't quite require a genius. But what do you guys think it's going to depend on? Yes? AUDIENCE: It's going to depend on this ratio of lifetimes, as well. PROFESSOR: Yes, right. So beta is going to appear in there. So I'll give you a hint, there are three things that determine it. AUDIENCE: Transcription, or the speed of transcription. PROFESSOR: Translation, yes. So the translation efficiency. So each mRNA, it's going to lead to some rate of protein synthesis. So yeah, the translation rate or efficiency is going to enter. There aren't that many other things it could be. But yeah, I mean, this is tricky. And it's OK if you can't just figure it out here because this, I think, is pretty subtle. It turns out it also depends on that k parameter because there's some sense that-- m equal to 1-- what it's saying is that that's the amount of mRNA that you need so that, if the protein concentration where 1, you would not get any change in the protein concentration. And given that now I had to invoke p in there and p is scaled by k, so then k also ends up being relevant for this mRNA. So you can, if you'd like, go ahead and start with a original, reasonable set of equations. And then, get back to this. But I think, once again, this just highlights that these non-dimensional versions of the equations are great. But you have to be careful. You don't know what means what. All right? Are there any questions about what we've said so far? OK. Now what we've done is we have now a protein concentration. We have mRNA concentration. And what I'm going to ask for now is, for these sets of equations, is it mathematically possible that they could, maybe, oscillate? Yes. I mean, we're going to find that the answer is that these actually don't oscillate. But have to actually do the calculation if you want to determine that. You can't just say that it's impossible based on the same argument here. And that's because, if you think about this in the case of there's some mRNA concentration. Some protein concentration. What we want to know is do things oscillate in this space. And they could. I mean, I could certainly draw a curve. It ends up not being true for these particular sets of equations. But you can't a priori, kind of, dismiss the possibility. Yes? AUDIENCE: That's like a differential equation. But if you write down the stochastic model of that, would that-- PROFESSOR: OK, this is a very good question. So this is the differential equation format of this and that we're assuming that there are no stochastic fluctuations. And indeed, there is a large area of excitement, recently, that is trying to understand cases in which you can have, so-called, noise induced oscillations. So you can have cases that the deterministic equations do not oscillate. But if you do the full stochastic treatment, then that could oscillate. In particular, if you do a master equation type formalism. And actually, I don't know, for this particular equations. Yeah, I don't know for this one. But towards the end of the semester, we will be talking about explicit models in which, predator prey systems, in which the differential equation format doesn't oscillate. But then, if you do the master equation stochastic treatment, then it does oscillate. Yeah, so we will be talking about this in other contexts. But I don't know the answer for this model. All right, so let's go and, maybe, try to analyze this a little bit. And this is useful to do, partly because some of the calculations are going to be very similar to what we're about to do next, which is look at stability analysis of a repressilator kind of system. All right. So this thing here is some function f of m and p. And this guy here is indeed, again, some other function g of m and p. And we're going to be taking derivatives of these functions around the fixed point. And maybe I will also say there's going to be some stable point. We should just calculate what it is. I'm sorry I'm making this go up and down. Don't get dizzy. So first of all, it's always good to know whether there are fixed points in any sort of equations that you ever look at. So let's go ahead and see that. First of all, is m equal to 0, p equal to 0? Is that a fixed point in the system? No. Right? So if m and p are 0, then this is a fixed point. But that one's not because we get expression of the mRNA in the absence of the protein. So the origin is not a fixed point. Now to figure out the fixed points, we just set these things equal to 0. So if m dot is equal to 0, we have 0. That's alpha 1 plus p to the n minus m. Again, 0 is this. So what you can see is that, at equilibrium, we have a condition here where m is equal to p. So from this, we get m equilibrium is equal to p equilibrium. So m equilibrium over here has to be equal to p equilibrium, we just said. And that's equal to this guy here. It's alpha 1 plus p equilibrium to the n. All right. And the condition for this equilibrium is then something that looks like this. Now this is maybe not so intuitive. But alpha is this non dimensional version of the strength of expression. And what this is saying is that, broadly, it's not obvious how to solve this explicitly. But as the strength of expression goes up, the equilibrium here-- and I'm saying equilibrium. And that's, maybe, a little bit dangerous. We might even want to just call it-- it's a fixed point in concentration, so it doesn't have to be stable. So if we don't want to bias our thinking, different people argue about whether equilibrium should be a stable or require a stable. We could just call it some p 0 if that makes you less likely to bias our thinking in terms of whether this concentration should be a stable or unstable fixed point. But for example, if we have that, in these units, if alpha is around 10, n might 2. Then, this thing gives us something. It's in the range of a couple or 2, 3. I mean, you can calculate what it should be. 2, 4, maybe even exactly 2. Did that-- yeah. All right, so yes. I'm just giving an example. If alpha were 10, then this equilibrium concentration or this fixed point concentration would be 2 if n were equal to 2 to give you, kind of, some sense of the numbers. And this is 2 in units of that binding affinity k, right. Now the question is, well, what does this mean? Why did we do this? Why do we care at all about the properties of that fix point? OK, so this might be some p 0. And this is, again, m 0 is equal to p 0 in these units. So there's some fixed point somewhere in the middle there. Now it turns out that the stability of that fixed point is very important in determining whether there are oscillations or not. Now the question of the generality or what can you say that's universally true about when you get oscillations and when you don't, this is, in general, a very hard mathematical problem, particularly in higher numbers of dimensions. But for two dimensions, there's a very nice statement that you can make based on the Poincare-Bendixson criterion. I cannot remember how to spell that. I'm probably mispronouncing it, as well. So Poincare-Bendixson, what they showed is that if, in two dimensions, you can draw some box here such that all of the trajectories are, kind of, coming in. And indeed, in this case, they do come in because the trajectories aren't going to cross 0. If you have some mRNA, then you're going to start making protein. If you have just protein, no mRNA, you're going to start making some mRNA. And we know that trajectories have to come in from out here because if the concentration of mRNA and the concentration of protein are very large then, eventually, the degradation is going to start pulling things in. So if you come out far enough, eventually, you're going to get trajectories coming in. So now we have there is some domain where all the trajectories are going to come in. Now you can imagine that, somehow, the stability of this thing is very important because in two dimensions here when you have a differential equation, trajectories cannot cross each other. So I'm not allowed in any sort of space like this to do something that looks like this because this would require that, at some concentration of m and p, I have different values for m dot and p dot. So it's similar to this argument we made for one dimension. But it's just generalized to two dimensions. So we're not allowed to cross trajectories. Well if you have a differential equation in any dimensions, that's true. But the thing is that this constraint is a very strong constraint in two dimensions. Whereas, in three dimensions, everything kind of goes out the window because in the three dimensions, you have another axis here. And then, these lines can do all sorts of crazy things. And that's actually, basically, why you need three dimensions in order to get chaos in differential equations because this thing about the absence of crossing is just such a strong constraint in two dimensions. Other questions about what I'm saying right now? I'm a little bit worried that I'm-- All right, so the trajectories are not allowed to cross. And that's really saying something very strong because we know that, here, trajectories are going to come out of the axis. And mRNA, we don't know which direction they're going to come. But let's figure out, if it were to oscillate, would the trajectories be going clockwise or counterclockwise? And actually, there's going to be some sense of the trajectories even in the absence of oscillations. But broadly, is there kind of a counterclockwise or clockwise kind of motion to the trajectories? Counterclockwise, right? And that's because mRNA leads to protein. So things are going to go like this. And the question is is it going to oscillate. And in two dimensions, actually-- Poincare-Bendixson-- what they say is that, if there's just one fixed point here, then the question of whether it oscillates is the same as the question of whether this is stable. So if it's stable, then there's no oscillations. If it's unstable, than there are. We'll just say no oscillations and oscillations. And that's because if it's a stable point and all the trajectories are coming in, then it just looks like this. So it spirals, maybe, into a state of coexistence. Well it spirals to this point of m and p. Whereas, if it's unstable, then those trajectories are, somehow, being pushed out. If it's unstable, then the trajectories are coming out of that fixed point. In which case, then that's actually precisely the situation in which you get a limit cycle oscillations. So if the fixed point were unstable, it looks like this because we have some box. The trajectories are all coming in, somehow, in here. But if we have one fixed point here and the trajectories are coming out, that means we have something that looks like this. It kind of comes out. And given that these trajectories can't cross, the question is, well, what can happen in between? And the answer is, basically, you have to get a limit cycle oscillation. There are these strange situations where you can get a path that is an oscillation that's, kind of, stable from one direction and unstable from another. We're not going to worry about that here. But broadly, if this thing is coming out, then you end up, in both directions, converging to a stable limit cycle oscillation. So it's a unstable fixed point, then this is the exact situation, which you get a limit cycle oscillation. OK. So that means that, what we really want to do if we want to ask-- let's try to back up again. We have this pair of differential equations. We want to know will this negative auto regulatory loop oscillate. Now what I'm telling you is that that question for two dimensions is analogous to the question of figuring out whether this fixed point is stable or not. If it's stable, then we don't get oscillations. If it's unstable, then we do. Any questions about this? So let's see what is is. On Tuesday, what we do is we talked about stability analysis for linear systems. We got what I hope is some intuition about that. And of course, what we need to do here is try to understand how to apply linear stability analysis to this non-linear pair of differential equations. And to do that, what we need to do is we need to linearize around that fixed point. So what we have is we have these two functions, f and g. And what we want to know is around that fixed point-- so we can define some m tilde, which is m minus this m 0. And some p tilde, which is p minus p 0. So when m tilde and p tilde are around 0, that's telling us that we're close to that fixed point. And we want to know, if we just go a little away from the fixed point, do we get pushed away or do we come back to where we started? Well we know that m tilde dot, which is actually equal to m dot, as well because m 0 and p 0 are the same. p tilde dot. We can linearize by taking derivatives around the fixed points. And in particular, what we want to do is we want to take the derivative of f with respect to m. Evaluate at the fixed point. That derivative is, indeed, just minus 1. So in general, in these situations, what we have is we have derivatives m, m dot p dot, and we have partial of this first function f with respect to m. Partial of g. Oh, no. So this is still f. Respect to p. Down here is derivative g with respect to m. Derivative g with respect to p. And this is all evaluated around the fixed point m 0 p 0. So we want to take these derivatives and evaluate at the fixed point. And if we do that, we get minus 1 here, derivative m with respect to m times m tilde. This other guy, when you take the derivative, you get a minus sign with respect to p. So we get a minus sign because this is in the denominator. And then, we have to take derivative inside. So we get n alpha p 0 to the n minus 1. And down, we get a 1 plus p 0 squared. So we took the derivative of this term with respect to p. And we evaluated at the fixed point p 0. Did I do that right? But we still have to add a p tilde because this is saying how sensitive is the function to changes in where you are times how far you've gone away from the fixed point. And then, again, over here, we take the derivatives down below. So derivative g with respect to m. That gives us a beta m tilde. And then, we have a minus beta p tilde. All right, so this is just an example of linearizing those equations around that fixed point. So ultimately, what we care about is really this matrix that's specifying deviations around the equilibrium. Right? So it's useful to just write it in matrix format because we get rid of some of the M's and P's. Indeed, so this matrix that we either call A or the Jacobean depending on-- so what we have is a minus 1. And we're going to call this thing x because it's going to pop up a lot is this minus n alpha p 0. So it's an x beta and minus beta. And then, we have our simple rules for determining whether this thing is going to be stable or not. It depends on the trace. And it depends on the determinant. So the trace should be negative. And is this trace negative? Yes. Yes because beta-- does anybody remember what beta was again. AUDIENCE: Ratio of lifetimes. PROFESSOR: Ratio of lifetimes. Lifetimes are positive. So beta is positive. All right, so the trace is equal to minus 1 minus beta. This is, indeed, less than 0. So this is consistent for stability. Does prove that it's stable? No. But we also need to know about the determinant of a, which is going to be beta, this times this, minus this times this. So that's minus. And this is a beta times what x was. So this gives us-- we can write this all down just so that it's clear that it has to be positive. So beta is positive. Positive, positive, positive, positive, positive. Everything's positive. So this thing has to be greater than 0. So what does this mean about the stability of Ethics Point? Stable. Fixed point stable. And what does that mean about oscillations? It means there are no oscillations. Fixed point stable. Therefore, no oscillations. So what this is saying is that the original, kind of simple, equation we wrote down for negative auto regulation, that thing was not allowed to oscillate mathematically. But that doesn't mean that, if you explicitly model the mRNA, it could go either way. But still, that's insufficient to generate oscillations. However, maybe if you included more steps, maybe it would oscillate. Question? AUDIENCE: So just to double check-- when you said, no oscillations, you mean stable oscillations? PROFESSOR: That's right, sorry. When I mean no oscillations, what I mean are indeed, no limit cycle oscillations. AUDIENCE: This is like a dampened-- PROFESSOR: Yeah. Yeah, so we, actually, have not solved exactly what it looks like. And I've drawn this is a pretty oscillatory thing. But it might just look like this, depending on the parameters and so forth. And indeed, we haven't even proven that this thing has complex eigenvalues. But certainly, there are no limit cycle oscillations. And I'd say it's really limit cycle oscillations that people find most exciting as because limit cycle oscillations have a characteristic amplitude. So it doesn't matter where you start. The oscillations go to some amplitude. And they have a characteristic period, again, independent of your starting condition. So a limit cycle oscillation has a feeling similar to a stable fixed point in the since that it doesn't matter where you start. You always end up there. So they're the ones that are really what you would call mathematically nice oscillations. And when I say this, I'm, in particular, comparing them to neutrally stable orbits. So there are cases in which, in two variables, you have a fixed point here. And at least in the case of linear stability, if you have purely imaginary eigenvalues, what that means is that you have orbits that go around your fixed point. And we'll see some cases that look like this later on. And this is, indeed, the nature of the oscillations in the Lotka-Volterra model for predator prey oscillations. They're not actually limit cycle oscillations. They're of this kind that are considered less interesting because they're less robust. Small changes in the model can cause these things to either go away, to turn into this kind of stable spiral, or to turn into limit cycle oscillations. So we'll talk about this more in a couple months. These are neutrally stable orbits. OK, but what I wanted to highlight, though, is that just because the original, simple, protein only model didn't oscillate and this protein mRNA together doesn't oscillate does not mean that it's impossible to get oscillations using negative auto regulation, either experimentally or computationally. And the question is, what might you need to do to get oscillations? AUDIENCE: So in the paper they talk about leakage in the negative-- PROFESSOR: OK, right. So in the paper, they talk about various things, including things such as leakage. It terms out that leakage in an expression only inhibits oscillations though. So in some sense, if you're trying to get oscillations, leakage is a problem, actually. And that's why they use this especially tight-- well we're going to talk about that in a few minutes. They use an especially tight version of these promoters to have low rates of leakage in a synthesis. But what might you need in order to get oscillations in negative autoregulation? Did you have-- have delay. Yes indeed. And that's something that they mentioned in the Elowitz paper is if you add explicit delay. So for example, if instead of having the repression depend on--OK, I already erased everything. But instead of having the protein, for example, being a function of the mRNA now, maybe if you said, oh, it's a function of the mRNA five minutes ago. And that's just because maybe it takes time to make the protein. Or it takes time for this or that. You could introduce an explicit delay like that. Or you could even, instead, have a model where you just have more steps. So what you do is you say, oh, well yeah, sure. What happens is that, first, the mRNA is made. But then, after the mRNA is made, then you have to make the peptide chain. Then, the that peptide chain has to fold. And then, maybe, those proteins have to multimerize. Indeed, if you right down such a model then, for some reasonable parameters, you can get oscillations just with negative auto regulation. And indeed, I would say that over the last 10 years, probably, the reigning king of oscillations in the field of system synthetic biology is Jeff Hasty at San Diego. And he's written a whole train of beautiful papers exploring how you can make these oscillators in simple G network. So he's been focusing in E. coli. There's also been great work in higher organisms in this regard. But let's say, Hasty's work stands out in terms of really being able to take these models and then implement them in cells and, kind of, going back and forth. And he's shown that you can generate oscillations just using negative auto regulation if you have enough delays in that negative feedback loop. Are there any questions about where we are right now? I know that we're supposed to be talking about the repressilator. But we first have to make sure we understand the negative auto regulation. So everything that we've said, so far, in terms of the models was all known. But what Michael wanted to do is ask whether he could really construct an oscillator. And he did this using these three mutual reppressors. We'll say x, y, and z just for now. x represses y, represses z, represses x. And has a nice model of this system that helped him guide the design of his circuits. So experiments-- as most of us who have done them know-- experiments are hard. So if you can do a week of thinking before you do a year of experimental biology, then you should do that. And what were the lessons that he learned from the modeling that guided his construction of this circuit? Yeah? AUDIENCE: Lifetime of mRNA. PROFESSOR: Right. So you want to have similar lifetime of the mRNA and the protein. And this is, somehow, similar to this idea that you need more delay elements because if you have very different lifetimes, then the more rapid process, somehow, doesn't count. It's very hard to increase the lifetime of the mRNA that much in bacteria. So instead, what he did is he decreased the lifetime of the proteins of the transcription factors. In this case, x, y, and z. And you mentioned the other thing that he maybe did. AUDIENCE: He introduced the leakage, but he didn't mention that that was PROFESSOR: That's right. So I guess, he knew that leakage was going to be a problem. I.e, that you want tight repression. So he used these synthetic promoters that both had high level of expression when on but then very low level of expression when being repressed. He made this thing. And in particular, he looked at it in a test tube. He was able to use, in this case, IPDG to synchronize them. And he looked at the fluorescence in the test tube. So the fluorescence is reporting on one of the proteins. We can call it x if we'd like. But fluorescence is kind of telling about the state. And if it starts out, say, here, he saw a single cycle. Damped oscillations, maybe. So the question is, why did this happen? So why is it that, in the test tube, he didn't see something that looked very nice? Oscillations. Noise. And in particular, what kind of noise? Or what's going on? Desynchronization, exactly. So the idea is that, even if you start out with them all synchronized-- you give it IPDG pulls, and they're synchronized in some way-- it may be that, at the beginning, all of them are oscillating in phase with each other. But over time, random noise, phase drift, and the different oscillators leads to some of them come down and come back up. And then, others are slower. You start averaging all these things together. And it leads to damped oscillations at the test tube level within the bulk. Yes? AUDIENCE: So what do you mean in the test tube? Like, you just take all these components and put it-- PROFESSOR: Sorry, when I say test tube, what I mean is that you have all the cells. So they still are intact cells. But it's just many cells. So then, the signal that you get the fluorescence is some average over all or sum overall. The fluorescence you get from all those cells. So there's a sense that this is really what you expect given the fact that they're going to desynchronize. Of course, the better the oscillator in the sense that the lower the phase drift, then maybe you can see a slower rate of this kind of desynchronization. But this is really what you, kind of, expect. All right. So that's what, maybe, led him to go and look at the single cell level where he put down single cells on this agar pad and just imaged as the cells oscillated and divided. Now there are a few features that are important to note from the data. The first is that they do oscillate. That's a big deal because this was, indeed, the first demonstration of being able to put these random components together like that and generate oscillation. But they didn't oscillate very well. So they said, oh, maybe 40% of the cells oscillated. And I have no idea what the rest of the cells were doing. But also, even the cells that were oscillating, there was a fair amount of noise to the oscillation. And the latter half of this paper has a fair amount of discussion of why that might be. And they allude to the ideas that had been bouncing around and from the theoretical computational side demonstrating that it may be that the low numbers of proteins, genes involved here could introduce stochastic noise into the system and, thus, lead to this kind of phase drift that was observed experimentally. I think that this basic observation that Michael had that he got oscillations, but they were noisy. That is probably what led him to start thinking more and more about the role of noise in G networks and so forth and led, later, to another hugely influential paper that is not going to be a required reading in this class but is listed under the optional reading, if you're interested. But we'll really get into this question of noise more a couple weeks from now. Were there any other questions about the experimental side of this paper? I wanted to analyze maybe a little bit of simple model of the repressilator. So the model that they used to help them design this experiment involved all three proteins, all three mRNAs. And what that means is that, when you go and you do a model, you're going to end up with a six by six matrix. And I don't have boards that are big enough. So what I'm going to do instead is I'm going to analyze just the protein only version model of the repressilator. All right. So what we have here is three proteins. p1 2 3 p1 dot. And we have degradation of this protein. And we're going to analyze the symmetric version, just like what Michael did. So that means we're assuming that all the proteins are equivalent. I'm sure that's not true because these are different promoters and different everything. But this gives us the intuition. So it's minus p1. And this is protein 1 is repressed by trajectory protein 3. Protein 2 is going to be repressed by protein 1. And then protein 3 is going to be repressed by protein 2. So this is what you would call the protein only model of the repressilator. Now just as before, the fixed points are when the pi dots are equal to 0. And we get the same equation that we, basically, had before where the equilibrium or the fixed point, again, is going to be given by something that looks like this. So it's the same requirement that we had before. Now the question is, how can we get the stability of that internal fixed point? It's worth mentioning here that now we have three proteins. So the trajectories are in this three dimensional space. So from a mathematical standpoint, determining the stability of that internal fixed point is actually not sufficient to tell you that there has to be oscillations or there cannot be oscillations because these trajectories are, in principal, allowed to do all sorts of crazy things in three dimensions. But it turns out that it still ends up being true here that when this internal fixed point is stable, you don't get oscillations. And when it's unstable, you do. But that, sort of, didn't have to be true from a mathematical standpoint. All right. Now since this is now going to be a 3 by 3 matrix, we're going to have to calculate those eigenvalues. Now how many eigenvalues are there going to be? Three? OK. So this thing I've written in the form of a matrix to help us out a little bit. But in particular, we're going to get the same thing that we had before, which is the p1 tilde. So these are deviations, again, from the fixed point. And we got this matrix that's going to look like this. Minus 1, again, 0. It's the same x that we had before conveniently still on the board. So this is just after we take these derivatives. And then, we have p1 tilde, p2 tilde, and p3 tilde. Now what we need to know is, for this Jacobian, what are going to be the eigenvalues? For this thing to be stable, it requires what? What's the requirement for stability of that fixed point? That p0? AUDIENCE: [INAUDIBLE]. PROFESSOR: OK. Right. For two dimensions, this trace and determinant condition works. It's important to say that that only works for two dimensions, actually, the rule about traces and determinants. So be careful. So what's the more general statement? Yeah. AUDIENCE: Negative eigenvalues. PROFESSOR: Exactly. So in order for that fixed point to be stable, it requires that all the eigenvalues have real parts less than 0. So in order to determine the stability of the fixed point, we need to ask what are the eigenvalues of this matrix. And to get the eigenvalues, what we do is we calculate this characteristic equation, this thing that we learned about in linear algebra and so forth. What we do is we take-- all right, this is the matrix A, we'll say. This is matrix A. And what we want to do is we want to ask whether the determinant of the matrix A minus some eigenvalue times the identity matrix. We want this thing to be equal to 0. So this is how we determine what the eigenvalues are. And this is not as bad as it could be for general three by three matrices because a lot of these things are 0. So this thing is just this is the determinant of the following matrix. So we have minus 1 minus lambda 0. This thing x that's, in principle, bad. Minus 1 minus lambda 0. Getting 0 x minus 1 minus lambda. Now to take the determinant three by three matrix, remember, you can say, well, this determinant is going to be equal to-- we have this term. So this is a minus 1 plus a lambda times the determinant of this matrix. And then, we just have that's this. The product of these minus the product of these. So this just gives us this thing again. So this is actually just minus 1 plus lambda cubed. Next term, this is 0. That's great. We don't need to worry about that. The next one, we get plus. We have an x. Determining here, we get, again, x squared. So this is just an x cubed. We want the same equal to 0. So we actually get a very simple requirement for the eigenvalues, which is that 1 plus the eigenvalues cubed is equal to this thing x cubed. Now be careful because, remember, x is actually a negative number. So watch out. So I think that the best way to get a sense of what this thing is is to plot it. Of course, it's a little bit tempting here to just say, all right, well, can we just say that 1 plus lambda is equal to x? No. So what's the matter with that? I mean, it's, sort of, true, maybe, possibly. Right. So the problem here is that we're supposed to be getting three different eigenvalues. Or at lease, it's possible to get three different eigenvalues. So this is really specifying the solution for 1 plus lambda on the complex plane. So the solution for 1 plus lambda we can get by thinking about this is the real part of 1 plus lambda. And this is the imaginary part of 1 plus lambda. And we know that one solution is going to be out here at x. This distance here is the magnitude of x. Now the others, however, are going to be around the complex plane similar distances where we get something that looks like this. So these are, like, 30, 60, 90 triangle. So this is 30 degrees here because what you see is that, for each of these three solutions for 1 plus lambda, if you cube them, you end up with x cubed. So this guy, you square it. Cube. You end up back here. This one, if you cube it, you start out here, squared, and then cubed comes back out here. Same thing and this goes around somehow. All right, so there are three solutions to this 1 plus lambda. And there are these points here. Now, of course, it's not 1 plus lambda that we actually wanted to know about. It was lambda. But if we know what 1 plus lambda is, then we can get what lambda is. What do we have to do? Right, we have to slide it to the left. So this is the real axis. This is the imaginary axis. 1. So we have to move everything over 1. Now remember, the requirement for stability was that all of the eigenvalues have real parts that were negative. That means the requirement for stability of that fixed point is that all three of these fixed points are in the left half of the plane. So what you can see is that, in this problem, the whole question of stability and whether we get oscillations boils down to how big this thing is. What's this distance? If this distance is more than 1, then we subtract 1, we don't get it into the left part of the plane. OK, I can't remember which case I just gave. But yeah, we need to know whether this thing is larger or smaller than 1. And that has to do with the magnitude of x. So if the magnitude of x-- do you guys remember your geometry for a 30, 60, 90 triangle? All right, so if the magnitude of x-- and this is indeed the magnitude of x. This short edge on a 30, 60, 90 is half the long edge, right? So what we can say is that this fixed point stable, state fixed point, is if and only if the magnitude of x is what? Lesson two. OK. That's nice. And if we want, we could plug in-- just to ride this out. This is n alpha p 0 n minus 1. So it's useful, once you get to something like this, to try to just ask, for various kind of values, how does this play out? What does the requirement end up being? And a useful limit is to think about what happens in the limit of very strong expression? So strong expression corresponds to what? AUDIENCE: Big alpha? PROFESSOR: Big alpha. Yes, perfect. And it turns out, big alpha is a little bit-- OK, and remember we have to remember what p0 was. p0 was this p0 times 1 plus p0. All right, so this is the requirement. And actually, if you play with these equations just a little bit, what you'll find is that, if alpha is much larger than 1, then this requirement is that n is less than 2 or less than around 2. This is saying, on the flip side, the fixed point is stable if you don't have very strong cooperativity and repression. And the flip side is, if you have strong cooperativity of repression, then you can get oscillations because this interior fixed point becomes unstable. So this is also saying that n greater than around 2 leads to oscillations. And this maybe makes sense because, when you have strong productivity in the repression there, what that's telling you is that it's a switch like response. And in that regime, it maybe becomes more like a simple Boolean kind of network where, if you just write down the ones and zeroes, you can convince yourself that this thing maybe, in principle, could oscillate. Now if you look at the Elowitz repressilator paper, you'll see that he gives some expression for what this thing should be like. And it looks vaguely similar. Of course, there he's including the mRNAs, as well. But if you think that this was painful to do in class, then including the mRNAs is more painful. Are there any questions about this idea? Yeah. AUDIENCE: So in the paper, did they also only do the stability analysis to determine the-- PROFESSOR: I think they did simulations, as well. So the nature of simulations is that you can convince yourself that their exist places that do oscillate or don't oscillate. Although, you'll notice that they have a very, kind of, enigmatic sentence in here, which is that it is possible that, in addition to simple oscillations, this and more realistic models may exhibit other complex types of dynamic behavior. And this is just a way of saying, well, you know, I don't know. Maybe someday because once you talk about six dimensional system, you never know if you've explored all of the parameter space. I mean, even for fixed parameters, you don't know if you started at all the right locations. You can kind develop some sense that, oh, this thing seems to oscillate or seems to not oscillate. And it does correspond to these conditions. But you don't know. I mean, it could be that, in some regions, you get chaos or other things. Right? So it's funny because I've read this paper many times. But it was only last night when I was re-reading it that I kind of thought about that sense like, yeah, I'm not sure either what this model could possibly do. Yes? AUDIENCE: In this linear analysis the three x's are the same. PROFESSOR: That's right. AUDIENCE: Because they're non-dimensional? PROFESSOR: All right. So the reason that the three X's are the same is because we've assumed that this really is the symmetric version of the repressilator because we're assuming that all of the alphas, all the ends, all the K's, everything's the same across all three of them. So given that symmetry, then you're always going to end up with a symmetric version of this. So I think if it were asymmetric and then you made the non-dimensional versions of things, I think you still won't end up getting the same X's just because, if it's asymmetric, then and something has to be asymmetric. Yes? AUDIENCE: [INAUDIBLE] so large alpha leads to-- PROFESSOR: Yes, OK. We can go ahead and do this. So for large alpha, this fixed point is going to be-- p0 is going to be much larger than 1. So this is about p0 to the n plus 1. We can neglect the 1 for large alpha. And then and then what we can say is that, over here, for example, if we multiply both sides by-- p0 squared, p0 squared, so multiply it by one. Then this down here is definitely alpha squared. And then, up here, what we have is p0 to the n plus 1, which we decided was around alpha for a strong alpha. So that gives us alpha times alpha divided by alpha squared. So this actually all goes away for large alpha. So then, you're just left with n less than 2. Did that-- AUDIENCE: Sorry. Where did that top right equation come from? PROFESSOR: OK, so this equation here is this is the solution for where that fixed point is. So in this space of the p0's, if you set the equations for p1, p2, p3, if you set that equal to 0, this is the expression always for a large alpha, small. So this is a need be location of that fixed point. And it's just, as alpha is large, then we get that p0 to the n plus 1 is approximately equal to alpha. And this is for alpha much greater than 1. And in that case, all of these things just go away. And you're just left with n less than 2. So for example, as alpha goes down in magnitude, then you end up getting a requirement that oscillations require a larger n. We'll give you practice on this. All right, so I think I wrote another-- if I can find my-- you can ask for alpha equal to 2. What n required for oscillations. I'll let you start playing with that. And I will make sure that I've given you the right alpha to use. So in this case, what we're asking is, instead of having really strong maximal expression, if instead expression is just not quite as strong, then what we'll find is that you actually need to have a more cooperative repression in order to get oscillations. And that's just because, if alpha is equal to 2, then we can, kind of, figure out what p0 is equal to. 1. Right. Great. So the fixed point is at one. That's great because this we can then figure out. Right? So this is 1 plus 1 square. That's a four. 1. 2. So this tells us that, in this case, we need to have very cooperative repression. We have to have an n greater than around 4 in order to get oscillations in this protein only model. Yes? AUDIENCE: It is kind of strange that even for a really big alpha you still need n greater than sum. PROFESSOR: Yeah, right, right. So this is an interesting question that you might think that for a very large expression that you wouldn't need to have cooperative repression at all. Right? And I can't say that I have any wonderful intuition about this because it, somehow, has to do with just the slopes of those curves around that fixed point. And it's in three dimensions. But I think that this highlights that it's a priori if you go and say, oh, I want to construct this repressilator, it's maybe not even obvious that you want it to be more or less. I mean, you might not even think about this idea of cooperative repression. You might be tempted to think that any chain of three proteins repressing each other just, kind of, has to oscillate. I mean, there's a little bit of a sense. And that's the logic that you get at if you just do 0's and 1's. If you say, oh, here's x. Here's y. Here's z. And they're repressing each other. Right? And you say, oh, OK, well if I start out at, say, 0 1 0 and you say, OK, that's all fine. But OK, so this is repressing. And it's OK, but this guy wasn't repressing this one. So now we get a 1, 1, 0, maybe. Then you say, oh, OK. Well now this guy starts repressing this one. So now it gives us a 1 0 0. And what you see is that, over these two steps, the on protein has shifted. And indeed, that's going to continue going all the way around. So from this Boolean logic kind of perspective, you might think that any three proteins mutually repressing each other just has to oscillate. And it's only by looking at things a little bit more carefully that you say, oh, well, we have to actually worry about this that you really have to think about you want to choose some transcription factors that are multimerizing and cooperatively repressing the next protein just to have some reasonable shot at having this thing actually oscillate. AUDIENCE: So in this, we might still be able-- I mean, oscillations like this might still [INAUDIBLE] but just not like, maybe, oscillations around some stable fix point or something. Like, they're just not limit cycle oscillations. Do you think that in a [INAUDIBLE] there would probably still be some kind of oscillations somewhere. Just not this beautiful limit cycle kind. PROFESSOR: Yeah, my understanding is that in, for example, this protein only model of the repressilator that if you do not have cooperative repression, then it really just goes to that stable fixed point. Of course, you have to worry about maybe these noise-induced oscillation ideas. But at least within the realm of the deterministic, differential equations, then the system just goes to that internal fixed point that's specified by this. Question? AUDIENCE: Can we think like that the cooperation, sort of, introduced delay? PROFESSOR: That's an interesting question. Whether cooperativity, maybe, is introducing a delay. And that's because, after the proteins are made, maybe it takes some extra time to dimer and so forth. So that statement may be true. But it's not relevant. OK? And I think this is very important. This model has certainly not taken that into account. So the mechanism that's here is not what you're saying. But it may be true that, for any experimental system, such delay from dimerization is relevant and helps you get oscillations. Right? But at least within the realm of this model, we have very much not included any sort of delay associated with dimerization or anything. So that is very much not the explanation for why dimerization leads to oscillations here. And I think this is a wider point that it's very important always to keep track of which effects you've included in any given analysis and which ones are not. And it's very, very common. There are many things that are true. But they may not actually be relevant for the discussion at hand. And I think, in those situations, it's easy to get mixed up because it still is true, even if it's not what's driving the effect that is being, in this case, analyzed. We're out of time. So we should quit. On Tuesday, we'll start by wrapping up the oscillation discussion by talking about other oscillator designs that allow for robustness and tunability. OK?
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https://ocw.mit.edu/courses/5-111-principles-of-chemical-science-fall-2008/5.111-fall-2008.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK. Let's get started. Can you go ahead and take 10 more seconds on this first clicker question here? OK. So it looks like most of you got that the electron configuration that we're writing here is for copper. So I'm actually going to give the benefit of the doubt that the people that didn't get it right were rushing to get out their clickers and didn't have time to think it all the way through. Remember that when we're talking about 4 s 1, 3 d 10, that's one of those exceptions where a completely filled d orbital is more stable than we would expect. So, that's actually the electron configuration we have when we're talking about copper and some other exceptions in the periodic table that you're going to be looking at. So, hopefully, if you were to go back and look you could see that this is, in fact, copper. We're actually going to do one more clicker question to get started with today, and as we do, I'll explain something we're going to be trying today, which is a little bit of a friendly competition in terms of answering the clicker questions correctly. So we've tagged each of your numbers to your actual recitation, so we can see today which recitation actually is going to be doing the best in terms of clicker questions, who's going to get the most correct today. So, you may or may not know this about your TAs, but this is a pretty competitive group of TAs we have this year, and they like to brag about how smart their recitation is, how good questions they're getting in the recitation section. So, do your TA proud today and see if you can be part of the recitation that gets the most correct in terms of a percentage. And at the end of class we'll announce which recitation that is, we'll also make sure to give you a little bit of a prize if you are, in fact, in that recitation. So we have extra incentive to get these clicker questions right. So, in this one we're selecting the correct electronic configuration for an ion. So, why don't you go ahead and take 10 more seconds on this second clicker question for our intro. OK. So, it looks like we have a little bit of a mixed consensus here. Let's go over this question. And I know there's a lot to talk about about this competition, but let's just get into listening mode here and talk about how we can figure out what the correct electron configuration is for this ion. Remember, ions are a little bit different. The first thing we need to do is write the electron configuration for the atom itself, and then we need to take an electron away. So here we're talking about v plus 1, so if we were to write it just for the neutral electron itself, we would find that the electron configuration is argon, that's the filled shell in front of it. Then 4 s 2 and 3 d 3. So this would be for the actual filled, the completely neutral atom. But remember what we said, which was when we talked about, this is at the end of class on Friday, we said that it turns out that even though 3 d is higher in energy when it's not filled, once we fill it with an electron, these 2 orbitals actually switch place in terms of energy. So if we were to write this in terms of energy, we would actually have to rewrite it has 3 d 3, and then 4 s 2. So, which orbital would we take an electron out of if we were ionizing this atom here? The s. So, we would actually take an electron out of the s, which gives us 3 d 3 and then 4 s 1. So, it's a little bit of a trick when you're dealing with ions. The best suggestion is just to write it out completely for the neutral atom, and then you want to take an electron out of the highest orbital. It makes sense that it's going to come out of the highest occupied atomic orbital, because that's going to be the lowest amount of energy that's required to actually eject an electron. All right. So let's go to today's notes. And actually before we start into today's topics, I want to remind everyone and hopefully you all do remember that our first exam is coming up and it's coming up in exactly a week, so it'll be a week from today, next Wednesday. And on Friday in class, at the beginning of class, I'll go over just in all the detail you could possibly imagine everything you need to know logistically for the exam -- where it is, what you do, what kind of calculators you can bring, which by the way are any calculator. So you'll get all of that information on Friday. So don't worry if you have some questions right now. I just want to let you know that. The other thing I want to let you know is that instead of having a new problem-set that you'll be assigned this Friday, what we'll do instead is we'll give you some practice problems, and these will be just more of the same type of problems that you saw before but that's another chance to try them out more. These won't be graded, you don't have to turn them in, it's just to give you some extra practice if you want while you're studying for the exam. We'll also post an exam from a previous year so you can actually see exactly what the format's going to look like. So when you go into the exam a week from today, it'll all look really familiar, you'll be comfortable with the format and you can just dive right in and start answering the questions. So you'll have all that information and we'll get it to you on Friday. The other quick thing I want to say is that I do have office hours today from 3 to 5, so feel free to stop by if you have questions about problem-set 3 that you're finishing up. And also, for those of you that did sign up for the pizza forum tonight, that's going to be at 5 o'clock, it's in room 56-502, so we'll see some of you tonight for that as well. All right. So, let's move on to what we're talking about today. What we're going to start with is discussing photoelectron spectroscopy, which is a spectroscopy technique that will give us some information about energy levels in multielectron atoms. We'll then take a turn to talking about the periodic table, we'll look at a bunch of periodic trends, including ionization energy, electron affinity, electronegativity and atomic radius. And then, if we have time at the end, we'll introduce one last topic, which is isoelectronic atoms and ions. I also want to note that the end of the material today, so this last topic here, that's the end of the material that's going to be on this first exam. So whether we finish it today, or more likely when we finish it up on Friday, once we get passed isoelectronic atoms, that's it, that's all you need to study for this first exam. So from that point on it'll be exam 2 material, so depending on how you like to come compartmentalize your information, you can separate that in your brain in terms of what you're trying to learn right now versus what you can put off until a little bit later. So, let's start with talking about photoelectron spectroscopy. This actually relates very closely to what we discussed in class on Friday before the long weekend, and what we were talking about is the energy levels of multielectron atoms. So what we'll start with today is talking about the technique that's primarily used to actually experimentally figure out what these different energy levels are. And this is called photoelectron spectroscopy, and essentially what it is is very similar conceptually to what we were talking about way back in the first couple lectures when we were talking about the photoelectric effect. Because here what we have is some atom that we're studying, in the case, it's going to be a gas, and we hit it with a photon that has some incident energy. So e sub i, some energy that the photon comes in with, and if it has sufficient energy to eject an electron, it will do that, and our electron will be ejected with a certain kinetic energy, which is going to be whatever energy is left over from the initial energy we put in minus what was taken up in order to actually ionize or eject the electron. So, you can see how this can directly give us different ionization energies for any atom that we're interested in studying. For example, with neon we can think about all of the different orbital energies we could be looking at. In the first case, so here is the electron configuration of neon. So we can think about what is our most loosely-bound electron, what's that highest energy orbital, and it's going to be the 2 p orbital, that's going to be what's highest in energy. So if we're going to eject an electron using a minimum amount of energy, that's where it's going to come from. So, you can imagine, that we'll actually probably have a lot of kinetic energy left over if we put a lot of energy in in the first place. We're only using up a little bit to eject the electron, then we'll have a lot left over. So, one difference between photoelectron spectroscopy and, for example, the photoelectric effect is that in this case, we're not just looking at one energy level, which is what we were looking at from the surface of a metal, now we're talking about this gaseous atom. So we can actually pop an electron or eject an electron from any single orbital that is occupied within the atom. So, for example, it's not just the 2 p that we could actually take an electron from, we could also think about ejecting an electron from the 2 s orbital. Now this, of course, is going to take more energy because a 2 s is lower, it has has a more negative binding energy than the 2 p, but that's OK as long as we put in enough energy, but what we're going to find is the kinetic energy coming out with the electron is actually going to be a little bit less, right, because we had to use up more energy to eject the electron, so we don't have as much left over. There's actually one more orbital that we could talk about if we're talking about this sample case of neon, which is a 1 s orbital, and if we're talking about a 1 s orbital, now we're going to be even lower in energy still, so that means the minimum energy required to eject an electron is going to be at its highest, so that means the energy that we have left over that turns into kinetic energy for the electron, is now going to be really quite small. And what happens when you irradiate one of these atoms that you're studying with this light is in photoelectron spectroscopy, you want to make sure that you put in enough energy to actually ionize any single electron that you have in the atom. So the way that we really make sure this is done is that we use x-rays. So you know that x-rays are higher frequency than UV light, for example, that means it's also higher energy than UV light, and if you think back to our photoelectric effect experiments, do you remember what type of light we were usually using for those? Does anyone remember? Yeah. It was UV light that we used. Well, we can't guarantee with UV light we'll have enough energy to eject every single electron, so that's why when we use x-rays, they're higher energy, you can pretty much be guaranteed we're going to eject all of those electrons there. So I said that this technique was used to experimentally determine what the different binding energies or the different ionization energies are for the different states in a multielectron atom. Another way to say states is to talk about different orbitals. So we can do this directly as long as we have certain types of information. The first that we need to know the energy of the photon that's incident on our gaseous atom. The second piece of information we need to know is what actually the kinetic energy is of the ejected electron, and that's something we can just measure by measuring its velocity. So we can use an equation to relate the incident energy and the kinetic energy to the ionization energy, or the energy that's required to eject an electron. This should all sound incredibly familiar, like I'm just repeating myself in terms of photoelectric effect, because essentially that's what I'm doing, and that's one reason we spent so much time and did so many problem-set problems on the photoelectric effect. So what we're saying here is the incident energy, so the energy coming in, is just equal to the minimum energy that's required to a eject an electron. When we talked about the photoelectric effect, that was called the work function. In this case, it's called the ionization energy, plus whatever kinetic energy we have left over in the electron. So if we want to solve for ionization energy, we can just rearrange this equation. Our ionization energy is going to be equal to the incident energy coming in, minus the kinetic energy of the electron. So, let's take a look at the different kinetic energies that would be observed in a spectrum for neon where we had this incident energy here. And it turns out that the first kinetic energy that we would see or the highest kinetic energy, would be 12 32 electron volts. So if that's the case doing a quick little calculation, what would the ionization energy be for a 2 p electron in neon? Yup, 22. So, basically all we did was take 12 54, subtract 12 32, and we got 22 electron volts. We can do the same thing for the other observed kinetic energy. So, for example, in the second case, we say that we see 12 06 in terms of the kinetic energy. Same sort of subtraction problem, what do we have for the ionization energy of the 2 s electron? Good, quick math. All right, so 48 electron volts. And let's look at the final kinetic energy that we'd observe in this spectrum, which is 384 electron volts, so what is that third corresponding ionization energy? I couldn't quite hear, but I have a feeling everyone said 870 electron volts. So, we can actually kind of visualize what we would see if we were looking at a photoelectron spectrum. And what we would see if we were graphing, for example, increasing kinetic energy, is we would see 1 line corresponding to each of these energies of electrons that we see coming out. And, of course, each of those electrons correspond to an electron coming out of a particular orbital. So in the case of 12 32, that is our highest kinetic energy, that means it's our lowest ionization energy -- it's the smallest amount of energy it takes to pop an electron out of that orbital. So that's why we see the 2 p here, the 2 s is 12 06, and it makes sense that what we see as the greatest ionization energy, which is also the smallest kinetic energy is that 1 s orbital. Remember, because that 1 s orbital is all the way down in terms of if we're thinking about an energy diagram, we're all the way down here, so we have a huge amount of energy we have to put into the system in order to eject an electron. So what I want to point out when you're kind of looking at these numbers here, what the significance is, look at that huge difference between what the ionization energies are for what we call those valence electrons, those outer shell electrons, versus the ionization energy for the 1 s orbital -- those are core electrons there. So we can think about something I mentioned last time, which is when we're thinking about chemistry and what's really interesting in terms of chemical reactions, it's mostly valence electrons we're talking about, those are the ones that tend to be involved in chemical reactions. It makes a lot of sense when we look at it energetically, because if we think about a 1 s core electron, that's going to be held really, really tightly to the nucleus. We see that we have to put this huge energy in to actually get a 1 s electron ejected, so it makes a lot of sense that we wouldn't want to pay that energy cost in a normal chemical reaction. And we don't -- we very rarely would see these core electrons actually being involved in any type of a reaction. All right, so one thing that I want to point out, which I said many, many times on Friday, and this is perhaps the last time I'll say it, but one last time is we can think about why we only see a line for the 2 p orbital, versus we don't see separate lines for a 2 p x, a 2 p y, and a 2 p z. Remember, we need those three quantum numbers to completely describe the orbital. Why do we just see one for all the p's? And the reason is that the energy of the orbitals, depend on two quantum numbers, and that's quantum number n, and quantum number l. M does not actually have an effect, in this case, on the energy of the orbital. So that's why we're not seeing separate lines in this spectrum. All right. So let's go ahead and try an example here in thinking about photoelectron spectroscopy. So, let's say we're looking at an element and we have an emission spectra, and we know that it has five distinct different kinetic energies in that spectrum. We might be asked, for example, to determine what all of the different elements could be that would produce a spectrum that gave us 5 different lines. So the first thing that we want to do, if we're thinking about something like this, is just to determine exactly what orbitals are causing the five different lines that we're seeing in the spectrum. So, if we're talking about five different orbitals and we're talking about a ground state atom, we know that we just need to start at the bottom and work our way out up. So, our first orbital that an electron must be coming from is the 1 s. What comes after that? 2 s. All right, then what? 2 p. After that? 3 s. Next? 3 p, and that's 1, 2, 3, 4 -- that gives us five different options, five different orbitals, five different energies right there. So, then all we need to do to determine which elements that corresponds to is take a look at our periodic table. So we want to look at any element that has a 3 p orbital filled, but that does not then go on and have a 4 s, because if it had the 4 s filled then we would actually see six lines in the spectrum. So that is relevant for all of these atoms here, so we actually have several different possibilities. It could be aluminum, silicone, phosphorous, sulfur, chlorine or argon. Any one of these different elements could actually produce a photoelectron spectroscopy spectrum that has five distinct lines. If I went on and told you what the different incident light was, and what the electrons were ejected with, and then you could look up the ionization energy for the particular different elements, you should be able to actually determine exactly which element it is, but just with the information given, we can only narrow it down to these choices here. So let's actually let you try another example of solving a problem that has to do with one of the spectrums. So, let's turn to another clicker question here. Remember, your answer holds great weight in terms of the state of the TA bragging for next week. So, how many distinct, so again, we're talking about distinct kinetic energies, would be displayed if you're talking about a spectrum for the element hafnium, and I'll tell you here that it has a z of 72, so you don't have to spend two minutes searching your periodic table. The period of table's on the back page of your notes if you don't see that there. All right. It looks like a lot of you are done, so let's take 10 more seconds here. Part of the challenge is speed, too, how quickly you can get these answers in terms of getting them in on time. So let's see what we say. All right. So I think I can safely say that most people had the right idea and were counting quickly, though I have a feeling that some people who wrote 13 might have forgotten about those 4 f, the 4 f electrons. So, remember when you're looking at your periodic table, don't forget about the lanthinides, sometimes they come into play. So it's actually 14, and the way that we got that answer was we just wrote out or just looked at your period table, figured out all of the different orbitals that you could have in terms of the principle quantum number, and then the l quantum number, and then write them all down -- it turns out to be 14, so that's what the answer is. So, it looks like this is good, because we'll have some separation in terms of not everyone's going to get 100% in terms of recitations here, which is what we're going for. All right. So let's turn our attention to a new topic, which is thinking a little bit about the periodic table, and also talking about periodic trends. And there's a lot we can explain by talking about what we see in the periodic table in terms of what different trends are in grouping different elements in different spots within the periodic table. So, here we have a picture of Dmitri Mendeleev, who is one of the scientists responsible for first compiling the periodic table. You'll notice I have what's a very flattering picture of him up here, and if you haven't done the reading yet you might not think this is particularly flattering, but if you look at the picture of him in the book, you'll notice I chose a very flattering picture of Dmitri up here, and here he's pondering putting these elements together in a periodic table. And he actually did this in the late 1800's, back before even all of the elements that we know today were discovered, really only about 60% or so, 70% were discovered then that we now know today. But still, he was able to put together a periodic table. And what he did what he actually grouped things in terms of their chemical properties. So the way that we like to think of things now is in terms of electron configurations, right, but at the time that wasn't really understood. So, instead, it was amazing he was able to group things in terms of the properties that he saw. So, for example, if he was talking about the group one metals, lithium, sodium, potassium -- he noticed these were all very soft reactive metals, those were grouped together. Versus looking at, for example, helium or neon or argon, these are all inert gases, inert meaning essentially do not react, those were grouped together in the periodic table. So basically, at the time he was just going on size and then traits, but what we actually know today is that we can also order things in the periodic table by electron configuration. In fact, that is the most logical way for us to look at it now. So, for example, if we're actually thinking about electron configuration and we look at lithium, sodium and potassium, these all have one valence electron. So basically, that means one electron in an s orbital in their outer-most most shell. So that explains why they're so reactive, they're all very willing to give up that 1 s orbital and then drop to a lower energy level. In contrast, helium, neon, and argon all have filled shells. That also explains why they're very stable. They're not going to want to add on another electron, because then it'll have to jump a very large energy level and start filling in another shell -- go from n equals 2, to n equals 3, and n equals 4, and so on. So it turns out that we can really know a lot just by looking at the periodic table. You will never in this class have to memorize anything about the periodic table. Depending on what kind of chemistry you go in to, you might accidentally memorize parts of the table, which is fine, but what you really want to know how to do is know how to use the periodic table. But you actually need to keep a few caveats in mind as you do this, which is the fact that trends predict a lot of chemical properties, but they can't predict everything in terms of biological properties. And after the periodic table was developed in the late 1800's, people didn't understand this quite as well, they took things a little more literally. They thought, for example, if you could do something with one element, if you looked at an element very close to it, it would be similar enough that you could maybe replace it with that. Today we know, for example, if you can put one certain kind of element in your mouth or eat that, it doesn't necessarily mean you want to put the element next to it and your mouth as well, that might not be safe. But this is things we've learned as the years have gone past. So, let's just take a quick example to show how not completely you can use these periodic trends, that there are limits. So if we consider lithium, potassium, and sodium, they're all together in the same group on the periodic table, knowing what we do about biology we can immediately think of sodium and potassium, or even just knowing what you know about table salt, for example, that these are two elements that we find, and particularly in the ion form in very high concentrations in our body. For example, sodium in our blood plasma is almost to the point sometimes of 100 millimol or that's very, very concentrated. Similarly, we find it in table salt, we're taking it in all the time, the same with potassium, think of bananas, were always eating potassium. Not so with lithium. I don't think too many people and here are probably taking lithium. It turns out there's actually no natural function known in the body for lithium. So there's nothing naturally going on unless we were to introduce it ourselves in our body that we know of, at least, that involves lithium. But this did not stop people, for example, in the late 1800's, early 1900's, and, in fact, in 1927 a new soft drink was put on to the market and they wanted to make a lemon-lime soft drink, these were very popular in the early 1900's, and to get sort of that lemony flavor, they decided to use citric acid, so that's a good idea, that gives that soury taste. And they wanted to use a soluble salt of citric acid, so they could have used sodium, they could have used potassium. But, you know why not do something a little special, little different, and they decided instead to use lithium. So, here we have this soda with lithium citrate, some of you might be familiar with this, soda is called 7-Up. So, 7-Up no longer has lithium in it, but from 1927 to 1950 it did, and, in fact, not only did they not try to hide the fact that there's lithium in the soda, this they used as a really special marketing technique, they really pointed out this is something that stands out about our soda, this is something special. There's a lot of good things about lithium. I don't know if you can see, probably not, what's written on here, so let me point out to you a few things. Lithium, slenderizing, that's great to see in a soda. Other nice things about lithium in your soda, it dispells hangovers, takes the ouch out of grouch. That's very nice. So basically, you get a lot of benefit supposedly from this 7-Up soda from the 1920's or so. And this went on and was unregulated for some time, but at some point the Food and Drug Administration did take a step in, so here's a case where they did do something important -- that's not what I mean at all -- where they did take the step, they do many things that are important, often not quickly enough. Here's a -- actually here it did take 25 years, but they did, they did eventually step on before we started drinking 7-Up. And what they said was, look, you can't put this in, we're starting to notice it does some strange things. Because it was in the 1950's or so, maybe the late 1940's, that people started to discover lithium, even though it had no natural function, it did do something in our bodies. Does anyone know what was lithium's used for? Yeah, it's an anti-psychotic drug, so, for example, some people with bipolar disorder even today still take it, it works really well for some people, for other people it doesn't work so well. But anyway, this isn't really something you want to have in your soda, so they did take it out eventually. Another side effect if you take too much lithium is death, so that's no good to have in sodas either, and it might not have been as big a deal back in the 1920's, but you can imagine with supersizing today, this might be a bigger problem. So anyway, when we talk about periodic trends, it doesn't always match up. This was eventually taken out, and actually just for your interest, there was no overlap between the time when cocaine was in Coca Cola and lithium was in 7-Up, so there was a few years difference between those two times, but it's amazing to think about what does go into processed foods. And the other thing to point out, which I don't know if this is true or not, but does anyone know -- well that's part's true, does anyone know what the atomic mass of lithium is? Yes, it's 7. So, I don't know if this is true or not, but I wonder if that's where the actual name 7-Up came from. So, even though we don't have the lithium anymore, we still keep that atomic number 7 around. All right. So that is an anti-example of using periodic trends. So let's go to some actual real examples, which might come more in handy for this class. So it's going to keep in mind the limitations, so let's start off with talking about ionization energy. Now this is a good place to start, because we are very familiar with ionization energy, we've been talking about it in a lot of different forms for quite a while -- it's that minimum energy required to remove an electron from an atom. And specifically, when we talk about ionization energy, it's assumed that what we mean is actually the first ionization energy. So, you can imagine, we could talk about any of the different electrons, or we could talk about taking out an electron and taking out second electron. Whenever you hear the term ionization energy, make sure you keep in mind that unless we say otherwise, we're talking about that first ionization energy. And we know what that's equal to, this is something we've been over and over, ionization energy is simply equal to the negative of the binding energy. So negative e, which is sub n l, because it's a function of n and l in terms of quantum numbers. So, let's think about kind of differentiating, however, between first ionization energy or just ionization energy, and other types such as second or third ionization energy, and let's take boron as an example here. So, if we want to think about what the first ionization energy is of boron, what you want to do is write out the electron configuration, because then you can think about where it is that the electron's coming out of. The electron's going to come out of that highest occupied atomic orbital, that one that's the highest in energy, because that's going to be the at least amount of energy it needs to eject something. So what we'll end up with is boron plus, 1 s 2, 2 s 2, and what we say is the delta energy or the change in energy as the same thing as saying the energy of the products minus the energy of our reactant here, and we just call that the ionization energy -- that's how much energy we have to put into the system to eject an electron. And again, this is just the negative, the binding energy, when we're talking about the 2 p orbital. So, this is first ionization energy, let's think about second ionization energy. So, second ionization energy simply means you've already taken one electron out, now how much energy does it take for you to take a second electron out. So in the case of boron here, what we're starting with is the ion, boron 1 s 2, 2 s 2, and now we're going to pull one more electron out. The highest occupied orbital is now the 2 s orbital, so we're going to end up with boron 2 plus 1 s 2, 2 s 1, plus the electron coming out of there. And what we say when we talk about the delta energy is that this is going to be equal to i e 2, or the second ionization energy, or we could say the negative of the binding energy of a 2 s electron in b plus. so it's important to note that it's not in b, now we're talking about b plus, because we've already taken an electron out here. So, similarly if we start talking about our third ionization energy, this is going to be going from b plus 2, to 1 s 2, 2 s 1. Now we're going to pull that second electron out of the 2 s, so we end up with boron plus 3, and then the configuration is just 1 s 2, plus our extra electron here. So, what we call this is the third ionization energy, or the negative of the binding energy, again of the 2 s orbital, but now it's in boron plus 2 to we're starting with. So, this raises kind of an interesting question in terms of what the difference is between these two cases, and we're talking about numbers of energy. So let's address this by considering another example, which should clarify what the difference is between these ionization energies. So let's think about the energy required now to remove a 2 s electron, let's say we're removing it from boron plus 1 versus neutral boron. So, in the case of boron plus 1, what we are starting with is the ion, so we're starting with a 2 s electron, and then we're going to 2 s 1 here. And what we call the binding energy is negative 2 s in b plus -- this is what we saw on the last slide. And the second case here looks a lot more like what we saw when we were talking about photoelectron spectroscopy, because here we want to remove a 2 s electron, but it's actually not the highest occupied orbital, so that's not the one that would naturally come out first, but let's say we're hitting it with high energy light sufficient to knock out all the different electrons, and one that we end up knocking out is this 2 s here. So if we think about what that delta energy is, we call that the ionization of the 2 s, that's different from saying second ionization energy. And that's going to be equal to the negative the binding energy of 2 s in b, in neutral boron. So, my question to you is are these two energies equal? No. All right, good answer. So, we can think about why is it that these are not equal. In both cases we're taking an electron out of the 2 s orbital. And it turns out that if we're talking about a 2 s orbital in an ion, that means it doesn't have as many electrons in it, so what we're going to see is less sheilding. There are fewer electrons around to shield some of that nuclear charge. So what we're going to see is less sheilding, which means that it will actually feel a higher z effective. So even though they're both 2 s electrons, in one case it's going to think its feeling more pull from the nucleus, and it, in fact, will be, than in the other case, and if its feeling a higher z effective, then it's actually going to require more energy to remove that electron, right, it's being pulled in closer and more tightly to the nucleus, you have to put in more energy to rip it away from that very close interaction. So, that's the difference in thinking about different types of ionization energy, so it can get a little bit confusing with terminology if you're just looking at something quickly, so make sure you look really carefully about what we're discussing here. If you see a problem that asks you, for example, the third ionization energy versus taking a second electron out of the 2 s in a photoelectron spectroscopy experiment, those are two very different things. So, let's make sure everyone kind of has this down, let's do another clicker question here. And in this case we're going to look at silicone, and we'll say if you can point out to me which requires the least amount of energy. So, which has the smallest energy that you have to put in in order to eject this electron? Will it be if you take a 3 s electron from neutral silicone, if you take a 3 p electron from the neutral atom, or if you take a 3 p from the ion? So this you should be able to know pretty quickly, so let's just take 10 seconds here. All right, great. So most of you see that, in fact, the energy that's going to be the least that we need to put in is in case 2 here. Let's compare case 2 and 3, since this where some people seem to have gotten confused. In case 2, we're taking it out of -- oh, it's kind of hard to compare case 2 and 3 when we can't see it anymore. In case 2, we're taking the 3 p out of the neutral atom, whereas in case 3, we're taking it out of the ion. Remember in the ion, we're going to have less electrons around to counteract the pull from the nucleus. So we're going to feel a higher z effective in the case of the ion compared to the neutral atom. If we have a higher z effective, it's pulled in tighter, we have to put in more energy in order to eject an electron, so it turns out that that's why case 2 is actually the lowest energy that we need to put in. The z effective is lower, so we have to put less energy in to get an ion out. So, let's take a look at this in terms of periodic trends -- that's our topic here, we're talking about periodic trends. So as we go across the row, and this is my beautiful picture of a periodic table here. As we go across the row what happens is that the ionization energy actually increases, and we can think about logically why it is that that's happening. As we go across the row, what we have happening is that the z or the atomic charge -- I'm not talking about z effective here, I'm just talking about z -- the z is increasing as we go across a row, that's easy to see. But we're still in the same shell, so we still have the same n value as we go all the way across a row in the periodic table. So, in general what we're going to see is that what happens to z effective if we have z increasing but we're in the same shell here. Would it increase or decrease z effective? All right. Kind of mixed thoughts here. So it turns out that it increases, and the reason is because the predominant thing that's going on here is that z is increasing. So the z is increasing, and what goes along with it is that the z effective is increasing, because it turns out that while we're in the same n, even though we know that energy depends on both the n and the l in terms of quantum numbers, while we're in the same n, the distance from the nucleus, it's pretty close, it's not hugely different. So the factor that predominates is that we're actually increasing z. That's why we see z effective increase, and that's why we see ionization energy increase. As we go down a column, what happens is that the ionization energy decreases. We can also think about this in terms of z effective. This is because even though z, the atomic number is still increasing, we are also getting further away from the nucleus. So, remember when we talk about Coulomb force, what's holding the nucleus and the electron together, there's 2 things we need to think about. The first is this the z effective, or how much charge is actually in the nucleus that's felt, or the I guess we would say the z, how much the charge is on the nucleus that holds it close together. But the second factor is how far away we are from the nucleus. So, if we're really close to the nucleus, that's when z effective is high, but if we get really far away, then z effective is going to get low, because even though we have the same charge in the nucleus, the z effective gets lower. So this is not even thinking about the other electron shielding, just if we think of the fact, all we need to think about is that the effect of going to a further away n actually dominates as we go down the table. We're getting further away from the nucleus because we're jumping, for example, from the n equals 2 to the n equals 3 shell, or from the n equals 3 to the n equals 4 shell. And when you're switching n's, you're actually getting quite a bit farther away. That's why in the earlier models of the atom, they're not horrible to sometimes think about just each n value as a little ring around. It's not complete and it's not accurate, but it's OK to kind of think of it in terms of how far we're getting away from the nucleus. So, as we go down a column, we see ionization energy's going to decrease. So, this means we have the general trends down, so we should be able to look at actual atoms in our periodic table and graph them and see that they match up with our trends. So here we have that graphed here, we have atomic number z graphed against ionization energy, so, let's fill in what the actual atoms are here, and we can see in general, yes, we're following the trend. For row one, we're increasing, as we should, across the row. Let's look at row two also. Well, we're generally increasing here, we'll talk about that more in a minute. And also in a row three, yeah, we're generally increasing, there's some glitches here, but the general trend holds true. Similarly we see as we go down the table, so as we're going from one row to the next row, so, for example, between helium and lithium, we see a drop; the same with neon to sodium, we see a drop here. So it looks like we're generally following our trend, that's a good thing. But hopefully, you will not be satisfied to just make a general statement here when we do have these glitches. So you can see, for example, we have several places where instead of going up as we go across the row, we actually go down in ionization energy a little bit. So between b e, and b, between n and o, magesium and aluminum, and then phosphorous and sulfur, what we see here is that we're kind of going down, or quite specifically, we are going down. So, let's take a look at one of these rows in more detail to think about why this might be happening, and it turns out the reason that these glitches occur are because the sub shell structure predominates in certain instances, and that's where these glitches take place. So I've sort of just spread out what we have as the second row here, graphed against the ionization energy. So, let's consider specifically where these glitches are taking place. So, let's look at the first one between beryllium and boron here. And the glitch that doesn't make sense just through periodic trends, is that it turns out that the ionization energy of boron is actually less than the ionization energy up beryllium. So I put the electron configurations and actually drew it on an energy diagram here, so we can actually think about why this might be happening. So what is this, which element did I chart here? Which one is that, the boron or the beryllium? I couldn't tell what you said, sorry. So, I'm going to assume that was beryllium, and then it turns out that if that's beryllium, the other one must be boron. So, we have beryllium in the first case here, it has four electrons, that's how we know it's beryllium, boron has five electrons. So, just looking at putting in the electrons, filling up the energy diagram here, we should be able to see a little bit why this is happening. And the reason is simply because the energy that we gain in terms of moving up in z, so from going to z equals 4 to z equals 5, is actually outweighed by the energy it takes to go into this new shell, to go into this new sub shell. So to jump from the 2 s to the 2 p, takes more energy than we can actually compensate with by increasing the pull from the nucleus. So, it turns out that in this case, and any time that we see we're going from a 2 s to 2 p, filling in of electrons, we actually see that little bit of glitch in ionization energy. So it's shown here for the second row, but it's actually also going to be true for the third row. The same thing when you're going from filling in the 2 s to putting that first electron into the 2 p. So that explains one of our glitches here, but we have another glitch, and that second glitch comes between nitrogen and oxygen. So, these sound more different, so I think I'll be able to distinguish. Which element is shown here? Yeah, nitrogen. So, nitrogen is shown here, we know that because it has 7 electrons. In this case, we're talking about 8 electrons, which is oxygen. So if we're comparing the difference between these 2 now, what you'll notice is that in nitrogen we have all half-filled 2 p orbitals, and now, once we move into oxygen, we actually have to add 1 more electron into 1 of the 2 p orbitals. There's no more 2 p orbitals to put it into, so we're going to actually have to double up. So now we're putting 2 electrons into the same p orbital, that's not a problem, we can do it, it's not a huge energy cost to do that. But actually there is a little bit of an energy cost into doubling up into a single orbital, because, of course, it takes energy when you create more electron repulsion, that's not something we want to do, but we have to do it here, and it turns out that that effect predominates over, again, the energy that we gain by increasing the atomic number by one. So, our two glitches we see when we go from the 2 p, or from 2 s to start filling the 2 p, and then we also get another glitch when we've half-filled the 2 p, and now we're adding and having to double up in one of those p orbitals. Again, we see the same effect as we go into different rows as well. So let's talk about another periodic trend, this one is called electron affinity. Electron affinity is actually the ability of an atom, or we could also talk about an ion to gain electrons. So it's the affinity it has for electrons, it's how much it likes to get an electron. We can write out what it is for any certain atom or ion x, so it's just x plus an electron gives us x minus. We have the minus because we're adding a negative charge from the electron. So, basically any time we have a really high positive number of electron affinity, it means that that atom or ion really wants to gain another electron, and it will be very stable and happy if it does so. So let's look at an example of chlorine here. So chlorine, if we talk about it in terms of electron affinity, we would be writing that we're actually gaining an electron here, and getting the ion, c l minus. And the change in energy for this reaction is negative 349 kilojoules per mole. So if we have a negative change in energy for any reaction as it's written, what that actually means is we're giving off energy as the reaction proceeds. So, in other words, this c l minus is actually lower in energy than the reactants were. So that's why we're giving off extra energy. We saw a similar thing as we saw electrons move from different levels. We can think of it in the same type of way when we're talking about actual reactions happening. So, if we have energy that's released, would you say that the chlorine ion is more or less stable than the chlorine atom? Who thinks it's more stable, show of hands? All right, who thinks it's less stable? Very good. So it turns out it is, in fact, more stable. It's more stable because you actually -- this happens spontaneously, you actually get energy out of the reaction as this happens. And we haven't talked about reactions at all yet, so you don't need to worry about the specifics of that exactly, but just that if you have this negative change in energy, you have a more stable product than you do reactant. So, we were talking, however, about energy in terms of electron affinity, so we can actually relate electron affinity to any reaction by saying if we have this reaction written as here where we're gaining an electron, we say that electron affinity is just equal to the negative of that change in energy. So, for example, for the chlorine case, we would say that the electron affinity for chlorine is actually positive 349 kilojoules per mole. That's a very large number, it's all relative, so you don't necessarily know it's large without me telling you or giving you other ions to compare to, but chlorine does have a very large affinity, meaning it's really likes getting an electron and becoming a chlorine ion. One major difference between electron affinity and ionization energy is that when we talked about ionization energy, remember ionization energy always has to be positive. We will never have a case where ionization energy is negative. Electron affinity, however, can be either negative or it can be positive. So let's look at a case where it's actually going to be negative. So, if we took the case of nitrogen, if we add an electron to nitrogen and go to n minus, we find that the change in energy is 7 kilojoules per mole. This means in order to do that we actually have to put 7 kilojoules per mole of energy into the reaction to make it happen. So this is not going to be a favorable process, we're going to find that the electron affinity is actually a negative 7 kilojoules per mole for nitrogen. So this means nitrogen has low electron affinity, it doesn't actually want to gain an electron. So, that also tells us that the n minus ion is less stable than the neutral atom itself. So, we can think about trends in electron affinity just like we did for ionization energy, and what we see is a similar trend. Electron affinity increases as we go across a row in the periodic table, and it decreases as we go down a column. I left out the noble gases here because they do something a little bit special, and actually, I'm going to give you one last clicker question today to see if you can tell me what you think noble gases do. To answer this question you just really want to think about what does electron affinity means. It means how much a certain atom actually wants to get an electron. So do you think noble gases would have a high positive electron affinity, a low positive, or negative electron affinity? So, let's take 10 seconds on that. All right. Great. So most of you recognize, if we switch back to the notes, that they do have a negative electron affinity. This should make sense to you, because they don't, in fact, want to gain another electron, because that would mean that electron would have to go into a new value of n, a new shell, and that's really going to increase the energy of the system. So they'd much rather just stay the way they are and not have another electron come on, and it turns out that halogens have the highest electron affinities.
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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Now we'd like to discuss angular acceleration for circular motion. So suppose we have our angle theta, radius r, and r hat and theta hat. Recall that we described the angular velocity as the derivative of d theta dt, and we made this perpendicular to our right-handed coordinate system, direction k hat. Now let's differentiate that to get our concept of angular acceleration. So alpha is the second derivative d theta dt squared k hat. And this quantity is what we call angular acceleration. Now we'll describe the component alpha z as d squared theta dt squared. So it's the second derivative of the angle. And also if we wrote this as omega z k hat, we can write that as the derivative of d omega z dt, as well. So this is the component. And now in circular motion, the quantities of omega and alpha z are very much like the linear quantities of the x component of the velocity and the x component of the acceleration. And again, when we've chosen a reference frame, let's look at what various components mean. Let's begin with the case 1 where omega z is positive. So when omega z is positive, that tells us that the angle d theta dt is increasing. And that corresponds to counterclockwise motion. Now given that case, let's look at what happens when alpha z is positive. Remember, that's the statement that d omega dt is positive, that omega z is increasing. So if an object is moving with a positive component of omega z and the angular acceleration component is positive, that corresponds to increasing. The linear example, if you had one dimensional motion, i hat, you had vx positive and a x positive, corresponds to an object increasing in its speed in the x direction. That's our first case. Now let's look at the second example when alpha z is less than 0. So now the derivative of d omega z dt is negative. What that corresponds to-- remember, omega z is the z component of the angular speed. And if that's slowing down, then, with alpha z less than 0, the object is slowing down. So in our linear case, if we had a x less than 0, this is the classic example of breaking. The object is moving in the x direction and slowing down. Now let's look at case 2. This is always a little bit complicated for circular motion where omega z is less than 0. In that case, the object is moving in the clockwise direction because the angle theta is decreasing, corresponding to clockwise motion. So in that case, once again, let's consider the two examples. Well, the first example is a positive component of angular acceleration. Now this is the one that can be a little bit confusing. The object is moving clockwise but it has a positive alpha z, which will correspond to slowing the object down. And if the alpha z remains positive, it will actually come to rest and then reverse its motion and start to speed up. So this is the case where d omega dz is increasing. And that's our first case. So something like that could correspond to, if we plotted omega z and we had an object that starts off with a negative omega z and increases. Notice that the slope here, which is alpha z positive, corresponds to a positive angular acceleration component. And the object slows down as omega gets closer to 0, stops, and now has a positive omega z, corresponding to motion in a counterclockwise direction. For our linear case, this corresponds to, again, with i hat, our object moving to the left, vx negative, and if a x is positive, it breaks in this direction, which means it's slowing down. And then eventually if alpha x, ax, stays positive, it continues in that direction. Now our final case, and I'll put it down here, b, this is again where omega z negative and alpha z negative. It's always helpful to see this immediately with the graph. Omega z is negative. Here, alpha z, which is the slope, is also negative. This corresponds to an object moving in the clockwise direction. And actually its speed is increasing because alpha z is negative. So it's going faster and faster in the clockwise direction, even though alpha z is negative. And for our linear case, again, this corresponds to an object moving in the negative x direction. And a x is negative, it's moving faster in the negative x direction. And so these are the cases of how we analyze the various cases for angular acceleration and angular velocity.
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https://ocw.mit.edu/courses/5-112-principles-of-chemical-science-fall-2005/5.112-fall-2005.zip
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The following content is provided by MIT OpenCourseWare under a Creative Commons license. Additional information about our license and MIT OpenCourseWare in general is available at ocw.mit.edu. Great. Well, let's get going. Last time we ended up by discovering the electron. We discovered the fact that the atom was not the most basic constituent of matter. But in 1911 there was another discovery concerning the atom, and this is by Ernest Rutherford in England. And what Rutherford was interested in doing was studying the emission from the newly discovered radioactive elements such as radium. And so he borrowed, or he got, from Marie Curie, some radium bromide. And radium bromide was known to emit something called alpha particles. And they didn't really know what these alpha particles were. Now, they did know that the alpha particles were heavy, they were charged and that they were pretty energetic. That is what was known. Of course, today we know these alpha particles to be nothing other than helium with two electrons removed from the helium, helium double plus. Rutherford is in the lab and has this radium bromide, alpha particles being emitted and has some kind of detector out here to detect those alpha particles. And he measures a rate at which the alpha particles touch his detector. And it is about 132,000 alpha particles per minute. That's nice. Then what he does is takes a piece of gold foil and puts it in between the radium bromide emitter and the detector. And that gold foil is actually really very thin. It is 2x10^-5 inches. Two orders of magnitude thinner than the diameter of your hair. I often wonder how he handled that, but he did it. He put it in the middle here and then went to count the count rate as a result of putting this foil there, and the count rate is 132,000 alpha particles per minute. It didn't seem like that gold foil did anything. The alpha particles were just going right through to the detector. It didn't even seem to matter that there was that gold foil. The post-doc that was working on it, Geiger, of the Geiger Counter, was actually disappointed. Gee, that is a boring experiment. But Geiger was even a little bit more unhappy because he had this undergraduate hanging around the lab, this undergraduate named Marsden. And Marsden was really enthusiastic about doing something in the lab. He really wanted to do something. And Geiger, you know, what am I going to do with this kid? Geiger goes to Rutherford, look, this kid really wants to do something. What should I have him do? And Rutherford said, well, what you should have him do is take this detector and have him build it so that it can be swung around. So that it can be positioned here. So that we can check to see whether or not any of these alpha particles are backscattered, scattered back into the direction from which they came. And Geiger went away and thought, good, this is something to give the undergraduate. This is a ridiculous experiment. We know all the particles are going right through the detector. Okay. But Marsden was real happy. He gets to build this detector. He swings it around and gets Geiger there to do the first experiment. He puts the radium bromide and they listen and hear tick, tick, tick, tick, tick, tick. Geiger says, "Oh, it must just be background. Let me do a control experiment. Let me take the gold foil out of here so that all the particles have to be going in this direction." They take the gold foil out of there and listen, and they hear nothing. They put the gold foil back and they hear tick, tick, tick, tick, tick, tick. And they put a platinum foil in there and they hear tick, tick, tick, tick, tick, tick. Whatever metal they put in there, there were some particles coming off. And they got Rutherford down in the lab. Rutherford looks them over their shoulder. They do this again and again. Hey, it is real. It is real. And what is coming off? Well, the count rate is about 20 particles per minute. Not large but not zero. And the probability here of this backscattering is simply the number of particles backscattered, which is 20, over the total number of particles, or actually the count rate that the particles backscattered over the total incident count rate. That is 2x10^-4. That is not zero. Wow. Rutherford was excited. Rutherford later wrote, "It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15 inch shell at a piece of tissue paper and it came back and hit you." What was the interpretation? The interpretation was the gold atoms that make up this foil, they must be mostly empty. Now, they knew that those atoms had some electrons in it because the electron had already been discovered. But these alpha particles seem to be going right through those gold atoms, for the most part. The atom, which he knew to be a diameter of about 10^-10 meters, most of that atom must be empty was the conclusion. But occasionally these helium double plus ions, these alpha particles, hit something massive. And that something massive then scatters those helium ions into the direction from which they came. And since that probability is small, well, the size of this massive part has to be really pretty small. And from knowing the probabilities and knowing roughly what the diameter of the atoms were and how many layers of atoms he had, he was able to back out of those experiments a diameter for this massive part of 10^-14 meters. And he called this massive part the nucleus. He called it the nucleus in analogy to the nucleus of a living cell. The heavy part, the dense part in a living cell-- that is where the name "nucleus" comes from. Now, Rutherford also realized that this nucleus here has to be positively charged. He knew about electrons and knew the atoms then were neutral, and so he reasoned this nucleus had to be positively charged. And then he did a bunch more experiments, more sophisticated experiments in which he actually measured here the angular distribution of the helium ion scattered from the nucleus. And from those very detailed measurements of the angular distribution, he was able to back out the fact that this nucleus, the charge on it was actually plus Z times e. Z is the atomic number. e is the unit charge. He did a bunch of different metals and was able to establish that the nucleus had a charge of plus Z times e. His model is that there is a very dense center, 10^-14 meters. This diameter of the nucleus is something that every MIT undergraduate should know. And he realized that then the electrons have to fill out the rest of this volume. That was his interpretation from these results. And think about Marsden, what a great UROP experiment. He discovered the nucleus. Isn't that great? Marsden had a long and successful career as a scientist also after that. Now, I should also tell you that this backscattering experiment is really the essence of how a quark was discovered. Quark are the fundamental elementary particles in protons and neutrons. Essentially, they took a high energy particle, scattered it through the proton or the neutron, and it backscatters. And, in that way, they discovered the quark and measured the diameter of the quark. And this was done by a couple of my colleagues in the Physics Department. Jerry Friedman and Henry Kendall, who has since passed away. Jerry Friedman is still around. He loves to talk to undergraduates, and many of you will get that opportunity. Now it is time for us to do our own Rutherford backscattering experiment. Yeah. [APPLAUSE] Here is our gold lattice. These Styrofoam balls are the gold nuclei. The space around them are the electrons. These things in the center here are just the posts on this frame. [LAUGHTER] This is a piece of equipment from my lab that I pressed into service, and so I couldn't cut these posts away because I would have trouble taking my manipulator out of my machine at a later time. So they are just there. But this is our one monolayer of gold nuclei. And so what are we going to do? Well, what we are going to do is try to measure the diameter of these Styrofoam balls in the same way that Rutherford did. And so we are going to need some alpha particles. What are we going to use for an alpha particle? Well, we have some ping-pong balls for alpha particles. Let's do that. We have 287 alpha particles, or ping-pong balls, and we are going to measure the probability of backscattering. The probability of backscattering will be the number that actually backscatter divided by the number that we throw, or the total number. That is what we are going to measure. But now I have to take this probability and I have to relate it to the diameter of these nuclei. How am I going to do that? Well, that probability is going to be equal to the total surface area of the crystal here. I have already measured the total area. I know that the total area is 2,148 square inches. That is in the denominator, but now the numerator is simply the total area of the nuclei. The total area of the nuclei is the area of one nucleus, A sub i, summed over the total number of nuclei, which I have already counted as 119. And so the total area is times the cross-sectional area here of any one of these nuclei. And that is pi d squared over 4. I can solve that equation, for the diameter, in terms of the probability. And when I solve that equation, d is equal to 4.79 times the probability to the one-half power. What we are going to do is measure this probability by throwing the ping-pong balls and calculating and determining how many backscatter. And then we are going to use that to get this diameter of the nuclei. The same experiment that was done to actually measure the diameter of the nucleus. Now you are going to do this experiment. Every one of you are going to get a ping-pong ball from the TAs. TAs, why don't you give out the ping-pong balls, and then I will give you some instructions. All right. The pi d squared over 4 is the cross-sectional area in terms of the diameter of these balls. I just wrote it in terms of d instead of r. Yes? That is correct. Good point. That balls that we are throwing actually have size compared to in the case of the Rutherford backscattering experiment where the projectile was almost a point compared to the size of the nucleus. In our experiment, you are quite right, our balls are about the diameter there. And so, if we were doing a more exact experiment, we would do a little different calculation. We would take into consideration the size of the actual ball that we were throwing. But we are not going to do that. Because we are not throwing that many balls, we don't really have the statistics to do a more exacting kind of calculation. But you are quite right. Yes? Well, he didn't know. Although, he knew the fact that it was backscattering, that it had to be much, much less massive than the nucleus. I think that he also measured the energy of the backscattered particle. And from that, you can back out the fact that it is much less massive than the nucleus. There are a few other details, you are quite right, that I have left out in this discussion that he had to know in order to get this number. Here is the thing. You have to aim your alpha particles at this lattice. And then you have to watch your ball. [LAUGHTER] You have to watch to see if it scatters back at you, because at the end I am going to ask you if your ball backscattered. And we need an accurate count. Now, if you hit one of these things and it backscatters, that doesn't count. Only if it hits the Styrofoam ball does it count. If it hits the Styrofoam ball and goes through, that doesn't count. It literally has to backscatter at you. Was there a question over here? If you miss you miss. [LAUGHTER] Now, I do invite you to come a little closer so that you can at least hit the crystal. Yes? That is correct. Well, you have got a defect. These are a little bit lighter. Oh, you have some more here. Oh, okay. You can have a regular one. Anybody need one yet? I have a couple. Oh, all right. You need one? Because I need them all thrown. Did you have a question? What is the mean free path? That I am going to have to give you an expression for at some other time, but there is certainly a decay pathway. I have another ball here. Now, are you ready? You can come down closer, but now I have one piece of advice for you. That is, only fools aim for their chemistry professor. [LAUGHTER] Go to it. Did you throw your balls? You missed the crystal. All right. Has our supply of alpha particles been exhausted? All done? All right. [APPLAUSE] Now comes the big test. How many of you had an alpha particle that backscattered? Let's keep your hand high because I have to count accurately. In this section I see one. Two? Cheater. No. Two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen. Did I get everybody? I got everybody? 13? Right, not deflection. If it hit and went through, that does not count. It has to come back at you. Yes. [LAUGHTER] That is right. All right. Does anybody want to change their count? 13 balls? I am sorry? If it just hit it and moved but did not backscatter, it does not count. The nuclei will move. They will move, certainly, because there is a momentum transfer. Well, not quite like that. No. We have 13 balls that backscattered? Okay. Let's see what we got. The probability, here, then, is 13 over 287. That probability is equal to If I now that this probability and plug it into here, what we are going to get is a diameter of 1.0 inches. And the diameter on the average of those particles is about 0.85 inches. You did a really pretty good job. You got the diameter of the nucleus. [APPLAUSE] That is great. And that is the way the nuclear diameter was, in fact, measured and discovered. But now we have the problem that the scientists had in 1912, and that is what is the structure of the atom? We now know it has a nucleus. It has an electron. How do they hang together? Where are they in the atom? We are going to talk about the classical description here of the atom. And the first question that we have to ask is, what is the force that keeps the electron and the nucleus together? What are the four fundamental forces? Gravity is one. And that is the strongest or the weakest? Weakest. Gravity. Next stronger force? Electromagnetic. I will just abbreviate it EM. Next stronger force? Weak force. And the next? Strong. Weak and strong are intranuclear forces. They are operable between the protons, the neutrons and the other elementary particles that make up the nucleus. It does not have a lot of effect, the weak and the strong force, on chemistry, except for beta emission for the radioactive elements. Gravity actually does have no known chemical significance to chemistry. And so all of chemistry is tied up here in the electromagnetic force, which I am, at the moment, going to simplify and just call the Coulomb force. Now, we know how to describe the Coulomb force between charged particles. We know what expression to write down. Let's do that. If we have the nucleus, which is positively charged, and the electron here, which is negatively charged, and they are at some distance r between each other, the expression that describes how that force of interaction changes with distance, this Coulomb's force law, it is just the magnitude of the charge of the electron times the magnitude of the charge on the nucleus over 4 pi epsilon nought times r squared. I am going to just treat the force as a scalar, just for simplicity purposes here. Epsilon nought is the permittivity of vacuum. It is a factor in there for unit conversation. r, then, is the distance between the electron and the nucleus. What does this say? Well, this says that when r goes to infinity, what is the force? Zero. The particles are infinitely far apart. There is no force between them. In this case, an attractive force between them. When r is equal to zero, what is the force? Infinite. And anywhere in between, that force is described by this one over r squared dependence. You can see that as the particles come closer and closer together, the force between them gets larger and larger. The closer they get, the larger the force, the more they want to be together. This expression is just telling me, if I held one particle and the other particle in my hand, and I held them at some distance from each other -- That expression is just telling me the force with which I have to kind of exert to keep them apart. But now, if I let them go, you know what is going to happen. They are going to come together. They are going to want to come together because of that force. And what is not in this expression? What is not in that expression is any information about how those particles move under influence of that force. Nowhere in this expression is there an r of t, how that distance changes with time. And so what we need to describe that is a force law. And in 1911, the force law that seemed to describe the motion of all bodies, including astronomical ones, of course, the equation of motion that described how bodies move are Newton's equations of motion. And, in particular, F equals ma. And, of course, I can write that acceleration as a time derivative of the velocity, dv over dt. And that velocity, of course, itself is a change in the position with respect to time. This is m, the second derivative of r with respect to time. If I know the force that is operation, which is this, I can take this and plug it in here, and I am going to have a differential equation. And that differential equation is going to allow me to solve for what r is as a function of time, the distance between the two particles. And it is going to allow me to solve for that distance in a way that we call deterministic, exactly. In other words, if I know where the particles are to start with, using this equation of motion, this force law, I can tell you where those particles are going to be for all future time exactly. It is deterministic, the classical mechanical approach. Now, in order to solve this differential equation, I am going to have to develop a model for the atom. All differential equations, for the most part, describing physical processes are going to need a model. They are going to need some boundary conditions or initial conditions. And the model, of course, that came to mind for the atom, is one in which the nucleus is in the center. And the electron moves around that nucleus with uniform circular motion and with a fixed radius. We are going to call that fixed radius r star. It is a planetary model. That seems like a good guess for the structure of the atom. Now, if you have a particle undergoing uniform circular motion at some well-defined radius here. That particle is being constantly accelerated. And I can write that acceleration a as the linear velocity squared over that radius of its orbit. It is being accelerated because the velocity vector. The direction is changing, so there is a constant acceleration. Now, this expression, for many of you, I pulled out of the air. Some of you have seen it before. It is an 8.01 topic. You are going to see it this semester, but later on and in You are not responsible for this right now here, but you will recall later on this semester that you have seen it here in 5.112. But, if this is the acceleration, I can take this expression for the acceleration and plug it into here. Plug in my operating force law. And, in so doing, I am going to get -- -- e squared over 4 pi epsilon nought r star squared. That is the F. Mass times the acceleration, m times v squared over r star. That is my equation of motion particular to this problem of a planetary model. And now I can solve that for v squared, the linear velocity of that electron going around the nucleus. That comes out to be e squared over 4 pi epsilon nought m r star. Now, the reason I wanted to calculate the velocity squared here is because I want to calculate kinetic energy. And that is easy to do. Kinetic energy, I will call K, is one-half m times v squared. If I plug in the v squared right in there, I get one-half e squared over 4 pi epsilon nought r star. So far, everything looks okay. We have a planetary model. Coulomb's law is operable. We know the acceleration. We just calculated the kinetic energy of this electron going around the nucleus. What I want to do now is I want to know the total energy of the system. I just calculated the kinetic energy of the system, but I want to know the total energy of the system. And the total energy of the system, I am going to call this capital E, total energy, is the kinetic energy plus the potential energy. And I want the total energy of the system for two reasons. One is I want to show you that the system is bound, that the total energy is going to be negative, that it is lower than the total energy when the electron and the nucleus are separated. I want to show you that within this classical model, the electron and the nucleus do look bound. To do that, I need to show you the total energy is negative. To do that, I need to calculate the potential energy. That is what I want to do. Secondly, I want to get an expression for the total energy. Because, using that expression, I am going to show you how this classical mechanics fails. How Newton's equations of motion won't work to describe this problem. Now, I have run out of time. I will do that on Monday, but that is where we are going. All right. See you on Monday.
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https://ocw.mit.edu/courses/10-34-numerical-methods-applied-to-chemical-engineering-fall-2015/10.34-fall-2015.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JAMES W. SWAN: So this is going to be our last lecture on linear algebra. The first three lectures covered basics. The next three lectures, we talked about different sorts of transformations of matrices. This final lecture is the last of those three. We're going to talk about in another sort of transformation called the singular value decomposition. OK, before we jump in, I'd like to do the usual recap business. I think it's always hopeful to recap or look at things from a different perspective. Early on, I told you that the infinite dimensional equivalent of vectors would be something like a function, which is a map, a unique map maybe from a point to x to some value f of x. And there is an equivalent representation of the eigenvalue eigenvector problem in function space. We call these eigenvalues and eigenfunctions. Here's a classic one where the function is y of x, OK? This is the equivalent of the vector, and equivalent of the transformation or the matrix that's this differential operator this time, the second derivative. So I take the second derivative of this particular function, and the function is stretched. It's multiplied by some fixed value at all points. And it becomes lambda times y. And that operator has to be closed with some boundary conditions as well. We have to say what the value of y is at the edges of some boundary. So there's a one-to-one correspondence between these things. What is the eigenfunction here, or what are the eigenfunctions? And what are the eigenvalues associated with this transformation or this operator? Can you work those out really quickly? You learned this at some point, right? Somebody taught you differential equations and you calculated these things. Take about 90 seconds. Work with the people around you. See if you can come to a conclusion about what the eigenfunction and eigenvalues are. That's enough time. You can work on this on your own later if you've run out of time. Don't worry about it. Does somebody want to volunteer a guess for what the eigenfunctions are in this case? What are they? Yeah? AUDIENCE: [INAUDIBLE] JAMES W. SWAN: OK, so you chose exponentials. That's an interesting choice. That's one possible choice you can make. OK, so we could say-- this is sort of a classical one that you think about when you first learn differential equation. They say, an equation of this sort has solutions that look like exponentials, and that's true. There's another representation for this, which is as trigonometric functions instead, right? Either of those is acceptable. [INAUDIBLE] the trigonometric functions, that representation is a little more useful for us here. We know that the boundary conditions tell us that y of 0 is supposed to be 0. That means that the C1 has to be 0, because cosine of 0 is 1. So C1 has 0 in this case. So that fixes one of these coefficients. And now we're left with a problem, right? Our solutions, our eigenfunctions, cannot be unique. So we don't get to specify C2, right? Any function that's a multiple of this sine should also be an eigenfunction. So instead the other boundary condition, this y of l equals 0, needs to be used to pin down with the eigenvalue is. So the second equation, y of l equals 0, which implies that the square root of minus lambda has to be equal to 2 pi over l, it has to be all the nodes of the sine where the sine is equal to 0. That's the equivalent of our secular characteristic polynomial that prescribes with the eigenvalues are associated with each of the eigenfunctions. So now we know what the eigenvalues are. The eigenvalues are the set of numbers minus 2 pi n over l squared. There's an infinite number of eigenvalues. It's an infinite dimensional space that we're in, so it's not a big surprise that it works out that way. And the eigenvectors then are different scalar multiples of sine of the eigenvalues, square root of the eigenvalues, minus x. There's a one-to-one correspondence between all the linear algebra we've done and linear differential equations or linear partial differential equations. You can think about these things in exactly the same way. I'm sure in 1050, you started to talk about orthogonal functions to represent solutions of differential equations. Or if you haven't, you're going to very soon. This is a part of the course you get to look at the analytical side of some of these things as opposed to the numerical side. But there's a one-to-one relationship between those things. So if you understand one, you understand the other, and you can come at them from either perspective. This sort of stuff is useful. Actually, the classical chemical engineering example comes from quantum mechanics where you think about wave functions and different energy levels corresponding to eigenvalues. That's cool. Sometimes, I like to think about a mechanical analog to that, which is the buckling of an elastic column. So you should do this at home. You should go get a piece of spaghetti and push on the ends of the piece of the spaghetti. And the spaghetti will buckle. Eventually it'll break, but it'll buckle first. It'll bend. And how does it bend? Well, a balance of linear momentum on this bar would tell you that the deflection in the bar at different points x along the bar multiplied by the pressure has to balance the bending moment in the bar itself. So this e is some elastic constant. I has a moment of inertia. And D squared y dx squared is something like the curvature of the bar. So it's the bending moments of the bar that balances the pressure that's being exerted on the bar. And sure enough, this bar will buckle when the pressure applied exceeds the first eigenvalue associated with this differential equation. We just worked that eigenvalue out. We said that that eigenvalue had to be the square root of 2 pi over l squared. And so when the pressure exceeds square root of 2 pi over l squared times the elastic modulus, this column will bend and deform continuously until it eventually breaks, right? It will undergo this linear elastic deformation, then plastic deformation later, and it will break. The Eiffel Tower, actually, is one of the first structures in the world to utilize this principle, right? It's got very narrow beams in it. The beams are engineered so that their elastic modulus is strong enough that they won't buckle. Gustave Eiffel is one of the first applied physicists, somebody who took the physics of elastic bars and applied them to building structures that weren't big and blocky, but used a minimal amount of material. Cool, right? OK, so that's recap. Any questions about that? You've seen these things before. You understood them well before too maybe? Give some thought to this, OK? We talked about eigendecomposition last time that, associated with the square matrix, was a particular eigenvalue or particular set of eigenvalues, stretches and corresponding eigenvectors directions. These were special solutions to the system of linear equations based on a matrix. It was a square matrix. And you might ask, well, what happens if the matrix isn't square? What if A is in the space of real matrices that are n by m, where n and m maybe aren't the same? Maybe they are the same, but maybe they're not. And there is an equivalent decomposition. It's called the singular value decomposition. It's like an eigendecomposition for non-square matrices. So rather than writing our matrix as some w lambda w inverse, we're going to write it as some product U times sigma times V with this dagger. The dagger here is conjugate transpose. Transpose the matrix, and take the complex conjugate of all the elements, OK? I mentioned last time that eigenvalues and eigenvectors could be complex, potentially, right? So whenever we have that case where things can be complex, usually the transposition operation is replaced with the conjugate transpose. What are these different matrices. Well, let me tell you. U is a complex matrix. It maps from the space N to R N to R N, so it's an n by n square matrix. Sigma is a real valued matrix, and it lives in the space of n by n matrices. V is a square matrix again, but it has dimensions m by m. Remember, A maps from R M to R N, so that's what the sequence of products says. B maps from m to m. Sigma maps from m to n. U maps from n to n. So this match from m to n as well. Sigma is like lambda from before. It's a diagonal matrix. It only has diagonal elements. It's just not square, but it only has diagonal elements, all of which will be positive. And then U and V are called the left and right singular vectors. And they have special properties associated with them, which I'll show you right now. Any questions about how this decomposition is composed or made up? It looks just like the eigendecomposition, but it can be applied to any matrix. Yes? AUDIENCE: Quick question. JAMES W. SWAN: Sure. AUDIENCE: Do all matrices have this thing, or is it like the eigenvalues where some do and some don't. JAMES W. SWAN: This is a great question. So all matrices are going to have a singular value decomposition. We saw with the eigenvalue decomposition that there could be a case where the eigenvectors are degenerate, and we can't write that full decomposition. All matrices are going to have this decomposition. So for some properties of this decomposition, U and V are what we call unitary matrices. I talked about these before. Unitary matrices are ones for whom, if they're real valued, their transpose is also their inverse. If they're complex valued, and they're conjugate transpose is the equivalent of their inverse. So U times U conjugate transpose will be identity. V times V conjugate transpose will be identity. Unitary matrices also have the property that they impart no stretch to a matrix-- or to vectors. So their maps don't stretch. They're kind of like rotational matrices, right? They change directions, but they don't stretch things out. If I were to take A conjugate transpose and multiply it by A, that would be the same as taking U sigma V conjugate transpose, and multiplying it by U sigma V. If I use the properties of matrix multiplications and complex conjugate transposes, and work out what this expression is, I'll find out that it's equivalent to V sigma conjugate transpose sigma V conjugate transpose. Well this has exactly the same form as an eigendecomposition. An eigendecomposition of A times A instead of an eigendecomposition of A. So V is the set of eigenvectors of A conjugate transpose A, and sigma squared are the eigenvalues of A conjugate transpose times A. And if I reverse the order of this multiplication-- so I do A times A conjugate transpose-- and work it out, that would be U sigma sigma U. And so U are the eigenvectors of A A conjugate transpose, and sigma squared are still the eigenvalues of A A conjugate transpose. So what are these things U and V? They relate to the eigenvectors of the product of A with itself, this particular product of A with itself, or this particular product of A with itself. Sigma are the singular values. And all matrices possess this sort of a decomposition. They all have a set of singular values and singular vectors. These sigmas are called the singular values of the A. They have a particular name. I'm going to show you how you can use this decomposition to do something you already know how to do, but how to do it formally. What are some properties of the singular value decomposition? So if we take a matrix A and we compute it's singular value decomposition, this is how you do it in Matlab. We'll find out, for this matrix, U is identity. Sigma is identity with an extra column pasted on it. And B is also identity. I mean, this is the simplest possible four by three matrix I can write down. You don't have to know how to compute the singular value decomposition, you just need to know that it can be computed in this way. You might be able to guess how to compute it based on what we did with eigenvalues earlier and eigenvectors. It'll turn out some of the columns of sigma will be non-zero right? There are three non-zero columns of sigma. And the columns of V, they correspond to those columns of sigma, spanned the null space of the matrix A. So the first three columns here are non-zero, the first three columns of V. I'm sorry, the first three columns here are non-zero. The last column is 0. The columns of sigma which are 0 correspond to a particular column in V, this last column here, which lives in the null space of A. So you can see, if I take A and I multiply it by any vector that's proportional to 0, 0, 0, 1, I'll get back 0. So the null space of A is spanned by all these vectors corresponding to the 0 columns of sigma. Some of the columns of sigma are non-zero. These first three columns. And the rows of U corresponding to those three columns span the range of A. So if I do the singular value decomposition of a matrix, and I look at U, V, and sigma and what they're composed of-- where sigma is 0 and non-zero, and the corresponding columns or rows of U and V-- then I can figure out what vectors span the range and null space of the matrix A. Here's another example. So here I have A. Now instead of being three rows by four columns, it's four rows by three columns. And here's the singular value decomposition that comes out of Matlab. There are no vectors that live in the null space of A, and there are no 0 columns in sigma. There's no corresponding columns in V. There are no vectors in the null space of A. The range of A is spanned by the rows corresponding to the non-zero-- the rows of U corresponding to the non-zero columns of sigma. So it's these three columns in the first three rows. And these first three rows, clearly they span-- they describe the same range as the three columns in A. So the singular value decomposition gives us direct access to the null space and the range of a matrix. That's handy. And it can be used in various ways. So here's one example where it can be used. Here I have a fingerprint. It's a bitmap. It's a square bit of data, like a matrix, and each of the elements of the matrix takes on a value describing how dark or light that pixel. Let's say it's grayscale, and it's value's between 0 and 255. That's pretty typical. So I have this matrix, and each element to the matrix corresponds to a pixel. And I do a singular value decomposition. Some of the singular values, the values of sigma, are bigger than others. They're all positive, but some are bigger than others. The ones that are biggest in magnitude carry the most information content about the matrix. So we can do data compression by neglecting singular values that are smaller than some threshold, and also neglecting the corresponding singular vectors. And that's what I've done here. So here's the original bitmap of the fingerprint. I did the singular value decomposition, and then I retained only the 50 biggest singular values and I left all the other singular values out. This bitmap was something like, I don't know, 300 pixels by 300 pixels, so there's like 300 singular values, but I got rid of 5/6 of the information content. I dropped 5/6 of the singular vectors, and then I reconstructed the matrix from the singular values and those singular vectors, and you get a faithful representation of the original fingerprint. So the singular value decomposition says something about the information content in the transformation that is the matrix, right? There are some transformations that are of lower power or importance than others. And the magnitude of these singular values tell you what they are. Does that makes sense? How else can it be used? Well, one way it can be used is finding the least square solution to the equation Ax equals b, where A is no longer a square matrix, OK? You've done this in other contexts before where the equations are overspecified. We have more equations than unknowns, like data fitting. You form the normal equations, you multiply both sides of Ax equals b by A transpose, and then invert A transpose A. You might not be too surprised, then, to think that singular value decomposition could be useful here too. Since we already saw the data in a singular value decomposition corresponds to eigenvectors and eigenvalues of this A transpose A, right? But there's a way to use this sort of decomposition formally to solve problems that are both overspecified and underspecified. Least squares means find the vector of solutions x that minimizes this function phi. Phi is the length of the vector given by the difference between Ax and b. It's one measure of how far an error our solution x is. So let's define the value x which is least in error. This is one definition of least squares. And I know the singular value decomposition of A. So A is U sigma times V. So I have U sigma V times x. I can factor out U, and I've got a factor of U transpose, or U conjugate transpose multiplying by b. So Ax minus b is the same as U times the quantity sigma V conjugate transpose x minus U conjugate transpose b. We want to know the x that minimizes this phi. It's an optimization problem. We'll talk in great detail about these sorts of problems later. This one is so easy to do, we can just work it out in a couple lines of text. We'll define a new set of unknowns, y, which is V transpose times x, and a new right-hand side for a system of equations p, which is U transpose times b. And then we can rewrite our function phi that we're trying to minimize. So phi then becomes U sigma y minus p. U is a unitary vector. It imparts no stretch in the two norms, so this sigma y minus p doesn't get elongated by multiplication with U. So it's length, the length of this, is the same as the length of sigma y minus p. You can prove this. It's not very difficult to show at all. You use the definition of the two norm to prove it. So phi is minimized by y's, which makes this norm smallest, make it closest to 0. Let r be the number of non-zero singular values, the number of those sigmas which are not equal to 0. That's also the rank of A. Then I can rewrite phi as the sum from i equals 1 to r of sigma i i time y i minus p i squared. That's parts of this length, this Euclidean length, for which sigma is non-zero. Plus the sum from r plus 1 to n, the sum over the rest of the values of p, for which the corresponding sigmas are 0. I want to minimize phi, and the only thing that I can change to minimize it is what? What am I free to pick in this equation in order to make phi as small as possible? Yeah? AUDIENCE: y. JAMES W. SWAN: y, so I need to choose the y's that make this phi as small as possible. What value should I choose for the y's? What do you think? AUDIENCE: [INAUDIBLE] JAMES W. SWAN: Perfect, right? Choose y equals p i over sigma i i. Right, y i is p i over sigma i i. Then all of these terms is 0. I can't make this sum any smaller than that. That fixes the value of y i up to r. I can't do anything about this left over bit here. There's no choice of y that's going to make this part and the smaller. It's just left over. It's some remainder that we can't make any smaller or minimize an smaller. There isn't an exact solution to this problem, in many cases. But one way this could be 0 is if r is equal to n. Then there are left over unspecified terms, and then this y i equals p i over sigma i is the exact solution to the problem. So this is what you told me. Choose y i is p i over sigma i i for i bigger than 1 and smaller than r. There are going to be values of y i that go between r plus 1 and m, because A was a vector that mapped from m to n, right? So I have extra values of y that could be specified potentially. If that's true, if r plus 1 is smaller than m, then there's some components of y that I don't get to-- I can't specify, right? My system of equations is somehow underdetermined. I need some external information to show me what values to pick for those y i. I don't know. I can't use them. Sometimes people just set y i equal to 0. That's sort of silly, but that's what's done. It's called the minimum norm least square solution. y has minimum length, when you set all these other components to 0. But the truth is, we can't specify those components, right? We need some external information in order to specify them. Once we know y, we can find x going back to our definition of what y is. So I multiply this equation by V on both sides, and I'll get V y equals x. So I can find my least square solution to the problem from the singular value decomposition. So I can find the least square solution to both overdetermined and underdetermined problems using singular value decomposition. It inherits all the properties you know of solving the normal equations, multiplying by A transpose the entire equation, and solving for a least square solution that way. But that's only good for overdetermined systems of equations. This can work for underdetermined equations as well. And maybe we do have extraneous information that lets us specify these other components somehow. Maybe we do a separate optimization that chooses from all possible solutions where these y i's are free, and picks the best one subject to some other constraint. Does it makes sense? OK, that's the last decomposition we're going to talk about. It's as expensive to compute the singular value decomposition as it is to solve a system of equations. You might have guessed that it's got an order N cubed flavor to it. It's kind of inescapable that we run up against those computational difficulties, order N cubed computational complexity. And there are many problems of practical interest, particularly solutions of PDEs, for which that's not going to cut it. Where you couldn't solve the problem with that sort of scaling in time. You couldn't compute the Gaussian elimination, or the singular value decomposition, or an eigenvalue decomposition. It won't work. And in those cases, we appeal to not exact solution methods, but approximate solution methods. So instead of trying to get an exact solution, we'll try to formulate one that's good enough. We already know the computer introduces numerical error anyways. Maybe we don't need machine precision in our solution or something close to machine precision in our solution. Maybe we're solving engineering problem, and we're willing to accept relative errors on the order of 10 to the minus 3 or 10 to the minus 5, some specified tolerance that we apply to the problem. And in those circumstances, we use iterative methods to solve systems of equations instead of exact methods, elimination methods, or metrics decomposition methods. These algorithms are all based on iterative refinement of an initial guess. So if we have some system of equations we're trying to solve, Ax equals b, we'll formulate some linear map, right? xi plus 1 will be some matrix C times x i plus some little vector c where x i is my last best guess for the solution to this problem, and x i plus 1 is my next best guess for the solution to this problem. And I'm hoping, as I apply this map more and more times, I'm creeping closer to the exact solution to the original system of equations. The map will converge when x i plus 1 approaches x i, when the map isn't making any changes to the vector anymore. And the converged value will be a solution when x i-- which is equal to i minus c inverse times c, if I replace x i was 1 with x i appear, so I say that my map has converged-- when this value is equivalent to A inverse times B, when it's a solution to the original problem, right? So my map may converge. It may not converge to a solution of the problem I like, but if it satisfies this condition, then has converged to be a solution of the problem that I like as well. And so it's all about using this C here and this little c here so that this map converges to solution of the problem I'm after. And there are lots of schemes for doing this. Some of them are kind of ad hoc. I'm going to show you one right now. And then when we do optimization, we'll talk about a more formal way of doing this for which you can guarantee very rapid convergence to a solution. So here's a system of equations I'd like to solve. It's not a very big one. It doesn't really make sense to solve this one iteratively, but it's a nice illustration. One way to go about formulating this map is to split this matrix into two parts. So I'll split it into a diagonal part and an off diagonal part. So I haven't changed the problem at all by doing that. And then I'm going to rename this x x i plus 1, and I'm going to rename this x x i. And then move this matrix vector product to the other side of the equation. And here's my map. Of course, this matrix multiplied doesn't make any-- it's not useful to write it out explicitly. This is just identity. So I can drop this entirely. This is just x i plus one. So here's my map. Take an initial guess, multiply it by this matrix, add the vector 1, 0, and repeat over and over and over again. Hopefully-- we don't really know-- but hopefully, it's going to converge to a solution of the original linear equations. I didn't make up that method. That's a method called Jacobi Iteration. And the strategy is to split the matrix A into two parts-- a sum of its diagonal elements, and it's off diagonal elements-- and rewrite the original equations as an iterative map. So D times x i plus 1 is equal to minus r times x i plus b. Or x i plus 1 is D inverse times minus r x i plus b. If the equations converge, then D plus r times x i has to be equal to b, we will have found a solution. If it converges, right? If these iterations approach a steady value. If they don't change from iteration to iteration. Is The nice thing about the Jacobi method is it turns the hard problem, the order N cubed problem of computing A inverse B, into a succession of easy problems, D inverse times some vector C. How many calculations does it take to compute that D inverse? N, that's right, order N. It's just a diagonal matrix. I invert each of its diagonal elements, and I'm done. So I went from order N cubed, which was going to be hard, into a succession of order N problems. So as long as it doesn't take me order N squared iterations to get to the solution that I want, I'm going to be OK. This is going to be a viable way to solve this problem faster than finding the exact solution. How do you know that it converges? That's the question. Is this thing actually going to converge or not, or are these iterations just going to run on and on forever? Well, one way to check whether it will converge or not is to go back up to this equation here, and substitute b equals Ax, where x is the exact solution to the problem. And you can transform, then, this equation into one that looks like x i plus 1 minus x equal to minus D inverse times r x i minus x. And if I take the norm of both sides and I apply our normal equality-- where the norm of a matrix vector product is smaller than the product of the norms of the matrices of the vectors-- then I can get a ratio like this. That the absolute error in iteration I plus 1 divided by the absolute error in iteration i is smaller than the norm of this matrix. So if I'm converging, then what I expect is this ratio should be smaller than 1. The error in my next approximation should be smaller than the error in my current approximation. That makes sense? So that means that I would hope that the norm of this matrix is also smaller than 1. If it is, then I'm going to be guaranteed to converge. So for a particular coefficient matrix, for a system of linear equations I'm trying to solve, I may be able to find-- I may find that this is true. And then I can apply this method, and I'll converge to a solution. We call this sort of convergence linear. Whatever this number is, it tells me the fraction by which the error is reduced from iteration to iteration. So suppose this is 1/10. Then the absolute error is going to be reduced by a factor of 10 in each iteration. It's not going to be 1/10 usually. It's going to be something that's a little bit bigger than that typically, but that's the idea. You can show-- I would encourage you to try to work this out on your own-- but you can show that the infinity norm of this product-- infinity norm of this product is equal to this. And if I ask that the infinity norm of this product be smaller than 1, that's guaranteed when the diagonal values of the matrix and absolute value are bigger than the sum of the off diagonal values in a particular row or a particular column. And that kind of matrix we call diagonally dominant. The diagonal values are bigger than the sum and absolute value of the off diagonal pieces. So diagonally dominant matrices, which come up quite often, can be-- those linear equations based on those matrices can be solved reasonable efficiency using the Jacobi method. There are better methods to choose. I'll show you one in a second. But you can guarantee that this is going to converge to a solution, and that the solution will be the right solution to the linear equations you were trying to solve. So if the goal is just to turn hard problems into easier to solve problems, then there are other natural ways to want to split a matrix. So maybe you want to split into A lower triangular part which contains the diagonal elements of A, and an upper triangular part which has no diagonal elements of A. We just split this thing apart. And then we could rewrite our system of equations is an iterative map like this, L times x i plus 1 is minus U times x i plus b. All I have to do is invert l to find my next iteration. And how expensive computationally is it to solve a system of equations which is triangular? This is a process we call back substitution. Its order-- AUDIENCE: N squared. JAMES W. SWAN: --N squared. So we still beat N cubed. One would hope that it doesn't require too many iterations to do this. But in principle, we can do this order N squared operations many times. And it'll turn out that this sort of a map converges to the solution that we're after. It converges when matrices are either diagonally dominant as before, or they're symmetric and they're positive definite. Positive definite means all the eigenvalues of the matrix are bigger than 0. So try the iterative method solving some equations and see how we convert. Yes? AUDIENCE: How do you justify ignoring the diagonal elements in that method? JAMES W. SWAN: So the question was, how do you justify ignoring the diagonal elements in this method. Maybe I was going too fast or I misspoke. So I'm going to split A into a lower triangular matrix that has all the diagonal elements, and U is the upper parts with none of those diagonal elements on it. Does that make sense? AUDIENCE: Yeah. JAMES W. SWAN: Thank you for asking that question. I hope that's clear. l holds onto the diagonal pieces and U takes those away. So let's try it. On a matrix like this, the exact solution to this system of equations is 3/4, 1/2, and 1/4. All right, we'll try Jacobi, we'll have to give it some initial guess for the solution, right? We're talking about places where you can derive those initial guesses from later on in the course, but we have to start the iterative process with some guess at the solutions. So here's an initial guess. We'll apply this map. Here's Gauss-Seidel with the same initial guess, and we'll apply this map. They're both linearly convergent, so the relative error will go down by a fixed factor after each iteration. Iteration one, the relative error in Jacobi will be 38%. In Gauss-Seidel, it'll be 40%. If we apply this all the way down to 10 iterations, the relative error Jacobi will be 1.7%, and the relative error in Gauss-Seidel 0.08%. And we can go on and on with these iterations if we want until we get sufficiently converged, we get to a point where the relative error is small enough that we're happy to accept this answer as a solution to our system of equations. So we traded the burden of doing all these calculations to do elimination for a faster, less computationally complex methodology. But the trade off was we don't get an exact solution anymore. We're going to have finite precision in the result, and we have to specify the tolerance that we want to converge to. We're going to see now-- this is the hook into the next part of that class-- we're going to talk about solutions of nonlinear equations next for which there are almost no non-linear equations that we can solve exactly. They all have to be solved using these iterative methods. You can use these iterative methods for linear equations. It's very common to do it this way. In my group, we solve lots of systems of linear equations associated with hydrodynamic problems. These come up when you're talking about, say, low Reynolds number flows, which are linear sorts of fluid flow problems. They're big. It's really hard to do Gaussian elimination, so you apply different iterative methods. You can do Gauss-Seidel. You can do Jacobi. We'll learn about more advanced ones like PCG, which you're applying on your homework now, and you should be seeing that it converges relatively quickly in cases where exact elimination doesn't work. We'll learn, actually, how to do that method. That's one that we apply in my own group. It's pretty common to use out there. Yes? AUDIENCE: One question, is that that Gauss, [INAUDIBLE] JAMES W. SWAN: Order N squared. AUDIENCE: Yeah, that's what I meant. So now we've got an [INAUDIBLE]. So we basically have [INAUDIBLE] iterations, right? JAMES W. SWAN: This is a wonderful question. So this is a pathological problem in the sense that it requires a lot of calculations to get an iterative solution here. We haven't gotten to an end that's big enough that the computational complexities crossover. So for small Ns, probably the factor in front of N-- whatever number that is-- and maybe even the smaller factors, order N squared factors on that order N cubed, play a big role in how long it takes to actually complete this thing. But modern problems are so big that we almost always are running out to Ns that are large enough that we see a crossover. You'll see this in your homework this week. You won't see this crossover at N equals 3. You're going to see it out at N equals 500 or 1,200, big problems. Then we're going to encounter this crossover. That's a wonderful question. So first small system sizes, iterative methods maybe don't buy you much. I suppose it depends on the application though, right? If you're doing something that involves solving problems on embedded hardware, in some sort of sensor or control valve, there may be very limited memory or computational capacity available to you. And you may actually apply an iterative method like this to a problem that that controller needs to solve, for example. It just may not have the capability of storing and inverting what we would consider, today, a relatively small matrix because the hardware doesn't have that sort of capability. So there could be cases where you might choose something that's slower but feasible, versus something that's faster and exact, because there are other constraints. They do exist, but modern computers are pretty efficient. Your cell phone is faster than the fastest computers in the world 20 years ago. We're doing OK. So we've got to get out to big system sizes, big problem sizes, before this starts to pay off. But it does for many practical problems. OK I'll close with this, because this is the hook into solving nonlinear equations. So I showed you these two iterative methods, and they kind of had stringent requirements for when they were actually going to converge, right? I had to have a diagonally dominant system of equations for Jacobi to converge. I had to have diagonal dominance or symmetric positive definite matrices. These things exist and they come up in lots of physical problems, but I had to have it in order for Gauss-Seidel to converge. What if I have a system of equations that doesn't work that way? Or what if I have an iterative map that I like for some reason, but it doesn't appear to converge? Maybe it converges under some circumstances, but not others. Well, there's a way to modify these iterative maps, called successive over-relaxation, which can help promote convergence. So suppose we have an iterative map like this, x i plus 1 is some function of the previous iteration value. Doesn't matter what it is. It could be linear, could be non-linear. We don't actually care. The sought after solution is found when x i plus 1 is equal to x i. So this map is one the convergence to the exact solution of the problem that we want. We've somehow guaranteed that that's the case, but it has to converge. One way to modify that map is to say x i plus 1 is 1 minus some scalar value omega times x i plus omega times f. You can confirm that if you substitute x i plus 1 equals x i into this equation, you'll come up with the same fixed points of this iterative map x i is equal to f of x i. So you haven't changed what value will converge here, but you've affected the rate at which it converges. Here you're saying x i plus 1 is some fraction of my previous solution plus some fraction of this f. And I get to control how big those different fractions. So if things aren't converging well for a map like this, then I could try successive over-relaxation, and I could adjust this relaxation parameter to be some fraction, some number between 0 and 1, until I start to observe convergence. And there are some rules one can use to try to promote convergence with this kind of successive over-relaxation. This is a very generic technique that one can apply. If you have any iterative map you're trying to apply, it should go to the solution you want but it doesn't converge for some reason, then you can use this relaxation technique to promote convergence to the solution. You may slow the convergence way down. It may be very slow to converge, but it will converge. And after all, an answer is better than no answer, no matter how long it takes to get it. So sometimes you've got to get these things by hook or by crook. So for example, you can apply this to Jacobi. This was the original Jacobi map. And we just take that. We add 1 minus omega times x i plus omega times this factor over here. And now we can choose omega so that this solution converges. We always make omega small enough so that the diagonal values of our matrix appear big enough that the matrix looks like it's diagonally dominated. You could go back to that same convergence analysis that I showed you before and try to apply it to this over-relaxation form of Jacobi and see that, while there's always going to be some value of omega that's small enough, that this thing will converge. It will look effectively diagonally dominant, because omega inverse times D will be big enough, or omega times D inverse will be small enough. Does that make sense? You can apply the same sort of damping method to Gauss-Seidel as well. It's very common to do this. The relaxation parameter acts like an effective increase in the eigenvalues of the matrix. So you can think about L. That's a lower triangular matrix. It's diagonal values are its eigenvalues. The diagonal values of L inverse-- well, 1 over those diagonal values are the eigenvalues of L inverse. And so if we make omega very small, then we make the eigenvalues of L inverse very small, or the eigenvalues or L very big. And again, the matrix starts to look diagonally dominated. And you can promote convergence in this way. So even though this may be slow, you can use it to guarantee convergence of some iterative procedures, not just for linear equations, but for non-linear equations as well. And we'll see, there are good ways of choosing omega for certain classes of non-linear equations. We'll apply Newton-Raphson method, and then will damp it using exactly the sort of procedure. And I'll show you how you can choose a nearly optimal value for omega to promote convergence to the solution. Any questions? No, let me address one more thing before you go. We've scheduled times for the quizzes. They are going to be in the evenings on the dates that are specified on the syllabus. We wanted to do them during the daytime. It was really difficult to schedule a room that was big enough for this class, so they have to be from 7:00 to 9:00 in the gymnasium. I apologize for that. We spent several days looking around trying to find a place where we could put everybody so you would all get the same experience in the quiz. I know that the November quiz comes back to back with the thermodynamics exam as well. That's frustrating. Thermodynamics is the next day. That week is tricky. That's AICHE, so most of the faculty have to travel. We won't be able to teach, but you won't have classes one of those days so you have extra time to study. And Columbus Day also falls in that week, so there's no way to put three exams in four days without having them come right back to back. Believe me, we thought about this and tried to get things scheduled as efficiently as we could for you, but sometimes there are constraints that are outside of our control. But the quiz times are set. There's going to be done in October and one in November. They'll be in the evening, and they'll be in the gymnasium. I'll give you directions to it before the exam, just say you know exactly where to go, OK? Thank you, guys.
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https://ocw.mit.edu/courses/3-091sc-introduction-to-solid-state-chemistry-fall-2010/3.091sc-fall-2010.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. Tuesday will be the first weekly quiz, celebration that is. That will be at the beginning of recitation. You'll have 10 minutes. It'll be a short one-pager. You just write on the page. All you bring is your periodic table, which you should have gotten in recitation yesterday. Periodic Table, table of constants, calculator, something to write with, but no aid sheet on the weeklies. Readings: Readings, I urge you to read before you come to class. And so if you go to the website, you can go to Schedule, and in the schedule, you'll see stuff like this that tells you what the readings are for the day. As I mentioned last day, the lectures are being videographed and posted probably within an hour on the website and any of the images that I show are also recorded, burned as PDFs and uploaded. So you don't have to be put in high-speed stenographic mode in order to attend class. What's the other thing I wanted to tell you? If you're new to the class or if you need to change your recitation section because your conditions have changed, do not simply go to the other class. We're trying to regulate enrollment, particularly on Tuesdays. If the TA shows up expecting 20 students and has 20 copies of the quiz and 25 people show up, that's not a recipe for success. So you must go to my administrative assistant, Hilary Sheldon in order to change recitation. And if you need any of the handouts and so on, it's just down the hall here in Building 8, Room 201. I think that's all that I had to say. If you go here the videos are all listed. So last day we started talking about taxonomy and that led us to the beginnings of atomic theory. We visited with Democritus, 400 BC. We had that detour with the idiocy of Aristotle, and then eventually got back to our senses, and we saw John Dalton with his table of the elements, and then ultimately onto Mendeleyev. And I wanted to pick up the thread there. But before doing so, draw attention the fact that John Dalton did more than simply develop a set of fonts for us. So he proposed the model of the atom and this goes back little over 200 years ago, and these are the features of the model. First of all, that matter is composed of atoms that are indivisible and indestructible. So that goes all the way back to Democritus, nothing new there. All atoms of an element are identical. Atoms of different elements have different weights and different chemical properties. This is the emergence of modern material science, the connection between properties and elements. So the weight arguably is one of the properties, but the only way they could distinguish elements at that time was by their atomic mass. Atoms of different elements combine in simple whole-number ratios to form compounds. Well, that makes sense because they're the elements. They are the elemental building blocks. If I told you you could build a structure made of blocks and part-way through your construction I say, why don't you cut that block in half, you'd say, well, then the block isn't the building block. It's the half-block that's the building block. So axiomatically if these are the elements they must combine in simple whole-number ratios to form compounds. And lastly, atoms cannot be created or destroyed. Well, he wasn't foretelling E equals mc squared. What he was saying was that if you take elements and you combine them to form a compound, if you subsequently decompose the compound you get the elements back as they were. So those are the features of John Dalton's model. And we fast forward to 1869. And this is the knowledge that was available at the time in terms of the elements that had been isolated and characterized. And it was with this set of elements that Mendeleyev operated. On file cards, in his breast pocket, he carried with him everywhere. And he wrote down the names of the elements and their atomic masses and their properties and whatever else he could use the way of characterizing them. And during the course of writing a textbook-- he had just finished a chapter on the alkaline metals and he was sitting in the railway station playing solitaire, and boom! The flash came to him that you don't put arsenic underneath aluminum even though it's next in mass to zinc. You move it over and you don't even put it under silicon. You put it under phosphorus. And furthermore, what he said was there's going to be an element here discovered under silicon and it will have these properties. And let's look a little bit more deeply at the properties. But before doing so I want to say that by announcing this prediction of what the element should be that's missing is that we start to see the evolution of principles of modern chemistry. First of all, he recognized the pattern. So did Lothar Meyer in Tuebingen. So they both proposed a periodic table of the elements, but where Mendeleyev pulled away from the pack and distinguished himself was that he developed a quantitative model. And I haven't shown you the quantitative aspect yet. That's coming next. That explains the observations, and that's good. You might say, well, that's just curve fitting if you're a cynic. But it makes predictions that can be tested by experiment, tested by experiment. So let's take a look. He said that under silicon, but above tin, there would be an element. He called it eka-silicon. Eka is a Sanskrit word, which means one after. So this is the element one after silicon. It was eventually isolated and given the name germanium. Mendeleyev said it would have an atomic mass of 72 grams per mole. In fact, it's 72.59. He said it would have a density of 5.5 grams per cubic centimeter. It's 5.36. This is 1869. He said that it would have a high melting point, whatever that means, and it melts at 958 Celsius. It's compounds. he said it would form a dioxide with a high melting point and a density of 4.7. It forms a dioxide and its density is 4.70. In fact, there's a story about a French mineralogist who came upon some of the stuff that ultimately became germanium dioxide, measured its density and reported it to Mendeleyev in a letter, saying, you know I measured the stuff and it's 5.3 grams per cubic centimeter. Mendeleyev wrote him back and he said, make the measurement again. You're wrong. He wrote back three months later and said, I measured it again. It's 4.7. That was the genius of Mendeleyev. To go way out on a limb and make those predictions. And so I've made the case for the table of the elements. Why do we call it the Periodic Table? What's the periodic about? Well, the periodic-- take a look here, if you go to the website there's a tab called Courseware and there's a tab called Periodic Table. And you can go to the Periodic Table and ask the software to plot property. So for example this is boiling point versus proton number or atomic number. And so you see the boiling point varies as you move from low atomic number to high atomic number. But it's not totally random. It's not a Gaussian distribution. There are features. It goes up and down, up, down, up and down. If you train your eye a little bit you'll actually see some regularity, a pattern there. Maybe that's not so good. Let's look at this one. This is electrical conductivity. And again, up, down, up, down and look at those red lines. There, there, there, there. Don't you see something? That's a pattern. And they're almost equally spaced. So that was where Mendeleyev announced his Periodic Law, where he said the properties are related to the identity of the atoms. And furthermore, he announced, that the properties are a periodic variation in atomic mass. So let's get that now Mendeleyev's Periodic Law, and the properties of the elements vary periodically with atomic mass. That was Mendeleyev. So now that we know that we can go forward, and here's now the full-blown Periodic Table according to the framework that Mendeleyev established. Now if you look at this carefully, you'll see down here things get whited out and there's these strange notations, uu, m, and all that stuff. What's that all about? This is where the super heavies lie. These are all synthetic elements. Transuranic, they're made by high-energy reactions, so-called high-energy physics in what you might call accelerators, atom smashers, what have you. And there's only three places on the planet where you can conduct such reactions. One of them is in Darmstadt in Germany. One is in Dubna, just outside of Moscow. And if you want to stay home-- and eschew the frequent flyer miles-- you can go to Berkeley, California. These are the three places where we have the accelerators capable of making such compounds. And so, take a look carefully at what the nomenclature is. The way you name them is by using these Latin ordinals. So un, bi, tri, quad and so on. So if you wanted to name element 115, it's ununpentium. You want the ium ending. And you can make these up. You could make up element 205 if you want to or whatever. My favorite is 111 because that's unununium. But there they are, so you can have fun with those. But with time, the elements are being named and these have been synthesized since your version of the table was printed. And so number 110 is named Darmstadtium in honor of the team at Darmstadt that first isolated it. And number 111 was just named two years ago and the name is roentgenium after Wilhelm Roentgen, who discovered x-rays. Now what is it about discovery? Well, here's an example of one such reaction that would give you an element. So if we had access to one of these devices we could take, for example, lead and nickel and accelerate them to very, very high energies. And then we could make 110, ununilium. Or now we'll call it Darmstadtium plus neutron. And in doing so we've generated the new element. But we can't just say we've made the element and publish. We have to be able to characterize it. Remember the reason that we gave Cavendish the credit for discovering hydrogen wasn't that he's the first to know that hydrogen exists, but he isolated it and gave it value. So if you look at the rest of the periodic table, you get things like boiling point, melting point, density, electronegativity, first ionization energy. There's a lot of information there. If you go down here there's nothing. It's all blanks. These things have very, very short lifetimes. Fractions of a second. But you have to isolate them. There's certain criteria before you can publish. And all this is regulated by this governing body called the International Union of Pure and Applied Chemistry, So UPAC, the organization that finally rules on the legitimacy of any of these. And actually there have been some retractions in recent years, where people published claiming-- I think there was a report out of Berkeley claiming that they'd synthesized 115 and then subsequently that was retracted because they couldn't support the property measurements Last thing is, if you're interested, want to do some extra reading, there's a fantastic book about Mendeleyev. He was the youngest of 14 children, came out of a very poor family in Siberia and rose to be a giant of his day. He was a polymath. He, among some of the other things he did, he worked for the Ministry of Weights and Measures under the czar. The czar was interested in taxation of alcohol. And if you mix equal volumes of water and vodka you don't get additivity. So 100 mL of water plus 100 mL of vodka doesn't give 200 mL. It gives less. And so Mendeleyev did a study to determine what the optimum ratio is so that people couldn't misrepresent the amount of alcohol in the beverage. And set the standard at 40% alcohol by volume, which is used the world over to this day. He also came to the United States in 1876 to go to Titusville, Pennsylvania, where the first oil well was drilled. And did an exhaustive study of what was the American petroleum industry at the time. And then went back to Imperial Russia and did the same survey for the Czar in Imperial Russia, including a report that recommended how to develop the natural resources of the time. He was really an amazing man. He wrote text books and so on. And nobody in science-- I would venture to say-- has not heard of the periodic table. Mendeleyev died in 1906. The Nobel Prizes were first offered in 1901. So there were five years where he was close to the top for winning the Noble Prize but was eked out by somebody else. When you look back at those other Nobel Prizes, they were deserved. But none more so than that for Mendeleyev. So ironically the man who gave us seminal knowledge of all chemistry was never awarded the Nobel Prize. And there's probably a lot of politics in there. And as I said last day, here's the typical picture of him. In this he sort of looks like a street person, disheveled and so on. But this was the man that gave us the periodic table. That's him at age 35 when he annunciated the Periodic Law. So good for him. Alright. So now, let's take a look a little deeper about the properties of the elements. How do we understand the properties of the elements? For the properties elements we're going to have to look inside the atom. If you did your reading you undoubtedly came across this table. Which at first pass, deconstructs the atom into three simple particles: the electron, the proton and the neutron. Here are their symbols, e, p and n. And they're distinguished by charge and mass. So the electron has charge, minus 1.6 times 10 to the minus 19 Coulombs. And a very low mass: 9.11 times 10 to the minus 31 kilograms. The electronic charge is balanced by the protonic charge. The atom is net neutral. So the proton has a charge of plus 1.6 times 10 to the minus 19 Coulombs. The neutron, as the name implies, has 0 charge. The proton and the neutron have very nearly equal masses, however. Right. And just a word about the units. The units here are given in terms of the Systeme Internationale. So when we use the term, C, capital C is for the Coulomb. And that's the unit of charge. And it has an uppercase letter because it's named after a scientist. In this case, the French scientist, Coulomb, whereas the gram is not named after a scientist and so it's lowercase. And then we can amplify by powers of three. So if I want 1,000 of these, I put a lowercase k here. If I put an uppercase K, I end up with the unit Kelvin, which is the unit of temperature named after Lord Kelvin. And all of this is known as SI units, which is the International System. And it's not because the scientists don't know how to develop an abbreviation. This was originally developed when French was the international language of science. So this is known as the Systeme Internationale and all of these units were defined at that time. And the term SI sticks that's the legacy. All right, so now if we go to the Periodic Table. When we start looking at the elements, we can look at any entry on the Periodic Table, and we have the chemical symbol that I'm designating here as uppercase X, and this was originally John Dalton with the I and the circle around it. And about 30 years later the Swedish scientist Berzelius suggested that we use neutral units and so therefore we have the Latin coming in for many of the elements, such as iron, Fe, ferrum, and gold Au, aurum. In the upper-left corner, we have the quality I'm representing here, A And A is the mass number. Some people call it the atomic weight. And it is the sum of the masses of all the constituents. So it's the sum of the mass of the protons, so it's the proton number plus the neutron number plus the electron number. But since the electron weighs 1/1800 of what these others weigh, you normally don't consider this. It doesn't matter. So just adding protons plus neutrons gets you to what we call the atomic weight. And then down in the lower left corner we have Z and Z is the proton number. And as the name implies, it's equal to the number of protons in the nucleus, which then equals the number of electrons outside the nucleus in the neutral atom. Now I'm specifying neutral atom, because it's not necessary for atoms to be neutral and we'll take a look at those in a moment. A point about redundancy here. We don't really need the proton number and the chemical symbol because the proton number really defines. The proton number is like the Social Security number. This is the identity number of the atom. If we change the the proton number, we change its identity. So for example, I could write sodium. Sodium 23 and 11. I don't need the 11. 11 means it's sodium or sodium means it's 11. So I could just write this as 23 sodium. So I know it's sodium, that means it's got 11 protons and 23 minus 11 must be neutrons. Or if I wanted to be a smart aleck, I could write this : 23 11. That's sodium. I don't need to put anything here. But there's no smart alecks here, of course. So for example, we could then show this reaction as-- this is what? 208. This is lead, 208. And nickel, 62 gives us Darmstadtium with a value of 269 and the neutron is 1. You can see how these reactions can be made to go. Now atoms don't necessarily have to be net neutral. We can have something that is net non-zero charge. Net non-zero charge on the atom gives it the term, ion. Ion is an atom with net non-zero charge. And we have two cases where the atom is net positive. If the atom is net positive that's the result of electron deficiency. The atom is electron deficient. And we term such an atom the cation. There's two types of ions. The cation. And then we have something that is net negative. If it's net negative, it means it's electron-rich. That is to say, there are more electrons than protons and the net negative ion is called the anion. And you can try to figure out ways. I sometimes think that cation has a t, which looks a little bit like a plus sign. Anion has five letters, minus has five letters. And they both end in n, but this has an n, which is negative or something. You'll figure something out. Now we've talked about varying charge at constant proton number. But the other thing we can do is we can look at-- you can vary the neutron number. Since the neutron has no charge you can vary the neutron number and not harm the identity and still have a neutral atom. So vary neutron number at constant proton number. And let's see what that is. That gives you something that looks like this. So for example, if you if you look at carbon, the atomic mass that's shown here is 12.011. And you'd say, well, gee, if it's got 6 neutrons and 6 protons, why isn't that 12 exactly? Well, this is the answer here. You can vary the neutron number at constant proton number. So let's take a look at how that plays out. The way that plays out is as following. Let's see I'm going to make a little table here. So we'll start with carbon 12. Carbon 12, so that means-- now I know what I'm going to do. I'm going to bring this down and make some headings for me. This will be my proton number and this will be my neutron number. And finally I'm going to talk about abundance, natural abundance. So carbon 12, since it's carbon, axiomatically it must have 6 protons. And 12 minus 6 is 6, so it's got 6 neutrons. And this is the dominant form of carbon. If you took a chemical analysis of the carbon you'd find that over 98%, 98.892% of the carbon atoms that you examined would be of this form, carbon 12. Now there's also carbon 13. Has to be 6, otherwise it's not carbon. That means it's got 7 neutrons. And it's a minority species. 1.108%. And then there's a third form of carbon and that's carbon 14. Again, has to be 6 and it's got 8 neutrons. And it's found in vanishingly small quantities, one part in 10 to the 12. Or we could call it ppt, parts per trillion. So this is same atomic number, same proton number, same Z but different mass numbers. Different A's. So all of these variants of carbon are found on the same place, the same spot on the Periodic Table. The Greek word for same is iso, and the word for place is topo, so these are called isotopes. The isotopes of carbon are species that have identical proton number but different neutron number. How about the units? What are the units here? Well, we have to give some kind of unit. I've been sort of freely going around and counting protons as one and so on. And here's the standard. The standard for mass is defined, and the definition goes like this. If you take carbon 12, which we just introduced to you, and we say that we're going to specify a mass of 12.000 grams exactly for a specified quantity, in other words, a specified number of these atoms. We have to say we'll take a certain number of these carbon atoms and specify the mass of that number is 12 exactly. And specified number of atoms being the mole. The mole. And it turns out that the mole has a value of 6.02 times 10 to the 23rd. How do they get that number? A little bit more in the way of definitions. It was a concept put forth by a professor. So we're going to take some time on it because we respect professors, in this class at least. And so this was a concept put forth by Professor Amadeo Avogadro. Professor Avogadro, who was a professor of physics at the University of Turin, Torino. And he was a contemporary of John Dalton's and they were both studying gases. And it was Avogadro who taught us that, when you keep the pressure constant equal volumes of different gases contain equal numbers of molecules. It doesn't matter if you have argon, which is by itself atomic, or we have oxygen, which is diatomic, or you have methane, which is CH4, five atoms making a compound. Equal pressure, equal volume, equal numbers of those species. So that was Avogadro's Law. So let's put that down. At constant pressure equal volumes of different gases, contain identical numbers of atoms. Equal volumes of different gases contain equal numbers of molecules. And here I'm using the term molecule as a counting unit. So it could be, strictly speaking, an atom or it could be diatomic and so on. That's what it was. And out of honor for Avogadro, we name the number of atoms in the mole the Avogadro number. Which I've written 6.02 times 10 to the 23rd. Now how do we determine Avogadro's number? That's an interesting story. So first of all, we need two pieces of information. Because we're going to do this by the noblest form of chemistry, electrochemistry. So the first thing we're going to do is we're going to look at the work Michael Faraday in England. And what Michael Faraday did is he studied the electrodeposition of metal. And specifically he passed current through a cell and he electrodeposited silver. So he starts with silver plus, that's silver a cation, and by the action of electric current attaches an electron to silver and renders it neutral. Silver, which now plates out onto an electrode and they measured the mass. They measured the mass of silver-plated and they compare it to the amount of charge that was passed. They measured the charge. And you can get charge, because you know current. So charge is simply equal to the integral of the current times the time. You know the current, that's easy. And what Faraday found was that to make what we now know to be 108 grams of silver, 108 grams of silver, which we're going to subsequently recognize as the mole, which is identical to the amount, the number of particles in 108 grams of silver, is equal to the number of particles in 12 grams of carbon. Sort of an Avogadro-type harkening. He found that that is-- the equivalent requires 96,485 Coulombs. So you can say 1 mole of electrons gives me 1 mole of silver, so that's the charge on 1 mole of electrons, where Coulomb is the elementary charge, because we know 1 electron per 1 silver atom deposited. So now if I know that's a mole of electrons, I need to find a charge on one electron, divide through and I get the Avogadro number. And to finish the story we have to wait about 50 years and come to the United States, where it's Robert Millikan, Robert Millikan at the University of Chicago doing the oil drop experiment through which we learn the elementary charge. And here's the cartoon of the oil drop experiment. I took this from a different text. It's not shown in your text. So I actually did this experiment as a sophomore at the University of Toronto. They had us repeat some of the great experiments of physics, the ones that were accessible, obviously. I couldn't do high-energy physics in an afternoon. That would have taken me a little bit longer. But we did this one. And so it consists of an atomizer, sort of a perfume atomizer, in which there's oil. And by the action of atomization we form a shower here, a very, very fine dispersion of tiny droplets of oil. And then-- this cartoon is hard to make sense of so I fixed this-- we shine high-energy radiation on this. And by the action of high energy radiation we take these neutral droplets and we turn them into ions. We eject electrons. And so now these are charged. And then we charge the plates. So if we have neutral species and they simply come out of the atomizer, they'll settle under gravity. But now if they're charged and I put a charge on the plates-- let's say as here the upper plate is positive-- if any of these particles is charged positive, the action of the electric field will accelerate the descent, because the bottom plate is negative attracting and a positive plate at the top is repelling. And vice versa. If I have a particle that's negative, the upper positive plate will actually cause it to slow down, and in the extreme, it may actually start to rise. And so what Millikan did is a set of experiments in which he studied all the different particle sizes. See this telescope? Right over here is Millikan. And Millikan's sitting there and he's squirting and he's watching. He's measuring the settling velocity. And he changes the magnitude of the electric field. He changes the intensity of radiation. He changes the nozzle. He changes everything he can. And what does he find? He finds that the distribution of velocities is not continuous. It's not continuous. You think, well, gee if you just keep dialing you should get every variation of velocity. Well, he doesn't. He finds that he gets variation down to a single value, below which he can't go. He determines that electric charge is quantized. That is to say there's a base unit. It's an element. I just talked to you about the elemental building block. That's an element in mass space. Now I'm going to go conceptually into charge space. There is an elemental building block of electric charge. Electric charge is quantized. And he found that the elementary charge, which we gave the symbol, e. e is not the symbol for electron. e is the symbol for elementary charge. It has a value, if you convert it to modern SI units, of 1.6 times 10 to the minus 19 Coulombs. So now I can take these two pieces of information, Faraday which is up here. This is known as the Faraday Constant. Script f, Faraday constant. So if I divide the Faraday constant, which is the charge on a mole of electrons, by the elementary charge, which is the charge on one electron, presumably I should end up with the Avogadro number. It should be the ratio of the Faraday to the elementary charge. And it gives us-- for the third time this morning-- 6.02 times 10 to the 23rd. If you like per mole, yes or no, doesn't matter. So now, what's the atomic mass unit? Now we can say the atomic mass unit is, 1 atomic mass unit then must equal what? It's going to equal 1/12 of the mass of carbon 12. 1/12 of the mass of carbon 12 divided by the Avogadro number, which gives us 1.661 times 10 to the minus 27 kilograms. Now be careful because the system is just a little bit rickety. You know we went SI, but look, this is still defined as 12 grams. And so sometimes if you look depending on where this is, 10 to minus 27 kilograms or 10 to the minus 24 grams. Just be careful. If you ignore this you'll be off only by factor of 1,000. That's a joke. But it's lost here. People are too serious. We'll lighten you up. All right, so enough of the history. Let's now do something dynamic. So far we've been studying static elements. But chemistry is really the action of elements in motion. So how do we describe a chemical reaction? Let's look at that. What are the rules to describe a chemical reaction? Write an equation. Write the equation of the chemical reaction subject to these rules. There are two simple rules. One is conservation of mass. We've been told the repeatedly since Democritus, conservation of mass. And the second thing, we use Dalton's Law of Molar Proportions. That is to say, the building blocks in integer ratios. And so I thought I'd do this in context. So I've got a specific example here. So this is something that I'm interested in. Some of my research is in metallurgical extraction by benign processes. What you're looking at is a billet of titanium. To give you a sense, you can see the stairwell back here. So this is about 4 feet, a little over a meter here. So you can see this is one honking big piece of titanium. This came out of the primary reactor, the Kroll reactor and this is subsequently swaged and hot worked and so on to form these billets. So this is the first step of turning dirt into metal. That's called titanium sponge. And titanium sponge occurs inside a Kroll reactor. It occurs inside a Kroll reactor, which was invented by a man of the surname Kroll in Luxembourg in the 1930s. And then with the advent of World War II, he decided to be smart, to get out. And he ended up in Oregon where he became a professor. So he's known as Professor Kroll, although the truth be told he really made his discovery before he became a professor. But he's still a professor and so we'll honor him. And so the Kroll process for making titanium centers around this reaction. Here's the reaction written according to the rules above. We take titanium dioxide, which is found in the Earth and by some prior chemistry convert it to titanium tetrachloride, and in a reactor that I'm going to show you in a moment, we react titanium tetrachloride with magnesium. And magnesium has a higher affinity for chlorine than does titanium and steals the chlorine from titanium to form magnesium chloride, leaving behind titanium metal. Now we have to have conservation of mass. So you can see, I've got 4 chlorines on the left but only 2 chlorines on the right. So I'm going to put a 2 here and double the magnesium chloride. But now I've got 2 magnesiums on the right and only 1 on the left. So I'll put a 2 in front of the magnesium and now we have a balanced equation. And here's what the reactor looks like. You can imagine a giant vessel with a pressure seal on the top and a couple of valves, big enough to make this. So this is about 15 feet by 30 feet. And so we introduce titanium tetrachloride, which is a gas, and magnesium as a solid and heat to 900 degrees C. And at 900 degrees C, if you look on your Periodic Table you'll know that magnesium melts at 650 degrees C. So we have a liquid sitting here, titanium tetrachloride here, and this thing is sealed. It's called a bomb reactor. Nothing can get in, nothing can get out. The pressure builds up here. And right at this interface the titanium tetrachloride reacts with the magnesium according to this reaction. Now this is very interesting. It's beautiful reaction because the titanium tetrachloride is a gas; magnesium is a liquid. Magnesium chloride is a liquid, but it is of different density, and it is insoluble in magnesium, and titanium melts at 1670 and it's a solid. So what happens over time is this. The magnesium chloride that forms pools underneath the magnesium liquid, gets out of the way so that we can continue to keep this interface clean and have the reaction proceed. You don't want to reaction where reactant A reacts with reactant B, makes a product that covers the interface and now the product is in the way of future reaction. So this is very elegant because I don't need any fans, I don't need any nose propellers, nothing. By density the magnesium chloride settles and the titanium settles. And it's sitting here at the bottom. And you can imagine if we do this long enough, this titanium at the bottom will continue to build until it looks like this. As long as you keep feeding TiCl and Mg. See I'm talking metallurgy now. TiCl and Mg, that's what you make. So that's how we make titanium, first step. And so suppose you get hired and it's your first day on the job and you're working at Cambridge Titanium and the boss says let's put in 200 kilograms of TiCl and we'll put in 25 kilograms of Mg. And the question is, what is the yield? What is the yield? How much titanium are we going to make? Well, you say, just multiply it out. But first you have to see if things are in balance. We have to study the stoichiometry of the reaction. Stoichiometry, what does this mean? It's from the Greek, stoicheia, which has to do with measurement proportions. So if these are not put in to the reactor in proportion to what they are in the equation we're not going to get the yield here. So first thing I gotta do, this is in moles, this is in kilograms. So I have to convert the kilograms to moles and then maybe I can make some sense of this. So if I divide by the atomic mass of titanium, four times the atomic mass of chlorine and convert; I will discover that I have 1,054 moles of TiCl. And I've got about 1,029 moles of magnesium. Well, this equation says I need 2 times the amount of titanium tetrachloride. Well, it's obvious to the naked eye, 1,029 isn't two times 1,054. So I've got a problem here. I'm not going to get as much titanium as I put in. Titanium chloride. This yield is going to be restricted. It's going to be restricted by-- this is sort of a chain is as strong as its weakest link-- the yield is restricted by the amount of limiting reagent. And in this case, magnesium is-- this is less than 2 times the mole number of titanium chloride. So this means this is the limiting reagent. Alright so now if we use that principle then I'm only going to get as much titanium as I had magnesium and you can see from the stoichiometry here, if I've got 1,029 moles of magnesium I'm going to have half of that number of titanium. So therefore the amount of titanium is equal to 515 moles of titanium. And you notice I'm not obsessed about a significant figures and so on. It's a metallurgical plant. Half of 1,029 is 515. Is it 514.5? If you wish. I don't care. So 515 moles and then I convert that, which gives me 24.7 kilograms of titanium when I use that amount of magnesium. And if you go to the text, Section 2.7, you'll see the nuts and bolts of how to run these reactions. For those of you who had a lot of chemistry in high school, I know this is review, but I want to bring everybody up to the same page. So we're starting with this. All right. I think that's a pretty good place to stop with the delivery. But I don't want you moving. You don't move yet. Because the last 5 minutes I'm going to still continue to talk. But on a slightly different topic. And so I don't want to hear the binders snapping and so on. We're here; you paid your money. Five more minutes. Five more minutes and then you're out there. Out there, then begins le weekend. But not until then. So a couple of things. First is, the music today. I try to link the music thematically. So the music playing today was Polovstian Dance number 17 from Prince Igor, by Borodin, Aleksandr Borodin. Why were we listening to this music? Well, because I insisted that we listen to it. Well, what about Borodin? Borodin lived in Saint Petersburg. He was a friend of Mendeleyev. OK, that's cute but more importantly Borodin wrote his music in his leisure time. He had a day job. His day job was professor of chemistry. And he worked at the Medical Surgical Academy in Saint Petersburg. He was an exceptional human being. In those days, women were forbidden to attend institutions of higher education. He set up an entire curriculum for women in a night school at the Medical Surgical Academy. He cavorted with artists and therefore obviously his politics were radical. And they were trying to reform the political scene in Czarist Russia at the time. And he was also quite a bon vivant. And he died on his feet dancing at a ball. So that's the way to go. Having a great time. That was Borodin. One other thing before you go. You were very very dour, so I thought I'd try to put you in a good mood to the extent this is possible with this group. And I wanted to share with you some news. There's been a new element discovered. You know these atoms smashers, they're always working. And so the discovery of the heaviest element known to science has been reported. The element, tentatively named administratium. I don't know if UPEC is going to go for this, but you can suggest names. So they're going to name is administratium, the discoverers. It has no protons or electrons. So that means its atomic number is 0. It does have one neutron, 125 assistants to the neutron. 75 vice-neutrons and a 111 assistants to the vice-neutrons. This gives it a mass number of 312. The 312 particles are held together in the nucleus by a force that involves the continuous exchange of meson-like particles called memo-ons. There's no electronic mail, because there's no electrons. There may be neutronic mail but we don't know yet. Now you've already learned something today. You know something. Since it has no electrons, what do we know about its chemical reactivity? It's inert. It has no electrons. It can't exchange. So this is chemically inert. So you say, how did they detect it? Because it seems to impede every reaction in which it is a present. According to the discoverers a few nanograms rendered a reaction that normally takes a fraction of a second, it took now four business days to conduct that same. There are a few other properties. We know so far that it's radioactive. And we're going to study radioactivity later, so there's a little bit of foreshadowing. It has a half-life of about three years, at which time it stops decaying and instead it undergoes a reorganization, in which the vice-neutrons, assistants to the neutrons and assistants to the vice-neutrons, exchange places. Some studies indicate that the mass actually increases after each reorganization. So you can imagine now we'll have something like this. See how this increased? So if they occupy the same place, they have the same proton number, but a different neutron number, in the case of administratium, they're called isodopes. So with that I will say, have a good weekend.
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https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/8.06-spring-2018.zip
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PROFESSOR: OK, so that's our equation. Three terms-- first term, spontaneous emission, second term, stimulated emission proportional to Nb, third term increases Nb's absorption. Our strategy now is, so what do you do with this? This looks like a good equation, but what are we to do? We haven't used three, really. So we can solve for U. And this time, the equation won't want to make U equal 0. It will give us something. And then we can compare with a thermal equilibrium relation to get something. So that's our goal. So how do we solve for U? Let's divide by Nb. So divide by Nb. So a 0 here, pass this to that side, and divide by Nb. So you get A on the left-hand side equals-- let's see, should I put the-- well, minus Bba Nb. Actually, I'll do it like this-- the second term first. Bab Na over Nb times U-- so I'll put it here, a U of omega Ba-- minus Bba. OK, so I moved the first term to the left-hand side, grouped the U. Why did I group the U? Because I said, let's find what U is, and try to compare with other things that we have. So now we can solve for the U. U of omega ba is equal to A divided by this quantity. So it's A divided by a big thing. Let me factor out the Bab. So now we have here Na over Nb minus Bba over Bab, or A over Bab 1 over Na over Nb is there, is e to the beta h bar omega ba-- It's a minus sign with respect to number 2-- minus Bba over Bab. OK, this is great. We're in very good shape. In fact, we're here. Let's compare with our thermal radiation. So here is what we got. And we have to compare with fact number three, which is the thermal radiation formula, which is h bar over pi squared C cubed omega cubed 1 over e to the beta h omega. And I'm comparing with the thermal radiation, U of omega ab. I canceled the D omegas, because this is U without the D omegas. So omega ba minus 1. All right. It's perfectly nice and ready. We need to compare this formula with this one. They have to be equal. Well, this factor, must be equal to this factor, given that here you have e to the beta h bar omega ba. These are constants. And the first thing that you discover is Bba equal Bab. And so yes, very nice. This is what we learn in perturbation theory. These two processes have identical rates. So the first statement that this equality requires is that Bab is equal to Bba. So there's just one B coefficient, just like there's one A coefficient. And the second thing that you learn is that A over Bab is this ratio-- is h bar omega ba cubed over pi squared C cubed. So we will be able to calculate the B coefficients, because they represent the familiar properties of harmonic perturbations transitions, and we've done already. Calculating A is harder, in principle. The process of spontaneous emission is a harder process. But this relation says that we don't have to worry about it. We already, if we know B, these are constants, we know A. So many times, even in problems in the homework, you will be asked, what is the spontaneous transition rate for this decay? You have an oscillator. It's in an excited state. What is the rate of spontaneous transition? And then you say, OK, this is complicated. But you calculate the stimulated transition rate, and then plug in this coefficient. So it will not be difficult. Now, spontaneous emission, as we discussed there in the top, doesn't include the factor having to do with the density of photons, the U that tells you how many photons there are. Because the photons play no deep role in producing the transition. But at some level, this transition-- while it's not stimulated by the photons that are flying around-- when you calculate it, in a serious, detailed calculation, you can think of them as stimulated by the vacuum fluctuations of the electromagnetic field. So if you were to quantize the electromagnetic field, there are vacuum fluctuations. And those vacuum fluctuations, you could say, they are stimulating what we call "spontaneous emission." So at the end of the day, the electromagnetic field is a quantum field. And therefore, it has zero point energies, Casimir effects, vacuum fluctuations. You can't get away from it. So in some sense, everything is stimulated emission, stimulated by a lot of photons or stimulated by the vacuum fields. Anyway, we'll still remain with the name "spontaneous emission," and keep those things very distinct. These rates have different effects at different temperatures, as well. So we'll consider that, for example, in that at low temperatures, the black body radiation has a very little number of photons. So most of the transitions, if they occur, are happening due to spontaneous transitions. On the other hand, as you increase the temperature, the number of photons that are available per unit volume increases, and stimulated emission takes over. So these are the different contributions to the rates. And it gives you a perspective-- at very low temperature, spontaneous dominates. At very high temperatures, stimulated emission dominates.
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https://ocw.mit.edu/courses/8-591j-systems-biology-fall-2014/8.591j-fall-2014.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality, educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Why don't we go ahead and get going. Today what we want to do is talk about predator/prey interactions. In particular, how a predator and its associated prey can in principal oscillate over time. Now in this area there's this very important model block, a Volterra model, that has been used as kind of the standard model. It's given us some intuition for how this might happen. But mathematical biologists really don't like this model very much, because the oscillations that it give are neutral, or neutrally stable cycles or orbits. What I mean is that it's not a limit cycle. It doesn't have a characteristic amplitude, nor a period. And associated with that-- and it's not a concidence-- and associated with it, what it means is that if you make sort of very small modifications to the model, then you can get qualitatively different outcomes. In particular, you can either abolish the cycles-- i.e., you can turn these neutrally stable orbits into stable limits-- or I'm sorry. Into a stable spiral. So it may have damped the oscillations. So it could be that the oscillations kind of go away over time. Or it could be that you turn them into true limit cycle oscillations. And this could just be from very small changes in the model. That's associated with this fact that the logical Volterra model was neutrally stable. Then we'll kind of switch gears a little bit and talk about these experiments-- that Yoshida paper that you guys read-- where they looked at this question of what happens in kind of a laboratory experiment where you have a predator and its prey. And in particular what they found is that if there is evolution in the prey population, that you can get qualitatively different oscillations in particular. Instead of having a 90 degree phase shift between predator and prey, you can end up with 180 degrees phase shift, so kind of anti-correlated oscillations. But also that the oscillations could be much longer than what you'd. Expect And finally we'll say something about these noise-induced oscillations, which I believe that you guys have been playing with a bit in the context of your homework. Is it the next one? Oh, I get it. Well, you will get a chance to play with a great deal. So pay attention then. All right. Any questions before we get going? Problem set's due tomorrow so that you can enjoy Thanksgiving with your family or friends. And then we'll have one more problem set. So this Lotka-Volterra model. There's kind of a fun history of this, which I'll tell you about. But I just want to highlight that there are two important things that you should be remembering. And I'm getting these two points from Mark Kot in his book Elements of Theoretical Ecology, where he says there are two things you need to know about this model. One is that it's a bad model. And of course you can argue what you mean by bad. I think it's maybe overstating it. But we'll try to be explicit about why he might say that. But then it's sort of profoundly important. He kind of says from a historical perspective. But I would say that maybe both of these statements are sort of true. It's a bad model mathematically in some ways, because it's somehow structurally unstable, but also profoundly important. The mark of intelligence is being able to keep two incompatible ideas in your mind at the same time. Somebody said something like that. More than two? OK. Well, we'll try to figure out what we mean by this. All right. So Kot, it's a nice book on mathematical ecology, if you're curious. The history of this is that it was in the mid-1920s. And there was a marine biologist named Humberto D'Ancona. Does anybody speak Italian? Well, he was a marine biologist. And he had been studying the prevalence of different fish species in fish markets, and kind of throughout Italy. So he went to fish markets. Over the course of about R-K. For about 10 years. Right? So kind of roughly from 1912 to 1923. Something like that. And he basically asked, well, what is the kind of the composition of fish being sold in all these fish markets across Italy. And he noticed something that was very interesting, which is that there was a marked change in the composition between more predatory fish as compared to more prey-like fish. And this was associated with something that was happening in Europe around that time. World War I. OK. So World War I in the middle. And what he found was that in the middle of this period there was an increase, and then later a decrease of what he called selachians, which is apparently a word for sharks and shark-like fish. But for our purposes, we'll just say they are typically kind of predators. So what he saw is that this number of selachians, a function of time, kind of went like this, right. Where this was kind of roughly World War I. And he wanted to try to understand why did this happen. Why is it that war favors predators? Probably not just having to do with the general zeitgeist at the time.So it's probably a more mechanistic explanation. And luckily, Humberto was engaged to the beautiful Luisa Volterra. OK. Oops. Engaged to Luisa Volterra. So one day mid-1920s, our hero Humberto was visiting his girlfriend. And he started chatting with his future father-in-law, who was Vito Volterra, and asked him this question. And luckily, Vito Volterra was a famous mathematician. So he could write down a fancy model and try to get some insight into this question. So this is a case of random personal interaction leading to something interesting. That's the Volterra-- Lotka had actually studied these equations almost 15 years earlier. First in the context of auto catalytic kind of chemical reactions and so forth. But then later indeed, in the context of population dynamics. So Lotka wrote a book in the mid-1920s. Volterra wrote an article analyzing D'Ancona's data. And then now it's called the Lotka-Volterra model. And we'll come back to this thing about why is it that war favors predators after we get a sense of the model. So what Volterra wrote down was the following. First of all, we should all make sure we can figure out which one is the predator and which one is the prey. So let's think about this for 15 seconds. And then we're going to yell it out verbally. Is y the predator or the prey? Ready, three, two, one. AUDIENCE: Predator. PROFESSOR: Predator. All right. So we have a predator here and prey. Is the total size of the population constant? i.e, is x plus y equal to a constant? Ready, yes or no. Verbally. Three, two, one. No. It would be a constant perhaps if these two terms-- if b were equal to d or so. No. That's not even true. Never mind. Yeah. OK. I guess really what I'm thinking about is this tells us about how many predators can be created from a prey. But then there's also the growth in death rate. All right. But even this act of converting prey into a predator, even on that level it's not conserved, right? And of course, vegetarians like to point out that you can create many corn burgers for the price of making a hamburger. Something like that. OK. Except in the case of salmon. This is not true. Well, all right. We can analyze the ratio of d over b for a variety of different newt sandwiches after class, perhaps. OK. So this is pretty much the simplest model you can kind of possibly write down that captures this basic idea that when these two species interact, we're assuming that there's this mass action kind of rate. That it's just proportional to the number of the density of x and y. Now I think that in the context of chemical reactions, this is going to be true over a huge range of densities, and perhaps almost even rigorously true if they are just single. There's just some x and y that bounce into each other. And at some rate they do something. Whereas I think in the context of predator and prey, this is on much less firm footing. That's OK. So it's at least the simplest thing we can write down. And of course, then we'll have to ask, how is it that the conclusions of the model might vary depending on the assumptions that go in here. One thing that we've spent a lot of time in this class doing is trying to look at a set of equations, and in words, be able to extract what the assumptions were that went into writing the equation. And associated with that, you can think about all these things about the nondimensionalized versions of these equations. You'll often see the predator/prey models written in some nondimensionalized version, where for example, you get rid of maybe the a and maybe the c. I think that in many cases I prefer to just leave these terms in here, because that way we can immediately see what happens if we go hunting, or if we do this or that. OK. Can we try to figure out what-- there are four assumptions, at least, that have gone into this that are relevant and are worth saying. Can you guys help me out? Yeah. AUDIENCE: It's well-mixed. The interaction in terms of [INAUDIBLE]. PROFESSOR: OK. Right. So it's well-mixed. And there's two aspects that we might-- So one thing, it is certainly worth saying that it's well-mixed. But associated with that, you're going to say that it's because of x times y. Right? Yeah. I guess that's what I feel you were about to say. AUDIENCE: I mean, I guess. PROFESSOR: Yeah. Right I guess. But I think we have to keep these things a little bit separate in the context of chemistry, those are maybe the same statements, right? Of course, we're writing these things. These are just quantities that are just functions of time. We have not included space explicitly. So in that sense, it's definitely a well-mixed model. But just because we're assuming that it's well-mixed, we're not keeping track of the state of density as a function of position does not mean that they have to interact with this term. And for two molecules that are bouncing against each other, then I think that is the case. But we have to keep these assumptions separate. AUDIENCE: You could imagine a colony of bacteria, for example, where they're all-- you can only prey on the outside. PROFESSOR: Yeah. Necessary. I guess what I would say is that if it were not well-mixed, it could still be that they interact like this. And you just keep track of the predator/prey density as a function of position. And they could still interact in this way, possibly. And in particular, in the context of Turing patterns, what we did is we allowed these things to vary as a function of position. But then we would have still typically written the interaction term as x times y. So it is true we're modeling this as a well-mixed population. But I think that's independent of this statement. Do you agree or disagree? Or? AUDIENCE: But if it's well-mixed, the encounter frequency will be proportional to [INAUDIBLE]. AUDIENCE: That's sort of the secret, right? PROFESSOR: Yeah. AUDIENCE: The assumption is the encounter frequency is proportionate, right? PROFESSOR: Yeah. OK. So what I would say is-- well, certainly in the context of predators and prey, it could be that if the prey is full, then it's not going to be so simple. In a sense, is that I guess it could be more complicated than that is all. But my statement-- AUDIENCE: So you're saying-- I'm sorry-- the interaction term would not necessarily be just a function of x times y. PROFESSOR: That's right. It could be more complicated than that. And the other thing is that even if it were not well-mixed, in the context of these reaction diffusion models, we would typically say that the rate that these two things hit each other is x times y. But they're each a function of position. Right. So it's not that there's-- it can either be well-mixed or not. And it can either be this or not. I guess. AUDIENCE: But it's not quite the same variable, because in that case it would be the density, whereas-- PROFESSOR: I agree. Although here, it's not obvious whether we're talking about number or density, frankly. Right. And we often talk about this as if it's the number of predator, number of prey. But then that's a little bit inconsistent with this in that if you were-- this should really be, if you're thinking about these terms being the product. And it should really be the density. And if you're looking at the population size in some fixed area or volume, then it doesn't matter. But even in the context of predator or prey, I'd say that the way in which this would make senses is via densities. Does this discussion make sense? Maybe? We are assuming it's well-mixed, because we're not thinking about these things as a function of position. But I would say that we cannot get that from just the interaction term. We are assuming it's well-mixed, though. Other things that we've assumed here? AUDIENCE: Something along the lines of the prey doesn't need the predator, for instance. PROFESSOR: Right. The prey does not need the predator. And in particular, what happens to the prey in the absence of predator? AUDIENCE: It grows. PROFESSOR: So the prey grows exponentially. AUDIENCE: Because it's not diet [INAUDIBLE]. PROFESSOR: Without predator. And the other thing that is perhaps worth mentioning in all this though is that this is a deterministic description of the world. So a here is telling us about-- the normal way that we would interpret a is it's the rate of division of the cells, or the rate of new deer being born, or whatnot. But of course, a could actually be the difference between this growth rate and a death rate. So just because we don't have an explicit death rate doesn't mean that there's no death. So I think you can't actually say that the prey does not die in the absence of predator. What you can say is that the prey population-- we're assuming that a is greater than 0. So what you can say is that the prey population grows exponentially without the predator at rate a. But this is really in principle the difference between the growth rate and the death rate of the prey in the absence of the predator. And in the context of a differential equation, this doesn't make any difference. But if we were to go and do a master equation type formalism, would this make a difference? Yeah. Right. So this is the whole thing about the Fokker-Planck approach. And you can get different variances at equilibrium, depending on whether the rate of forward back reactions and so forth. Does that sound familiar to you guys? No. What else have we assumed in this model? AUDIENCE: The predator dies exponentially. PROFESSOR: Right. So the predator dies exponentially when? So the predator dies exponentially without the prey. So it's a nice simple assumption. Other things that you might like to point out about this? AUDIENCE: A single predator is [INAUDIBLE]. PROFESSOR: OK. Right. So there is a sense that it's just x times y. So the most simple way to think about this is that it somehow is a single predator eating a single prey. There's maybe no group hunting type behavior. So if we wanted to try to understand how a pack of wolves can bring down a buffalo, then you might not want to use this equation. So this is that the rate of predation is proportional-- we'll say goes as x times y. And that embodies many different kinds of assumptions. These are four findings. OK. So since it's such a simple model, we can go ahead and we can just solve it. Right. How many fixed points are there going to be? Two. OK. It's always good to remember these things. And so what we can think about, this as x. This is y. There's indeed going to be a fixed point somewhere in here. But then the other one is going to be at zero, zero. Because the top equation has x's everywhere. Bottom equation has y's everywhere. So indeed, there's going to be one fixed point at zero,zero. So-called the trivial fixed point. Is this a stable, or is this an unstable fixed point? Let's think about this for a moment. And then we'll vote verbally. Stable or unstable? OK. Ready, three, two, one. AUDIENCE: [INAUDIBLE] PROFESSOR: So I think there were some disagreements there probably. It was a little hard to tell from the verbal. And one way that we can think about this as that we can ask, well, the predator in the absence of the prey, what happens to it? So it kind of comes down here. So along this axis it's stable. However, in the absence of predator, if you think about the prey, that'll grow. So this is actually an unstable fixed point. Stable along one axis and unstable on the other. Incidentally, what does this tell us about the eigenvectors around this fixed point? AUDIENCE: [INAUDIBLE] PROFESSOR: That's right. The eigenvectors are really just 0, 1, 1, 0. Because you can see that if you start along one of these, you come straight down. Start along the other one, you come straight down. And the other one we can just see by-- well, we can solve it directly. Just we pull out an x. It's when a minus by is equal to zero. And also when minus c plus dx is equal to zero. So the other fixed point, this x star, y star, and this is maybe the important one. It's going to be just c over d and a over b. Super simple model. Right? You can calculate where the fixed points are kind of immediately. But it's actually a model that has many weird properties. Some of which you might think of as features. Some of which you might think of as bugs. But in particular, the location of this fixed point, I think, is really sort of surprising in that its dependence on this abcd. This is part of why I really think that when analyzing this model, I very much prefer to keep the a's, b's, c's, d's, rather than using the nondimensionalized version of it. Because when you do that, you lose track of what's going on. So this model for example, makes very clear predictions of what should happen if you hunt the predator. There are various contexts in which wildlife managers have been interested in trying to help a prey population. So if a prey population is suffering in some way, then you can reasonably think what you should want to do is kill the predator. And this is all these debates where you have the wildlife managers, where they buy automatic rifles in order to shoot wolves from helicopters. We have a Canadian in the room, right? I mean, wasn't this a Canadian proposal? AUDIENCE: I'm very American. PROFESSOR: All right. We'll have to do some more research to figure out exactly. OK. Right. So the question is, well, what happens at least in this model? And people apparently have seen this sort of thing occurring in natural populations as well, right? If you hunt the predator by going out and shooting them, what does this do to the predator and the prey populations. In particular, we think about x star, y star. Well maybe this-- am I confused? Maybe this is the example that I'm confused by. Well we have to complete it now that I started it. But now I might have gone backwards. We'll see what happens, and then discuss. All right. Ready. Three-- Oh wait. I don't have enough options. OK. Sorry. We should do one, then the other. Because sometimes things don't change, right? So how do I do this? OK. Sorry. OK. We should just do one, then the other, right? No change. This is down. OK. There's gonna be a problem [INAUDIBLE]. Let's first do y star. Cause that's the most direct one. Sorry. I think that my story I got backwards. But we'll figure this out. OK. Ready? OK. So the question is, there's a population that you want to-- all right. Now I'm going to redo my story to make it-- but there's a population you want to keep in check somehow. So you might reasonably go out and shoot it. Right? So the question is, if you go out and you do this on the predator, at least in this model, what happens? All right. Ready. Three, two, one. Right. So although we actually have a fair number of disagreements on this. So we're actually kind of 50-50 at this stage. What is it if you go out and from the helicopter, you start shooting this animal, what does that do from the standpoint of this model? Right. So we're shooting the predator. a, b, c, d, what you think? What should it change? c. And does c go up or go down? C goes up. OK. Well, how does that affect y star? AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah. AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah. Right. You know, I mean, I'm thinking daily helicopter rides where you shoot wolves as you see them. AUDIENCE: This is Canada. PROFESSOR: This is Canada. This is Canada. Yeah, those Canadians. AUDIENCE: So effectively, you're changing the death rate of-- that's what you're saying. PROFESSOR: Well, OK. So that's my claim. Of course, all of these things can be more complicated. But it certainly is change in the death rate when you go and shoot them. So I guess I would say that in the context of this model, that's the simplest way to think about it. So the statement is that if you hunt this predator, you're increasing the death rate for the predator. You're increasing c. Hunting predator. This causes c to go up. And the striking thing is that y star does not change. OK. AUDIENCE: I believe you. I think it's funny. PROFESSOR: OK. [LAUGHS] You believe me. OK, I'm glad. One thing is believing me in the context of this model. Another thing is asking whether this is happening in real life. But I think people have seen such weird phenomenon in the context. But then of course, it's a matter of is this a dominant source? What's going on? Like many, many things. Right. But certainly in this model, there's this weird phenomenon where the death rate of the predator does not alter this y star. OK. We have to figure out what the y star means here in a moment. What is the effect on this fixed point number concentration of the prey? It goes up. So my original story, I think I was getting confused. So at least in this model there's a sense that if you hunt the predator, it's true that you don't bring down the steady state, or the time average-- we'll see-- time averaged number of the predator population. But you do increase the prey population. Because in this case you can help the prey. But you don't bring down the predator. Now I want to say something more about justifying this thing about why we might care about x star and y star so much. So this thing is x star/y star. Well, maybe all. Yes. AUDIENCE: As soon as you stop shooting, this is a very temporary solution. When you stop shooting, c goes back to its original value. PROFESSOR: That's right. When you stop shooting, c goes back to its original model, or original value. And then the prey population will come back down. So this is something you have to continue to do. Yes. What do I wanna say? If you go ahead and you calculate the-- if we linearize around this fixed point x star/y star, that's our standard thing that we did early on the semester, what we find is that this Jacobian, this matrix A, the linearized matrix you get is-- all right. Well, we come here and we take the derivative with respect to x. So we get a minus by. Derivative of this guy up top with respect to y. And we get minus bx. And derivative of this g function with respect to x. We get dy. Now aspect to y, we have a minus c plus dx. Now we want to evaluate this around that fixed point at x star/y star, which is given here. We plug this in. So evaluate at y star. It's a over b 0. Evaluate at x star. This is just minus bc over d. Now y star, this is ab over b. And here we can get 0. So this is telling us about the linearized dynamics around a fixed point x star/y star. Now for this sort of matrix, what are the eigenvalues? It's not that the eigenvalues are 0. It's a slightly different statement. What are the eigenvalues for a matrix like this? What's that? AUDIENCE: They're imaginary. PROFESSOR: Right. They're purely imaginary. So the real part is equal to zero. So indeed, we can figure out this. Cause remember, to figure out what the eigenvalues are, you take the determinant of this vector a minus this lambda times the identity matrix. So we end up with, we want to take the determinant of-- we have a minus lambda, a minus lambda, and then minus bc over d. This is lambda squared. And then this is a plus, we have bc over d, ab over b. So we end up that the eigenvalues are equal to plus minus square root of a times ci. So they're purely imaginary eigenvalues. And what does this mean again? What can you say when you do this sort of analysis and you have purely imaginary eigenvalues? AUDIENCE: The orbits look like ellipses or something like that. PROFESSOR: Right. So the statement is that-- OK. And then we have to be careful in all of this business. So what we've done is we've taken a set of non-linear, a pair of non-linear differential equations. We've done the linear stability analysis. And we get purely imaginary. So one thing that you can say is that if you started with a linear system, and you got purely imaginary eigenvalues, that would tell you that indeed, you have these neutrally stable orbits, they go around. They can have a variety of different shapes. But given that the order that we did things in is that we took a nonlinear system, and we linearized, and then got this. What does that allow you to say? Does that prove that you actually have these neutrally stable [INAUDIBLE]? Yeah. Because this is one of those border cases, it unfortunately does not actually allow you to say that you have neutrally stable orbits. Because it turns out that the slight nonlinearities in the equations could cause problems, and cause it to go either to a stable limit cycle or to a stable spiral. The confusing thing is that in this case it is true that you have neutrally stable orbits. But that did not have to be true. And guys will get the chance to do this proof and so forth. Because there some conserved quantities. Of course, people have analyzed this thing in more depth. And it is true that you have neutrally stable orbits. But what is very hard to remember is that it didn't have to be the case just based on what we've said so far. Yeah. AUDIENCE: So you mentioned, it's not a coincidence. There is a deeper reason why? PROFESSOR: Well, coincidence. I guess what I'm saying is that you can prove that this thing does have these neutrally stable orbits. But we have not proven that. And we're not going to. Do these orbits go around clockwise or counterclockwise? Ready. Three, two, one. AUDIENCE: Counterclockwise. PROFESSOR: Counterclockwise. Right. And you can get that from thinking about what predator and prey should do. But also I've already drawn some arrows here. And that kind of helps provide some guidance. So they kind of come around like this. Now is this the only orbit that I should be drawing here? No. Right. Because it turns out that in this case there are an infinitely large number of orbits that could possibly come around. Yes. AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah. Well, in the context of our simulation, the simplest thing that you guys would probably do is you would simulate and see what comes around. But yeah. I think even in the reading, did they prove it? Different authors in different books either prove it or don't. And there are a number of different-- you'll see in the problems, you can find that there is a quantity that is conserved along the equations of motion. But the quantity's different for each of these orbits. Right. So it's really saying that as you go, there's some quantities conserved. And it still has the same value when you go around. And so that means you had to kind of come back to where you were. So there is a conserved quantity. You'll see it. Now these neutrally stable orbits are kind of funny in several ways. Well, first of all, this eigenvalue tells you about the period of the orbits when you're close to the fixed point. And you can see that it doesn't depend on everything in the model. It depends on a and c. Does that kind of make sense, maybe? Yeah. Sort of. Cause a is telling us about how rapidly we grow up here. c is talking about how rapidly we die over here. So you know, it has something that at least has units of 1 over time. It makes sense that a and c should appear here. But I would submit that it's not totally obvious that b and d should not appear at all. So neutrally stable orbits, there's no characteristic amplitude to the oscillations, nor period. And this is at least true far from the fixed point. This tells you about maybe these orbits. But then it doesn't tell you about what happens far away. And these are not the kind of oscillations that from a mathematical standpoint we like the most, which are limit cycles. And remember, a limit cycle would look like this. If it were a limit cycle, it would be that there's some orbit that is stable in the sense that if you start inside of it, then you would approach it over time. If you start outside of it, then again, you would approach it over time. And this orbit then would have a characteristic amplitude, and a characteristic period, because it's one orbit. Whereas here, this has kind of an infinite number of different orbits. What this means is that just any sort of noise-- demographic fluctuations and so forth-- would cause the system to drift over time in terms of this amplitude of period. So it's true that if you got it started in the absence of any noise, it would keep on doing that oscillation forever with that amplitude. But any random noise will cause the thing to drift. The other thing that is worth stressing in this is that if you do a time average of the predator concentration, or the prey concentration, so if you do the average or the mean, x as a function of time, or y is a function of time is indeed x star y star. That's also something that you'd have to show. Not at all obvious from this. It's not a surprise for these orbits that are out here that the time average is indeed x star /y star. But for these big orbits, they probably already didn't have to be. So this is telling us that the analysis that we were talking about-- well, what happens if you do one thing or another thing-- that's actually kind of a reasonable quantity to be trying to calculate, because that is the average or the mean number of predator/prey in this model. Yes. AUDIENCE: So in a zombie apocalypse, how do we thin down the zombie [INAUDIBLE]. PROFESSOR: Yes. That's a very important question. In the event of a zombie apocalypse, what do we have to do in order to thin out the zombie population? All right. Well, the zombie movies typically are not yet at equilibrium. But certainly at equilibrium, what do you want to do? This is not a formal recommendation, by the way. But in this model, if you want to decrease the number of the predator, what is it that we want to do? Right. Well, OK. So in this model, if you want to decrease the number of the zombies at equilibrium, you want to decrease a, which corresponds to-- AUDIENCE: So it's a is equal to 0. So the birth rate is equal to the death rate, then there aren't new zombies. PROFESSOR: Yeah. There are probably also no people. AUDIENCE: But there are. The birth rate is equal to the-- so you have a stable population. PROFESSOR: Right. So in the limit, as a goes to 0, it's true that the number of zombies goes to 0. AUDIENCE: So you just have to beat [INAUDIBLE]. PROFESSOR: Yes. This is a good situation where it's important to ask about whether your assumptions in the model are good before making public policy based on them. OK? Indeed. But can we go back just for a moment before we switch gears, to ask about this question of why it was that war might have favored these predator fish? And by favored, what he really measured, he measured the number of the frequency of these kind of predator fish at the markets. So it's somehow is maybe the ratio of those two, or so. Yeah. That's right. Yes, it could that the fisherman just-- yeah. So they could have fished in a different region. And so the explanation could be something-- AUDIENCE: Because it's not very well controlled. PROFESSOR: It's not a controlled experiment. That's why you need to have many different wars in order to average the importance of large data. AUDIENCE: That's going to be a great [INAUDIBLE]. PROFESSOR: Right. And then of course, in all this business you have to ask, well, which thing are you varying more or less, et cetera, et cetera. But let's say that to first order, the fishermen are going out there and they're just catching all the fish in an unbiased fashion. But the war is making it more difficult to fish. So not as many fishermen go out. And then what does that do to these parameters? Right. So it might increase a, because the prey are not getting caught as much. And it might do something also to the predator. What? So the predator fish are also being caught, cause they're showing up at the market. Right. So which other parameters does it change? So what's that? AUDIENCE: [INAUDIBLE] PROFESSOR: Right. So lets be clear. It's a war makes a go down. All right. And what other parameter might change? AUDIENCE: Fewer fishermen. PROFESSOR: Oh. I'm sorry. I'm sorry. You're right. You're right. I was starting to think about the next one. OK. So a goes up. And what also happens? And c goes down. And this is just because a is a growth rate. c is a death rate. So it's not that the fishermen are preferentially changing what they're doing between the prey fish and the predator fish, but it's just that these are defined different ways. Right? And in particular, what this means is that in times of war, Voltaire's argument to his future son-in-law was that what you expect is that the predator population should go up. And the prey population should go down. So the ratio of them should certainly shift in favor of the predator. Or maybe it's just that the fishermen didn't go into the deep regions of water or something, where predator fish like to hang out. But it's striking at least that a simple model like this can actually provide some insight. Any questions about where we are before we modify the model in some way? Now, the problem with this model is that it's making lots of assumptions that we very much think are not true. So we have a few of the assumptions up there. Which ones are you guys least happy about among those assumptions? Yeah. The second one, the prey grows exponentially, i.e., without bound. That's not physical. We should fix that. So let's go ahead and do that. All right. So real prey populations-- and all populations-- populations saturate. Don't go to infinity, right. We can just make a quick little fix to our model to capture this, right? X dot. So now it's going to be this ax. But what we can do is we can add a logistic term here. And we can leave everything else constant. So we just add this logistic row term. So now in the absence of a predator, what will happen is that the prey population saturates at some point, at some value k. Right? And k could be rather large, if you like. And I think this is the kind of situation where you say, oh well, that could be just a really modest change. Because it's primarily telling us about what happens kind of in the absence of the predator. But really, the dynamics around here could be unchanged. But what's striking is that because of this neutral stability of the model, any little change can lead to a qualitative change in the outcome at equilibrium. And in particular, adding a carrying capacity causes the sustained oscillations to disappear. Instead you get damped oscillations. So in this case, xy, you still have this fixed point. But what happens is that you get-- it orbits the look. Kind of like this. Yes. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah OK. So that depends upon [INAUDIBLE]. And indeed, as k goes to infinity, the timescale of this decay goes to infinity as well. AUDIENCE: [INAUDIBLE] clean scaling. PROFESSOR: A clean scaling. AUDIENCE: But I mean just [INAUDIBLE]. PROFESSOR: I mean, I don't have nothing intelligent to say. But what I can say is that if you put in a k at values that are close to this fixed point, then the oscillations really disappear. It's really only as k goes way, way out. And I maybe won't do this calculation. But you can do it quickly. It's the kind of thing you can do in five minutes on an exam or so, right? But then the question you might ask me, well what can you do in order to get these oscillations that we really like? These sustained limit cycle oscillations? And there are two ways you can think about this. One is by a kind of taking this sort of model and adding features and playing with it and so forth. Or you can ask a mathematician who kind of derives everything in kind of totality. And then you can see it. I'll say that one thing you can do on a concrete basis to change the outcome of this is you can add something that is kind of like some saturation effect at the level of the interaction between predator and prey. So instead of scaling purely as x times y, if instead there some sort of Michaelis-Menten type behavior as a function of the prey, then this model is actually converted to one that has a stable limit cycle. And in this case, can you guys remember how you show that something has a stable limit cycle? Yes. AUDIENCE: [INAUDIBLE] use the Poincare-- PROFESSOR: That's right. There's this Poincare-Bendixson theorem, which tells us that if we can draw some box out here where the trajectories are all kind of coming in, indeed, here the trajectories are kind of coming in. Now in this case the trajectories have to come in everywhere. But this is typically true in any of these reasonable systems. Because they should, if there's just not that much food, they should be coming in. Now in this case, there's a single fixed point. So then the question of whether there's a limit cycle is equivalent to the question of the stability of this interior fixed point. It's a stable fixed point, then all of these trajectories will spiral in. However, if it's an unstable fixed point, then you have to have a limit cycle oscillation somewhere in between. Yes. AUDIENCE: It could be that it's a stable midpoint. And then you have two limit cycles. Could that be? PROFESSOR: Right. So the question is whether there can be a stable limit cycle and two limit cycles. I think that that requires more fixed points. Because let's say you had a limit cycle here. That this is saying that the trajectories come in. But it's also saying the trajectories are coming out. I think that you have to have a fixed point inside here in order to have these trajectories going-- AUDIENCE: So what I'm saying [INAUDIBLE] two circles, and way outside, all the trajectory is [INAUDIBLE]. But between the two circles, the trajectories go from the outer circle to the inner circle. And then from the inner circle in [INAUDIBLE]. PROFESSOR: Are the circles encircling each other? AUDIENCE: Yeah. PROFESSOR: OK. So here's one circle. Here's another circle. And the trajectories are coming into this one and what are they doing? AUDIENCE: So each of these limit cycles are only semi-stable. There are some-- PROFESSOR: Yeah. All right. All right. So you're talking about these limit cycles where the trajectories come to it from one side, but then leave from the other side. I'd have to think about this more. Everything that I'm talking about are the limit cycles that are stable from both sides. And I think that in the presence of any amount of noise, those are the only ones that we care about. Because you can write these models where you have orbits that are stable from one side but not the other. But then they don't matter in any real system that is subject to noise. Because it's not a stable orbit in that case, right? AUDIENCE: Then we can make the outer circle stable from both sides, and the inner circle [INAUDIBLE] from both sides, and we could get it [INAUDIBLE]. PROFESSOR: OK So it could go like this. So you're saying that this outer orbit is now stable from both sides. And then this one is unstable from here. And oh, OK. Now you're saying this thing is unstable from-- so now it's stable. AUDIENCE: So we can have-- PROFESSOR: Yes. OK. So I can certainly draw the trajectories, as you pointed out. And it works. Yeah. I don't know if this is a loophole in the wording of the theorem, or if this is just not possible. AUDIENCE: [INAUDIBLE] There's like one way in which it applies. And then the other way you kind of usually think that it applies, but then you're wrong because there are these counter examples. PROFESSOR: I see. Well, how about this. I will look it up. And then I will tell you what I think the answer is. I'm certainly not going to be able to derive it on the spot. But You're right, though. I can draw the orbits and they do it. I don't know what that means. AUDIENCE: So I think the theorem is if on this large circle on the outside everything is coming in, then if your fixed point is unstable, then you must have at least one stable [INAUDIBLE] outside. Or you must have at least one that is [INAUDIBLE] or something like that. PROFESSOR: OK. But you're saying that it-- AUDIENCE: But if you flip it so the fixed point is stable, then you could have-- PROFESSOR: I see. That may be right. I can't remember which direction. But I'll look it up. But if you are curious about the situations in which these predator/prey functions will have limit cycles, you can look up Kolmogorov's conditions. So he basically has these four conditions in which you will get a stable limit cycle in the predator/prey populations. And maybe I will. And they're surprisingly simple. They were in the reading. So you can look at them. But they're basically just these questions of the derivatives. And they're derivatives of the per capita growth rates. So you take those functions, you divide by x. So it's x dot over x. And then it's some f, y dot over y, sub g. And the derivatives of those functions withe respect to x and y have to be something, at least in some limits. And then you get limit cycles. But I want to talk about the experiments. Because I think that they were quite pretty. So this is a paper by Yoshida, et al. This is Nature 2003. And the title was "Rapid Evolution Drives Ecological Dynamics in a Predator/ Prey System." Could somebody kind of say what they think is the big point of this paper? Do you guys like the paper, dislike the paper? Too long? Was it three pages? Although it's really only-- I mean, one of the pages is figures. And one of the pages is essentially methods. So it's really basically a one-page paper, so it's not such heavy lifting, maybe. All right. Maybe I'll be concrete. What features of their predator/prey oscillations were different from a normal predator/prey oscillation? What were the features they were trying to explain? AUDIENCE: There's a longer phase life between the maximum. PROFESSOR: And this is an interesting history, actually, that these are authors-- they took what I think are basically like samples from the Great Lakes, or something like that. So they took a predator and prey. So it's a rotifer algal system. So the rotifer eats algae. They had previously published a paper just a few years before, where it was called something like "Crossing the Hopf Bifurcation in a Predator/Prey System." And so what they did is they had a chemostat where there was some constant rate of dilution in their chemostat. And then they measured the dynamics between the predator and the prey. And they showed that depending upon, for example, the nutrients or this dilution rate in their chemostat, they could go from a situation where you have a stable coexistence-- so a stable fixed point between the predator/prey-- but at increasing dilution rate, they started getting these oscillations. So they went from stable to oscillations. And then at higher dilution rates, they got collapse of this predator/prey system. So this is crossing the so-called Hopf bifurcation, where you go from stable to unstable. So they could actually see that they could use a simple model to try to get some insight into why it is that a predator/prey system might oscillate or it might not oscillate. And so that was great. But there were some features in their data that they didn't understand. So the features were that the oscillations had a really long period. Long period oscillations. And in all this data, it's a little bit hard to get the exact phase lag, but in many of the cases they saw that the predator and prey were not at this 90 degrees out of phase like you would expect. And in particular, for the predator/prey, given that you have x and y, and it's going around some circle, the idea is that first the prey population peaks. And then kind of 90 degrees later phase, you get a peak of the predator. And you can imagine that in any predator/prey model that you write down, not just Lotka-Volterra, but if you write down a differential equation with some x and some y, it's hard to get something where it looks much different. Of course, you can imagine maybe funny orbits or something like that. But generically you really expect this 90 degree phase lag of the predator relative to the prey. Whereas in the experiments, they often saw 180 degree lag, lack of predator relative to the prey. Yes. AUDIENCE: So when you say [INAUDIBLE] theory of oscillation, what does that mean? What's the time scale? PROFESSOR: Yeah. AUDIENCE: What would make this suprisingly [INAUDIBLE]? PROFESSOR: Right. Yeah. OK. So this is really in the context of-- I mean, they wrote down a model where they did basically have something like this, where they said there's some-- AUDIENCE: [INAUDIBLE] PROFESSOR: That's right. And in particular the predator, I think, was not really dying on it's own. Except for the dilution. So it's really the dilution rate from the chemostat gave them c. And they could see that indeed, the rotifer wasn't dying much. And then a is telling you about what happens if you don't have the rotifer, you don't have the predator. And then at that dilution rate, you ask what the division rate is. And that's essentially the growth rate of that prey population minus the dilution rate of the chemostat. I'd say that they had these parameters pretty well. Yes. And what they saw experimentally was that it took much longer. I mean, the period was 5 times, 10 times longer than what they would have expected based on the measured parameters for the predator and the prey. And of course, that was annoying for them, because they're actually doing the experiments. So they had to do these chemostat experiments that lasted six months or something like that, in order to see a couple oscillations. And so this longer period wasn't just a mathematical-- I'm sure that they were disappointed to see these really long time-scale oscillations, because it was something that made it very difficult for them to do the measurements. But this is what they saw experimentally. And they didn't know what was causing it. But they later then did more modeling where they asked, which of the assumptions in our original model might not be true. And the nature of these things is that there are an infinite number of things are true that are not incorporated in the model. And this makes it quite challenging to isolate what the effects might be. But at the very least what they can do is they can go in. And they have some sense of their system. And they can make guesses of what might have been the primary things missing. And then they kind of went through the models and they asked, well, if we kind of change this assumption or that assumption, what does it do in terms of the oscillations. And then what they found in a modeling paper-- so this is their previous experimental paper before this. What they found in their models was that prey evolution was kind of the one thing that if they allowed prey evolution, i.e. If they allowed different types of prey in their prey population that might be oscillating independently, then they could get both of these effects. So from the models they concluded that prey evolution could give this, could maybe explain those two things. These are the two experimental features. Yeah. AUDIENCE: It might be that you have [INAUDIBLE] symantic point. If you have two types of species, pre-species-- [INAUDIBLE] they have these characteristics and they're not evolving. They're just fixed. PROFESSOR: Yes. AUDIENCE: You would still be able to-- PROFESSOR: Exactly. Right. This is a major-- AUDIENCE: [INAUDIBLE] PROFESSOR: Yes. You can argue about what you want to call evolution and not revolution. Over these time scales what they are not looking at is the de novo kind of emergence of new mutants spreading in the population. What they're looking at is variations in say, different types of prey. AUDIENCE: So is that what we would call evolution? PROFESSOR: Well, OK. This is a major question. I think different people call evolution different things. And in particular there's a distinction between the ecologists and the evolutionary biologists, or evolutionary microbiologists, or whatnot. And this is something that I certainly encounter. For the experimental microbial evolution guys, I'd say that for them, if you use the word "evolution," what they really want to see, or what they're expecting to see is kind of new mutants arising in the population, spreading, fixing, and that changes the character of the population. Whereas from that standpoint of ecologists-- and indeed, if you look up the definition of evolution, it's some change in allele frequency over time. And this could certainly be a change allele frequency over time in the sense that you could imagine this arising just from, for example, there could be a point mutation in some gene that leads it to do something or another. And in this paper they didn't identify exactly what was going on. I'll tell you a little bit about a later paper where they get a better sense of it. But at least you could imagine it being the case that there's just a point mutation, some gene that has some fitness defect in the absence of the predator, but then allows the algae to avoid the predator in some way. AUDIENCE: [INAUDIBLE] PROFESSOR: That's right. That's right. As far as the model goes, it could be that the prey are really-- one species was just a point mutation. Then you'd say, oh yeah. That's kind of evolution. Or it could be two different prey populations and so forth. Right? And depending upon what information you have access to, you would either be aware of these things or not. In this case, they just look under the microscope and say, oh yeah, this looks like an algae, or whatnot. And maybe they know that it's the same species. But they could be different in lots of different ways. And I think this discussion is highlighting that a lot of the same effects that you see in ecology also you see in evolution and vice versa. And depending upon your focus, you might want to put it more in the bin of evolution or in terms of population dynamics. I think that's a matter of taste at some point, actually. Yeah. Because they say rapid evolution. And I think-- AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, right. So I think that different people will have different takes on this. I think, once again, this is really very much from the standpoint of an ecologist, this is rapid evolution. Because every cycle, the allele frequency is changing. And that's what's maybe leading to this. But the thing is, it's very easy for you to read certainly the title, and come away thinking it's something different from what it was. I agree. But one of the things I like about this whole line of research that took place over the course of say, 15 years, was that they did these nice measurements. They're guided by models. They saw some things that they were expecting, some things they couldn't explain. And then they went and did more modeling. And then they could explain it. And that guided this experiment. Because before what they did is they just took kind of the sludge, or whatever, from the lake. So there were many different types of algae and of the rotifer, for that matter. And then they looked at the predator/prey oscillations between those. And then in this paper what they did is they took individual prey populations that came from a single prey. So they were kind of isogenic. Of course eventually they would evolve, and blah, blah, blah. But in this case they took individual isolates from the prey. And then what they saw is that these two features went away. The oscillation period was more what they were expecting. And they got the 90 degree phase lag. So it's was really a case of these models led to the experiments where they took kind of clonal prey. And then they recovered their classical predictions. And indeed, they've recently done some other measurements that I think were quite nice. They had an ecology letters paper just a couple years ago where what they did is they took two different algae that had different types embodying these trade-offs that we were talking about. In particular, one of the algae can divide rapidly. But it kind of exists as singles. Another algae is kind of clumpy. So this algal type was able to divide more rapidly than this algal type. But this type was harder to eat just because it was clumpy. OK? It's one of things that once you hear it, you say, oh yeah. Sure. But then in the abstract, when you read this 2003 paper, you think, oh, I can't imagine what kind of behavior could possibly help the algae avoid these rotifers. But then in this ecology letters paper, what they did is they actually tracked population densities of this type, this type, and of the rotifer over time. So then they could see all three sub-populations oscillating. So then you can really kind of see how this evolution, if you want, or you could just say it's ecology. Because it could be different species for all we care. But in any case, it's certainly prey heterogeneity anyway you look at it. Whether it's evolution or ecology. It's heterogeneity in the prey population that leads to these very qualitatively different population oscillations. Any other questions on that before we-- And so just for the last few minutes I'll say something about this question of noise-induced oscillations. This is going to be something you're going to be playing with over the next week. So you'll be experts eventually. And the discussion in the homework is really guided by this paper by McKane and Newman. All right. So McKane. Newman. McKane is a professor at Manchester. And what they showed is that if you take this model here, did this model have limit cycle oscillations, sustained oscillations? No. So this model where we include the carrying capacity of the prey, this does not have sustained oscillations. So you take a model like this that just has a stable spiral. Now this is a model in the context of-- it's a differential equation model. Now the question is, how do you incorporate noise? Just the fact that individuals give birth, they die at random times. The first order way that we often think to do this is we just add some noise on to the differential equations. So what we do is, you might say, oh well, you could add some a to x, and some a to y. Maybe this noise should be proportional to x or something. So you could think about the strength of this might be some pre-factor. So you could just take kind of a [INAUDIBLE] approach where you add noise onto the differential equations. And then what you get is a noisy kind of path to that fixed point. The perhaps surprising thing is that instead of adding noise to a differential equation, if instead you start with the individual based approach, and you take kind of a master equation approach where you say they're individuals and they're doing something. Individual predators are eating individual prey, et cetera. So if instead you take a master equation, just like what we did for the chemical equations modeling the cells and so forth that we talked about at the beginning of the class, if you formulate this predator/prey system as an underlying set of kind of individual based interactions, then what you actually find is that you get surprisingly large sustained oscillations. So you really end up with a situation where this thing comes. And it's noisy, of course. But it kind of comes around here in some way that looks like this. And you'll see this in your simulations next week where-- and I don't know if you can see this at all. But this is some plots of predator and prey at reasonably large numbers, where you can actually have 1,000 predator or prey. Somehow you can get some sort of resonant enhancement of these oscillations. And kind of what's going on is that the demographic fluctuations of the demographic noise excites the system at all frequencies. But then there is a characteristic frequency given basically by this frequency of the inherent oscillation there that is somehow amplified. So you end up with a situation where you get these oscillations. And they're noisy and whatnot. But it's really because it's this particular frequency that was amplified. And just because it takes a long time for those oscillations to go away. So you can imagine that the amplitude of the oscillation falls off as kind of a root end scaling, because it's kind of a demographic type noise. So it is true that the relative amplitude of these oscillations, it gets larger as you to smaller population sizes. But it still ends up being a surprisingly large effect. And you'll see how this plays out in this model that's basically guided by this paper. And this was a PRL in 2005. Incidentally, other people have since studied how these sorts of ideas can result in noise-induced pattern formation, as well. All right, with that, I will let you guys go. Have a good Thanksgiving. I'll see you guys on Tuesday.
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https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/8.04-spring-2013.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Hi everyone. Spring has regressed. So, we have-- we're going to have a guest at the end of lecture today, which should kind of entertaining. Just as a warning, if you see someone come in. So questions, before we get started? No questions about anything? At all? Math? Nothing? Yeah? AUDIENCE: Can you explain the physical significance of the crystal momentum? PROFESSOR: Yeah. OK. Let me go over that. That's a good question. So the question is what again is the significance of the crystal momentum? So let me answer that in a slightly backward way. So this is a form of the explanation I haven't given you. It's going to be a slightly different one. Let's step back and think about the momentum, and ask what the momentum is. Now you guys showed on a problem set, the following fact. That if you have a wave function, sine of x, such that, the expectation value in the state SI of x in the state SI is equal to x naught, and the expectation value of p in the state SI is p naught. Hat, hat. Then if you want to change the momentum, increase momentum by h bar k, the way to do that is to take SI and build a new wave function, SI tilda, is equal to e to the i, k x, SI of x. And then the expectation value of x is the same, SI tilda, still equal to x naught, because this phase goes away from the two complex, from the wave function is complex content we give the inner product. But the expectation value, the momentum is shifted in state SI tilda, is shifted by each h bar k, p naught plus h bar k. So all the intuition you have about momentum, you can translate into intuition about the spatial variation of the phase of the wave function. Yeah? AUDIENCE: [INAUDIBLE] PROFESSOR: OK, good. OK, good, So we have a sneaky [INAUDIBLE]. So, the information about the momentum can be encoded in these spatial variation of the phase of the wave function. So another way to answer the question of what is momentum, apart from it's the thing that-- so what are ways to answer the question, what is momentum, you could ask well what is momentum? It's the thing that commutes with p or with x by i h bar. That's one way to answer it. Another way to answer is to say that translations by l can be expressed in terms of momentum as e to the minus i upon h bar p l. So these are both ways of describing what the momentum is. But another way of talking about the momentum is the momentum p governance the spatial variation, the x dependence of the phase of the wave function. So these are always talking about what the momentum is. So now let's turn this around, and let's ask about the crystal momentum. Oh, and one last thing, a last defining property of the momentum, a central property from the Schrodinger equation is at the time variation d dt of p is equal to the expectation value of minus d the potential of x d x. Also known as the force. So this is the Ehrenfest Theorem Statement that the classical equation of motion, p dot, is equal to the minus d v d x is equal to the force, Ehrenfest's Theorem tells us that the classical equations of motion are realized as expectation values. And equivantly, if there's no potential, the potential is constant, this tells us that the momentum expectation value is time independent. Right? A familiar fact. So these are all true lovely and things about the momentum. So let's turn all these facts around into the crystal momentum. So let's talk about crystal momentum. Which was the question, what is the crystal momentum? So the crystal momentum is defined from beginning, from the following property. If we have a potential v of x, which is invariant under shifting, by one lattice spacing, by some l, v of x, then this tells us that the energy operator is invariant if we shift by l. If we translate by l equals zero. And from this fact, we deduced via block or a la block, that the wave functions are really the energy eigenfunctions, can be written in the form e cubed is equal to e to the i q x, u of x, where u, we're going to take to be a periodic function. So what is this parameter q doing? Q is governing the spatial variation of the phase of the wave function. Cool? So in precisely this sense, the momentum difference is space of the wave function. Here, in the case of a periodic potential, the crystal momentum q is governing the spatial variation of the phase of the wave function. So q is the thing the governs the phase as a function of x. Well what about-- another fact about the crystal momentum which you show in your problems set, is that if you impose an external force d q d t, and really d h bar q. d t is equal to-- d dt of the expectation value of h bar q, is equal to the expectation value of the force. I'll just write-- OK? So again, this is a quantity, and this was assuming that we had a sharply peaked wave packet. So this is for a wave packet sharply peaked at q naught. And so let me just write this as h bar q naught. So the central value of your wave packet-- so this is what you've shown on the problem set that the central value of your wave packet, the peak of your wave packet varies in time according to the external force. And so in particular, if the force is zero, we turn no external driving force, your wave packet maintains its crystal momentum. It's time independent. So the crystal momentum is something that time independent, unless an external force is applied, just like the momentum. And it's something that governs the phase of the wave function just like the momentum. However, it's different in a crucial way. It is not the eigenvalue p on five sub e q is not equal to a constant p naught times 5 sub e q. Because when we take-- when we active p or we active the derivative, you pick up a term from here, which gives us a constant, but we also have this overall periodic piece. And its spatial variation is generically non-zero. And if the potential is nontrivial, it's always non constant. So when the momentum operator hits this guy, it will generically not give us zero. It'll get two terms and we will not get an eigenvalue equation. So q is not the eigenvalue h bar q is not the eigenvalue of p. And what's the last important property of q that's different from the momentum? It comes from the commutator, which tells us that the thing that's conserved is the expectation value of p l is really the precise statement. And in particular, what this tells us is that the eigenfunction, or the eigenvalue of our wave function, under translations by l, is a quantity that can be determined simultaneously with knowing the energy. However, the eigenvalue of t sub l, on this state, is equal to e to the i q l. Which means that q is only defined for determining the eigenvalue up to 2 pi over l. If you have q, which is 0, and you increase it to pi over l, that value, pi over l, is effectively the same as the value minus pi over l. Because at least they're the same eigenvalue. But that's really strange because that means that q itself, it's not strictly conserved. It's conserved mod 2 pi over l. When you have momentum conservation, momentum is strictly conserved if there's no force. And even if there is a force, it's increasing control by the force as you turn on the force, it just constantly increases. For the crystal momentum, that's not the case. You turn on a force, it increases according to the conservation law. But it's not increasing constantly. It's periodic. It's periodically defined. So it increases then it ends up at a smaller value. It increases and ends up at a smaller value. OK? So it carries many of the same properties. It governs the phase. It's time independent unless there's an external force applied. It's the eigenvalue. Controls the eigenvalue of an operator that commutes with the energy when you have a periodic potential, in the same way that the momentum commutes with the energy when you have no external force, when you have a constant potential. Does that help? Good. OK. So developing an intuition for the crystal momentum, I think, is best done by just playing with examples. And you'll do that more in the course on solids, which I encourage you all to take. Because it's really beautiful stuff. But for our purposes, this is going to be the full set of ideas we'll need for 8.04. Yeah? AUDIENCE: [INAUDIBLE] PROFESSOR: Ah. So good. So thank you. So this involves a slight subtlety, which I've been glossing over in the entire story here. Which of the following. So, is u of x a real function? Well, so when we started out asking what are the eigenfunctions of the transit by l operator, all we showed was that, and I'm going to do this on a separate board just to make it clearer. Tell me if this turns off, because it kept bumping. OK. So when we started with translate by l, and we constructed it's eigenfunctions, we said that translate by l q Phi sub cubed is equal to some phase, and this is unitary, so we're talking about they must be an actual phase in the i alpha of Phi sub q of x. And let's just suppose that this is true. Then this tells us that Phi sub q times e to the minus i q l equals u. So, I'm just going to use this to define a new function, u sub q. Or just u. I'll use sub q. Fine. of x. So this defines a new function, sub q. I take an eigenfunction, I multiply it by some phase. Sorry, minus i q x. If we choose q l to be equal to alpha, then acting on u sub q, by translate by l, on u sub q, of x, is equal to-- well, if we act on Phi sub q with translate by l, what happens to Phi sub q we pick up a phase e d i alpha. What happens to e to the minus i q x? x goes to x plus l. We pick up a phase e to the minus i q l. So if q l is equal to alpha, those two phases cancel, and we just get u back. u sub q of x. But translate by l, if u sub q, by definition, is equal to u sub q of x plus l. So we've determined is that if we take q l is equal to alpha, then Phi sub q if eigenvalue label by its eigenvalue, q, can be written in the form e to the i q x u sub q of x, where this is periodic. Everybody agree with that? OK. So that's step one. Step two is to say well look, since the eigenvalue of this guy, under t sub l, e d i alpha is equal to e to the i q l. Since this is periodic under shifts of q, by 2 pi upon l, I can just choose to define q up to 2 pi over l. So 2 q, I will take to be equivalent to q plus 2 pi over l. And the reason I'm going to do that is because it gives the same eigenvalue, and if I want to label things by eigenvalues, it's sort of redundant to give multiple values to the same eigenvalue. Now there's a subtlety, here though. And this little thing here is this. Suppose we have a free particle. Does a free particle respect translation by l? So if we have a free particle, the potential is zero. That constant function is also periodic under shifts by l. Right? Because it's just zero. So it's stupidly periodic, but it's periodic nonetheless. So now I'm going to ask the following question. What are the common eigenfunctions of the energy and translate by l for the free particle? We did this last time. So the common eigenfunctions of translate by l and the energy are the wave functions Phi sub q, comma e, are equal to e to the i q x times some function u of x, on general grounds. But we know what these eigenfunctions are. They're just e to the i k x. Where k squared upon 2 m is e. [INAUDIBLE] So we know that these are the correct eigenfunctions, but we're writing them in the form e v i q x u. Now you say that's fine. There's nothing wrong with this. We just say u is constant and q is equal to k. These functions are of this form, but they're of this form with e v i q x being e d i k x and u of x being constant. Right? There's nothing wrong with that. Everyone agree? Perfectly consistent. However, I thought we said that q is periodic by 2 pi? If q is periodic by 2 pi, then that would seem to imply that k is periodic by 2 pi, and we know that's not true because any k is allowed for a free particle. So if we want to think about q is periodic by 2 pi upon l, then we cannot require that u is real. Because it must be the phase that makes this up. It must be, so I can always write this as e to the i q x where q is less than 2 pi upon l. I'm sorry, where q is between 2 pi or pi upon l and minus pi upon l. So that it's defined only after this periodically thing. But times some additional phase, e to the i k minus q x This is trivially equal to e to the i k x. But now u is not a real function. On the other hand, if we hadn't imposed the requirement that q is periodic, we wouldn't have needed to make u real. We could just taken q to be equal to k, for any value k, and then u would be constant. u would be real. So this is important for answering the excellent question that our fearless restation instructor provoked me to answer. Which is that so what-- we'll come back to the question in just a second. But what I want to emphasize this, that if we're going to take q to be not periodic, Sorry. If we're going to take q to be defined only up to shifts by 2 pi over l, it's important that we allow u to be not real. It must be able to be an overall phase. But if we want u to be always real, we can do that. We just can't impose this periodicity. Different values of q mean different wave functions. And this is really what's going on when you see those plots, sometimes you see the plots as parabolas. The bands are represented by parabolas with wiggles, and sometimes they're folded up. And that's the difference. The difference is that when you fold them up, you're imposing this periodicity and you're labeling the eigenfunctions by q, and the overall amount of the number effectively of k phases that you're subtracting off. Yeah? AUDIENCE: So is this an arbitrary choice? [INAUDIBLE] PROFESSOR: Yeah. I mean, how to say? It's exactly akin to a choice of variables. In describing the position of this particle, should we use Cartesian coordinates, or should we use Spherical coordinates? Well it can't possibly matter. And so you'd better make sure in any description of your system, that changing your coordinates doesn't change your results. And here, that's exactly what's going on. Do we want to define our variable to be periodic by 2 pi upon l? Well, OK then. But u can't be real. Or we could take q to be not periodic by 2 pi l and impose that u is real. It's just a choice of variables. But it can't possibly give different answers. The point is, this is a subtle little distinction it we gloss over, and is glossed over into my knowledge every book on intro to quantum mechanics that even covers periodic potentials. It can be very confusing. Anyway, the reason that I had to go through all this, is that in order to answer the very, very good question professor Evans posed, I'm going to need to deal with this fact. So for the moment, let me deal with-- let's work with u real. And q, q an unconstrained, real number. OK. So not periodic. Are we cool with that for the moment? So if we do that, then notice the falling nice property of our wave function. Our wave function, Phi sub q, is equal e to the i q x times u of q, or u of x. Which is real. So when we can construct the current-- remember that j boils down to the imaginary part, h bar over 2 m i. Well, h bar over m times the imaginary part of SI complex conjugate derivative, with respect to x, which is the current, in the x direction of SI. And we need this to be imaginary, or we will get no current. You show this in a problem set, if you have a pure, real wave function, for example. A single real exponential, that's decaying, as on the wrong side of a barrier. Then you get no current. Nothing flows. And that make sense. It's exponentially decaying. Nothing gets across. So we need the wave function to be real. So if q were zero we would get zero. And what you can immediately do from this, compute from this, is that while the derivative, if the derivative doesn't hit e to the i q x, if it hits u, than the phase e to the i q x cancels. And so the contribution from that term vanishes. So the only term that's going to contribute in here is when the derivative hits the e to the i q x. But then this is going to be equal to h bar q. And we want the imaginary parts, that's going to be e to the I over m. And then we're left with u squared of x. So this is the current, but we have to do it-- we had take advantage in order for this to be sort of clean, we had to take advantage of u being real. Everybody cool with that? Now there's one last twist on this, which is that if I have k-- if I have q. So this is a side note. Going back up here, to this logic. If I have q, and I want, I can always write it as some q naught plus n pi over l. And so now what I want to do is I want to take sort of a hybrid of these two pictures. And I want to say Phi sub q is going to be equal to e to the i q naught x. Where this is the value that's periodic by 2 pi. e to the I n pi over l x u. And so now really what's going to happen, what I'm doing here is I'm labeling q, not by a single number. I'm labeling my wave function not by single number q, but by q naught and an integer n. Comma n. So q naught and n. So now q naught is periodic. It's defined up to shifts by 2 pi. n is an additional integer, and what it's telling you is how many times did you have to shift back to get into that fundamental zone between pi and minus pi. And this fits nicely into this story, because now all we're going to get here is q, which is q naught plus n pi. So the current depends on both the part defined mod 2 pi over l, and the integer, which tells you how many factors of 2 pi over l did you have to subtract off to get into that fundamental domain. So let's think back to our band structure. So what is this n quantity? Let's think back to our band structure. In our band structure, we had something that looks like this. And here's the value of q. But am I plotting q? No. I'm plotting here q naught. I'm plotting the part that's periodically defined up to 2 pi over l. So this is pi over l. This is minus 2 pi over l. Or minus pi over l. OK. And what we see is that there isn't a single energy. Because this is the energy the vertical direction for the band pictures. There isn't a single energy for a given value of q. In fact, the set of energy eigenvalue-- or the set of allowed states or energy eigenvalues for an allowed value of q would say this particular value of q naught, how many of them are there. Well, there are as many as there are integers. One, two, three, four, count. So to specify a state, I don't just have to specify q NAUGHT, I also have to specify N. Which one of these guys I'm hitting. And when you unfold this into the parabola picture, remember where these came from. These came from these curves. Came from shifting over. And the higher up you go, the more you had to shift over. And that's exactly the integer piece in n pi over l. And so we can write the current now, in terms of h bar q naught upon m, u squared-- I'm sorry. h bar q naught upon m plus n pi h bar upon m u squared of x. So we get a contribution from the crystal momentum and from which we're in. OK? So sort of an elaborate story to answer the phase question. Yeah? AUDIENCE: [INAUDIBLE] PROFESSOR: Good. So here we had SI-- so SI-- I'm sorry. I should have done this for Phi. But I meant this wave function, right. This is Phi, this is Phi q. So from here we're going to get the imaginary part. So we get the imaginary part of this wave function which is u to the minus i q x u of x derivative of e to the i q x u of x. Now the term that contributes is when the derivative hits the e to the i q. x pulls down a factor of i q, and the two phases cancel from these guys, leaving us with a u of x here, and a u of x here. AUDIENCE: [INAUDIBLE] PROFESSOR: Oh sorry. This is a potential. Good. That's the point. So this is the potential. So in this statement that what we have this translation by x. So this is just some function. It has nothing to the potential. It's defined in terms of the wave function. The eigenfunction of translate by l. So the logic here goes, if we know we have a function of translate by l, then I construct a new function u. Nothing to do with the potential, just a new function. Which is e to the minus i q x times it. You can't stop me. You hand me a function, I will hand you a different function. And then we pick q felicitously, to show that u is periodic. So u is just some periodic function which is contained which is defined from the wave function. From the energy. From eigenfunction of t l. Did that answer your question? OK. So here, it just came from the fact that u is Phi is the e to the i q x u, x and then a factor of u for each of these. Other questions. Yeah. AUDIENCE: [INAUDIBLE] PROFESSOR: So this picture, when it's unfolded, first off, you know what it is for a free particle. So we want the energy as a function of q. So what is it for a free particle? Parabola. Yeah, exactly. And now let's add in-- let's make this a function of q, not q naught, but so here's pi over l. Here's 2 pi over l. Here's 3 pi over l. And I need to do this carefully, because it's incredibly difficult to get the straight. OK. My artistic skills are not exactly the thing of legend. OK. So here's the parabola that would have been, if we had not turned on a periodic potential. As we turn on the periodic potential, we know that the energies change. And so in the first band it's easy to see, because for minus pi over l, it's pi over l. We don't have to do anything. So it look exactly the same as the lowest band over here. So in particular-- OK? So what about this second band? Well what I want to know what's the allowed, the other allowed energy that's say, plus pi over l. Plus pi over l, it's going to be something greater than this value. But plus pi over l, we already know the answer from that diagram, because plus pi over l is the same as minus pi over l, so what's the value over here? Well, the value over there for the second band is slightly above, and then it increases and decreases. So slightly above, and then it increases. Shift by pi over l. Whoops. Did I shift by pi over l for this guy? That's one, two. Yes. I did. Good. And it goes the other way. So just noting that it goes away from the top. I have a hard time drawing these things. So for every value of q, there's an allowed energy. But it's different than it would have been for the free particle. And then we do the same thing for the next state. And it looks like this. So now imagine what happens when we take this, and we it over one two. What we get is a band the looks-- that should look like this. That's what the second band should look like. And indeed, when we put it in the fundamental domain, this is what we get. This is what the first band and the second band together look like. And then the third band, we'll move this over once, and then twice, it's going to look like this. And this guy, move it over once, twice, looks like whoops. Yeah? AUDIENCE: If we wanted to plot u with respect to k instead, would that just be a parabola dotted line? If so, why do we not have really-- PROFESSOR: If we just wanted-- sorry. Say it again? AUDIENCE: E as a function of k instead of q. PROFESSOR: Oh. Yeah. E as a function of k is always going to look like that. But k is not a well-- so what is k? K is just defined as h bar squared, k squared upon 2 m is equal to e. So this doesn't tell you anything. Right. Because any allowed k. Sure any allowed k is some valid value of e. But this didn't tell you which values of e are allowed. Only some values of e are allowed, right? There are no values of e-- there are no energy eigenstates with energy in between here and here, right? And so that tells you they're no allowed k's because k is just defined, it's just completely defined by e. So this doesn't tell you anything about which states you're at. It just that given an e, there's some quantity that could define k. This is a definition of k, in terms of e. What this diagram is telling you is which e's are allowed. AUDIENCE: [INAUDIBLE] PROFESSOR: Yes. Yes. There should be. Let's see. What's AUDIENCE: [INAUDIBLE] PROFESSOR: Oh, here. Yes. Yes, you're absolutely right. Over out. Thank you. Excellent. That's exactly right. Yeah. Oh man. I made a dimensional mistake. Thank you. Jesus. OK. Good. Yeah. AUDIENCE: Could you like re-explain how imperfections and a lattice leads to actual conduction? PROFESSOR: Yeah. I'm going to do that. So that's an excellent question. The question is could you explain again how imperfections and a lattice leads to actual conduction. As we talked about last time, when you have a perfect lattice, there is actually no current flowing in response to an applied electromagnetic field. If you put on a capacitor, played across your perfect lattice, you don't get any current. So the particle, the charged particle in your lattice, just oscillates back and forth in a block oscillation, running up the band, and down the band, and up the band, and down the band. So, let me slightly change your question, and turn it into two other questions. The first question is given that that's obviously not what happens in real materials, why don't we just give up on quantum mechanics and say it totally failed? And so this is a totally reasonable question, and I want to emphasize something important to you. Which is the following. That model led to a prediction, which is that if you put a capacitor plate across a perfect crystal, then you would get no current flowing across, you would just see that the electron wave packets oscillate. Or block oscillations as we discussed last time. And that is manifestly what happens with copper. But the experimentalist comes back to you and says look dude. That is a ridiculous model because the copper isn't in fact perfect, it's messy. So how do you test the model? Well there are two ways to test-- to deal with the situation. One is you improve the model to incorporate properties that copper actually has. And see if you can actually get the same conductivity that you see. But the other is you could improve the material, instead of improving the theory. So let's make up what-- can we actually build a perfect crystal? This is actually something that I'm doing research on right now. Not on the building side, but on the theory side, because I'm a theorist and you should not let me in a lab. But I collaborate with experimentalists, so they're nice people. They're very good physicists. So here's something you can do. You can build a system that has exactly a periodic potential. It turns out it's very difficult to do this with quantum systems. But what you can do is you can do it with lattices not of atoms, but lattices of dielectric. So the equation. Here's a cool fact, the equation for light going through a dielectric, where the dielectric has different constants, like wave guides. You've got glass, you got air. You've got glass, you got air. That equation can be put in exactly the same form as the Schrodinger equation for the time evolution of a wave function. They're both waves. And so it's not so surprising these two wave equations are related to each other a nice way. Meanwhile, the index of the dielectric turns into the potential for the quantum mechanical problem. So if you have a periodic potential, what do you want? You want a periodic dielectric constant. Yeah. And so you can build a system which incredibly, cleanly, has a periodic dielectric constant and no disorder. And then you can put light into the system, and you can ask what happens to this system. So here's the idea, I take a system which is a periodic-- I'm going to draw the potential here. So I'm going to draw the dielectric constant. So small, large, small, large, small, large, small, large, et cetera. But instead of having it be a one dimensional lattice, I'm going to make it a two dimensional lattice. So now, basically, I've got a set of wave guides. Let me draw this differently. So does everyone get the picture here? So literally what you have, is you have glass, glass with a different index, glass, glass with a different-- if you can think of those as a line of glass fibers. Optical fibers. And you shine your light that's reasonably well localized, in both position, and in phase variation, or crystal momentum. Because you can control the phase of the light. So you send this wave packet in and you ask what happens. Well not a whole lot happens. It's a wave packet. It's going through a wave guide, but we haven't implemented an electric field. To handle an electric field, you need the potential to be constantly varying. Uh huh. So it's at a linear ramp into the potential. Instead of making it just perfectly periodic, let's make the index ramp just a little bit. And this experiment has been done. In this experiment, so as the wave packet moves along, what's discovered is that the position-- if I draw the x as a function of t, so now the role of t is being played by the distance it's moved along the wave guide, what you find is that it does this. It exhibits beautiful block oscillations. And this has been proved in a very small number of real honest quantum mechanical systems. The most elegant experiment that I know of was done by Wolfgang Ketterle, who's here at MIT. And he got three data points because it was preposterously difficult and declared victory. So I talked to him about this in the hallway one day. And he said yes, this was ridiculous, but we got three data points. We got small, we got large. Victory. We declared victory. But it really needs to be done well. So one of the interesting questions in this part of the field right now is we know that it's true. But we want to see it. We want to feel it, so various people around the world are working on making a truly beautiful demonstration of this bit of physics. Yeah. AUDIENCE: [INAUDIBLE] PROFESSOR: It's totally impractical, because any interference is just going to kill you. Unfortunately. So, you have to work ridiculously hard to make systems clean. So the question is really a question about quantum computation, which we'll come to next week. But, the basic question is how robust is this. And the answer is it's not robust at all. But which you can tell because everything in the real world has enough impurity that it conducts. Or as an insulator. Yeah. AUDIENCE: What place sort of like the larger role in sort of like the perfection of a lattice like temperature or impurities. PROFESSOR: That's a very good question. So the question is what's the most important property? What's most important disordering property that leads to conduction? And there's temperature fluctuations, there are impurities in the lattice. There are decohereing effects which is a more complicated story. And that's actually, it depends on the situation, it depends on the system. And exactly how it depends is something that is an active area of research. Now there are many, many ways to probe this physics. So we know that these block oscillations are true. We see them in all sorts of different systems that are analogous. So there's lots of [INAUDIBLE], it's not like this is an ambiguous bit of physics. But it's one that turns out to be surprisingly difficult to tease apart. The reason I bring all this up is to emphasize the following, our model made a prediction that disagreed explicitly with the connectivity property of copper and other materials. So don't throw away the model. Observe that you've modeled the wrong system. If you find a system that fits your-- that is-- that shares the assumptions of your model, that's when you ask did it work. And it worked like a champ. OK. So now let's talk about real materials. This is going to close up our discussion bands and solids. And this is actually what I wanted to get to at the beginning of the lecture. But that's OK. There are lots of questions and they were good questions. So this is an extremely brief. But I want to ask you the following question. What happens in the following three systems? So first, imagine we take why don't we take a system with built out of single wells, which have some set of energy eigenstates, and then we build the periodic array out of them. What do we expect? And let me draw this bigger. What do we expect to see when we build a lattice? We expect that this is going to-- that these states are going to spread out into bands a funny way Yeah and let's just talk about the 1 d potential. So what we'll find is that this band turns into-- I'm sorry. This state, this single state turns into a band of allowed energy eigenstates. There's now a plot of the energy. And similarly, this state is going to lead to another band with some width. And this state is going to lead to another band, which is even wider. Everyone cool with that? Quick question? In 1 d, do these bands ever overlap? No. By the node theorem. Right? OK. Now let's take a single electron, and let's put in-- let's take a single electron, and let's put it in the system. What will happen? Well if we put it in the system, what state will this single electron fall into? Yeah one event. But which state? AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah, if you kick the system around, you let it relax a little bit. It's going to fall down to the ground state. You have to couple to something else like hydrogen has to be coupled with an electromagnetic field to decay. But couple it, kick it, and let it decay. It'll settle down to its ground state. So you get an electron down here in the ground state, and looking back at that band, we know that the band for that ground state looks like this. So, here it is. There's our electron. It's sitting in the lowest energy eigenstate. Is it moving? Well, it's in a stationary state. Is the expectation value of the position changing in time? No. The expectation values don't change in time, in the stationary state. That's part of what it is to be a stationery state, to be an energy eigenstate. OK. Great. it's not moving. Now, in order to make it move, what do you have to do? What kind of state corresponds to the position changing in time? Yes. Superposition. Right? From the superpositions we'll get interference terms. So if we put in a superposition of say, this state, and this state, which corresponds to different energies. If we put it in a superposition of these guys, then it's meaningfully moving. It has some meaningful, well defined time variation of its position expectation value. So in order to induce a current, in order to induce a current of this system where the electron wave packet carries a little bit of momentum is changing in time it's position, what do I have to do to the electron in the ground state? I have to excite it, so that it's in a superposition of the grounds state and some excited state. Or more generally, into a superposition of other states. Yes? In order to induce the current, I must put the electron into a higher energy state and in a particular superposition of higher energy states. Everyone down with that? Here's why this is so important. Imagine each one of these wells is actually not some square well, but it's an atom. And let's say the atom is hydrogen, just for-- this doesn't actually happen, but just imagine-- in particular what it means is it has the ion, the nucleus is charge plus 1. And so in order for the system to be neutral, I must have one electron for every well. So if I have n wells, I must have n electrons in the system. Everybody agree with that? In order to be neutral. Otherwise, the thing's charged and all sorts of terrible things-- electrons will get ripped off from nearby cad. So we must have an electron per well. How many states are in this band? For n wells? n. Right? OK. So if I put in the n electrons I need to neutralize a system, where do those n electrons go? Yeah, they fill up the first band. And if we let the system relax with lowest energy configuration, every state in this lowest band will be filled, and none of these states will be filled. Everyone down with that? So here's my question. When I've got that ground state configuration of this lattice of atoms with one electron per well, in these distributed wave functions, filling out these bands, is anything moving? Wow, you guys are so quiet today. Is anything moving? This system is in an energy eigenstate. In particular, it's in a completely antisymmetrized configuration, because they're identical fermions. So, nothing is moving. If we want to induce a current, what do we have to do? Yeah. We have put them in a superposition. But where's the next allowed energy eigenstate? Next band. So it's in the next band. The next allowed energy eigenstate. So the configuration we have now is that these guys are all filled, these guys are all empty, but in order to take an electron from here and put it into this excited state, we have to put in a minimum amount of energy, which is the gap between those two bands. Right? So now think about it this way. Suppose I take light and I send my light at this crystal. In order for the light to scatter off the crystal, you must have electrons in superposition states so that they can have a dipole and absorb and radiate that energy. Yeah. But in order for that to happen, the light has to excite an electron across the gap. It has to give it this macroscopic amount of energy. Well, it's not macroscopic. it's large. It's not infinitesimally small. That means that there's a minimum amount of energy that that incident light must have in order to excite the electron in the first place. So very long wavelength light will never do that. Light along wavelength will not have enough energy to excite an electron across this gap into the next band to allow there to be a current, which could oppose the electric field. So the only for light to scatter off of this crystal, is if the energy, h bar omega, of the light is greater than or equal to, let's say greater than approximately, the band gap delta e. That cool? We've just discovered something. Crystals are transparent unless you look at sufficiently high frequencies. That's cool. Right? A crystal is transparent unless you look at sufficiently high frequencies. If you look at low frequencies, your crystal should be transparent. Well that's really interesting. In particular, we immediately learn something cool about two different materials. Consider diamond and copper. These are both crystals. They're solids made out of a regular array, perhaps not perfect, but extraordinarily good, regular array of atoms of the same time. Array in a particular structure. Diamond, anything and I think face inner cubic. I don't remember. I really should know that. Anyway, copper. It's a lattice. That's embarrassing. I really should know that. So we have these two materials which one has the larger band gap? Diamond, because it's transparent. At in the visible. So the band gap, delta e of diamond is much larger than the band gap for copper. But in fact, this is a little more subtle, because copper in fact, doesn't even have a band gap. We made an important assumption here. So I want to think about-- we're going to come back to copper in a second, but I want to point out the nice thing here. Which is that diamond has to have a band gap. It's transparent. It must have band gap. It must be such that when you fill up all the electrons you need for it to be neutral, there is a gap to the next energy states. And that gap must be larger than visible wavelengths of light. Yeah. That's cool. And that must be true of all the transparent crystals that you see. Otherwise, they wouldn't be transparent. They would respond by having free electrons that could respond like a metal. Yeah. AUDIENCE: So, [INAUDIBLE] PROFESSOR: Yeah. Are diamonds good conductors? No. They're terrible conductors. In fact, there preposterously-- if you compare the number of-- I'll get into this later. But yes, they're terrible. AUDIENCE: [INAUDIBLE] PROFESSOR: Uh, that's a slightly more complicated story, which let me come back to. Hold on to that, and if I don't answer today, ask me after in office hours because it's a little more-- what? Really? Wow. Well, MIT. It's all about the intellect. And everything else has to-- OK. So, this is pretty good, but here's the thing. In one dimensional crystals, the only thing that can happen is, look if you have each band come from allowed energy state and each energy state, each well comes with one electron, or two electrons, or three electrons, you will always have filled bands, and then a gap and filled bands and then a gap. Does everybody agree with that? You can't have a partially filled band if each band comes from a bouncy, in a single well, and each well comes with an integer number of electrons. You just-- you're stuck. Yeah. AUDIENCE: [INAUDIBLE] PROFESSOR: Oh. I'm lying about spin. But spin in one dimension is little-- I'm lying about spin. But do you really want me to get in spin? Man. OK. So if we include spin, and we have splitting, then it becomes a more subtle story. If we include spin, then there are two states for every allowed energy eigenstate of the potential. However, there are generically going to be interactions between the-- there are generically going to be magnetic interactions which split the energy of those two spin states. Electrons spin up, and electrons spin down, will generically have different energies. Now in 3D, this isn't such a big deal, because those splittings are tiny, and so the states can sort of overlap. But in 1D they can't. So I mean, that's also not exactly true, but it depends on exactly the details. It depends on the details of the system, is what I wanted to get to. Curse you. So let me talk about the same phenomena in an easier context, where we don't have to worry about spin, which we haven't discussed in detail, in the class. Which is in three dimensions. Where the story changes in a dramatic way. Yeah. AUDIENCE: [INAUDIBLE] PROFESSOR: Oh. It's not. Generically, no, it's not. It will depend. AUDIENCE: [INAUDIBLE] PROFESSOR: You say it happens to be the salient one. Yeah, exactly. That is exactly right. There the gaps are not the same. That they do not remain constant. OK. So let's talk more about this system, but let's talk about it in three dimensions. So in three dimensions, you guys did an interesting thing, when you studied, you didn't know this was about the structure of solids, but it really was. When you studied the rigid rotor. And when you studied the rigid rotor, you found that you had energy eigenstates and they were degenerate with degeneracy 2 l plus 1. The various different l z eigenstates. Yeah. And then we turned on an interaction which was the energy costs, the energy penalty for having angle momentum in z direction. Which added an l z term to the energy. And what you found is that as a function of the coefficient, which I think we called epsilon, of that perturbation of the energies of the energy was equal to l squared over 2 i plus epsilon l z. What you found is that these guys split. So this remained constant. And this split into, so this is the [INAUDIBLE] l equals 1. So l equals zero. So this is one, this is three, this is five. So the l equals zero state, nothing happens. l equals 1. There's one that changes, one that doesn't. And then this guy has five. One, two, three, four, five. OK. And what we found here is that these guys could cross. States from different multiplates, with different values of l, had energies that could cross as a function of the strength of the deformation of your system. Right? The deformation is where you have a sphere and you stick out your arm. So it's no longer symmetric top. So here we can have states crossing. There's no nodes here in three dimensions. So as a consequence, when you have a three dimensional material built out of atoms. So here's my sort of pictorial description of three dimensional system built out of atoms. You have a potential well, potential, potential well. Now, if the energy in one particular potential well, is like this, and like this, and like this, then when we add in a lattice we get bands again. The structure's a little more intricate because it depends on the momentum. But these bands now can overlap. OK. Everybody see that? Because there's nothing preventing states from different-- in different multiplates from having the same energy in three dimensions. There's no nodes here that tells you have to keep the ordering constant as you turn on the potential. Now we turn on the multiple particle potential, and they can interact, they can overlap. As a consequence, when we fill up, let's say we two electrons per potential, or per well, when we filled those first two bands, well, there is the first-- so the first band is now filled. The second band and part of the first-- part of the third band and most of the second band are going to be filled. But part of the second band is now available, and much of the third band is now available. We filled in 2n electrons, but we haven't filled up this band, because it's really two bands jammed together. Or really bands from two different orbitals jammed together. They happened to overlap. So as a consequence here, if this is the length of the energy of the last electron that you put in, how much energy do you have to give the system, do you have to add the system, to excite the energy-- or to excite the electrons into excited states, in particular into superpositions so that the electrons can move? AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah. Preposterously small amount. An amount that goes like one over the number of particles. So in the continuum limit, it's zero. There's an arbitrarily nearby energy. So how much energy does it take to excite an electron and cause a current that opposes the induced electric field? Nothing. Any electric field that you send in will be opposed by an induced current. So this behaves like a classical conductor. You turn on an electric field, and the charges will flow to oppose that externally imposed electric field. You get charges then building up on the walls of your capacitor plates. So, this is where we have a conductor. Because there's an unfilled band. And back here , we had an insulator because we had filled bands separated by gap. The gap between the filled band and the next available band. This is actually called a band insulator. Because there are other ways of being an insulator. So from this so far, just from the basic quantum mechanics of a particle and a periodic potential, we now understand why some crystals are transparent. Why some materials conduct. Why the materials that are transparency are also insulators. And the things that conduct are not transparent, generally. Yeah. AUDIENCE: [INAUDIBLE] PROFESSOR: Excellent. Excellent question. So what's so special about diamond and differ from copper? And so the answer goes like this. So what determined the exact band structure in for a 1D periodic potential? Two properties. One was l, the periodicity. And that came in the q l and k l. And the second is the detailed shape of the potential. Now in three dimensions, the story's going be a little more complicated. In three dimensions, the things that are going to determine the potential are not just the distance between atoms, but you have a three dimensional lattice. And the three dimensional lattice could have different shapes. It could be cubic, it could be hexagonal, could be complicating in all sorts of different ways. Right? It could be bent, it could be rhomboidal, and it could have all sorts of different crystallographic structures. So that's going to go into it, in the same way that l went into it, which is the only parameter in one dimension. In the same way that l goes into it. So the crystal structure, the shape of the lattice, is going to determine it. Secondly, the structure of the orbitals is different. Different atoms are different wells, so they'll give you different band structure. So different materials for example, diamond versus copper, are going to give you different bands allowed energies, because the potential is different. It has different shape. And so when you solve the problem for the energy eigenvalues is a function of now the three different components of the crystal momentum, you'll just get a different set of equations. And working those out is not terribly hard. But it's a computation that must be done, and it is not trivial. And so one of the sort of, I don't know if I'd say exciting, but one of the things that one does when one takes a course in solids, is you go through a bunch of materials. And you understand the relationship between the potential, at the atomic orbital structure of the individual atom, the crystal structure, and the resulting band structure. And there's some sort of nice mnemonics, and there are calculations you do to get the answer. AUDIENCE: [INAUDIBLE] PROFESSOR: You will almost always find overlapping bands in three dimensions in sufficiently high energy. I can't off the top of my head give you a theorem about that, but yeah, it's generic. Yeah. AUDIENCE: --analog to conductor in one dimension? You have these like, non-zero band depths? PROFESSOR: Yeah. And this is why Matt was barfing at me. So the answer to that is yeah. There aren't [INAUDIBLE]. But what would we need? What we need is one of two things. We need either the band gap coincidentally is ridiculously small. What's a good example of that? A free particle. In the case of a free particle, these band gaps go to 0. Right? And so that's a conductor. Just an electron. Right. It conducts, right? OK. So that can certainly happen. But that's sort of stupid. I mean, it's not totally stupid. But it's sort of stupid. But a better answer would be, well, can you have a system where there are bands but you didn't have one electron per potential well? And yeah. You could orchestrate that in lots of ways. Now it involves orchestration. So it's not the generic system that we were talking about here. But you can't orchestrate it. So spin is a useful thing that gives you an extra handle. If you have twice as many states per well then you can have half a band filled. So that's one way to do it. Then it becomes dependent on details of the system, which is what I didn't want to get into. But yeah, you can orchestrate it. It's just not a generic thing from what we've done. And it's really not for spin-less systems. On the other hand, accidental small gaps. Easy. That happens. That certainly happens. So that brings me to the last thing I wanted to talk about before getting to entanglement, which is accidental small gaps. So what happens to a system which is-- so there are some systems that are neither conductors nor insulators. They are reasonably good conductors and reasonably bad insulators. But they're not perfect. And these materials are called semiconductors. I want to talk about why they're called semiconductors and what that means. So this is going to be very brief. Then I'm going to give you-- we're going to get into entanglement. So consider a system exactly using the same logic we've used so far which has the following property. We have two bands. And the bottom band is filled because we've got just the right number of charged particles. Bottom band is filled. And this guy is empty, but the gap is tiny. OK. Delta e is very small. Now delta e has dimensions. It has units, right? So when I say small, that doesn't mean anything. I need to tell you small compared to what. So what's a salient thing that controls an energy scale for a real material? Well the temperature. If you have a hot piece of copper, then the lattice is wiggling around. And every once in a while, an ion can hit one of the electrons and excite it, give it some momentum. And so there's an available reservoir of energy for exciting individual electrons. You have it really hot, what happens is every once in a while an electron will get nailed by a little thermal fluctuation in the system and get excited above the gap. And now it's in a super-- and generically, it's going to be in a superposition state of one of these excited states. So it's in general going to be moving. It can radiate. It will eventually fall back down. But you're constantly being buffeted. The sea of electrons is constantly being buffeted by this thermal fluctuation. And as a result, you constantly have electrons being excited up, cruising around, falling back down. So you end up with some population of electrons. And they can ask-- and both when asked, although not quite in this language, how likely are you to get an electron up here? How likely is an electron to be excited up thermally? And those of you taking 8.04 will know the answer to this. The probability goes as e to the minus delta e over kt. So let's think of this where this is the Boltzmann constant. So what does this mean? At very low temperatures, if the gap isn't 0, then this is 0. It doesn't happen. But at large temperatures, the denominator here is large. If the temperature is large compared to the width of the gap, then this is a small number. And e to the minus of a small number is close to 1. So at high temperature, you're very likely to excite electrons up here. And now if you have electrons up here, you have a bunch of available states down here-- also known as holes-- and you have a bunch of available electrons up here with lots of available states. So at a high temperature, a material with a small gap-- or at least at temperatures high compared to the size of the gap-- it's basically a conductor. And at low temperatures, it's basically an insulator. This is called a semiconductor. And there are notes on the Stellar web page that discuss in a little more detail what I just went through and show you how you build a transistor out of a semiconductor. And the important bit of physics is just this. OK. So that finishes us up for the band gap systems for periodic potentials. We've done something kind of cool. We've explained why diamonds are transparent. We've explained why they don't conduct. We've explained why copper does and it's opaque. And that's pretty good for 15 minutes of work. It's not bad. But along the way, we also talked about the analogous system of what are called photonic crystals. Systems of periodic arrays of dielectrics. Like wave guides. And those have the same structure. They have bands of allowed energy and gaps of disallowed energies where no waves propagate through. So you might think that's a little bit of a ridiculous example. So just to close this off, you've all seen a good example of a photonic crystal flying past you. You know that highly reflective at very specific frequency structure on the surface of a butterfly wing that makes it shiny and blue? It looks metallic. It looks like it's a crystal reflecting in a specific frequency. At some sharp blue. And the reason is, it's a photonic crystal. It is exactly this form. If you look at it under a microscope, you see little rays of protein which have different dielectric than air. And they form exact crystals-- or not exact, but very good crystals-- that reflect at very specific wavelengths. And as a consequence, they have a metallic sheen. So why would a butterfly put a photonic crystal on its surface? Well it's extremely light. It's fairly rigid. It looks shiny and metallic without actually being shiny and metallic. And it's not a pigment, so it doesn't absorb light and decay over time. It's like the best thing you could ever do if you wanted to be a shiny, fluttery, flying thing. Anyway. So there's an incredible amount of physics in this story of the band gaps. And consider this an introduction to the topic. OK. So that's it for band gaps. And I want to move on to the remainder, the last topic of our course. Which is going to be entanglement and quantum computation. And here I need to give you one quick observation and then move on to the punchline of today. The one quick observation is this. We've talked about identical particles before. And we've talked about identical particles in funny states. So for example, imagine I have two particles described by a wave function where the first particle could be in the state a and the second particles in the state b. And I can build a wave function for the first particle being in state a and the second particle in state b in the following way. So let's say position a and position b. I could take a single particle wave function, chi of a, and a single particle wave functions phi of b. And we've talked about what this tells us. And you've studied this on your problem set. What this tells you is that the probability of finding the particle at point A is given by chi a squared. And this is normalized, so when we integrate against it, we get 1. And similarly, the probability that we find the second particle at b is this thing norm squared. And it's independent of what a is. But we also studied-- and so this was called the distinguishable. We also studied the symmetric configuration, which was equal to 1 over root phi, root 2. Chi of a. Phi of b. Symmetric, plus chi of b phi of a. And this tells us something totally awesome. What's the probability that I find the first particle at a? It's the norm squared of chi of a phi of b, right? If we integrate over all phi b, this is the norm squared integrates to 1. So it's fine. So there's a factor of one half. We either find it at chi of a or chi of b. However if I tell you that I've measured the first particle and I find it in the state chi, what can you say about the second particle? It's in the state phi. If you know the first particle's in the state chi, the second part is in the state phi. Because we measured it and it's not in the state-- the first particle's not in the state phi. So measuring one particle tells you something about the second particle. And this is deeply disconcerting, because I could've taken these particles, put them in this entangled state, and sent one particle off to a distant planet and the second particle to my sister in DC. And my sister measures this second particle and determines what state it's in and is immediately determined what state the first particle is in over in this distant planet Zorg, right? So that's deeply disconcerting. And to those of us who have studied quantum mechanics up to this point-- which we all in this room have-- to those of us who have studied quantum mechanics to this level of development and understand that it is a correct description of many experiments, this should be yet another moment of serious discomfort. We've run into a bunch of these over the semester. But this one should be troubling to you. Because look. How can something here dramatically change the state, the configuration, the initial configuration, of a particle arbitrarily far away? Isn't that deeply concerning? And if you think about relativity, this should be all the more deeply disconcerting. Because how does relativistic causality fit into this? So there was a person that roughly this time, a little earlier, who was troubled by this problem. And his name was Einstein. And so one of the things that's kind of amazing is that he created a thought experiment which we're going to study in detail next week called the EPR experiment. And there's a beautiful historical story about the setting and the meaning and the particular person. And unfortunately, I'm not a historian so I can't tell you that story. It sure would be nice if we had someone who wrote a biography of Einstein to tell you a little bit about that story. Oh look, it's Tom Levenson who wrote a biography about Einstein. So Tom is-- TOM LEVENSON: Oh, I need a microphone. Those of who have taken courses in [INAUDIBLE]-- and I'm sure that's all of you because of the GIRs-- know this is larger than the usual [INAUDIBLE] class. So I'm very used to microphones, but not in this context. OK. Is this-- yeah, it's on. Can you hear me? All right. So there are lots of ways to slice the story of Einstein by the time he reaches the EPR experiment, which is Einstein, Podolsky, and Rosen for the three people who actually wrote the paper. Just to dot the I's and cross the T's on the paper itself, Rosen is apparently for person who first came to Einstein. Podolsky and Rosen were two young physicists in Princeton after Einstein moved to Princeton. Einstein moved to Princeton in 1933. About three weeks before-- I'm sorry, he moved to Princeton '33. He left Germany in 1932 December, about three weeks before Hitler took power. And he did so with decisiveness and dispatch and a head of almost all of his-- in fact, I think all of his German-Jewish physicist colleagues and those German physicists for whom the Hitler regime was unacceptable. Which shows that Einstein really was smarter than most of his peers. That's one of many different ways you can ascertain that. And so he came to Princeton in '33. He actually went to Caltech before we went to Princeton. As part of an ongoing visitor-ship he had there. Came back to Europe, hung with the queen of Belgium who was a friend of his. Went to England. And then headed across the Atlantic and took up residency in Princeton at the Institute for Advanced Studies where he stayed for the rest of his life. And over the course of the-- that was '33, he died in '55, I think. I should know that, but I think that's right. 22 years. He worked with a lot of different, mostly younger physicists. And Podolsky and Rosen were early members of that chain. So Rosen was talking with him some day and starts to frame this experiment. Einstein develops it. The three of them talk about it. They write the paper and they put it out. And I want to share with you, actually, a really lovely description of the way the problem was represented in a way by-- this is from a book that I recommend to all of you. It's actually really hard to find. It's really sweet. Jeremy Bernstein, who is a physicist. He's sort of been around. A physicist and writer. He's in his eighties now. He lives in Aspen. He worked with CERN for a number of years. He's always been independent. He wrote for the New Yorker. Anyway. So you've all heard of the physicist Bell, I assume? Bell's inequality? OK. So Bell had a lovely way to describe-- I'm trying to find. I had this marked and then I lost my piece of paper. I have already lost it. That's terrible. So Bell has this wonderful way of describing this problem of entanglement. And it's based on his description of an actual person. I was going to read you his actual quote. Now I'm just going to paraphrase it for you. He had a friend named, I think, Bartelstein. Or at least someone known to him. Who had two quirks. An unusual color sense and a taste for mismatched socks. And so Bell used to say, if you saw Bartelstein and you could only see one leg and that sock was pink, you knew to a certainty that the other sock was not pink. He comes up I think-- I'm trying to remember who this is originally attributed to. Same thing. If you have a coin and you cut it in half down the-- so you've got two coin shape disks. You cut the disk in half, not-- and you have one side that's the head and the other side that's the tail. And they're separated. They get handed to two different gamblers. And one gambler tries to cheat the gambling establishment by tossing in his half coin. And you see the head that you know somebody-- somebody at some other casino is cheating by tossing in the half coin that only has a tail on it. So there are lots of ways to represent this. And many physicists being very witty indeed have come up with different metaphors for it. So Allan just described for you the basic claim in EPR. Its weirdness. That you have two particles that are entangled in some way and then go their separate ways. And thus you have-- if you have knowledge of what's the state of one, you have certain knowledge of the state of the other, violating relativistic ideas of locality. And just kind of making you queasy if you're sort of approaching it naively. What Einstein, Podolsky, and Rosen argued was actually something a little bit-- in fact, the paper comes to an end on that note of queasiness. But what they argue is a little bit more subtle. Because what they said is, OK. You perform this thought experiment. You send the two particles off. You measure position of one, you know absolutely the position of the other. You've conferred-- and the paper turns on a discussion of the connection between a measurement-- a physical measurement-- and a property of physical reality. And they have definition for what reality is. And that is something whose-- if you can perform a measurement, you know that quantity absolutely. I don't have the mathematics to express that properly. But that'll do for this hand waving. You can then do another experiment and measure a complementary property. And you know that piece of reality. But you can't do the-- so on the one hand, quantum mechanics says you can't know physical reality to this level of precision. And on the other hand, the fact that you can do that measurement violates the relativistic picture of reality. So you have what they claimed was a paradox. And this paper was published. And it received a range of reactions from indifference by younger physicists who said, we don't care that it's weird. We're going to keep on doing quantum mechanics and performing experiments and making measurements. And just see where this leads us. Remember, this is happening in the mid '30s. 1935. One of these three books will tell me precisely in a moment. And the quantum theory, as it turned into quantum mechanics, developed in its first period between '23 and '27. And by '35, you have enormous numbers of productive results and unexpected things and the prediction of the positron and then its observation. And I mean, the theory is enormously, dramatically, excitingly productive. So those who are really heads down doing the work are, for the most part, saying, this is fine. We'll get back to it when we're old and retired and bored. But that wasn't the uniform case. And most notably Niels Bohr found this paper really troubling. And spent about six weeks, apparently, discussing this and trying to come up with a response to it. And what he responded was essentially that-- in some ways, it was the same reaction as his younger colleagues. Get over it. But more precisely, it was he said, there's no description of reality that excludes the measuring apparatus anymore. You can't make statements about physical reality unless you include a description of the measuring apparatus. And you've said that we can measure this one quantity with precision and know the other thing. And then we can subsequently, in a separate observation, measure a complimentary quality and know the other one. You still can't know much them at the same time. It's still true that the complementarity in essence means that once you know one part of the picture, you know some other part the picture. And that's just the nature of the quantum world. Einstein had argued that the EPR paradox suggested that quantum mechanics was incomplete. And Bohr essentially responded in effect that Einstein's description of quantum mechanical explanation was inadequate. The important thing to remember-- and I want to just spend a couple minutes going back into the pre-history of all this, and then a couple minutes speculating on why Einstein reached the position he did. And what that might tell you about the practice of science as a lived experience as opposed to one reflected in your textbooks. But the thing to remember is that there's nothing logically wrong with the EPR paper. Right? You know. It does what it says it does and there's no overt error in it. And there's nothing wrong with Bohr's response. And in fact, when the experiments were-- Bell formalized the-- what Bell's inequality really does is it formalized the two arguments. It says, if Bohr is right, you will observe this in the experiment. And if Einstein is right, you would observe something different. The experiments were done, and I imagine are still being done, as sort of demonstrations. And they showed that Bohr's interpretation was correct and that yes, quantum mechanics produces results that are non-local just as Allan described to you. And that the world really is as strange as people first glimpsed in 1925, '26, and '27. And the question of whether or not that strangeness is adequately explained without the explanations that you're going to learn in this class and subsequent ones are quote "complete" or not. And completeness is a very funny, very, very tricky concept. But the question whether or not the framework of quantum mechanics is somehow unsatisfactory in any kind of formal a technical sense is one that's at least partly dependent on your scientific temperament, I think. So that's the cartoon version of what happened in '35. Einstein with his two young colleagues proposes-- really you should understand the EPR paper as a description in detail of a consequence of quantum theory as it was then expressed with the conclusion-- and I just want to read you this thing-- no reasonable definition of reality could be expected to permit such a result. In fact, it's called a paradox, but it isn't. It's a complaint. You know, it's a memo to the Flying Spaghetti Monster that the universe shouldn't be this way if, in fact, experiments turn out to show that it is and they have. The oddity here for a biographer of Einstein opposed to a physicist is given what you know about Einstein between 1879 when he's born and say, 1925 or so when he completes the last of his really great physics. How could-- I mean, I actually keep-- I've been working on Einstein off and on for years and years. I keep finding out new ways in which he's just inconceivably bright and on target and with a nose for the right problem and insightful. And yet by the 1935, 10 years later, he's still a relatively young man. He's in his '50s. Which being in my '50s I think is an extremely young man. Just 10 years after doing work that's right on the edge of modern quantum mechanics that is essential to its foundation. That's really extraordinary. 10 years after that, he's saying, no reasonable definition of reality should be permitted to behave this way. Where does that come from? Well the first thing I want to tell you-- again, this is all going to be a really cartoon version. Because there's not much time, I understand. Is that Einstein-- I mean, how much are you aware of Einstein's role in the creation of the quantum theory? A lot. I mean, a lot? None? OK. I mean, I'm going to make a claim that except for Heisenberg, Schrodinger, maybe Bohr. Maybe Born. Maybe a couple of others. There's no one more important to the quantum theory than Einstein. And you could maybe even argue that from a sort of foundational point of view that without Einstein, rigorous thinking about quantum mechanics would have taken much, much longer. I mean, he's really central to it. Planck in 1900 publishes as an ad hoc solution to the black body problem the first quantum theory. In 1905, Einstein says it's not an ad hoc thing. If you look at the photoelectric effect is the particular problem he's dealing with in explanation. But that's really-- the behavior of the photoelectric effect is really presented as the confirmation of this idea that light exists as quanta with particular kinds of behavior. And from 1905 on, he spends probably more time on quantum problems than he did on any other physics problems. Certainly more than on relativity. Though he spent enormous energy on special and general relativity. One of most amazing things about Einstein, in fact, is that despite the fact that he's seen and appears by 1935 to be this hidebound old guy who can't accommodate himself to the new world is he had an extraordinary capacity to do what the Red Queen did in Alice in Wonderland. and believe two impossible things before breakfast. Just think in 1905. April, he publishes The Quantum Theory Of Light. June he publishes Special Relativity, which treats light as a wave. And makes no mention of his revolutionary-- I mean, he called it revolutionary in a private letter in 1905. So he knew what he had in the quantum theory of light. But in special relativity, go read the special relativity paper. It's actually lovely reading. And you'll see he doesn't even nod in that direction. He doesn't say, you know this is a hero-- he says nothing. So he's capable of doing excellent-- and there's a reason that year is called the annus mirabilis, the year of miracles. And in part it's because Einstein is able to actually really focus on these things. And I realize class is almost over. So there's several more steps in Einstein's quantum journey. But you know, what you should take away is that Einstein's ability to deal with the problems of quantum pictures extends to the point-- he's the first person to suggest there might be a problem with causality in quantum mechanics. He does this in 1917, eight years before quantum mechanics is invented. When he starts looking at what the quote "classical" quantum theory tells you about the emission of radiation from an excited atom. He realized you can't predict it precisely. Radioactive decay has the same problem. He says, well-- he writes in a letter. I don't want to give up causality, but we may have to. So he's aware of these things. So I see class is over now at 12:30. OK, sorry. So juts to finish off, the question here is why does Einstein give up on this. And the answer, I think, is because in addition to his-- as he started at the beginning of his career, he says with the quantum theory of light and with special relativity, ignore your physical pictures. Try and look at the phenomena and explain those. And by 1935, that becomes very difficult for him. Because the phenomenology becomes too strange. One of the things that quantum mechanics does is it takes away the immediate ability to visualize physical systems. There. English is my first language sometimes. And that's an aesthetic failure on Einstein's part. He had the intellectual capacity and explicitly said, quantum mechanics is a logically consistent theory that incredibly powerfully describes lots of problems. He said that in print. He nominated Schrodinger and Heisenberg for Nobel Prizes twice. I mean, he wasn't stupid. He was Albert Einstein. But he was aesthetically incapable of pursuing this new physics in ways that were possible under the research possibilities of the time. And that is what I would leave you with. Physics is an aesthetic as well as an intellectual pursuit. So thank you all. [APPLAUSE]
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https://ocw.mit.edu/courses/7-014-introductory-biology-spring-2005/7.014-spring-2005.zip
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And I just wanted to mention a few more things about community ecology before I move onto the final lecture which I've forecasted to you where I'm going to try to tie everything together through a research story that I want to tell you about. But before we go to that, I just want to present one very famous ecological experiment from your textbook. Because what we're trying to do here in this lecture is bring together the first set of lectures where we talked about biogeochemical cycles and productivity which we think of as the function of ecosystems with the last set of lectures where we talked about population biology, community structure where we actually talk about different species and how they interact. And the structure of the community effects productivity in biogeochemistry. And these processes in turn feed back and affect the structure of the community. And this is something ecologists have known, but it's not easy to demonstrate this experimentally. So one of the famous experiments was done by David Tillman of the University of Minnesota because these are very long-term experiments. And, of course, it's easier to do these kinds of experiments with plants because they actually stay put. So he asked the question is the species diversity of plants in a living community related to the productivity of that community, and also the resistance and the resilience of that community to stress? So what he did is he set up plots. And he went through generations of graduate students monitoring these over many years. This would be a plot that would have a single species in the plot. And here's a plot that has 24 different species in the plot. If we mixed them together they were all indigenous species that would grow there. And he showed first that the total plant cover here as a percent was a function of the number of species per plot. So establishing that indeed productivity, the amount of plant biomass produced was a function of species diversity. And when you think about this it makes sense because the more diverse species you have the more likely they're able to exploit the full suite of resources in the soil and are probably more resistant to predation. So the more diverse the plot it has a greater biomass. The next thing he looked at, and this is in your textbook, was the effect of the biomass to the resistance to disturbance. So this is a change in biomass one year before a drought and then to the peak of the drought showing that this ratio increased with the number of species. And finally the resilience, that is how long it takes for the community to recover after it's been stressed, he was also able to show increase as the number of species increased. So there is a relationship between community structure, and indeed productivity, resistance and resilience increase if there is more diversity, which is, of course, one of the motivators for preserving species diversity on the planet globally. OK. Now we're going to try to tie all this together through this story. Now, going back, many of the slides that I'm going to show you you've already seen. Parts of this story you've already learned from my previous lectures, so I'm trying to tie this together and remind what you know and help you to think about how you can apply what you know to current issues in global ecology. So you remember this, our global carbon cycle showing the photosynthesis of the plants balanced by the respiration of the plants and the animals. You learned this. And superimposed on that is our burning of fossil fuels and land use changes, i.e., cutting down trees increasing CO2 in the atmosphere. And as we showed last time, I mean in the first set of lectures that this sudden excavation of this fossil photosynthate is causing a dramatic increase in CO2 in the atmosphere relative to historical concentrations of CO2. This is thousands of years before present. And we're worried about that. This is all reminding you, getting you in the train of thought here. We're worried about that because there's good evidence that these increases in CO2 are already increasing the temperature of the planet. This is average temperature over the last thousand years. And if it isn't already there is certainty that it will in the very near future. So let's look here in the ocean, my favorite ecosystem, where the phytoplankton, that you know all about now, play a critical role in drawing CO2 out of the atmosphere. And this draw down is referred to as the ìbiological pumpî of the oceans. And we talked about this very briefly. Here's the phytoplankton community photosynthesizing, drawing CO2 into the surface ocean. And then because of the food web, that you've learned all about, most of this phytoplankton productivity is eaten by zooplankton and by fish going through the marine food web. As it's eaten they are respiring and CO2 is released and it goes right out of the system. So CO2 in, CO2 out. But some of that carbon, some of that photosynthetic product finds its way to the deep ocean through fecal pellets of zooplankton, through aggregates of dead cells, through just basic mucus that fluffs off of jellyfish. It's all carbon that came from photosynthesis, but some of it settles down to the deep ocean where it's chewed upon. Now, here is where all those deep consumers that I showed you in the DVDs last time are. There are fish and squid and jellyfish in the deep oceans that feed on this carbon that rains down, because there is no photosynthesis down there, and bacteria that then regenerates that organic carbon into CO2. So this functions as a pump. And the concentration of carbon dioxide in the deep ocean is very high if you look at a depth profile of CO2. This is depth and this is CO2. And at the surface, of course, it's in equilibrium with the atmosphere. And it's very high in the deep ocean because at low temperatures and pressures it can hold. So this is a huge reservoir of CO2, so much that if you did a thought experiment and you killed all the life in the oceans and just shut off this biological pump and then you let the oceans mix all the way to the bottom, which they won't do, but this is just a thought experiment. And you let that all equilibrate with the atmosphere, all of that CO2, the concentration in the atmosphere would double to triple what it is today. That's how much CO2 is in the deep ocean. And so this is a natural function of the ocean ecosystem, is to maintain this pump and maintain this gradient of CO2. So this you also learned, remember? That the oceans are not this homogeneous soup of phytoplankton, but there are areas of very high phytoplankton biomass and productivity and low biomass and medium in the green here. And we talked about nutrient limitation of that primary production, which is the availability of nutrients. And what I told you was that the sort of standard understanding of this system was that nitrogen and phosphorus were the primary limiting nutrients, this is all review for your final exam, in aquatic ecosystems. But I told you that I was going to tell you in the last lecture that there was more to it than that. And there is indeed much more to it than that. So for years we thought that nitrogen and phosphorus were the nutrients that were regulating this differential productivity, but we knew that something was wrong with our understanding. Because these satellite images, which I just showed you, of the distribution of productivity showed this distribution. But if you took a simulation model and modeled the global productivity based on the availability of nitrogen and phosphorus to the phytoplankton this is what the model showed. They said this is what the ocean should look like, not that. There should be much more productivity here in the equator and down here in the Southern Ocean than there actually is. And, in fact, people wrote papers why isn't the equator greener? And they had all these different hypothesis for why that might be, a lot of them having to do with grazing with the food web. Well, it turns out that iron is a really important limiting factor in the ocean. And this is a story that has just unfolded in the last 15 years. And I'm not telling you in exactly the order that it unfolded, but more or less in the order that it unfolded. A fellow named John Martin who is an oceanographer out at Moss Landing Marine Labs had been studying iron for quite a long time in the ocean. And he had a hypothesis that iron was limiting. But most people wouldn't believe him because most people, when they went out to measure iron in the oceans, got very high concentrations. So they said how could that be limiting? Well, it turned out that most people were measuring contamination in their iron samples. And if any of you have ever been on a marine ship, if you look around the deck you notice there is always rust. They are constantly painting marine ships, right, because the seawater is very corrosive. And so there is rust everywhere. And it turns out that you have to be heroically clean in order to measure the concentration of iron in seawater. And John Martin and his group realized this and went out and developed these techniques to collect the sample and to have it never see air before it went into the sample bottle. And they acid washed the sample bottles through this process which takes weeks and weeks and weeks. I mean you had to really believe that iron was limiting in order to go through all this agony to measure the level. So his group was able to measure really low levels of iron in seawater, but still they weren't able to convince people because this was such a totally different way of thinking about the oceans. So there was a lot of pushback. He also argued that iron is introduced to the oceans through atmospheric dust. Those people did believe. And that if you looked at the atmospheric dust flux, which is proportional to the iron flux, you'd see that in the Atlantic it is relatively high because you have wind patterns coming off these deserts in Africa. And also over here in the Western Pacific it's relatively high. It's low down here around the Antarctic because there's no land source there. So these patterns of dust delivery map on pretty well to this discrepancy to what we see and what we model if nitrogen was the limiting nutrient. So what John Martin argued was that there was lots of nitrogen and phosphorus in these regions but there's not enough iron for the phytoplanktons to actually utilize that. Remember the Redfield ratio we talked about, how it's the availability of nutrients relative to what's required by the plant that determines what's limiting? So he did some experiments where he went out in the boat and took samples. In the control sample he would add nothing and another sample he would phosphorus and nitrogen and another sample would add iron. And he was able to show that the addition of iron caused phytoplankton to bloom. He said iron is limiting in these regions of the ocean, but everybody said bah, bah, bah, no it's not, we don't believe you. They made up reasons why these experiments couldn't be true. There are not zooplankton in the bottle so this and that and the other thing. And so he persevered. And he said OK. If you don't believe my bottle experiment, I'm going to go out and I'm going to add iron to the ocean, then you'll believe me. So he said we're going to go out with a boat. And what they did was, and my lab was involved in these experiments. We had a small role, that's the boat, in measuring a certain component of the phytoplankton. But they pumped iron into the propeller wash of the boat and made a zigzag path in the ocean where there is about ten kilometers by ten kilometers. And the natural mixing in the surface ocean in about a day would mix that iron through that patch. And, of course, meanwhile the patch is moving. The oceans are moving, the patch is moving, they've got these location buoys, the captain of the ship is going crazy trying to navigate relative to these buoys rather than relative to the solid earth. But we were able to actually follow the patch. And, oh, here's John Martin who was also a friend of mine. And sadly enough he died of cancer before the first experiment showed unequivocally that iron is limiting. But he knew it was so that was good. Anyway, he threw out this line. ìGive me a half a tanker of iron and I'll give you the next Ice Age. This was before the experiment because he was trying to drum up enthusiasm for the experiment because he wanted to do the science. But you see the connection here. So what's the connection between a tanker of iron and an Ice Age? Exactly. If you fertilize with iron, the phytoplankton photosynthesized more. It could be argued that they draw more CO2 out of the air and will cool the planet. OK, so he got a lot of attention because of that, even though this was a scientific experiment. So now I'm going to, and we'll come back to that, of course, later. I'm going to take you on an oceanographic cruise so you know what it's like. And this is one that my post-doc went on. This was not the first iron fertilization experiment. There have been about five of them now, but this was one of the actually one of the more recent ones that was done in the Southern Ocean. They've been done in the subArctic Pacific, in the Equatorial Pacific. I went on the one in the Equatorial Pacific because I'm a cruising light-weight, but my post-doc went on the one in the Southern Ocean and that's where these slides come from. But just to show you what goes into doing this. These are the vats where they mixed the iron which they mixed with an acid solution. And they also put in sulfur hexafluoride which is an inert tracer so they can use that to trace the patch. Here's the oceanographic ship. What you do when you launch a cruise is you patch things in these vans. And sometimes you have your whole lab in your van. And, in fact, for this trace metal clean work they have special trace metal clean vans where they're all Teflon lined and positive air pressure and all of that. So you ship this out to the Antarctic, and it gets loaded onto the ship, the van, and then gets tied down. This is what the lab looks like before the scientists arrive. It's just a bunch of tables that are tied. And these labs get broken down and rebuilt for each cruise because different types of scientists have different needs. And this is what it looks like when everybody is settled in. It's a very crude makeshift thing very crowded with equipment and wires and is all set up temporarily for a month's worth of work. So here's the shift leaving port. They're going out of New Zealand which is where most of these cruises leave from for the Antarctic. Here it is in the rough seas of the Southern Ocean. You cannot see this very well. But that's an iceberg which is a big problem down there, because that's just the tip of the iceberg. So navigation is very tricky. So here's a radar showing these icebergs scattered around that they have to look out for. Here's another one. Here's how the samples are taken. These little plastic PVC pipes are all electronically wired. And this ball, when they're open, is drawn into it. So when these are lowered in the water they fill up with water, and then you trigger it and the ball shuts it and then you bring it up. So you can set it, say this one goes off at 5 meters, that one 20, that one 50, that one 100, whatever. Wherever you want them you set them, and then it comes up and you have your water sample. Working in the Antarctic is particularly difficult. In this case this person is out there taking the snow off of these incubators that have sample water in bottles trying to make the phytoplankton think that they are still in the ocean, but they are in controlled experiments here. And they're taking the ice off so that the light intensity stays the same. And this is to summarize the results of the iron fertilization experiment just in a picture. This is the water without iron added and this is the water with iron added. The addition of iron to these waters causes major algal blooms. And here is just some of the data. We can just look at chlorophyll ìaî, which you know is a measure of phytoplankton biomass in the patch versus outside of the patch. And this is actually a satellite image taken off of the NASA satellite of this iron-enriched patch. So there is no question now that the availability of iron limits primary productivity over vast regions of the ocean. I didn't think I was going to use the board but I will use the board. And because the Redfield ratio, which you guys now know, of carbon, some nitrogen, some phosphorus. Remember? We talked about this, 106:16:1. If we add iron to this, iron is about 0.005. Anyway, tiny amounts of iron are required relative to nitrogen and phosphorus in order for a phytoplankton cell to grow. So you can leverage, if there's abundance of nitrogen and phosphorus, it just takes a little bit of iron to get this big bloom. And that was very appealing to people. Any time it takes a little bit of something to get a lot of something, I think people get interested. And I think what motivated this whole interest in ocean fertilization, I think, is motivated subconsciously a lot by man's ability to manipulate a system so large with so little effort. But that's sort of a [subset? . So the success of these scientific experiments, were really just to go out there and understand what regulated productivity in the oceans, were picked up very rapidly by entrepreneurs. And the proposal that was put forward was to develop a commercial ocean fertilization industry where you'd fertilize the surface oceans with iron. And I'm kind of joking here that money comes out the bottom, but I'll show you how this works in a minute. And turn this into a business. And there are a lot of unknown questions here. First the experiment showed that you could increase the phytoplankton growth here, but they didn't show that you could increase this export because the timescale of this is much longer than you can stay out on a ship to follow it. So it could be that this is increased and then this arrow is increased and you have no net flux. There is a little data on this flux now, but it's not compelling yet. Then the next question is if this was an industry, could you actually verify how much carbon was exported, if it could be exported? And then the collateral effects, which we're going to talk about in a minute, could it be made profitable? And the biggest question is what would the unintended consequences be of such an intentional fertilization? And this is where you guys come in because you have learned in this class a lot of things that could help you assess what the unintended consequences are, and we're going to talk about that. But how could you make money doing this? And it depends on a lot of things. Right now you couldn't, but there are people depending on a future in which you could, and this is the way it works. There is an emerging market in carbon trading credits, especially for countries that did sign the Kyoto Accord where there's a commitment to reducing CO2 emissions. And so these carbon offset credits are worth money. So the way this would work is, if it works, was you'd have this industry, you'd fertilize with iron, you'd be able to claim that you buried X amount of carbon in the deep ocean. And with that claim that would give you these carbon offset credits that are worth money. Utilities companies could then buy those, and if there was a cap on how much carbon emissions they could have this would increase their cap. So it's just like, you've probably heard about tree plantations generating carbon offset credits. And that's a going industry now. So that's how it would work. And there are companies, these are some websites. Planktos.com, they have patents filed on this process. How they're able to do that is beyond me since it's published in the open literature, in the scientific literature, but they do. Their mission is to develop formulations to [manage? phytoplankton productivity in carbon export. There's another one that you don't have in your slides. I just slipped this in this morning. But here's the website, Green Sea Venture, if you're interested. But it's another company that is marketing this idea. And here's another. I just saw this ad last month in [EOS?] which is a publication of the American Geophysical Union. And it seeks professionals to work on an ocean nourishment demonstration. This is a new word. The oceans need to be ìnourishedî. So there's a psychology here. They'll talk about ocean deserts that are nutrient poor, need to be nourished. And this is not only for sequestering CO2, but the claim is to increase wild fish stocks by fertilization. And so some of these outfits are really on the edge of credibility, and I will let you figure that out yourselves. If you did enough research you could find out. That's not part of your assignment for the class. But some of them are actually really rational-thinking scientists and engineers behind them. So there's a whole spectrum of people interested in this. So why am I concerned about it? In fact, one of my good friends, who is a very good scientist, really thinks we should explore the idea of fertilizing the Southern Oceans to bring the whales back. Because he thinks that the whales are gone because the krill are gone, the krill are gone because the phytoplankton are gone and the phytoplankton are gone because the whales aren't there to recycle the iron. And we don't know any of that, but that's the hypothesis. He thinks the ecosystem needs to be ìjumpstartedî by an iron fertilization. So just to give you an idea the way people are starting to think about ecosystems and our ability to manipulate them. So why am I worried about this? And why should you be worried? Because you know now from taking this class that it's not that simple, that ecosystems are complex. And you know that if you fertilize with iron and you create a lot of organic carbon, phytoplankton, and some of that settles, a lot of that settles down to the deep ocean where there is no productivity, if it's consumed and digested by bacteria oxygen is going to be consumed, right? And respiration, heterotrophic bacteria, that's what you learned in the first set of lectures, oxygen will be consumed and CO2 will be regenerated. But if you have enough of this it will actually, the oxygen in the deep ocean waters will decrease and can even go anoxic if you do it long enough. And when you change -- That's the function of the system, the oxygen concentration. You change the community structure of the system and you'll have a different assemblage of microbes there than you had before. And one of the things that could happen, for example, is that you would increase the ammonia concentration by this reminerialization, you could stimulate nitrification and denitrification, which you learned about in my second lecture, I think. And you remember that a byproduct of that is nitrous oxide. And nitrous oxide is also a greenhouse gas that is 300 times more effective molecule per molecule than CO2. In terms of its greenhouse capability, in terms of its absorption of heat. So you're doing this whole thing to suck CO2 out of the atmosphere, but if a side effect is creating nitrous oxide, the amounts of which are impossible to predict at this point, you could be worse off than you started. And none of these proposals takes this downstream effect into account. The other thing that can happen in low oxygen waters is the stimulation of methanogenic bacteria which produce methane that is 22 times more effective molecule per molecule CO2 as a greenhouse gas. So there are ecosystem consequences to this. You cannot just say I'm going to add this and make carbon and that's the end of it, because you make carbon and things happen to that carbon. The other thing that is overlooked that you guys know about, remember this diagram of global ocean circulation? If you fertilize the Southern Ocean with iron and utilize the nitrogen and phosphorus here, when those waters upwell over here along the equation that nitrogen and phosphorus isn't in them. Now, that nitrogen and phosphorus is fueling the productivity of these ecosystems upon which fisheries are based. People are fishing the fish from those systems. So those people should be able to say to these people, hey, you took my nitrogen and phosphorus. That has to be factored into your balance sheet. My loss of fish, my loss of income from the fish needs to be factored into your balance sheet for your carbon credit. And if you do that the profitability is really marginal. Finally, there are models from a group at Princeton that show if you do this in a sustained way that after many, many years, this is 1500 which is the extreme, but even after 100 years you create, this is latitude. Here's the equator 40 degrees, 40 degrees south. So this is a broad swath of the ocean and this is depth, OK? So this shows a huge, huge anoxic zone in the oceans that would be causes by sustained fertilization in this way. Not surprising, you're making a lot of organic carbon. It's going to be consumed by bacteria. So that is all a story, a scenario to get you thinking about your future and your relationship with the earth's ecosystems in the future. Because our relationship with these systems is changing dramatically. We are now in charge. We've been in charge for a while, but we haven't taken the responsibility of being in charge for a while of these systems. And your generation is going to be making these kinds of decisions. As we move forward and we experience the aftermath of some of the manipulations we've done in the past like burning fossil fuels and increasing CO2 in the atmosphere there will be choices to make. Do we just adapt to this global warming or do we know enough about how the earth works to try to counteract it in ways like fertilizing the oceans. And then you'll have to decide are the risks of fertilizing the oceans much greater than the risks of adapting to climate change? And a new trend in thinking about the earth is thinking about nature and ecosystems as not simply as something that we value because they're part of our world and we should set up reserves so that we can enjoy them and see nature and so generations in the future will know what natural ecosystems look like, but starting to think of ecosystems as things that provide services for humans and actually have a monetary value that is not part of our economy, it's not part of our economic system but they provide functions to our world for free. And so as we destroy them we're losing those functions. And this is a very well known paper. There's a journal called Ecological Economics that tries to talk about factoring in ecosystems into our economy. And this is a very well known paper on the value of the world's ecosystem services and natural capital. And course this is impossible to do but you've got to try. So they evaluated the ecosystem services as worth $33 trillion. And this is just a brief list of some of the services that they analyze in this. One that's pretty easy to wrap your brain around is the pollination of crops by insects. We rely on insects to pollinate crops. They do it for free and we rely on that. And just this is estimated to be worth $6 billion. That if you had to hire somebody to manually pollinate your crops if there were no bees and all of the things that are doing it, it would cost $6 billion globally. There is decomposition of waste, recycling nutrients, dispersion of seeds, control of pests, purification of the air and water, ecosystems to do this. And this is compared to the GNP of $18 trillion per year, the global gross national product. So this is sort of a shift in our thinking about how we think about ecosystems. And just in the last -- This slide was in one of your handouts about, I don't know, five lectures ago, but I didn't get to it so I thought I would bring it in now. Just in the last, well, April 14th was announced this Millennium Ecosystem Assessment Report, which was over a thousand scientists worldwide assessing the state of the earth's global ecosystems and from the point of view of strengthening capacity to manage ecosystem's sustainability for human well-being. So the focus is on how do we manage these things for ourselves and the future generation? And you can go to the website if you want to get depressed. Well, you can get depressed about what we've done. But you can be optimistic that we're really facing up to this challenge in a very systematic way. But the bottom line is that two-thirds of the natural machinery of the earth has already been degraded by humans. And water use is dramatic. Major rivers are dry before they reach the oceans. We are mining groundwater basically. We're taking water out of the ground much faster than it's being recharged. One-quarter of all fish stocks are already over-harvested. I mean we're in a non-sustainable mode. I don't need to tell you. You read the newspaper and you know this, and so I'll skip over this. Oh, just this last part I think is the most important. The argument is that more and more people are going to be living in the cities in the future, so less and less in touch with nature. So nature is going to be more and more of an abstraction to us. And people are really worried about this because conservation of natural spaces is not just a luxury. This is a dangerous illusion that ignores our dependency on these systems. And we have to really strengthen that understanding. OK, so now I want to lead you, this slide several students have told me is the only thing they remember from this class and my other ecology class. And that makes me very happy. If this is the only thing you remember that's great because this is really the take-home message for you and your generation. And this is how we have changed our relationship to the earth and this is where we are right now. So you learned in the first lecture about the biosphere, autotrophs producing organic carbon, heterotrophs consuming organic carbon and with a little bit of input from the earth's crust, nitrogen and phosphorus, and this system running pretty well before humans. And before the Industrial Revolution societies fit into this system of tight recycling taking a little bit off of the autotrophic productivity and putting a little bit of waste into the system. But then we had the Industrial Revolution. We've cut down massive amounts of trees. we've changed the very landscape of the autotrophic system, and we've dramatically increased the waste stream. And we're mining the lithosphere in a huge way, elements, mining for metals and things used in manufacturing, etc., which is, of course, increasing this waste stream. So this is where we are now. And we know that this is not sustainable the way we're operating. So this is your generation's decision. Are you going to go this way or are you going to go that way? Very simple. You've got to point your compass in the right direction. So this way just increases these streams, the waste stream and the erosion of the natural ecosystems and the mining from the lithosphere. This stream works toward recycling within the societies that we've already built, restoring natural ecosystems so they can do their functions properly for cleaning air and water, and leaving enough productivity for the rest of the heterotrophs on land. You know we're not the only heterotrophs. There are all these other species that rely on this primary productivity, the birds and, well, all the species. So we need to leave some of that for them so the ecosystems can sustain themselves. And so this is where we should be. And we just need to find the will to get there. So easily said, not easily done. And there's an organization called The Natural Step. I put this website, which you might want to write down because I don't think this was on your slide, if you're interested you can write it down, that I think has a very, very creative approach to working with industry to try to direct things in the right direction and where they talk about the compass. And they say there are basic system conditions for sustainability that if we don't maintain we will be going in the wrong direction. And it's just a total no-brainer. It's very simple. Substances from the earth's crust must not systematically increase in nature. Obviously, that's not sustainable if you do that and substances produced by society must not systematically increase in nature. We cannot keep pouring waste into nature. The physical basis for the productivity and diversity of nature, i.e., the green part, the autotrophic cycle cannot be systematically deteriorated. And then they add to this we must be efficient enough to meet basic human needs. We have to work toward efficiency which is working toward tightening this recycling here. And they've been very effective in many countries in pointing industries in the right direction. Every time that they make a decision about what metal to use in a particular manufacturing process, they look at how much of that metal has already been mined. Is there an alternative metal that they could, etc. They're constantly using at this compass. OK. Obviously, the compass is not pointed in the right direction. A no-brainer. Building cars that use more fuel and get less miles per gallon is not the right direction. So some of these are very easy to answer. And I don't want to cast dispersion on SUVs, but in case you didn't recognize it that's what that is. I mean fuel efficiency is here, we can have it if we want to, and we need to work on that. And finally, well, this isn't finally. This is second to finally. I love this cartoon. This is the [doctor Saturn? or whatever looking at the earth and diagnosing our planet at this stage in its evolution. And finally I'll leave you with this image which is my favorite image of the earth because it has no national boundaries and it really does remind us that it's a living planet and that time is now. I mean, I know this sounds overdramatic but it's not. I mean we've changed this planet so much in the last 200 years relative to all the years before. And the next 50 years, your time to make a difference is absolutely critical, so I hope you guys will go off and save the planet for us. Have fun and have a great summer.
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https://ocw.mit.edu/courses/5-08j-biological-chemistry-ii-spring-2016/5.08j-spring-2016.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ELIZABETH NOLAN: We're going to move on with GroEL/GroES and a few more comments about where we closed yesterday and then talk about experiments that were done to determine what polypeptides are folded by this machinery. So I'm just curious. Has anyone stuck trigger factor or GroEL into PubMed to see how many hits you get? Yeah, so Rebecca's question yesterday, or on Monday, about trigger factor and active versus passive folding motivated me to take a look. So just to give you some scope, if you put trigger factor in PubMed, as of last night, there's 11,810 hits there. GroEL is closer to 2,000 to 3,000-- in that range. If you put trigger factor active folding, you end up with 34 hits. Most of those are about using trigger factor in protein overexpression. So if you also express trigger factor, does that help? And it looked like there was one paper in those 34 that suggests an active folding role for one of the domains. But that is just looking at an abstract. And so, the point there is there are many, many studies that consider these chaperones and a huge literature to search. So what we're able to cover here is really just the tip of the iceberg for that. There's also a new review out on GroEL/GroES, which is not required reading, but we're posting it on Stellar. So it just came out last month, and I really enjoyed reading this review. I thought they did a very good job of talking about current questions that are unanswered yet in terms of models and presenting different models for how this folding chamber works-- so passive versus active, for instance. And they also give a summary of the substrate scope-- so the experiment we'll talk about today. So where we left off last time, we went over the structure of this folding chamber and here's just another depiction of the overview. So effectively, we have to back-to-back heptamer rings as shown here. Some polypeptide in its non-native state can bind. It initially binds up at the top by these apical domains, and there are some hydrophobic interactions. OK, ATP also binds, and we have all seven ATPs found within one ring, the ring that has the polypeptide. OK, we see the lid come on, and then this polypeptide has some time, a residency time, in this chamber to fold. And then after the residency time, which is generally quoted on the order of 6 to 10 seconds, the lid comes off, and it gets ejected. And during that time, the ATPs are hydrolyzed. So somehow, this ATP hydrolysis gives conformational changes that drive this cycle. OK, and then we see, again, we flip to having function in the other ring. So one point to make involved cooperativity, so I hope you've all seen cooperativity before, probably in the context of hemoglobin. We have examples here of positive cooperativity and negative cooperativity. So within one heptamer ring, ATP binds to all seven subunits. So that's positive cooperativity. And then we can think about negative cooperativity between the two rings, where we only have ATPs bound to one ring. So the other heptamer ring will not have ATP bound here. So what is happening inside this chamber? The polypeptide enters the chamber, and it's given this protected environment to fold. And we saw that when the GroES lid comes in that the hydrophilic nature, hydrophobic nature of the interior changes. And it becomes more hydrophilic. So I just want to point out-- and this also builds upon Rebecca's question from last time-- is this passive folding in the chamber so effectively in Anfinsen's cage, where the primary sequence dictates the trajectory? Or does the actual chamber itself play a role? So that would be active folding. And effectively, is there forced unfolding or refolding by GroEL itself? So perhaps the apical domains can force unfolding before polypeptide is released into the chamber. And the cartoon that was just up indicated that to some degree. Maybe the cavity walls are involved. And what I would say is that the pendulum on this has swayed quite a bit over the years in terms of whether or not GroEL is a passive folding cage or actively involved in folding. And some of the debates in the literature have resulted from experimental set-up that may bias results to indicate one thing or the other. And that's something the community is striving to work out these days. And I'll talk about that a bit more on the next slide. But I'll just note-- these questions are still there, and the recent review I just noted discusses these questions. There was a study just a few years ago that was performed with very dilute polypeptide substrate-- so below one nanomolar. And what they conclude from this study is that GroEL is involved in active folding of a maltose-binding protein mutant. One question I'll just spring up with this is, maltose-binding protein is a nice model polypeptide, but what happens for a native GroEL substrate? And is there utility in studying those? So why have I emphasized this dilute protein sample point here? So what happened in some early work, in terms of studies that were done to try to differentiate active or passive folding, is that there were some complexities in in vitro studies. So, here, I just have a cartoon of folding in the chamber. And if we think about only one polypeptide within the GroEL chamber, it's folding in isolation. So there's no possibility for it to form an aggregate or a ligamer with other polypeptides. It's all alone here. So this folding in the chamber avoids the complications of the folding landscape we talked about in the introductory lecture to this module. So what happens in aqueous solution, right? There's the possibility that, depending on your conditions, maybe there's some sort of aggregate that forms. And if this aggregate forms, what does that mean in terms of what you see? And so, in earlier work, there were some in vitro kinetic studies that indicated GroEL accelerates folding relative to folding in dilute aqueous solution. But some of these comparisons weren't appropriate, because as it turns out, oligomerization might compete with what you're watching for. And so, if there's some oligomerization happening, it might indicate that the rate is slower than you think. So there's ways to monitor for this. And it's just a point in terms of what control studies do you need to do to make sure your experimental setup is appropriate there. I think it'll be exciting to see what's to come in future years about this question and what kinds of biophysical techniques are applied, including single-molecule studies here. So where we're going to go, moving on, is to think about what actually are the substrates for GroEL. So what polypeptides get folded in this chamber? And how do we begin to address that question from the standpoint of what's happening in the cell? OK, so first, we're just going to consider some observations. And then we're going to go into the experiments here. So here are some observations. So the first one is that polypeptides, up to 60 kilodaltons, can fold in this chamber. So that's quite big-- 60 kilodaltons. Some proteins or polypeptides need to enter the GroEL multiple times to be folded. So that means the chaperone has the ability to bind and release and re-bind the polypeptide here. So when studies are done in vitro, what's found is that almost all polypeptides interact with GroEL. So you just saw even an example of that in terms of this non-native maltose-binding protein. So many polypeptides will interact. And this really contrasts what's observed in the cell, where, in vivo GroEL is involved in only folding about 10% of E. coli proteins here. OK, so what observations three and four suggest is that GroEL has some preference for particular endogenous polypeptides. And what we want to answer is, what are these polypeptides, and what are their properties here? OK, so Hartl's group did some nice studies to look at this, what needs to be done. First of all, there needs to be a way to isolate the polypeptides that are interacting with GroEL in the cell. And then, once these polypeptides are isolated, they need to be analyzed in order to learn about their identity and properties. OK, so we're going to look at experiments that were done to address this. And they involve pulse-chase labeling of newly synthesized proteins, amino precipitation, and analysis here. So in terms of addressing what are these substrates, we're going to begin with pulse-chase labeling. OK, so basically, the goal of this experiment and why we're starting here is we want to determine which proteins interact with GroEL. And, in addition to which proteins, we want to determine how long they interact. OK, so what is the experiment? These experiments are going to be done with like E. coli cells. So we want to know what's happening in the cell. So imagine we have an E. coli. And so these bacteria are grown in some culture medium. And the trick here is that they're going to be grown in medium that's depleted in methionine. So incubate, or grow, in medium with no methionine. OK, so effectively, we're depleting them of that amino acid. OK, so then after some period of growth, what are we going to do? We're going to spike the culture with radiolabeled methionine. And this is the pulse. So we're going to add 35S methionine. And we're then going to incubate for 15 seconds. OK, and so that's the pulse with a radiolabeled amino acid. Then what are we going to do? And after we go through the steps, we'll go through why. After this stage, we're going to add excess unlabeled methionine. And we're going to then continue this culture for 10 minutes. OK, this is the chase here. And during this chase period, basically, samples will be taken at varying time points. OK, and then, at some point, we're just going to stop this. OK, so just say, stop culture and experiment. So what's happening in each of these steps? And why are we doing this? So what we want to do is think about newly translated polypeptides. OK, so we have a living E. coli. It has ribosomes. And these ribosomes are going to be synthesizing polypeptides over the course of this experiment. So during the pulse period, all proteins, or all polypeptides, synthesized are radiolabeled. Right, because the methionine has been depleted from the culture medium. And so effectively, the methionine that these organisms are seeing are the S35-labeled methionine. And all polypeptides have an informal methionine from the initiator tRNA and what other methionines are in the sequence. So, if we think about doing this for 15 seconds, and we think about the translation rate, which I gave as 6 to 20 amino acids per second when we were discussing the ribosome, we want to think about how long are these polypeptides going to be? So we have a translation rate of 6 to 20 amino acids per second. OK, so, if we think about 15 seconds of a pulse, we're getting polypeptides on the order of 90 to 300 amino acids synthesized during that time. So newly synthesized polypeptides in these 15 seconds are radiolabeled. What happens next? OK, we have this chase period where we flood the system with unlabeled methionine here. Why are we doing this? So certainly, there are some polypeptides that are longer than 300 amino acids that still need time to be synthesized. And if there's new peptides being synthesized that start in this stage, we won't see them, because this unlabeled methionine is in vast access over the radiolabeled methionine that was added early. So here, we have, the synthesis of larger polypeptides can be completed. And we have, no longer producing radiolabeled new peptides. OK, so this allows us to only see the peptides that were radiolabeled during this pulse period here. So what are we going to do in terms of the sampling at various time points? So let's say we want to sample at one minute, five minutes, ten minutes. What do we need to do? So can we just aliquot some of these E. coli and put them on our bench? We could, but that's not going to be very helpful to us, because what we want to do is stop the translation machinery and all of the cellular machinery here. AUDIENCE: You need some kind of clench? ELIZABETH NOLAN: Yeah, we need a clench. And not only do we need a clench, we're dealing with a living organism too, right? So we need to break open the E. coli in whatever this condition is to stop the reaction. OK, so we're going to take aliquots at varying time points. And basically, we care about time, so you have to immediately lyse, or break open, the cells. And this was done in the presence of EDTA. So what is EDTA? AUDIENCE: Ethylenediaminetetraacetic acid. ELIZABETH NOLAN: Yeah, ethylenediaminetetraacetic acid. So it's the chelator. And why might this lysis be done in the presence of this metal chelator? AUDIENCE: [INAUDIBLE] processes like-- [INAUDIBLE] magnesium, which would help [INAUDIBLE] AUDIENCE: Are the proteases that are not binding? ELIZABETH NOLAN: There certainly are zinc proteases. So that that's one class of protease. So EDTA will chelate many, many different metals. The main point here is we want to stop stop translation, shut down processes here. OK, so we have these samples. What do we need to do next? We need to detect these newly synthesized proteins that interact with GroEL. And we want to do this at each time point. So how are we going to do this? We have a very complex mixture that has all of the cellular components. So the next step in this will be immunoprecipitation. And so, what will happen in immunoprecipitation in these experiments is that the researchers had an antibody that binds to GroEL. And this antibody was put on a bead and used to fish out GroEL from this complex mixture. And we need to talk about these antibodies a little more. But just in starting, I imagine there's a bead. And we think about antibodies as being Y-shaped biomolecules. So here, we have a GroEL. And imagine that, in this mixture, we have GroEL that has some polypeptide bound. That's one of its endogenous substrates. So, if these are mixed together, then the antibody binds GroEL with the polypeptide attached. OK, here, we can imagine "capture" of this species here and using the bead to separate, say, by centrifugation. So let's think about this a little bit and a little background to have everyone up to speed. If you need to learn more about antibodies, please see a basic biology textbook for further details. But these are Y-shaped molecules that are produced by a type of immune cell called B cells. And they're used by the immune system to detect foreign biomolecules and help to neutralize them. And so, in these, the tip of the Y contains the paratope that ideally binds specifically to a particular epitope-- in this case, GroEL here. And so, we often think about a lock-and-key model with antibody and think about the antibody binding its target with precision here. So for these experiments that were done, just realize the researchers had to come up with an antibody to GroEL. How is that done? They may have immunized, say, a rabbit or given a rabbit GroEL and allowed that rabbit to produce antibodies. And then they isolate the antibodies here. So something we want you to take home from this course is, yes, the antibodies should bind the target with precision. But there's huge problems in terms of use of antibodies in research. This is just the start of an article that was published last year around this time. And it's focused on pharma and clinical trials. But this is much more broad. And often, antibodies aren't as specific as indicated by the label on the container from the supplier here. And it's pretty dismal what they quote in this terms of how difficult it is to reproduce data here. So if you're going to use an antibody, you always need to test it to see whether it is selective or not for the species of interest that you want to detect there and have that information on hand so you don't misinterpret your data here for that. So what are the steps for this immunoprecipitation? Basically, as shown on the board, beads will be functionalized with the antibody and then just added to the cell lysate. And the antibody can recognize GroEL. And the goal and hope are that whatever polypeptides are associated with GroEL are pulled down together. So that's something a bit incredible here that these polypeptides remain bound to GroEL during the steps of this process. You can imagine, if there's a low-affinity binder, it could be lost. So the sample can be centrifuged. And then, you can isolate these beads here. So, in cartoon form, a complex cell lysate in your microcentrifuge tube. You can add the antibody, centrifuge. And see, down here, we've pelleted the beads with GroEL attached. And then some sort of workup needs to be done to dissociate the protein, or polypeptide, substrates here. And then they can be analyzed. AUDIENCE: How long do they do that for? Do you know how many-- ELIZABETH NOLAN: How long do they centrifuge for? AUDIENCE: No, for the immunoprecipitation. Is it 30 minutes? Is it-- ELIZABETH NOLAN: I don't know how long the incubation time is. Need to go back to the experimental, but that's getting right back to this question as to how do they stay bound. AUDIENCE: How do they stay bound? ELIZABETH NOLAN: Yeah. So, see the point here. If you have a high-affinity complex, that's one thing. If you have low-affinity association between GroEL and the polypeptide, you can imagine it might get lost during this workup. And how much do we know about those affinities there? AUDIENCE: You said that they would just give rabbits GroEL, and hopefully antibodies would just happen. But if a rabbit's immune system encountered GroEL, would it actually see it as an antigen that it had to develop antibodies against? ELIZABETH NOLAN: So, yeah. So here's the point-- would it? So, if it's E. coli GroEL, would the rabbit recognize this, yes or no? And if no, then what can you do to provoke an antibody response? And so, what can be done is, say, you could take a GroEL subunit and attach that to something immunogenic. So there are carrier proteins that will mount an immune response. So one of the subunits of cholera toxin is an example that can be used. And then the idea is you're mounting an immune response against that carrier protein. But you'll also get antibodies to whatever is attached. So that's another strategy for doing it if direct injection doesn't work. And too, not going off on a big tangent, but there are some decisions that need to be made. So would they use the full-length GroEL? Or maybe they would just use a polypeptide region, like some shorter polypeptide that's a portion of GroEL. So there's a lot of possibilities there in terms of what you use to generate the antibody for that there. And it's something that a lot of companies do these days. You can send them your protein or your polypeptide fragment. And they'll conjugate it to one of these carriers and treat the rabbits or whatever animal and then isolate those antibodies. And then they need to be characterized there for that. OK, so how are these samples going to be analyzed? That's the next step. So, for the analysis, effectively, we're going to have some mixture. And, at the onset, we don't really know how complicated this mixture will be. I told you initially that about 10% of E. coli polypeptides are thought to be substrates for GroEL, which is quite a large number if we think about the total number of proteins in E. coli. And the other point is we have this radiolabel, which we're going to use for detection there. OK, so, for analysis-- OK, there's two things. We need to separate these various polypeptides in each sample. And then we need to determine what their identities are here. So-- that were bound to GroEL from one another. OK, and then, we need to determine identities. And once we know the identities, we can think about their properties. And this needs to be done in every sample that was collected along this time course, which is also going to give some temporal information. So what are the methods that have been used? So, in order to separate the proteins in this complex sample, the method is a 2-D gel-- so 2-D gel electrophoresis. OK, and in terms of determining the identities, what's done, once these polypeptides are separated, is to do a protease digest and then mass spectrometry. Has anyone here ever run a 2-D gel or seen the equipment? One person. Has anyone heard of 2-D gels? Fair number. OK, so, we'll go over this briefly in terms of 2-D gel. So, in terms of 2-D gel electrophoresis, we talk about running these gels in two dimensions. And, in each dimension, we separate based on a different property. So, in the first dimension, the separation is based on charge. And effectively, we can talk about the pI of a protein. So the pI is the isoelectric point. And it's the pH where the net charge on the protein is zero. And so, the type of gel we use here is called isoelectric focusing, or IEF. And effectively, what's done is that the gel electrophoresis is done through a continuous and stable pH gradient. And, in this gel, the protein will migrate to a position where the pH corresponds to the pI. Then the anode is low pH and the cathode high pH. So that's quite different than SDS, where, in an SDS-PAGE gel, we're coating the protein with negative charge. So then, the second dimension is something most of us are familiar with, is SDS-PAGE. And so, what happens in SDS-PAGE? We have separation based on size here-- on molecular weight. So has anyone not run an SDS-PAGE gel? And this is totally fine. I never ran one till I was a postdoc. So it's not something to be ashamed about if you haven't. OK, so everyone has. So what's the ratio of SDS molecules to amino acids? So if you take your protein sample and you put it in your loading buffer and run your SDS-PAGE, what is the ratio of binding? What is SDS? AUDIENCE: Sodium dodecyl sulfate. ELIZABETH NOLAN: And what does it do? What happens to your protein in SDS? AUDIENCE: Denatures it. ELIZABETH NOLAN: OK, what else? So it's a denaturant. So it denatures the protein. So why does SDS-PAGE let you separate based on molecular weight, more or less? AUDIENCE: It coats the protein, more or less, uniformly with negative charge. ELIZABETH NOLAN: Yeah. AUDIENCE: Do we know the exact ratio of binding? ELIZABETH NOLAN: Yeah, so what's the ratio of binding that can be done in terms of grams of SDS per grams of protein or number of SDS molecules per amino acid. What is it? And there'll be some error, but there's approximates. But it's something to think about, right? You're putting your sample into this. So it's about 1.4 grams of SDS per gram of protein. That's the ratio there. And as said, the idea is that SDS is giving the protein a large net negative charge. So it's going to override whatever the intrinsic charge is of the protein. And so, it gives all proteins a similar mass-to-charge ratio here. With that said, sometimes, there are proteins that migrate in the gel in a manner that's not reflective of their molecular weight. That's just something to keep an eye out on. So within the slides that will be posted on Stellar, there'll be some background information about both of these methods-- the IEF gel and SDS-PAGE, which I encourage you to take a look there. OK, so back to the 2-D gel-- how is this actually going to be run? So it's one gel. First, it needs to run the IEF gel. And you need a special apparatus for. This it's called a cylinder, or tube, gel-- so not flat like what you're all accustomed to for SDS-PAGE. Then, this gel needs to be equilibrated in the SDS-PAGE buffer. And then, you run the SDS-PAGE separation. And, in this step, just to note, the gel is rotated 90 degrees. OK, so what you get-- you get a gel where we have molecular weight here. We have pI here. And if it's a cell lysate, there's going to be many, many spots. These should all be spots unless you did a poor job running the gel. So this 2-D gel is being used, because it's going to provide better separation than a standard 1-D gel. Imagine trying to separate peptides out of some cell lysate using just a 1-D gel. Even after this immunoprecipitation, we'll see that these samples are very complicated here for that. So what we need is some way to detect the spots that indicate different polypeptides. So what are methods? Maybe Coomassie stain for total protein. We can use the radiolabel-- autoradiography, for instance, which is what's done here. We're looking at the S35 radiolabel-- or maybe Western blot here. So how are we going to get from this gel to knowing the identity of each of these spots? AUDIENCE: You have to identify your spot, excise it, extract the protein from the gel, adjust it, and then run NS and line it up with known protein for evidence. ELIZABETH NOLAN: Exactly. So what will be done is that each spot of interest will be cut out of the gel. So you need a way to mark them. You'll see they're numbered in the data that we'll look at. The protein needs to be extracted out of the gel. Then the protein will be incubated with a protease that will give some number of fragments. Trypsin was used in this work. And then that digest can be analyzed by mass spec. And so, for each sample, you get all of the m over z values for the different polypeptides that resulted from the digest. And then, effectively, you can compare that to some database of E. coli protein sequences. So further details are provided throughout here. So what are the major questions? And what are we going to look for answers for in the data here? So first, how many proteins interact with GroEL? We can imagine getting an answer to this by counting the number of spots. What are the identities and structural features and properties of the proteins that interact with GroEL? We're going to get that from the mass spec analysis and then literature studies. And then another question we can get at is asking, how long do proteins interact with GroEL? Because we're calling the pulse-chase samples were taken at various time points over that 10-minute period. So, at two minutes, do we see the same polypeptides associated as we see at 10 minutes? Or if we monitor one given polypeptide, when does it show up and potentially disappear from the gels? So all of these samples can be addressed with these methods. And where we'll begin on Friday is going through the data in some detail. But just as a prelude to that in the last minute, here's the data from the paper for these gels. So this is looking at the 2-D gels for, on the top, total soluble cytoplasmic proteins at zero minutes and then total cytoplasmic proteins at 10 minutes. So this is without the immunoprecipitation. And then, at the bottom here, what we're looking at are the polypeptides that we're isolated from the immunoprecipitation with the anti-GroEL antibody at zero minutes and 10 minutes. And so, before we meet next time, what I encourage you to do is take a close look at these gels and see what information can you pull out just from a qualitative look. So simple questions, like, we see a lot of proteins here. And please don't go and try and count all the spots. I'll give you the numbers next time. How do these gels here from the immunoprecipitation differ from these up top? And it's not just the total number of proteins. There's some additional subtleties in these data. OK, so next time we'll begin examining these data, looking at what polypeptides were pulled down. And then we'll move into looking at the chaperone DnaK, DnaKJ system there.
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https://ocw.mit.edu/courses/5-112-principles-of-chemical-science-fall-2005/5.112-fall-2005.zip
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The following content is provided by MIT OpenCourseWare under a Creative Commons license. Additional information about our license and MIT OpenCourseWare in general is available at ocw.mit.edu. Today is a very big day for all of you. Your first semester at MIT is half over. And, in addition to that, you have a lot of challenges remaining ahead of you for the rest of this semester. I am going to help make the chemical challenges fun for you. That is what I am here to do. I love chemistry. I love teaching 5.112. This is the best part of my week. So, here we are. Do you guys know that chemistry can be incredibly fun? Let me just show you how fun chemistry can be. And that is just the gravy. The fun part is actually making the molecules. Any of you see this picture in U.S.A. Today recently? Not reading that important journal, I see. Well, this picture did appear very recently in U.S.A. Today in a story that talked about the recent Nobel Prize in chemistry that was awarded to Professor Richard R. Schrock, of our department. And, in fact, I don't know if you realize this, but Professor Schrock was the first person ever awarded the Nobel Prize in Chemistry at MIT for work conducted while at MIT. That is a pretty amazing accomplishment because we have so many tremendous chemists on the faculty here, and he is the first one to break through in that manner. And maybe you are here at a very special time, and we will see a number more of these in the next few years. May with you contributing, so much the better because you are here. And let me just explain something, too. Chemistry is fun. And only on very rare occasions like this would Professor Schrock be drinking champagne at 8:30 in the morning, when this picture was taken. The Reuters reporter, Brian, who took this photo, is actually a pretty good friend of Professor Nocera here. And so he rode over from the Schrock household. And Dick and Dan came in together in the same car, as they often do, from Winchester on that morning, which was a very exciting day. If you come to see me for office hours maybe I will tell you more about it in detail, but it was an exciting day. And I am in here, too. My name is Kit Cummins. Kit is my nickname. My name is Christopher. You can call me Christopher or Kit, or you can call me Professor Cummins, or you can just say hey, chemistry guy. That will be fine. I am in this picture primarily because it is part of being in the right place at the right time, just like I am here now in front of you. But also I am in it because we three are three in one of the areas in our department that really focus our research on inorganic chemistry. And what is inorganic chemistry? I will say more about this during the semester. But inorganic chemistry is really the chemistry of the elements of the Periodic Table inclusively. So, a pretty broad topic. And I like doing chemistry with lots of elements. And so that is why these guys and I share an association with each other. And, in fact, I was a graduate student with Professor Schrock. And I am very, very proud to be able to say that. Hopefully, my computer will be staying on, here. And I want to point out, too, that people who love science want to know something about its history. And part of all of that is knowing the academic family from which you derive, depending on the person you've studied with in learning about chemistry. And I studied with Schrock. And Schrock studied with Osborn, a fantastic chemist in his own right, no longer alive, sadly, to see his progeny, Professor Schrock, win the Nobel Prize. And Osborn studied with Wilkinson, and Wilkinson was also a Nobel Prize winner. So, you do not have to go very far back in that lineage to encounter that recognition, the Nobel Prize, Schrock, and then it skipped Osborne, and if you go back to Wilkinson, there is another one. And you can keep going back. And I am not going to go through that whole history today because I have another history that I am going to need to go through with you today. And I am going to bring this back over so it does not distract you. And this other history that I need to impart to you today has to do with some of the contributions of Gilbert Newton Lewis. This is one of the most famous of all chemists ever. And if you like chemistry, I encourage you to think of the word Newton, as it is found engraved in one of the pillars in Killian Court at MIT as the middle name of Gilbert Newton Lewis. It is conceivable that having that middle name weighed heavily upon Gilbert during his lifetime. But it remains a good moniker. And in the future, when you have kids, I encourage you to consider using it as a middle name. It worked well in this particular case. When you get the notes for today, which incidentally will be available on the website after class today, so you don't need to write down everything I am writing down. You are actually going to get a little bit more in the online version than what I am writing down here. And if you find that you forgot something that I said during class, just email me. I will tell you what I said. It is not a problem. But if you want to write down notes while I am speaking just to make notes that you can refer to later to refresh your memory, that is fine. I think your taking in information in lots of different ways. We are going to use the computer and we are going to use the blackboard today. And you are going to have available stuff from the Internet a little bit later, too. And all of that is good. I think you should try to reinforce and not worry if some of the information is a little redundant. And never be afraid to ask questions, especially, I like to take questions by email or if you come knock on my office door. In 1916, Gilbert Lewis published a landmark paper in chemistry in the journal of the American Chemical Society which remains the flagship journal of the American Chemical Society. And this paper was entitled "The Atom and the Molecule." Besides being an incredibly important and influential paper in the history of chemistry, "The Atom and the Molecule" was also the basis for -- Like I said, besides being a hugely influential manuscript, "The Atom and the Molecule" also laid the foundations for Lewis's thinking about electronic structure theory that shaped the way we still today talk about valence and chemical bonding. So, an incredibly important paper from 1916. In the lecture notes today, you will find the specific volume and page number for that reference in the journal of the American Chemical Society. And I encourage you in your copious free time to download that paper. It is really great. All the American Chemical Society journal articles are available, from the very beginning. And you can download it and read it. And, having had this introduction today, I think it will probably prove to be a very fascinating exercise for you, if you choose to do so. It also was the basis of a book that Gilbert Newton Lewis later wrote on valance and on atoms and molecules. That is also a classic in chemistry. And, should you be shopping for chemistry books on e-bay or something, you might find a rare copy of it. And if you do and choose not to buy it, please alert me and I will buy it. But Gilbert Newton Lewis, "The Atom and the Molecule." In this paper you are going to see, if you do choose to download it, that he was thinking about polarity. And he was also thinking about the relationship between the polarity of a molecule relates to the polarity of a substance. And I mean, when I say a polarity of a substance, that that would be a sample, you could say a water molecule down here and liquid water up here. And you will see this kind of sigmoidal graph that Lewis made in thinking about how polar condensed phase substances were as compared with the structure of the molecules that comprise the substance. Down here, in this region, you will find that there are substances that are not terribly polar. And what I have written here is the organic shorthand for the hexane molecule. Liquid hexane is an alkane, akin to methane, which is natural gas. And so, if you are not familiar with this way of writing organic molecules, I will just tell you briefly that this terminus is CH three or a methyl group, and then it is CH two, CH two, CH two, CH two, and then CH three, another methyl group. So, that is some organic shorthand. And I will go through that kind of shorthand occasionally so that you become familiar with it. And that is the hexane molecule, a very non-polar molecule composed of elements, carbon and hydrogen, that have very similar electronegativity. And then he goes on up the curve here, and you find this familiar molecule, benzene, considered next, as being somehow more polar than hexane, and liquid benzene accordingly in its solvating properties in the condensed phase, being a more polar medium, as it were. And then, as we continue progress up this scale, he writes next molecules like the diethyl ether molecule. No doubt you are familiar with ether and its applications as an anesthetic. The first use of ether as an anesthesia in an operation in medicine was carried out just across the river in Boston. There is this historical Ether Dome there that you can go and look at. But, nonetheless, we are just getting up a little bit more polar than benzene. And then next you find molecules like this one. That contains an organic functional group called an ester. That is ethyl acetate, which is a substance that you might have smelled if you ever built models, because some of the glues that you may have used would contain it, or the paints. And so, it is a system that, in terms of its molecular structure, is more polar. And, when you use it in the condensed phase, it behaves in a manner that you would describe as more polar. And then, finally, you get up to some very polar systems, like the water molecule. Like the ammonia molecule. And, finally, Lewis talks about a species like this, which is sodium twice sulfur. That is actually called sodium sulfide. And it is a salt analogous to table salt, sodium chloride. And being a salt, it is obviously very polar because there is quite a large degree of charge separation between the atoms in the molecule. And, accordingly, if you take that salt and heat it up until you have got it as hot as its melting point and make it liquid, that molten salt has individually charged ions moving around in solution. And that provides a very polar medium, which would be excellent at dissolving polar things. You will notice here that on the non-polar side we have elements, carbon and hydrogen, that are very close to each other in electronegativity. And then, as we examine this relationship, we start putting in elements like oxygen or nitrogen, which are very electronegative close to the upper right-hand side of the Periodic Table. That introduces polarity both in the molecule and in substances comprised of the molecule. And then all the way over to the involved atoms, sodium and sulfur are very different in electronegativity and, therefore, have very strongly separated charges. And, in thinking about this, Lewis proposed the notion of a continuum. And his musing on this are really quite lucid in that paper, but the idea was that a molecule, or a part of a molecule even, you can analyze molecules in terms of smaller pieces of them that can be identified within the larger molecule, can lie on either side of a continuum in reference to a number of different properties. And the section of class that we are launching into today deals with acid-base properties. Let's mention that first, acid and base. These are two ends of a continuum. But he also recognized that you could have a molecule that would be an oxidant. And the counterpart to that is a reductant. And then, here is a term that maybe those of you who have studied organic chemistry in more detail than most would recognize. This is electrophile. Electrophile has a counterpart called a nucleophile. And then, the fourth one that I would like to mention is acceptor. And, of course, the counterpart to that is donor. Please remember those four pairs because when you can understand and make predictions about those four properties as a function of molecular structure, then you will be a chemist. I mean that is a large part of what it is all about, understanding all the different chemical properties that will make it interested in terms of acid-base, oxidant-reductant, electrophile-nucleophile, and acceptor and donor. And the currency of chemistry can be seen, with reference to this, to be the electron. That is what chemistry is all about. And so, we will be talking a lot about atoms and molecules with reference to their properties and how the properties stem from knowing something about the electrons in the molecule. And I think you are going to find this to be quite fun. With respect to the properties of the medium, let's talk about a substance like hydrogen chloride. This is just to illustrate what I was saying over there. If you take a molecule like hydrogen chloride, and it is a gas, of course. And if you make a solution of it in hexane, hexane is going to be our solvent, and then what you find is that in this low polarity medium hexane solution, liquid hexane, the HCl molecule remains intact and floats around in solution, interacting weakly with hexane molecules that surround it. But if you take this HCl molecule and put it into a solvent like water, then something on the other end of the continuum transpires because we get ionization to produce hydrogen ions. And they go off and are separated in solution from chloride ions. And we will talk more about just what is going on when you separate H plus from Cl minus and let them diffuse apart in aqueous solution. But you can see, I think, the dichotomy. Either the proton shares in the electrons of the chloride, or it just pops off and ionizes and goes out into solution. And Lewis wanted to understand this. And, to understand it, he developed a theory. And his theory was the "cube theory." In this part we get to draw some cubes. And he decided that if you thought of the nucleus, which he calls the kernel, of an atom as being located at the center of the system, then if you had an atom like neon, which is a noble gas with eight valance electrons, that these valance electrons would want to get away from each other as far as possible because they repel each other. They are all negatively charged. This could be, here, the neon atom. Here, on that cube, is how he would represent the electronic structure of the neon atom, with the eight electrons here represented by peach-colored circles, each the vertex of a cube. So, they got far away from each other. And you can see that in the historical archives, some of his original sketch books containing these ideas are there, and now you will know what they mean. You will see that he progressed beyond understanding the atom to understanding the molecule. And here is one of the molecules that he talks about. Here we have a molecule containing 14 valance electrons, all arranged in that manner. And this could be, for example, the I two molecule. And let me draw it in terms of a dot structure, -- -- like that. And there was known at the time, partly because of the contributions of Lewis himself, something called the Rule of Eight. And that was a reference to this notion that most atoms want to have eight electrons in their valance shell, and it is also something that we call today the Octet Rule. So, when you see the Rule of Eight mentioned in his paper, it is the familiar Octet Rule. And you can see here that one of the key features is two electrons shared equally. And when electrons are shared equally like that, you have a molecule with symmetrical charge distribution, charge referring to the electrons where they are in space. And then, he showed that you could think about this another way. You could consider some kind of a geometric distortion of the I two molecule, like this. Distorted, not only because of my drawing, but because of the way the atoms themselves are thought to be rearranging in this process. Notice that what has happened here is that the I two molecule is doing something electronically such that one iodine is starting to pull away from the other. This one over here is carrying with it the electron that came from over here. And, in that case, what we have is a molecule in which we are developing partial negative charge on the right-hand side and partial positive charge on the left-hand side. As I was talking up here, I was thinking about what I was going to say here, because the ideal is that molecules, even if they are inherently non-polar like the I two molecule, in their equilibrium structure they can have electron fluctuations that lead to partial charges in the manner shown here. And Lewis was working beautifully toward the development of a theory like this that would allow the description of electronic structure to be made graphically apparent. And this is something that chemists really like to do, is we like to be able to see stuff. I hope I am going to be able to show you some really cool stuff this semester. And, in fact, even some really cool stuff before we finish here today. So, that is the development of charge as represented using the Lewis cube theory in an I two molecule. Next, one thing you could do is say, let's just continue this process, and the right-hand iodine with its eight electrons pops off. That would give you an ionized situation in which you had an I minus floating away from an I plus molecule. And that situation is drawn in the notes, and because I am running out of space on this board I am not going to draw that one up. Instead I am going to go onto another triumph of the Lewis cube theory, which was its ability to explain different chemical bond orders. By the way, Lewis never got the Nobel Prize. He certainly should have. A dramatic oversight. He was born in Massachusetts. He was a Harvard professor for a while and did not like it, and came to MIT to be a professor for a while. And, although he liked that, that was not far enough away from Harvard, so he went to Berkeley, where he became the department head and was very successful over a long period of time in learning about chemistry and in teaching chemistry. I will invite you, and you will see this in the notes, too, to go ahead and see if you can find a good biography of Lewis because it is very instructive. Anyway, here is another molecule considered by Lewis's cube theory. How many electrons are being shared by the two nuclei or atoms involved? Four electrons shared. We can circle them here. That is four electrons shared. Therefore, that is not a single bond. We call that a double bond. And so, that might be a representation of, for example, the O two molecule. Here we have I two, here we have distorted I two with charges developing, and here we have O two. O two is a very interesting diatomic molecule. We are going to talk about that later in the semester, but for now let's just leave it at that. But then, even though this was quite a successful theory, the cube theory, it has certain problems, certain shortcomings, such as, we can do single bonds with it quite well, and now we can do double bonds with the cube theory, but what do you do if you have to make a triple bond? That was a problem for Lewis. That should be a problem for you, too. Cubes just don't fit together that way. They have edges and faces, and that is it. And they have vertices. You can do one vertex. So, triple bond is a problem. And we know that there are molecules that have triple bonds. And we cannot handle them with the cube theory, so we are going to have to something a little different, for that reason. But there is also another reason that had to do with just considering the chemistry of light elements like hydrogen and helium. Those elements tend to associate themselves with only two electrons. And so, Lewis decided, well, maybe the rule of two is more fundamental than the rule of eight, and that we should be thinking about pairs of electrons somehow. Because he looked at what was known about chemistry and realized that in most molecules, the number of electrons is always even, is always divisible by two. That is something interesting about two electrons at a time being important. You take that together with the triple bond problem, and Lewis started thinking in terms of electron pairs, and this is where this "electron pair theory" comes from. Although he spent a lot of time on the cube theory, he was perfectly willing to cast that theory aside in favor of something that maybe was going to work a little better, the electron pair theory. And how does that look? Well, of course, his initial pictures still had the cube, which just shows that once the human mind fixes on something, it fixes on it with great tenacity and it can be hard to shake your way of thinking about things. But hopefully you will be able to do that. He still decided, let's arrange these electron pairs on a cube with the nucleus of the atom supposed to be at the center, and let's keep these electrons, now, nicely collected in pairs. And let's get these pairs as far away from each other as possible, for the same reasons we used to be putting eight electrons as far away from each other as possible, namely, electron repulsion. Now, the problem of electron repulsion between the pairs was not really solved. And Lewis decided that the pair theory worked so well that he was not going to worry about that, and so we won't worry about it yet, either. You have your four pairs of electrons arranged at the center of four of the edges of the cube, and that geometry is what we call a tetrahedron, which I am drawing the edges of here in pink. You have a tetrahedral disposition of four pairs of electrons, now, for an atom. This might again be the neon atom. But the nice thing about this choice is that if you want to you can make single bonds with this theory, and you can make double bonds -- -- with this theory, like that. And then, due to the properties of the tetrahedron, you can also make triple bonds. This is marvelous. You can see that the famous chemist Gilbert Newton Lewis was talking about electronic structure in terms of symmetrical things, like tetrahedra and cubes, as a way of arranging electrons reasonably in space. This could, again, be our iodine molecule. This could be, now, our O two molecule. And, if we continue that progression in the Periodic Table, what is this molecule? Dinitrogen, N two, the major component of our atmosphere, another interesting molecule that we will be talking about as the semester progresses forward. But chemical language and electronic structure theory are very much still permeated with this affection for symmetry and the problems that you can solve using symmetry to analyze electronic structure theory. Let's talk next about another molecule because we want to talk about Lewis's development of the electron pair theory, but also we want to understand how that relates to acid-base properties. And so, let me draw here another molecule. And when you look at this molecule, which is the aluminum trichloride molecule, in which I am drawing out explicitly all of the electrons, not forgetting these two. And having drawn the aluminum trichloride molecule and looked at it like that, you can ask yourself is the number of electrons divisible by eight? Yes it is. We have a 24 valance electron system here. It is divisible by eight. Of course, it is also divisible by two. You can also ask yourself is this molecule, as written, one that satisfies the rule of eight for all four of the atoms? No. We have a problem there because this thing is electron deficient. And if you were Lewis, you would want to know, well, if it is electron deficient, this molecule must be doing something about that to get more electrons. Which atom is missing electrons, here? The aluminum. How can we give aluminum more electrons, in a structure like this? Can we get it up to the rule of eight? Double bonds, awesome. Beautiful. We can make a double bond. And, if we do that, we can allow aluminum to satisfy the rule of eight, or allow the rule of eight to satisfy aluminum. Whichever way you want to look at it. And see what we have done, here? We have taken one of the electron pairs, say, this one, and we have moved it in between the aluminum and the chlorine, so it joins the other one in there. And we get a double bond. And now, everybody has an octet. But what is the problem with that? The problem with that is that chlorine is very electronegative, and it does not want its electrons stolen away by the very electropositive aluminum center. There are other ways that aluminum can ultimately satisfy its octet, and this is how we get into the realm of "Lewis acid-base theory." And I want to show you one of the other ways that aluminum can satisfy its octet, and then, we are going to move to the computer so we can visualize some of these properties. If I put my aluminum here -- and do you remember this molecule up here, the ammonia molecule? I am going to say that if I draw that ammonia molecule in place, here -- The ammonia molecule comes in, and it has its eight electrons. It is sharing six of them with three hydrogens. The idea is that it has an extra pair of electrons on that nitrogen in the ammonia molecule, that it can share with the aluminum. And now we go ahead and complete the picture. And we can have eight electrons around everybody. And so the rule of eight is now satisfied for the three chlorines, for the aluminum, and for the nitrogen of the ammonia molecule. The ammonia molecule is what we call a "Lewis base." And the aluminum trichloride molecule is what we call a "Lewis acid," for reasons that should now be quite clear. And I want to show you how we can look at these things. And if you will give me just a second here, I am going to load the aluminum trichloride molecule up onto the screen. What we are going to do is remember that there are lots of different ways in which you can visualize molecules. And I want to show you some of the more interesting ways. All that is is a picture of the aluminum trichloride molecule with spheres drawn at kind of arbitrary radii. The size of these spheres is not representative of any particular physical quantity, but we can change that. And so now, what we are going to do is draw the thing a slightly different way. And what you can see is that the positions of those nuclei are in this flat box. I made the box kind of flat. It is kind of a slab because this AlCl three molecule is conveniently kind of flat in its structure. And what are we going to do now? We are going to draw a contour surface. And I like this. We are going to use the solid surface method, first. And can you see that now? These curves that enclose the four nuclei of the aluminum trichloride molecule do have some physical significance, because what I have done here is taken results, in this case from the quantum chemistry calculation, but you could also take it from an X-ray crystallographic structure determination study. And I have drawn a three-dimensional contour map. The value of the electron density at every point on the surface of one of those ovaloid shapes is the same. Just like when you have a contour map to describe in two dimensions an interesting complicated surface, we can do this in three dimensions as well. And one of the things you notice, when you look at a surface like this, is that the value of the electron density actually goes down to some low value in between the nuclei, for this. When you see molecules represented as balls and sticks, or if you represent molecules this way, just showing a pair of dots between the aluminum and chloride, you may not be getting a physically complete picture of the situation. You can get a much more complete picture if you have access to the actual electron density. And then, you can even go a little further, and I want to show you something very cool. First I am going to need to load in a second molecule. And that will be the product of the first "Lewis acid-base reaction" that we have discussed. Because, I didn't say it over there, but that was your first exposure to a Lewis acid-base reaction, in which the ammonia molecule comes in and binds to the aluminum trichloride molecule, in order to satisfy the valence requirements of the respective atom types. And so, here we are going to look at that molecule, and we are going to look at its electron density. And I think you are going to like this very much. Here it comes. And then I am going to represent this as a solid surface. The value of the electron density everywhere on this surface is 0.1 electron density units. I want you to start thinking about how that shape appears there because I am going to underscore that a little bit more in just a second. But one thing you should notice is that at the NH three molecule, which is pictured at the top of that representation, what you are seeing is that the value of the electron density does not get very small as you go between the nitrogen and hydrogen nuclei because both nitrogen and hydrogen are quite similar in electronegativity as compared with the aluminum atom in this system. The fascinating thing about illustrating a molecule in this way is that we can actually associate another property with the color on the surface. Now what we are doing is looking at the aluminum trichloride ammonia system. And let me see if I can make this thing gradually spin, so that I can talk about it without being pinned to my computer. What we have on the surface on each of these electron density isosurface enclosed regions is a color that corresponds with a value of a function that tells us about the propensity of electrons to be paired at that point in space. The Lewis theory is now colored onto the electron density isosurface of this Lewis acid, Lewis base complex molecule. And the values that correspond to those regions in space where electrons are most likely to be found paired up are colored blue. Where they are least likely to be found paired up are colored kind of red-orange, here. And then an intermediate color is this green. And so, you can see that you have electron pairs associated with each of the three NH bonds, up at the top. You have another electron pair here in blue, which is very proximate to this aluminum center, that is very electron deficient as we had found right over here because it lacks an octet. And then, each of these chlorines is able to attract electron pairs. Because chlorine is very electronegative, it attracts the electron pairs to itself and sort of withholds them from the aluminum center, but yet, that development of charge separation keeps everything together in what we can call an ionic interaction, an electrostatic interaction. You can see here covalent bonding, donor-acceptor complex formation, and electrostatic contributions to bonding. And all of this is just an illustration of the brilliant ideas of Gilbert Newton Lewis.
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https://ocw.mit.edu/courses/22-67j-principles-of-plasma-diagnostics-fall-2023/22.67j-fall-2023.zip
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PROFESSOR: So we're very fortunate today to have Dr. Maria Gatu Johnson join us to give a guest lecture on neutron diagnostics. Maria is a principal research scientist at the PSFC. She did her PhD working on neutron spectroscopy on jet, on magnetic confinement. But now she works on inertial confinement fusion. And in particular, her work on the magnetic recoil spectrometer on NIF was a key to understanding the record neutron yield that we got last year in the ignition shots. So we're very fortunate to have Maria here, and looking forward to hearing her talk. MARIA GATU JOHNSON: OK, thanks, Jack. So an hour and a half on nuclear diagnostics. For ICF, that's kind of a lot to squeeze in. So I did a bit of a sampling. I want you guys to ask questions as we go through. If there's anything you're particularly interested in, stop me and we'll talk a little bit more about it, try to touch on the key things. Also, there's a lot of familiar faces today. So a lot of you know details of what we'll be talking about today, in some cases, better than I do. So then feel free to chime in as well. So with that, we'll get started. So yeah, as Jack said, we're going to be talking about nuclear diagnostics for ICF plasmas. This, in particular, illustrates three facilities where we do ICF work-- the National Ignition Facility in Livermore in California, the Z facility at Sandia National Labs in Albuquerque, New Mexico, and the OMEGA laser in Rochester, upstate New York. And this is actually a picture of the instrument that Jack just talked about in his introduction, the magnetic recoil neutron spectrometer installed on the NIF target chamber, which is the blue, circular, spherical part that you can see there in the background. But I thought, as part of this, we'll touch a little bit on-- if I can get this to move forward-- some comparisons to diagnostics for magnetic confinement fusion as well. So these pictures here, that's the inside of the jet [INAUDIBLE] time of flight neutron spectrometer at the [INAUDIBLE] which I did build for my graduate work. So we'll talk about that one a little bit as well when we get to that point. Brief outline-- find this a little weird between switching the computer and being behind the camera. We'll start by talking about implosion parameters and nuclear signatures. What are we actually looking for in the ICF implosions? What do we get from the nuclear diagnostics? So that's kind of the broad overview background. And then we'll go into a little bit more technical stuff-- the nuclear diagnostics that we use. And I've divided it into neutron activation, neutron spectrometry, neutron imaging, charged-particle spectrometry, touching a little bit on other charged-particle magnets as well, and then finally, reaction-rate history. And finally, if we have time, I'll just spend a few slides discussing the impact of nuclear measurements on the ICF program that is in particular. So let's start with the implosion parameters and nuclear signatures. So the nuclear emission from an ICF experiment carries information about the state of the fusion fuel. And this is actually what really excites me about nuclear diagnostics is that they carry information directly about what's happening in their reactions. They're the products of the reaction. So they know exactly what's going on. We can count the number of nuclear products, and that gives us a measure of the number of fusion reactions that happen. We can look at the energy spread of the fusion products, and that gives us a measure of the plasma ion temperature. We can look at the energy upshift of the fusion products to fuel velocity. Was there a comment online? Can you guys hear me OK? AUDIENCE: We can hear you great, thanks. MARIA GATU JOHNSON: Great. And we can look at a scatter or downshift of the nuclear products to study area of density, which I'll discuss what that is, of the compressed fuel and shell. We can also take an image of the nuclear emission to study the spatial burn profile, and we can look at the temporal evolution to determine how nuclear burn evolves in time. And actually, if you look at this, quite a few of these are also relevant for magnetic confinement fusion. We can obtain similar information by looking at the nuclear emission from a tokamak, for example. The exception is the area of density, which is a specific quantity to ICF. So I want to spend a few slides talking about why that's important. As I think all of you know already, ICF uses the inertia of a dense shell to confine the plasma before it blows apart under its own pressure. We can express a confinement time in terms of the sounds, cs here. This is kind of illustration of how it works. And if we take a-- [INTERPOSING VOICES] AUDIENCE: Should I turn this one round? And then you've got one fewer computer to look at. MARIA GATU JOHNSON: Trying to switch slides here. OK, that's good. AUDIENCE: There we go. MARIA GATU JOHNSON: We can take a mass average of the local confinement time. And that gives us the confinement time as the radius over the sound speed with a 1/4 of r to the 4 in that. So I'm not going to go through that integral in detail, but it's a very simple one. Yeah, so this is the confinement time. OK, so then if we look at the standard, number density times confinement, and plug this expression from the previous slide in for the confinement time, it gives us this expression. So we see that number density times confinement time is a direct function of [INAUDIBLE], which is this area of density that we keep talking about, which is essentially a confinement parameter in inertial confinement fusion. And that's why we care about it so much. Oh yeah, even highlighting it there. High areal density-- or rho R, which you should refer to it as-- is required for a significant fraction of the fuel to burn before it disassembles under its own pressure. We can express the burn fraction, fb, as rho R divided by rho R times 6 grams per centimeter squared. That's at an ion temperature of 30 K. You can derive it at different [INAUDIBLE] temperatures. So that's kind of high compared to what we usually operate at. And this expression actually is derived from the fusion burn rate integrated over the confinement time. If we throw in some numbers on this, we can throw in that we want a burnup fraction of 25%, which really would be required for high gain. That means we need a rho R of 2 grams per centimeter squared. OK, just to put this in perspective, the best performing NIF implosions to date have had a burnup fraction of about 5%. So this is really high performance. And then if we assume solid D-T, D-T ice has a density of 0.25 grams per cubic centimeters. Then we find that for ignition, we need a fuel mass about 1/2 a kilogram. Again, this is with this expression, solid D-T ice. So then the question is, as I'm sure you've seen this 1,000 times before in ICF presentations, can we really work with 0.5 kilograms of D-T fuel in a laboratory? Of course, the answer is no. That gives us a little too much yield, which is not quite what we want. And that motivates-- that in order to achieve the required areal density, our confinement parameter, i temperature and confinement time without destroying the lab, we need to compress the capsule. We actually have to compress it quite a lot. We get these rough parameters, starting from about a 2 millimeter size capsule. Get down the radius of 30 to 50 microns, density like 700 grams per cubic centimeters from the 0.25 that we started with. And an areal density of 2 grams per centimeter squared, temperatures from 5 to 40 keV, and confinement times from about 20 to 200 picoseconds. So looking at these numbers, these inclusion parameters really set the requirements on the different measurements. Typically, we want to achieve a 5% to 10% accuracy on these kinds of numbers. So that really tells you how well we need to be able to make these measurements. OK, so then we have a lot of different products that we can work with. And they really carry a wealth of information about ICF implosions. First of all, we have our primary products. All of you know that we primarily work with the D-T reaction, which gives us the alpha particle and the neutron [INAUDIBLE]. PROFESSOR: I don't think so unfortunately. MARIA GATU JOHNSON: OK, keep jumping. PROFESSOR: Yeah, sorry. MARIA GATU JOHNSON: You could use a board eraser, but that's only going to give you an extra foot or so. So it's probably not worth. MARIA GATU JOHNSON: OK, so that's the primary one we work with. If you look over here, that gives us the yield, the ion temperature, the areal density. We can also use it for yield versus time and use it to infer the confinement time and the radius of the capsule. There's also another branch, which gives us a gamma particle, which is actually quite useful. And we'll look a little bit at that as well later in the class. In addition to D-T, we can also look at primary products from the D-D reactions. We have two, one that gives a [INAUDIBLE] proton, one that gives a helium 3 and neutron. And actually, a lot of-- many confinement experiments have been working primarily with D-D to date. So you can use that neutron for a lot of measurements as well. And then the helium 3, we also work with in a lot of surrogate experiments where we don't want to do D-T and it's a different thing. And then actually, it's quite similar to D-T. We also get alpha particle, get a proton instead of a neutron with pretty high energy. OK, so those are the primaries. Counting the number of primaries to generate gives us a direct measurement of how many fusion reactions we had, obviously. So that's a quick way of getting the yield for inflation. Not necessarily always easy, but basically that's how it works. We can also have secondary products, which I actually won't be spending much time on today. But if you're interested in that, I'm happy to answer questions afterwards. So that's when one of the fast products produced in a primary reaction goes on to react with a thermal fuel and to give a broader energy spectrum of the reaction products. And then we have knock-on reactions. We'll spend a little bit of time looking at how this works. So that's when the fast neutrons born in the D-T reaction hit one of the fuel ions to give faster fuel ion and a scattered neutron. We actually use this quite a lot. And we can also have similar reactions for the alpha particles scatter on the fuel ion and give [INAUDIBLE] fuel ions. I won't spend much time on this today, but this is the signature that they can use to look at what the alpha particles are doing in the plasma, in particular, when these fast neutrons react again, in a tertiary reaction, which I think the example's in here-- yeah, to give these fast neutrons, which we call alpha knock-on neutrons. Think some of you have heard about that before. OK, yeah, so this is actually pretty cool, these knock-on reactions are the easiest way of measuring rho R. So we look at some examples of that as we go through. OK, we already talked about this-- counting the number of emitted primary fusion products in a set. Solid angle and scaling it up to 4 pi gives us a measure of the total yield. So this is basically the yield reaction. I forgot to put the y in front of it. But if you do the integral over volume and time, the densities of the reactants-- typically, it could be [INAUDIBLE] and [INAUDIBLE] for D-T plasma-- and the reactivity-- and this is just a Kronecker delta. So if it's 2D, you get a factor of 1/2. That gives you the [INAUDIBLE]. And the reactivity is the integral over the reactants of the fuel ion distributions, the cross section. And then you can get-- this is actually an example. Reactivity is calculated according to that formula. That's written up in this paper, which if you haven't seen this paper before, this is really a key reference to go to if you want to look at how likely a reaction is. Obviously, the key is the most probable. Then we have the 2D reactions. We have a probability. And you can go to lower probability, [INAUDIBLE]. OK, yes. Go ahead, John. AUDIENCE: How do you deal with the fact that you're technically not-- so, when you do this calculation of integrating over 4 pi, you're assuming some spherical symmetry. But we know for a fact-- MARIA GATU JOHNSON: Great question. So in ICF, actually, you can typically assume 4 pi. There are some variations, which we can look at-- will look at later in the talk. But in magnetic confinement fusion, you don't have 4 pi symmetry. So then you have to correct for it. And actually, this leads to the work that you are doing. This is an example from a paper in 2010 by Sjostrand, where he's using a neutron spectrometer to infer the total yield of neutrons from the [INAUDIBLE] tokamak. But he needs to correct for the emission profile by using the profile monitor so he can scale that single line of sight [INAUDIBLE] neutrons to what the [INAUDIBLE] emission would look like. Great question. OK. OK, so, thinking about what happens to the particles after they're born, most nutrients escape an implosion and can be counted. But some of them scatter. We'll talk more about that later. But for charged fusion products, like from the helium 3 reaction, stopping in the assembled fuel has to be considered. So the nutrients, they might scatter on their way out, lose some of their energy. But most of them escape directly. Scattering probability is relatively low. Any charged particles are going to lose energy as they traverse the assembled field. And in fact, an example of what that can look like, we have the helium 3 critical spectrum be born in 14.7 MeV. This is the lower [INAUDIBLE] implosion, lose just a little bit of energy on its way out of the capsule. For high convergence implosions with high rho R, the ions will be fully stopped and can't be counted. So then we really can't rely on measuring primary fusion products, charged fusion products, for those experiments. And actually, it turns out that ion stopping plasmas is a rich research topic that our group has been working on. I have a few example references there, and there are a lot of other references. OK, let's see. Yes, so coming back a little bit to the knock-on reactions. So the neutrons, when they scatter off the fuel ions, they will also up scatter the fuel ions and energy. So fuel ions that are starting the thermal distribution, or basically, coal, can be up scattered in much higher energy by these 14 MeV neutrons. The energy of the ion can be calculated according to that equation, where A is the mass number of the scattering nucleus. And theta is the scattering angle. En is the neutron energy. And this is the examples of what it can look like. So this is the scattered spectrum for tritons, deuterons, and protons. And basically, the energy of a deuteron tritonal proton that's scattered in this way is going to depend on the scattering angle. So this is what you get if you integrate over all scattered [INAUDIBLE]. OK, and we can use this fact. We often measure these products. And we use a number of those products to infer areal density because the areal density depends on the ratio or scale, basically the opposite. The ratio of the number of knock-on ions to the neutron yield will be a function of the areal density. Does that make sense? The more fuel that's assembled, the more of the neutrons are going to scatter. So we can use that relationship to infer what the areal density of an implosion is. But this only works up to a certain areal density. Like, the knock-on deuterons, for example, would be fully ranged out [INAUDIBLE] 200 milligrams. And we talked about before, it might be at the order 2 grams per centimeter squared. So this is really relatively low performing implosions that we're looking at. So we often look at the neutrons instead. And we can derive the energy of the scattered neutrons to see that the neutrons carry information about the symmetry of the assembled plasma. Because again, the scattered neutron energy will depend on the mass number of the scattering nucleus and the scattering angle. So we find, if we do that math, that for a detector at a set angle [INAUDIBLE] implosion, neutrons that end up at the detector in a certain energy range are going to be sampling a certain part of the shell of the implosion. So at the NIF in particular, we often infer that compression by looking at energy range 10 to 12 MeV, which is a really clean energy range of the neutron spectrum. I think I might have an example later where you can see. There are no other sources of neutrons that might contribute in that range. So by looking at the number of neutrons in that range, you really get a measurement of only the neutrons that are scattered. OK, so if you do that, if you look at the range from 10 to 12 MeV, the neutrons are scattered out from tritons. It's part of the shell. Neutrons are scared of the [INAUDIBLE] part of the shell. And it's a little bit different because deuterons and tritons have a different mass. And you can also broaden the range, and actually, we'll look a little bit at neutron imaging. Neutron imaging typically looks over a broader neutron energy range in order to get enough statistics in the sample the broader part of the shell. So OK, this looks nice and simple. In reality, typically, the source of [INAUDIBLE] is significantly broader. So this smears out. You're not going to have all the neutrons coming exactly from the center. But this is the basic idea. The unscattered energy spectrum, the neutrons that go straight out, like the green arrow here that don't lose energy on the way out, will carry information about ion temperature and velocity of the assembled fuel. And actually, in principle, this will be true of any fusion products. When they're born, they will be born with this information. The problem is, if it's a charged fusion product, it's going to lose a lot of that information as it's losing energy on the way out of the capsule. It's going to be a lot harder to infer that information. This is the expression for the energy of a single fusion product. Then we can make the moments of this energy over the distribution of reactant plans to find that the width of the neutron spectrum is proportional to the ion temperature. There's a small ion temperature related peak up shift of the spectrum. And the peak will be shifted to the direction of flow of the emitting plasma. There's actually one piece of key information that we've gotten from neutron data at NIF. They found early on that the capsule, when it was pushed through the lasers, it ran off in one direction, basically, based on the neutron spectrum being shifted, which we had to correct for because that prevented efficient conversion of the compression energy into thermal energy of the capsule. So this is a non-relativistic expression. Really recommend you read this paper to get the relativistic math for how this works. I'm not going to go through it today, but this is an excellent reference which everyone that does neutron diagnostics for ICF uses all the time. And it's actually, originally, comes from the magnetic confinement community. So it's another example of the connections. OK, so I think by now you've understood that the neutron spectrum provides information on areal density, iron temperature, and yield. And this is an example of what a neutron spectrum can look like. So the primary [INAUDIBLE] here that are unscattered [INAUDIBLE], the width of that primary spectrum is related to the ion temperature of the plasma. We can also-- not illustrated here-- but we can have an upshift that's related to the velocity. And then by counting them, we get the yield, scaling up to 4 pi. And then by taking the ratio of the neutrons in the down scattered range, the neutrons in the primary range, we get a measure of the areal density. And actually, we often talk about a down scatter ratio rather than an areal density because we can measure the number of neutrons in this range compared to the number of neutrons in this range. And, yeah, go ahead. AUDIENCE: How do you distinguish between neutrons that have down scattered within the fuel versus neutrons that have down scattered in the lab? MARIA GATU JOHNSON: OK, so that becomes a technicality of the instruments. You have to collimate them really well in order to look at that. And it turns out the way that ICF is set up, the capsule is at the center of a large chamber. So the room return from the back wall of the chamber becomes a much, much smaller fraction of the neutrons that go down your line of sight. So basically, the way it's set up for-- Chris is actually looking at exactly this problem-- for the magnetic recoil spectrometer, which we'll be talking about later, the foil is 26 centimeters from target chamber center, back almost 5 meters away. So the solid angle for scattered neutrons coming down the same line of sight is just so much smaller. It becomes negligible. For the [INAUDIBLE], I don't think it's been [INAUDIBLE] I don't think it's been looked at in detail. But what they do is they take reference implosions so that they know if there's no assembled rho R, they know what the background in that range is. Yeah. And then Chris is looking at the concept of putting numerous foil really far away. And I'm making him do simulations to see if that's going to work, or if we're going to have a problem with the returns [INAUDIBLE]. AUDIENCE: Why do you cut off the down scattered region at 4 1/2 MeV? MARIA GATU JOHNSON: OK, so that's actually just kind of random. What we typically do when we do this is we [INAUDIBLE] This is not the best spectrum to look at. We're gonna see if you can see this on the camera. We look at DSR, or down scattered ratios, we call it. And that's the integral in the [INAUDIBLE] sort of see it? Integral in the 10 to 12 MeV range divided by the 13 to 50 MeV range. [INAUDIBLE] just get a quantitative number. So that's this range here divided by that range here. And the reason we look at that range is because we have contributions from neutrons from the T2 reaction contributing up to 9 1/2 MeV. We have the D-D neutrons contributing here. You can kind of see that peak here. And there's also multiple scatters that kind of break the correlation between rho R and the number of neutrons that you get. So it's the cleanest reading to look at in the spectrum. So that's how that works. Any other questions? OK, so I mentioned that we can also use the fusion products to look at the spatial emission. So we can take images of primary and scattered neutrons, for example, to provide information on the burned region, size R, and also the thickness of the high density shell. So in this case, this is actually a reconstruction, taking primary images in the 10 to 12 MeV range, down scatter images in the 6 to 12 MeV range, and then doing a fluence-compensated image, which gives us this artifact here, which gives us the picture of the neutron source, which is the primary neutrons and the high density shell, which is scattered neutrons. And we can measure the nuclear reaction rate to get information about the confinement time and the bang time of the implosion. So what this is here is-- we call it the Lagrangian plot where you follow this simulation. You follow the same fluid element as a function of time. And then you see that the red is that interface between the capsule shell and the gas on the inside. This is for a gas-filled implosion example rather than the [INAUDIBLE] ETIs. You drive it with the lasers. You get ablation of the surface material, which is why some curves are going off, and the other curves are compressing inwards until you get convergence. The shift you will show in particular moves inwards. The rest of it is converging. Get a little bit of burn here when the shocks hit the center, and you get more burn here where the capsule is at peak convergence, when it's maximally heated. And then you can measure the emission history as a function of time. And you can see this shock burn and this compression burn. And this particular example [INAUDIBLE] implosion. So gas [INAUDIBLE], so then you often get both of these components. And then with the implosion, you're going to have very little shock. And you can have a lot more compression where we would be completely dominated by the [INAUDIBLE]. OK, so those are some examples of the parameters we're looking for. So with that, I plan to go into more about the technical detector details. So any questions before we move on? Actually, I have no idea how I'm doing on time. PROFESSOR: Oh, you've got half an hour-ish. MARIA GATU JOHNSON: That should be good. PROFESSOR: Yeah. MARIA GATU JOHNSON: OK, then let's jump into it. So the first one I thought we'd talk about is nuclear activation diagnostics. So they're typically based on indium 115, copper 63, or zirconium 90 isotopes for measurement of primary D-D or D-T neutron yields. If you look at D-D first, that's what we use indium. When a D-D neutron hits the [INAUDIBLE] indium, we get a isomer and the scattered neutron. The threshold for that reaction is about 1 1/2 MeV. And this isomer state will decay, emitting gamma. But it has to have about 4.5 hours. And this is the gamma that we count to ensure how many reactions happen. On omega, we use copper to measure the D-T yield. And again, it's copper 63. And then the neutron [INAUDIBLE] neutron means copper will be an end-to-end reaction. So with copper 62 and two neutrons, threshold is about 11 MeV. We'll look at the shape of the cross-section. I think it's on the next slide. And then what happens is that copper 62 is radioactive, will decay to nickel 62. And the half-life for this is 9.8 minutes. And what we actually count are the gammas here as well. Zirconium [INAUDIBLE] zirconium 90. We again get an end-to-end reaction. Threshold is about 12 MeV in this case, which means we're really narrowing in on the primary neutrons at 14 MeV. This is what we use at the NIF. And again, zirconium 89 is not stable. The end product that we get is gamma 909 keV, which is what we're counting. And if you look at-- OK, so this first plot has the indium and zirconium reaction cross-sections. So you can clearly see why we use zirconium for D-T. The threshold is at about 12. Really covers our primary D-T [INAUDIBLE]. It's also really sharp though, which is actually a useful tool because if the peak is shifted up or down, it's going to impact what you're counting, which means you get an impact of velocity [INAUDIBLE]. You so you can see differences around the implosion. And then indium, on the other hand, is a really broad cross-section, which actually makes it a really blunt tool. If you want to use it to measure D-D, it's by far the easiest. In a pure D2 implosion, we don't have the down scattered D-T neutrons [INAUDIBLE] In principle, you can use a cocktail of different nuclear activation detectors to piece together information about the full neutron spectrum. And here, we have some examples of parts of the spectrum that can be of interest. So in the D-T implosion, this is what the D-D spectrum look like. You actually have some of those secondary neutrons that we talked about before that are at 14 MeV. You have the primary D-Ts at 2 1/2 MeV, down scattered D-Ts, and then just a little bit of scattered in between. And that we can get at with the indium in principle. We have the T-T neutrons, which have a peak at about 9 MeV and go from 9 1/2 all the way down to 0. Those you can also attack a little bit. And then, really, the primary thing we're looking at is the D-Ts, which some of the up scattered then-- and we looked at this before-- where the neutron has hit a fuel ion, giving it a lot of energy. That energy in turn reacts to produce another neutron. That's when we get these really high MeV tertiary neutrons, 15 to 30 MeV. And that's actually, in many cases, also really interesting measurements [INAUDIBLE]. Oh yeah, that completely fell off. There's another reaction here that has this cross-section here. It's an isotope of carbon, but honestly, don't remember which one. So then you can really focus in on just those highest energy neutrons. And this actually also shows you-- we have-- it's carbon-12. [INAUDIBLE] Copper you kind of see is this orange line here. And we have zirconium as the red line. So zirconium is a much sharper threshold of 12 MeV. Copper starts already at 11. So they have a little bit different sensitivity to [INAUDIBLE] neutrons. At omega, copper activation is used for measurements of the primary DT neutron yield. We have, basically, a little retractor tube that allows a puck to be inserted and then dropped after a shot by pushing a button. But it's still very manual. So many times, I've been up there for a shot, and this old Russian guy, Vladimir Glebov, pushes the button, gets the black disk, and then runs it over to the counting detector in a different lab. And it's using sodium iodide detectors to detect the gammas in coincidence [INAUDIBLE]. And it's actually quite useful because then you go down to really low neutron thresholds. So cryogenic implosions at OMEGA produce upwards of 10 to the 14 neutrons, but you can measure neutrons down to 10 to the 7, which means you can look at experiments where you're not producing many neutrons at all and still know what you produce. On the NIF, zirconium is the primary activation element used. And it's used routinely for measurements of primary DT neutron yield. In fact, from the high-performing implosions, there are two measurements that provide the yield that's then reported out. One is the zirconium nuclear activation, the other one is MRS. So it's implemented in a number of different versions. We have the Well-NADs, which is kind of the go-to reference. It's inserted to 4 meters from target chamber center. There's three different pucks that sit very close to each other so you can compare the numbers from the three and make sure you're not making any mistakes. And then you have the Snout-NAD, which you can insert much closer. And actually, it's more common to vary the elements in these packets and have the cocktails to look at different neutron interactions. And then finally, you have the Flange-NADs, which sit on the outside of the chamber. And there's a large number of those attached in different positions around the chamber to look at symmetries. The zirconium detectors are transported-- well, actually this is kind of modified to the Flange-NADs-- we'll talk more about later-- are now counted in situ at the NIF, but they used to be transported. Zirconium detectors are transported to the Lawrence Livermore National Lab NAD Data Analysis Facility, which looks like this with some really old hardware but still does its job. And then yeah, so the Flange-NADs has actually been converted, fairly recently, to 48 real-time zirconium nuclear activation diagnostics, or RT-NADs, that are permanently installed-- semi-permanently installed on the chamber with a lanthanum bromide detector counting the activation from those continuously. And you can see the peaks when there is an implosion. But recently, it used to be 48, which is great. You can really look at the symmetry of the emission, which-- I think I'll get to this on the next slide, but the symmetry emission tells us about the aereal density symmetry. So we could look at the symmetry emission. The reason high-yield implosions have started killing these detectors, so we're now down to 21 version, and we actually have to remove them before every high shot. So they're no longer permanently installed on the chamber. Yeah. So this is what I was trying to get to. The nuclear activation diagnostics often show large low-mode aereal density asymmetries in NIF implosions. So this is when you have this network of 48 detectors that all provide a measurement of the yield above the threshold of 12 MeV. If you're starting to see variations in that above-12-MeV threshold, that means that the birth distribution is uniform in 4 pi. So that means something must have happened on the way out, where some of them, more than others, were scattered on the way out, which means that the aereal density is not symmetric around the implosion. And actually, turns out the typical scattered neutron fraction is about 20%. You can look at them, the number density times the cross-section times the shell thickness, you get just a rough measurement of how many neutrons will scatter on the way out. It's about 20%. [INAUDIBLE] frequently see variations of plus/minus 8% in the unscattered neutron yield, which means it's a large aereal density variation from one side of the capsule to the other. And this has also been a really useful diagnostic tool in figuring out what's going on with these implosions as we're trying to improve them further, make them perform better. And yeah so I've mentioned before that you also have an impact on peak shifts here because the cross-section is higher and higher energy. So if you have a flow where the [INAUDIBLE] runs off in one direction, you can have upshift of the peak in that direction, downshift of the peak in the other direction. It turns out it's a smaller effect than the rho R asymmetries, but it's significant enough where it has to be corrected. First, you have to measure that directional flow and correct the distribution for that effect as well. And we use low aereal density gas-filled DT exploding pushers to set the baseline variations, basically, as-- losing the word-- baselining. There's another word. Maybe it'll come to me later. OK. Did that all make sense, nuclear activation detectors? PROFESSOR: So why is it that you use copper on OMEGA and zirconium on NIF? MARIA GATU JOHNSON: Just historical reasons. PROFESSOR: Oh, OK. Is one better than the other or that can [INAUDIBLE]? MARIA GATU JOHNSON: So I actually-- the guy who runs the neutron diagnostics at OMEGA now would like to start using zirconium instead. And if I remember correctly-- yeah, the reason for that is the longer half life. It's really hard to work with a 10-minute half life. You have to really run to get to that detector fast enough, whereas zirconium, with the three days, is a little bit easier. And also, the threshold's actually better at 12 compared to 11. Yeah? AUDIENCE: For looking at the upshift from zirconium, do you just compare that to a baseline where you have a more uniform emission profile versus a non-uniform one, where that steepness of the cross-section actually matters? Or do you compare it to a baseline cross-section that's flatter? MARIA GATU JOHNSON: I'm not sure if I fully understand the question. But so what you're doing is you're looking in many different directions, and you compare the results in different directions. But you know-- AUDIENCE: So but you just-- you know your baseline just by assuming a 14.1 MeV uniform profile? MARIA GATU JOHNSON: OK, so there's a couple steps to this. If you get a map kind of like this one, I mentioned you have to correct for the velocity, which is the peak shift. We actually don't get the velocity from this diagnostic. We get it from the neutron spectrometers. So if you have neutron spectrometers in six lines of sight, you can measure there. You don't just measure a number above the threshold, you actually measure the neutrons because you know what the option is, which means you can infer the actual 4 pi velocity vector. And then you can correct this for that. Yeah. Neutron spectrometers. Any other activation question? OK. Then with that, let's go into neutron spectrometry. This is kind of touching on exactly that point. We do have a large suite of neutron spectrometers on the NIF. These five, the blue ones here, are all based on the same technology. They're neutron time-of-flight spectrometers, which we'll discuss in detail. And then this one in red is the magnetic recoil spectrometer, which I already mentioned a couple of times. There's actually one more that's not included on this cartoon, which is also a neutron time-of-flight spectrometer based on a different detector technology that's fielded together with the neutron imager on roughly this line of sight. It's been a few years since it was working, but we're trying to bring it back to resolve some [INAUDIBLE] everything. So total of seven neutron spectrometers on the NIF that provide good implosion coverage together. And this similar setup exists in other ICF facilities. Like on OMEGA, for example, think there are six now that run on DT in different lines of sight. So you can compare the results in those six lines of sight to, again, infer the flow vector in addition to measuring the ion temperature and the ion temperature variations. And fewer of them work on DT but still enough to get a good coverage. OK. So let's start with the magnetic recoil spectrometer. So again, this is what I've been working with since 2010, so happy to take any questions on this one. And Chris is working on a very similar concept. Sean's kind of working on a similar concept, too, for Spark. So you guys are very familiar with this already. But for those who have not heard about it before, the way it works, you will have the neutrons emitted from target chamber center, that little blue dot. A fraction of those neutrons will reach a conversion by the plastic conversion foil, 26 centimeters, in the case of NIF, from target chamber center. And this is actually deuterated foil. So the neutrons that interact with the foil, some of them will knock out deuterons. Forward-scattered neutrons will reach this magnet, which is outside of the target chain wall. It's just a vacuum in between. So they all reach the magnet. And then there is momentum separated in the magnets [INAUDIBLE] different physical location of the detector, right, depending on their energy. Then you use that to reconstruct a recoil deuteron energy spectrum. And then from that spectrum, you can infer what the incident neutron spectrum must have looked like. We use deuterated plastic, in particular because the detector we use in this instrument is CR-39, and it turns out the deuteron tracks are much, much easier to distinguish above background compared to using protons. Yeah. And we also-- there's a number of detectors with [INAUDIBLE] the sodium hydroxide, scan them in microscopes after the shot, and then stitch the data together. Looking at-- zooming in on the foil here, so the neutron will hit a deuteron in the foil. And then the recoil deuteron energy will depend on the incident neutron energy, and then also the energy loss that the deuteron has from its place of birth until it hits the back of the foil. So a neutron born at the start of the foil would come out with a lower energy than a neutron born at the end of the foil. And that has to be considered in the analysis of the data. We can look at, also, a couple of other aspects of this. So when I have-- we used to want a high efficiency to be able to count all the neutrons that came out. Today, we're actually running into saturation problems instead, so we don't really want this to be so high anymore. But the efficiency of the MRS can be back of the envelope calculated as the foil solid angle times the number density for deuterons in the foil times the foil thickness the n, D differential cross-section for forward scatter, and the aperture solid angle where the aperture is opening in front of the magnet. And you can throw some numbers on that. This is an example from the NIF. The foil solid angle will be the area of the foil divided by the total sphere at that distance where the foil is sitting. And then number density, we calculate based on manufacturer's specifications. Foil thickness is measured. This differential cross-section in the forward scatter direction is roughly this number. Aperture solid angle take the area of the aperture and just divided by the foil aperture distance squared. And this is a correction for the fact that the aperture actually isn't sitting straight. It's tilted in front of the magnet. And then this is the [INAUDIBLE] of the [INAUDIBLE] foil which you also have to add as the correction. This gives us a rough number. In reality, this is not what we use to get the yield number out. We use MCP simulation. Actually, I thought on the way over here, I should have included a slide on MCP because we use MCP a lot as a tool in understanding the response of the detectors that we're looking at. Even for the nuclear activation detectors that we looked at before, to know how many neutrons that they see and how many might be scattered before they hit the nuclear activation detector, we also have to use Monte Carlo neutron transport tools such as MCMP. So it's not just building detectors. There's a lot of modeling that goes into this as well. OK. We can also look at what we expect for the resolution. What that is, basically, is if you have monogenic neutrons emitted from target chamber center going through that whole system, then we're going to end up with a wider spectrum than just monogenic on the MRS. So we look at, how wide would that spectrum be assuming we had monogenic neutrons to understand the broadening-- the instrument of broadening. So we can look at that as three components. We have a broadening effect due to the foil thickness. And this is, again, where deuterons are born on one end or the other, and they're going to lose energy as they go through, which gives you a broadening. We're going to have broadening based on the scattering geometry. We have neutrons that hit the foil head on, deuterons that go straight out. But then we also have neutrons that hit one edge of the foil and go at an angle. And that gives us a broadening effect. And then finally, we have some broadening depending on the ion-optical properties of the magnet. And that leads us to a total broadening. And it actually turns out that these are counter-related. So if you want higher efficiency, you have to add more foil material, which also enhances the broadening. You want a narrow broadening, you want high efficiency, so it becomes an optimization problem. In the design of a magnetic recoil spectrometer, you have to balance efficiency and resolution against each other. This is an example for the thin foil magnetic [INAUDIBLE] spectrometer at JET, which is very similar to MRS and maybe even more similar to what Sean is working on for Spark. So this is looking at proton collimator radius and foil thickness, and seeing how varying those two will impact the efficiency and resolution, trying to find some set points that would be ideal operation for getting the signal you need and still having a good enough resolution to make the measurements you want. And that, actually, we can take a look at what that looks like at JET. This is the MPR magnetic recoil proton spectrometer. In this case, we do use protons not deuterons, because they use scintillator detectors they can count protons without any problems. This is what it looked like before the shielding was added. This is the magnet housing. And we add a lot of shielding to prevent contributions from scattering neutrons that you don't want to look at. And it's the same concept as for MRS. You have the neutrons emitted from the plasma over here. You have collimators, so they only look at a certain fraction of them. The foil is inside the magnet housing here. The second collimators [INAUDIBLE] the protons are born in foil here. And then the protons are momentum analyzed in the magnet [INAUDIBLE] in a different physical location on the detector array. In this case, it's an electromagnet rather than a permanent-- maybe I probably forgot to mention that for MRS. But for MRS, it's a permanent magnet. An advantage with an electromagnet is that you can tune it so you can operate at different energies. The MPR, in particular, can be tuned to operate either for 14 MeV DT neutrons or 2 and 1/2 MeV neutrons. And we've used them with the JET tokamak with an oblique angle like that. If you look from the top, you can see how it traverses all the way through the plasma. PROFESSOR: Why is the oblique angle used? MARIA GATU JOHNSON: I'm not sure. That's an interesting choice. I think it might have actually been to maximize efficiency because you see more of the plasma that way. AUDIENCE: OK. MARIA GATU JOHNSON: It's a question for, Johan, though. He was involved in the actual building of this system. OK. And then so just look at what a recoil deuteron energy spectrum can look like, this is an example from the NIF. It's a really old example. But you have the primary DT peak, and then you have the down-scattered neutrons. And then from this, you infer the total neutron yield by scaling for height, the aereal density by comparing the number of neutrons here-- deuterons here and deuterons here and the ion temperature from the width. And what you do when you analyze this kind of data is you take a model neutron spectrum and fold it with the instrument response function simulated using GM-4 or MCP or a combination thereof and get it on the deuteron energy scale, and then adjust the ion temperature the peak position, which is related to velocity, the amplitude, which is related to yield, and the rho R, which is related to the down-scatter relative to primary diffraction. So you get those numbers out of the analysis. Yeah? AUDIENCE: You're going past just fitting a Gaussian to this peak. You're using a more sophisticated analytic model? MARIA GATU JOHNSON: So actually, in most cases, I fit a Gaussian to the peak and then just have a second component that accounts for the scattering. But yeah, there are some slightly more advanced models as well. And for magnetic confinement fusion, you'd typically have to have more advanced models because you have fast ions due to heating that contribute to broadening of the peak. And to resolve those, you have to have a model for the beam-thermal reactions or the beam-beam reactions that have slightly different shapes. And you can find the relative contributions of those by fitting those different shapes to the peak. Other questions? OK. So with that, let's look at the neutron time-of-flight technique. So for ICF, this is actually simple because all the particles are assumed to be emitted at the same time. The burn time is so short, order of 100 picoseconds, so you can make that assumption. So then you really only have to measure the neutrons as they arrive on a scintillator at a set distance, d, from the implosion. And you use that time to infer the neutron spectrum. Yeah. And I mean, this in particular, it's already been converted to temperature expression. But really, what you're looking at is the neutron energy just on a time scale. Yeah. And the same as for MRS, the ion temperature is determined from the width. And actually, so on OMEGA, the nTOFs are still used as the yield measurement, too. On the NIF, we gave up on that a while ago because it's so hard to know how the gain of the electronics [INAUDIBLE] drifts in time. So what we do is we calibrate it relatively routinely to the nuclear activation detectors and MRS. And since it's cross-calibrated, we no longer use it as the absolute yield number. nTOF detectors are also used to diagnose aereal density. And I think already mentioned at some point that we use that by comparing to what we call zero rho R implosion, where we really just put DT gas in the really thin shell, drive it really hard directly with a laser, so we know there's no aereal density or negligible aereal density, and then we can look at the difference between zero rho R implosion and one with significant rho R to [INAUDIBLE] the rho R's. And yeah, this is identifying on the time scale what the 10 to 12 MeV neutron energy range will be. The detectors at the NIF-- I think the closest one is 18 meters from the implosion, and the furthest one is at 27 meters from the implosion. This is what the original equatorial nTOF looked like. It's since been upgraded to look more like that. That actually looks like this, but it's kind of hard to tell. The way it works now is instead of having these PM tubes directly attached to the scintillator, which is in this volume here, like [INAUDIBLE] you have the photomultiplier tubes facing. So implosion-- I'm the implosion. Detector is over here. The photomultipliers are facing this way so that neutrons hit the scintillator, and the light is collected in this direction, so you have [INAUDIBLE] contributions from scattering in the detectors themselves. And there's four photomultiplier tubes at each scintillator, so you can have different settings on different photomultiplier tubes to optimize them to look in different parts of the neutron energy spectrum. There's collimators on the way. This is an example of how neutrons come through a wall collimator, with this detector. In this case, you can also see that both neutrons and gammas coming through the collimator hit this high-sensitivity and fast detectors, which we call a spec detector, which is used to measure rho R. Also recently installed these quartz Cherenkov detectors to just a really thin rod in the same line of sight, which you can use to look at both the neutron and gammas. And these are actually more optimal for the velocity measurement because you get a really precise measurement of the primary structure from those. Yeah. And it's similar at OMEGA. So this is at OMEGA. It's below the target chamber center in kind of a basement, which we call LaCave. It's this large detector which is a liquid scintillator material. It's quenched xylene, which allows you to have a really fast time response. So it falls off as quickly as possible after a primary peak, which makes it easier to measure the down-scattered neutrons. Actually, another detail-- on OMEGA, the nTOFs times do not measure down-scattered neutrons in this energy range, because the rho R is much lower at OMEGA than on the NIF's much lower-power laser. So instead, we're using the backscatter edge of n, D backscattering. So neutrons that hit the back of the implosion scatter off of tritium and reach the detector on this side, which gives us an edge at 3.4 MeV, which is much easier to distinguish on OMEGA. And so that's done with this detector. Again, for photomultiplier tubes, they're optimized for different ranges of spectrum. And look closely, you can see there's two detectors in front here-- one Cherenkov detector as well-- that thin rod-- and this pattern detector, which is one of the primary ion temperature detectors. So it has much better resolution than with large xylene detector. Any questions about that? Ben? AUDIENCE: Is the thickness of the detector a significant source of uncertainty? I guess I'm guessing that the thickness of these detectors versus the length of these beam lines is really small to be negligible. But is that a source of uncertainty? MARIA GATU JOHNSON: So it depends on what you mean with "uncertainty." I mean, so definitely, you get a broader spectrum from this large detector than from those thinner ones in front, which reduces your resolution, so it's harder to measure ion temperature. So that's why, like in this case, this detector is optimized for the rho R measurement, and this one is optimized for the ion temperature measurement. This one needs high efficiency to get that weak component of down-scattered neutrons. This one needs high resolution to measure the peak accurately. AUDIENCE: So similar to MRS, it's a trade-off between efficiency and resolution? MARIA GATU JOHNSON: Yeah, yeah. Mm-hmm? AUDIENCE: Someone asked at APS, and I didn't know why-- what are the benefits of MRS over the nTOFs if they all give temperature, rho R, [INAUDIBLE]? MARIA GATU JOHNSON: So OK. My perspective, the primary benefit is having more than one technique because you really need to know independently what you're measuring. You can compare the results from both. And many times over the years, as one technique started drifting, and then we figure out what's going wrong by comparing with the other technique. So I think that's the primary advantage. You can also say that one is that the MRS gives you the absolute yield. It's calibrated from first principles compared to cross-calibrated to other detectors. Yeah. But I think it's really important to have both. PROFESSOR: But this data is available directly after the shot, whereas the MRS is a while. MARIA GATU JOHNSON: That's true, that's true, which is a huge, huge advantage. And yeah. And this could be scaled up to rep rate as well because you can make sure you can analyze it quickly after a shot, whereas MRS would CR-39 indefinitely. [LAUGHTER] Well, Chris is working on electronic detection, so we'll get there. Other questions? OK. So then the magnetic confinement fusion equivalent-- we do have a neutron time-of-flight system here, too. But here, it becomes more complicated because we can no longer assume that all the neutrons are emitted at the same time. So here, we have to have two sets of scintillators. We have a start scintillator, which we call S1 here, and a stop scintillator, which is S2. This is what it actually looks like in real life. There's a collimator through the floor here. The detector sitting in the roof lab above the JET tokamak. So the neutrons come through the collimator in the floor, hit the start detector first and then the stop detector. We have the start detectors layered to allow us to count at a higher rate. There's five layers in there. And then the stop detector is divided into 32 segments to actually-- that's more of a resolution [INAUDIBLE] thing because you want to know where the light is coming from in order to be able to measure the [INAUDIBLE]. And replacing them on the constant time-of-flight sphere so that you can compare the data from all the different detectors and stitch to make one spectrum. Yes, Kai? AUDIENCE: How do you know if a neutron hits the S2 is the same one that you just measured at S1? MARIA GATU JOHNSON: Great question. So you don't. And that's where this comes in. AUDIENCE: Oh, OK. MARIA GATU JOHNSON: So we look at data from the different scintillators in coincidence. So you take-- what this example is, is all its events that you can get in the S1 detector on the top. You get a lot more events in S1 because it's closer and it's directly in the beam of neutrons from there. The S2, the beam actually goes through in the hole in the center, so it doesn't hit the S2 directly. The S2 only sees scattered neutrons. But that means you get a lot fewer events in S2's. And what you do is you go through and look at coincidences between the two detectors. And actually, what you get is you get all the coincidences. You get the true coincidences, and you get background random coincidences. And you have to subtract that back out. But when you do that, the peak will appear because that's then the true correlation that it will be the same between all neutrons [INAUDIBLE]. You were saying? So in this case, the flight time for a 2 and 1/2 MeV neutron between S1 and S2 is about 65 nanoseconds. AUDIENCE: But the beam drift time or the amount of time that the neutron spends moving between S1 and S2 is very short, because looking like these detectors, they're very-- MARIA GATU JOHNSON: 65 nanoseconds. AUDIENCE: So does this function for DT neutrons, or is this [INAUDIBLE] MARIA GATU JOHNSON: So it's best for DB. So this, it will [INAUDIBLE] they show up at 27 nanoseconds. The time resolution isn't anywhere near as good simply because the flight path is shorter. If you wanted really good resolution, you'd have to make the flight path really long [INAUDIBLE]. But of course, another difference here is here, we didn't really make that point with the inertial confinement fusion nTOFs. But what we're looking at here is running the scintillators in current mode. We're just opening them up, looking at the signal current as a function of time. In this case, we're looking at individual pulses from single neutrons interacting with the scintillator. So we can divide it into-- we recorded over the entire duration of the pulse and we can reconstruct neutron spectra for any time interval we want where we get enough statistics. So we can actually look at the time evolution of the neutron spectra this way. And at ICF, we simply get too many neutrons at the same time so it becomes complicated. But Chris has spent quite a bit of time trying to figure out how to do that, too. OK. So that's all I plan to say about neutron spectrometer. Any more questions before we move on? OK. OK, I think I have just a very short section on neutron imaging. So what we do in neutron imaging is we use a pinhole or aperture close to the implosion and the detector really far away to obtain good magnification. This is actually very-- we have the source again here, which is a target chamber center, which is where the neutrons are emitted. You have a lined aperture, which can either have penumbra, which will encode the signal, or a simple pinhole, where you have a direct correlation, basically, between the neutron emission and opposite, kind of inverted to the detector. The magnification will depend on the pinhole standoff distance and the detector distance. OK, so I took this slide from somewhere else, and I don't know what numbers they actually threw in there. Oh, assuming a magnification of 200, which you would get depending on what L1 and L2 values you have. If you have 5 microns at a source, it's going to be 1 millimeter at the detector, which magnifies your radius. I talked about before, that implosion would be 30- to 50-micron radius. You magnify them so you can separate individual features much easier on your detector on the outside of the chamber. And actually, for the NIF in particular, I think typically the aperture is about 20 centimeters from target chamber center. The detector is 28 meters away. Yeah, this is what-- ha, actually, those are the exact numbers. So this is what it looks like. You have the NIF target chamber over here. That's target chamber center. The aperture is fielded right here, 20 centimeters from target chamber center. And then the neutrons that are selected by the aperture will travel through this collimator structure all the way to the detector back here 28 meters away. And so the neutrons-- this is just a [INAUDIBLE] it looks like. The neutrons come out here through the line-of-sight collimator. This is that other nTOF detector I talked about. You can kind of see that it's a different technology. It's flat plastic scintillators with photomultiplier tubes. But then the primary-- so that that's just another neutron spectrometer. The primary for the imaging system is the scintillating fiber array, which is fiber coupled to a camera and it's done this way in order for you to be able to gate the camera and get two snapshots. So you can get the primary neutrons and then the scattered neutrons at a later time. There's actually also-- this is the original NIF neutron imaging system. There's two more now, so we can look at symmetries around the chamber. One of them only has image plate detectors, which I planned to bring image plate, but I forgot. But it's-- PROFESSOR: We discussed it, actually. We talked about X-ray diagnostics, so we talked about it a little bit. Yeah. MARIA GATU JOHNSON: Yeah, so you know you can't time gate on image plates. So then you just get one image. OK. And I wish I had a better picture. But these pinhole apertures are actually extremely complicated. So they're made of gold. They're about that long. And you have to make pinholes that are precise all the way through. It's too hard to drill them circular, so they make them triangular instead. And then they're tapered to minimize scatter. So basically, you have an opening, and then the neutrons that go through that opening are all going to be captured at the back end. None of them are going to stop because of the taper inside. It's still a really hard machining problem to get those even triangular pinholes precise all the way through. We need gold because, as we talked about, neutrons have a pretty low likelihood for interacting in matter. And you want to only select the ones that go through the pinhole. You don't want all the ones around to also make their way all the way back to the detector. Yeah. So that's fun. And these are examples of the penumbral apertures where the information will be decoded in the penumbra. You'll also get straight through neutrons in [INAUDIBLE] that you can't use to infer anything about the shape of the implosion. So this kind of aperture array is used on all three lines of sight now, but there's development going on to make it coded aperture, which is supposedly going to be simpler and thinner and easier to manufacture. We'll see if that actually works out. See. Oh, yeah, so these are examples of primary and scattered neutron images. And again, they're routinely obtained. And all DT NIF implosions in these days, it's even in three lines of sight. And there's a lot of work going into tomographic reconstructions to make sure we understand the full emission region. And we have our equivalent in magnetic confinement fusion, which John is working on. And we call them neutron profile cameras. So it's, again-- the idea, again, is to probe the shape of the neutron source distribution. But it's much bigger here and more complicated. So instead of that pinhole array that's trying to reconstruct the 50-micron spot, we're looking at a much larger emission. And we do it by using a number of different lines of sight and counting particles along this line of sight, and then try to reconstruct the full emission [INAUDIBLE]. And John can answer a lot more questions here. OK. So with that, I'm going to jump right into charged-particle spectrometry. And I think I touched on this in the beginning, but it's routinely used to diagnose low to medium rho R implosions. And they can be filled with deuterium, DT, or D helium-3 fuel. And the reason we don't typically use it for high rho R implosions is, again, because the charged particles stop in the assembled fuel, so they don't become indicators on the outside. OK. This example is showing the proton from the interaction for 14.7 MeV downshifted to about 11 MeV You can look at this downshift to infer the aereal density, which in this case was about 84 milligrams. And you can-- actually, let's see. Yeah. I have an example, one with nice [INAUDIBLE] spectrometer. The cool thing is you can do this thing in a number of different locations around the target chamber to look at symmetry again to see if things are compressed uniformly or if there's non-uniform issues. So this is a super simple spectrometer. We call it wedge range filter spectrometer. It's just an aluminum filter that's shaped like a wedge. Pass this around. You can see it. So the way this is fielded, it's got a bunch of holes in front. And you know the holes are used to register where the detector is fielded behind that wedge. You know the thickness of the function of precision on this. And based on, actually, the hole size of the CR-39 detector as a function of precision relative to those holes, you infer the proton energy spectrum. And it's a little bit more complicated than it sounds because you need to know the diameter versus energy response of CR-39. And that varies from piece of CR-39 to piece of CR-39. So you have to come up with a pretty intricate method for inferring that from the data. But that's done, and these are routinely used to measure rho R from the helium-3 gas-filled implosions in many different locations around the target chamber. Pass that around. It's a small 5-centimeter round packet. So you really can get it in many locations. And this is also the detector material that's used for MRS, as we talked about before, CR-39 plastic. So the wedge range filter spectrometer is one example that you can see here, too. We also have charged-particle spectrometers which are very similar to MRS. It's a magnet outside of the target chamber wall. The difference is we don't have a conversion coil, so we're just looking at charged particles directly from the implosion. And you can actually-- this is another advantage of MRS. You can also run MRS in charged-particle mode for experiments where we're not interested in the neutron spectrum. And then you can look at the charged particles that come directly from the implosion. Yeah. And this is an example of looking at that symmetry and how it can vary around the implosion. This is an insertion module on the NIF, where we can field, actually, up to six of these wedge range filter spectrometers on a single insertion module. There's four insertion modules that have capability of fielding these. So you understand we can field a lot from one implosion. On OMEGA, we can field up to seven on one implosion in different directions. And in each direction, you can add a few more if you want. So a lot. Yeah. So in this case, these are fielded at 50 centimeters from implosion. These are fielded at 10 centimeters from implosion [INAUDIBLE]. So we can look in different directions. And this is actually a really old example from the paper by Johan in 2004, where he's fielded protons with [INAUDIBLE] in different locations around the OMEGA target chamber [INAUDIBLE] look in different [INAUDIBLE] spectrum. And if you think back to this, for D3, you often get these two peaks in time, a sharp peak and a compression peak. So you can also say something about the evolution of the experiment. Here, you have the time evolution. You see the small, sharp peak and the larger compression peak. And then you can also see that in the energy spectra, you have less range down-- more range down compression profiles. So you can tell the difference in aereal density between those two types of implosion. It's kind of neat. OK. I feel like I'm running out of time, so I've got to speed up. We already talked about image plates, so don't need to talk about that. CR-39, I kind of touched on. This is an example of what it can look like after we've etched it in sodium hydroxide at 80 degrees Celsius for of order hours. If we put it on one of these microscopes, step over, and take pictures for roughly 400-micron frames, the microscope automatically picks up tracks which are due to particles interacting in the CR-39, and records their roundness or eccentricity and the track size. And then we use that to reconstruct whatever information they wanted from that diagnostic. OK. So then, spend a few minutes on reaction-rate history. So there's a number of ways to do this. And again, so we're looking at really short burn, order of 100 picoseconds. We can field a plastic scintillator really close to the implosion and combine it with the streak camera to measure their reaction-rate history. This is done at OMEGA. So this picture is of the OMEGA target chamber. This is the laser. So this is not part of the diagnostic. This is how we drive the actual implosion. This is what the detector will look like. So it will be fairly close to the target chamber. There is the plastic scintillator. The light from the scintillator will be coupled through an optical light path [INAUDIBLE] camera to record a streak image. And this is where the burn history is encoded. And that scintillator will have a rise time of about 20 picoseconds but a fall time of 1 and 1/2 nanoseconds. So the information, really, is encoded in the rising edge of the signal. So we have to unfold it. But when you do that, you can get the burn history as a function of time for [INAUDIBLE] compression. And this is used a lot for neutrons on OMEGA in the neutron temporal diagnostic. There's also the particle temporary diagnostic or the particle and X-ray temporal diagnostic-- similar concept, where you can tweak that scintillator setup in the center to have a number of different channels, some of them optimized for X-rays, some optimized for protons, some optimized for neutrons, depending on how you filter them and what neutron density focus you put behind [INAUDIBLE] on the streak camera and reconstruct it after. So that's actually a really neat diagnostic that's useful for a lot of things. I told you, promised you, early on that we'd get back to what we used to gammas for. So one cool thing about the gammas is they don't have the same time dispersion that neutrons do. So the neutrons, when they're emitted from target chamber center, they're going to disperse in time, which is really why we can use just the neutron time dispersion for neutron time-of-flight. So if you put the nTOF detector 20 meters away, look at the neutrons, we measure the energy spectrum, not the emission history. For the gammas, they don't disperse in time, so we can have a gamma detector relatively far away and we still retain that time history. So that's the cool part. We have the lower probability for getting gammas. I think we saw the branching ratio is about 10 to the minus 5 of the neutrons. But there is enough of them to count. And by counting the gammas-- in this example, we're using the gamma reaction history detector, which is based on converter gamma rays that are converted to electrons. And electrons generate Cherenkov light in this gas cell, and then it's detected as a function of time. And that's how you infer the bank time or burn history. You get both from this measurement. This is what this detector looks like on OMEGA, and this is what it looks like on the NIF. And on the NIF, there are four different channels, which you can vary in gas pressure set at different gamma detection threshold. You don't get spectral information, but you can set a threshold and then compare the results from the four. And actually, it even made it into a movie, this detector. This is the gamma reaction history detector. Hans Hartmann, who built this detector, was really proud to be able to take his kids to this movie at the movie theater. [LAUGHTER] OK. So coming back to magnetic confinement fusion, again, here we typically use fission chambers to measure the nuclear reaction rate. And the way this works is you have the fissile material, you get the fission products going into the fuel gas. They ionize-- or, sorry-- yeah, they ionize the fuel gas, and then you get an electric pulse that goes out. And you measure that as a function of time. And here again, you need MCP model in order to determine what your measured signal actually means in terms of nutrition. These are often also in-situ calibrated, where you move the source around inside the chamber and see what the signal looks like on efficient chambers on the outside. OK. And I think have five more minutes, Jack, right? PROFESSOR: Go for it. MARIA GATU JOHNSON: So we'll take a few minutes on the impact of nuclear measurements on the ICF program at the NIF. And I think we've really touched on this throughout the talk today, right? The nuclear data have been essential for guiding the initial experiments to ignition. This is the time axis of the yield from the experiments from 2010 through now to 2024. And you can see that this is a logarithmic scale. It's obviously increased a lot over that time frame. And MRS has been part of it from the beginning. I started here at MIT in August 2010. I think the first data from the NIF came back from MRS a week-- two weeks after I started. They shipped it back in this huge moon lander looking container. It was, like, octagonal box with lots of cool packs to keep this [INAUDIBLE] cold. And we had to work day and night to etch and scan it and turn it around. And that's when we're at this yield level, right? Not registering on the scale. And then we've been working our way through up to the regions where we actually have target gain. And we've looked at many of these diagnostics today that have been essential for [INAUDIBLE] experiments to ignition. We looked at how we get ion temperature, hotspot velocity, fuel density or aereal density, and yield from the neutron spectrometers. We looked at how we get the burn width and bang time from the gamma reaction in-situ detector, which is related to the confining time. We also get neutron yield from the activation detectors as well as the map of the fuel uniformity from the real-time activation detectors. And we use neutron images to get the hotspot and fuel shield shape. And really in particular, these two have been essential for identifying those asymmetries. And seeds to asymmetries have been really hard to eliminate along the way to get there. OK. This is an example that Johan put together two years ago. On August 8, 2021, an implosion experiment at the NIF ignited and generated a then record neutron yield of 4.5 10 to the 17 1.35 megajoules. This is the MRS spectrum from that particular experiment. We were so excited about that experiment, which really, internally in the community, that was ignition. And I'll explain why in the next slide. But so this explains what I was talking about before. We have a model neutron energy spectrum, which is a Gaussian with a width governed by the ion temperature, the mean energy determined by the birth energy plus the peak shift, which is related to the velocity. And then we have this component, which is related to the aereal density. We vary those parameters to fit into our mesh and reconfigure an energy spectrum, and then we get a best fit neutron energy spectrum which explains what we actually had. And so if we look, in particular, at this implosion from 210808, many of the key nuclear observables point to this implosion being in a fundamentally new regime. We saw how the ion temperature took off. Earlier implosion under the same campaign had 5 keV. Now we measured neutron average ion temperature 10 keV-- a dramatic step up. We saw how-- the burn history is a little bit hard to see, but what happened actually is the burn history peaked later and got narrower. And what this is saying is that the yield took off during compression. So where the yield would have previously tanked, you start having it climb more and more instead. And then it becomes a really narrow burn because it's burning on as the explosion is exploding rather than as you're compressing. And that, you can see in the neutron images as well. These are neutron images and as well as one X-ray image on the top row from two predecessor implosions. This one, you can see how it gets a lot bigger. And this is, again, because it's going on the expansion. Yeah. And then the other one-- well, you're actually measuring-- this is not the best spot to show it, but what we're finding is we're actually measuring a lower down-scatter ratio. And the reason for that is we're now probing-- the density ratio that we're measuring is probing the timing the implosion after peak compression. So even though the actual down-scatter ratio-- the actual compression is the same, we're seeing fewer scattered neutrons because the peak compression is before the probing. Does that make sense? OK. And yeah, this is just another illustration of the same point, where you really see the temperature climb or jump. This is the fusion yield on the y-axis. This is an inferred hotspot mass and hotspot energy, also jumped for this one implosion. Neutron radius increased and burn rate decreased, really showing that we're in a new regime. And that was this one couple years ago. So definitely done some better ones since then. And there's another one from October 29 that's not yet on this chart, too, which is the second best performing ever, so falls right between these two. And we're at 4 with gain over 1 at this point. OK, I think that's all I had for today. PROFESSOR: Thank you very much. Any other last questions? MARIA GATU JOHNSON: Yeah? AUDIENCE: What's next to try to go even higher on the gain? Because I think a lot of the changes were capsule quality, et cetera. And is there anything that you're seeing in your new neutron data from this new regime that's guiding further changes? MARIA GATU JOHNSON: Good question. I don't think there's anything super obvious right now. Part of it is pushing for bigger implosions, which is not directly related to the neutron data. Yeah. Yeah, no, I don't think there's any defect signatures right now that we're going after. AUDIENCE: Are there any open questions that you see in the neutron data [INAUDIBLE]? MARIA GATU JOHNSON: Yes, there's a big one. Great question. So actually, I talked a lot about peak upshifts and how we infer velocity from that. So there's two aspects to that. There is the peak shift that's different in each of the different lines of sight that's showing us in which direction the implosion's taking off in. But also turns out that there is a uniform [INAUDIBLE] that's the same in all lines of sight that's anomalous, that's not explained by the ion temperature and not explained by the direction of velocity. And right now, it looks like the only way to explain that upshift is by non-Maxwellian effects in the fuel line velocity distributions. It's not clear why those would arise. So that's definitely a big outstanding question that we're looking at, which excites me a lot. I like puzzles.
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https://ocw.mit.edu/courses/3-60-symmetry-structure-and-tensor-properties-of-materials-fall-2005/3.60-fall-2005.zip
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PROFESSOR: I think it's time that we got started. I haven't given a problem set out for a couple of days. So I have one that will cover some material that we'll go over with and develop today and some new material that we're not going to introduce for a while. AUDIENCE: [INAUDIBLE]? PROFESSOR: Yes, this is Thursday. And this is for the near future. I've given you two-- two a week before. So it's the adjective you're objecting to not the problem set, of course. Today we're going undertake another major step in our development of three dimensional crystallographic symmetries. We have just completed derivation of the crystallographic point groups, so these are the macroscopic symmetries that crystals can have. And then, by analogy, to what we did in two dimensions, the next step would be to take each of these point groups and hang it at a lattice point of some three dimensional lattice that can accommodate them. We have not as yet developed the three dimensional lattices, and that will be our agenda for today. Seems like we've already done that, haven't we? We showed that in two dimensions, a cell can be oblique, it can be rectangular, it can be centered rectangular, it can be square, or it can be hexagonal. So these are the different kinds of bases that one can have for the cell, right? So there should be the same number of space lattices. Actually it's more complicated than that. And let me remind you by another handout what the two dimensional space groups, the plane groups, look like assembled in all of their glory on one sheet. And there you see the different ways symmetry can be placed in these five different kinds of lattices-- oblique, rectangular, centered rectangular, square, and hexagonal. Now a lattice in two dimensions, a potential base for a unit cell, can be rectangular if and only if there is symmetry within that net. That demands that it be exactly rectangular. In other words, if a lattice is to be rectangular in the base of the cell, it has to have in it one of the symmetries that correspond to those in the plane groups that give and require, indeed demand, a rectangular lattice. Similarly a lattice can be truly square, identically square, only if there's a fourfold axis in it. That demands that it be square. The same could be said for either threefold or sixfold symmetry. One of the groups, P3, [? P3, ?] mP3, 1m, P6, and P6mm, has to be in the base of the cell if that cell is to have the specialization. So let's give an example with a twofold axis. Let's suppose we have a base of the cell that has an unspecialized shape-- two translations that are unequal in length and an angle between them which is a general obtuse angle. That's about as general as you can get. But suppose that this net has as well a two dimensional-- a twofold axis in it. And suppose we pick our third translation such that it picks up this net which is going to be the base of our three dimensional lattice and translates it over to this point in the base. Well that translation picks up everything. It picks up not only the two translations that we've been calling T1 and T2 to this point, but it also picks up all the twofold axes that are hanging on these lattice points. So if this is the terminus of T3 and projection, it will have picked up a twofold axis and plopped down a twofold axis here where no twofold axis exists. And we can take one of the theorems that we've seen, namely that the operation A pi, followed by translation, is equal to a new rotation operation through pi located halfway along the translation. And turn this around and ask what happens when we combine two rotations through 180 degrees about different points. And let's just draw it out to see what happens. Let's say there's a twofold axis here. It takes motif number 1, moves it to number 2. And then here's a second twofold axis and this takes number 2, rotates it over here to number 3. I is number 1 related to 3? But in terms of this first theorem, not surprisingly, we have introduced a translation into the space which is equal to twice the spacing between the twofold axes, which I"ll label as delta. So in other words, if we put a twofold axis down in this place in addition to the ones that we already have, we have created, in the base of the cell, a new twofold axis in this location. That just mucks everything up because it's going to rotate the translations that we have, it's going to rotate the twofold axes that we have. And we will not any longer have a group. We will not have a set of operations that closes upon itself. So the conclusion then is that if we're going to safely pick a third translation, T3, and combine it with a net that has twofold axes in it, in order to get a group, we are going to have to ensure that the twofold axes line up in every net in the stack. And there could be several ways of doing that. We could pick T3 so that it goes straight up. And then the twofold axes are just all slit up parallel to themselves, and they remain in coincidence. Or we could pick T3 so that it terminates over the point that's halfway along T2. And then this twofold axis gets picked up and put down over one that's already there. This one gets picked up and put down over one that's already there and so on. So that's an OK choice. That's a safe choice as well. Similarly we could pick T3 such that, within the plane of the net, it had a component that was one half of T1, so that it picked up this net and slid it over to this location. So once again all the twofold axes lined up. Or, in the same fashion, let it terminate over the center of the cell. So those are the four ways we can combine an oblique net that contains a twofold axis in such a way that the lattice points in the existing twofold axes remain invariant and are not moved into locations that create new twofold axes. OK so let's back off a little bit and start with plane group P1-- no symmetry at all. So this is T1 and T2. If there's no symmetry whatsoever in the base of the cell, we are free to pick the third translation in any orientation that we choose. So trying to sketch it in three dimensions. If this is T1, and this is T2, T3 can be in any orientation such that not only is this angle general, but this angle is general, and this angle general as well. And if I would try to carefully complete the parallelepiped so that its geometry looks reasonable. This is going to be the general oblique parallelepiped with three translations that are unequal in magnitude, and three angles between these translations. Let me write them as the angle between T1 and T2-- not equal to the angle between T2 and T3 necessarily, not equal to the angle between T3 and T1. And all of these angles are completely general and can assume any values that they like. So this then is the lowest common denominator. This is a totally unspecialized space lattice. It has the shape of the general parallelepiped-- no special relations between any of the translations whatsoever. AUDIENCE: [INAUDIBLE] PROFESSOR: I say they're-- Oh I'm sorry, magnitude are not equal. [INAUDIBLE] I meant to put not equal signs there. Yeah, they can have any length they wish. They're not equivalent. Any other questions? OK let's go back then to the case of an oblique net to which we've decided to add a twofold axis. And that gives twofold axes in the locations that we've become familiar with in the plane group. And if I pick the translation so that it goes directly normal to the base of the cell, I'm going to have a new kind of lattice. The lattice is going to have two angles between axes that are exactly 90 degrees, not approximately so, but exactly so, because only if that translation, T3, is exactly normal to the base of the cell will the twofold axes in all of these nets lineup exactly. So here we have a degree of specialization. The translations remain unequal in magnitude. They can be any length they wish and not change things, but two of the angles-- axial angles, T1 to T2, is identically 90 degrees. The angle between T2 and T3 is identically 90 degrees. And the third angle, the angle-- I'm sorry, T1 and T2 is my general angle. So this can be anything. So that can be any general angle. T2 and T3 and T1 and T3, want to be exactly 90 degrees. So we're getting space lattices that have a degree of precise specialization. And the reason is that the geometry of the base of the cell is inseparable from symmetry in the base of the cell that demands the degree of specialization that makes the space lattice unique. So let us look at a couple of the other choices for the third translation that will leave the twofold axes in coincidence. Another choice would be to pick the third translation such that it terminated exactly over the midpoint of T2. So here's T1. Here's T2. Now what I'm going to do is something I have not done to this point. Having this translation T3 terminate exactly over the midpoint of a translation [INAUDIBLE] was clearly a specialization. If I were to go up two translations, T3, I would be directly over the end of T2, and that is a much more convenient way of demonstrating that special feature. So let me pick a new T3 prime. It goes up exactly normal to the base of the cell, and in so doing, I will have defined a cell which is a double cell. And the catch is an extra lattice point in the middle of one of the pair of faces. So I'll refer to this general sort of situation-- this is a double cell. And I'll call this, in words, a side-centered cell. And it has, with this redefinition of translations, exactly the same degree of specialization as before-- magnitude of T1 not equal to the magnitude of T2, not equal to the magnitude of T3. And it has the same specialization of the angles. The angle between T1 and T3 is exactly 90 degrees. The angle between T2 and T3 is exactly 90 degrees. And the angle between T1 and T2 is general. General, but by convention we will assume the obtuse angle rather than the acute angle. OK so by picking this double cell, we've defined a cell that has a twofold redundancy. But in doing so, we have gained the advantage of showing that this specialized geometry is exactly the same as this one-- in the same sense that in two dimensions, the primitive rectangular cell and the centered rectangular cell were both cousins. They were both had orthogonal translations. But the other thing that we gain is that even though we can't have an orthogonal coordinate system, if we use the three translations as the basis of a coordinate system, at least we have two 90 degree angles on our coordinate system. And operationally, that is going to be an enormous advantage. Yes, sir. AUDIENCE: So is T3 [INAUDIBLE] PROFESSOR: T3 was selected so that we picked it as 2 T2. I'm sorry, 2 T3 minus T2. That was my new T3 prime. I went up two translations in T2 and then back T2, and that put me directly over the lattice point from which my translations emanate. AUDIENCE: So your T3 is actually [INAUDIBLE] PROFESSOR: It goes to the next level up in the stack of nets. If I would draw this in projection, which is not nearly as clear in some respects, this would be the base of the cell. Let me use x's for the next level up. So here's the net offset by half of T2. And then if I go up two of these T3's and then back by T2, my next layer up is directly over the first one-- the third layer up is directly over the first. AUDIENCE: Shouldn't it be T3 prime then? PROFESSOR: Yes if I were to be consistent, I should call this T3 prime, which I did here. But now I should call that T3 as well. I got a little bit careless because we're going to forget that we ever had this T3-- to find this selected as a stacking vector. OK well in P2 there are four different kinds of twofold axes. So we could have the third translation terminate over either of the remaining pair of twofold axes as well. I think that it's apparent that if I pick T3, the original T3, as one half of T1, so that the next layer up terminates in lattice points in this orientation. And then I pick a T3 prime, which is equal to 2 T1 minus-- 2 T3 minus T1, I will have gotten again a side-centered lattice. The only difference is that it's going to be a different pair of faces. It's going to be the O1Oa faces that have the extra lattice point. So that is not fundamentally a different sort of lattice. It is also side-centered. And differs from our second result only in whether it is the side face with the shortest translation or the longest translations of the base that bears the extra lattice point. One final choice though does result in a new lattice with special features. And this would be one where I pick the third translation such that it terminates over the center of the net below. So let's let this be T1, this be T2, and pick T3 so that it moves the origin lattice point directly over the center of the parallelogram face below. And then if I go up two translations, T3, and then subtract off T1, and subtract off T2, I will again have a T3 prime, which is exactly normal to the base of the cell. OK this is a type of lattice that you've come to know and love. In the cubic system, this is referred to as a body-centered lattice. AUDIENCE: So then, T3 is just [INAUDIBLE] PROFESSOR: T3, the original choice, would pick this original net up and move it so that that net was moved-- such that it's corner was over the center of the original net. So if we go up two translations, we have lattice point over lattice point, and we redefine T3 by going back 1 T1 and minus 1 T2. And that makes it exactly normal to the base. AUDIENCE: Not on the sides? PROFESSOR: Nothing on the sides . Now in fact, if we try to do that, we would have something that's not a lattice. If we tried to have some additional lattice points in the center and the edges of this upper level, we would have destroyed the original T1 and T2 in the base of the cell. Let me do one or two more and then I think the way in which one proceeds should be quite clear. If you take this list of all of the plane groups, if we would like to have a three dimensional cell that has a rectangular base, the base can be rectangular only if there is a mirror plane in that net or a twofold axis with the mirror plane. The two situations are different. So let's examine first the case of a rectangular net that has either a mirror plane or a glide plane in it. And they're both going to have the same sort of constraint. If I have a mirror plane, that-- let me use a slightly wiggly line so the mirror line is not confused with the edges of the cell. So here are two translations, T1 and T2. And if I pick a T3 that moves this rectangular net up and over itself, we have to do so in such a way that the mirror planes coincide. Because if I have a mirror plane in space that passes in proximity to a lattice point, I can turn my theorem around and say that a translation, T1, combined with a mirror plane that is a reflection operation that's removed from the lattice point, I would have to introduce a new translation to delta which is going to be incommensurate with T1. So I have to pick my third translation in one of two ways. Let me indicate that by drawing a circle. T3 can terminate anywhere over this mirror plane. I could pick this net up, slide it parallel to T2, and as long as I put it down so that this is the lattice point in the next layer up, that will be a legitimate placement of the net. Because all the mirror planes are simply slid into coincidence with one another. So if I sketch that thing in three dimensions-- this is the base of the cell and it has a rectangular shape. And the third translation goes up at an angle. That's going to be exactly something that I've made before. And the only difference is that the oblique angle is now on the vertical plane but that's a right angle. And this stays a right angle. So that's exactly a lattice of the shape that we had last time with twofold axes, except this cell is sitting on one of its rectangular sides rather than sitting on its oblique base. So that is something that's not new. And if I would pick the translations so that T3 had a component that was one half of T1, plus some amount z that is perpendicular to the base, I can redefine that in terms of a cell that has one pair of faces inclined to one another. And the difference is going to be that the lattice point would be caught in the center of the cell. This would be my original translation T3 and I'll redefine that as a T3 prime. And that gives me two right angles and one general angle. So those are exactly the same lattices that I obtained from stacking a net with a twofold axis in it. Let me introduce now another piece of jargon as we go along. The relative angles between the translations are going to determine the shape of the coordinate system that we pick along those translations to specify the geometry of features within the lattice. So we've had a first case, and this was just a general oblique net. So we had a T1 not equal to a T2, not equal to T3, and all three angles were general. All three pair of axes are inclined. And this coordinate system, if we want to refer to it in terms of words, is called a triclinic. This is the triclinic system. And the triclinic system is short for coordinate system. So this is as general as things get. And if this is the shape of the lattice there is no symmetry. Which is the same as saying that the point group, the only point group that can fit into such a lattice, is point group 1. And then we hit another sort of geometry. And that is where the three translations had a general angle-- a third translation that was exactly normal to the plane of the first two. And one pair of axes is inclined. And you can see that this notation is a little strange, perhaps, but consistent. And it says something about the nature of the arrangement of cell edges. So one pair of axes inclined is called the monoclinic system. And this can contain lattices that are either primitive, side-centered-- with either the longest or shortest translation in the parallelogram that's coming out of the face that's centered. Or we saw there was one final possibility. And that was body-centered. So we'll refer to these lattices subsequently as monoclinic primitive, monoclinic side-centered, or monoclinic body-centered. These lattices were compatible with point group 2 or point group m. We obtained a lattice of the same shape with point group m or with point group 2. If it works for 2 and it works for m, why not both at the same time. It would also work for 2 over m-- twofold axis perpendicular to a mirror plane. Any question at this point? Yeah? AUDIENCE: Also 2m, right? [INAUDIBLE] Yeah PROFESSOR: No,because as soon as you got a mirror plane, then you have to have a rectangular net. So it's going to be-- it's going to fall into this family. You're right, a mirror plane by itself and we've got that here. That's where the mirror plane falls in the side of the thing. You're right. But if it has a rectangular shape, then we have to have either orthogonal mirror planes and a twofold axis. The base, if the base-- yeah. OK. Part of the amusing part of this game is that you can, upon appropriate redefinition of cells, come up with a given coordinate system that defines more than one lattice. And I'll give you-- because you get a little space doing this-- I'll give you a two dimensional example. Here is one cell. This is T1, this is T2. This is a primitive cell. Nobody should bother to write this down. Here's a T1, here's a T2. So anybody know what that is? This is a primitive cell. This is a jail cell. Couple sniggers, thank you. Still another one-- this is T1, this is T2. Anybody guess what that is? That's a soft sell. I got a million of them. Know what that is? It's an underground cell. Here's one that's kind of dated-- I don't think anybody will get this one. Does anybody know what that is? That was a misadventure of the Ford Motor Company that was named the Ed cell. Well, I'll give you equal time. Anybody who, after intermission, wants to respond in kind, I'll let you have an opportunity to do so. But obviously these are not standard crystallographic cells, but they make for a smile in an otherwise dreary business. I don't know how to proceed from here. I have a set of notes for you which carries all this out in incredible thoroughness. So let me pass this around. Let you take a copy of this and we won't go over every single thing, but I think just to outline quickly the sort of choices that we have. We haven't yet finished with the rectangular cell, but if it's got something like 2mm in it. with twofold axes here and mirror planes running this way-- P2mm in the base, P2gm, P2gg, can be stacked only in such a fashion that T3 brings twofold axes and mirror planes into coincidence with one another. So T3 could be 0T1 plus 0T2 plus some amount z which is straight up. Or it could be equal to one half of T1, 0T2, plus z. That would be making the origin twofold axis fall over this location. Or T3 could be equal to 0 plus one half of T2 plus z. And they're both going to, upon redefinition, give the same result. Or T2 could be equal to one half of T1 plus one half of T2 plus z straight up. And upon redefinition, the first one that would be a brick-shaped unit cell. That's going to be a primitive cell. Picking T3 with a component of either one half of T1 or one half of T2 is going to be something that gives us a cell, which can again be redefined. This was the original T3 in terms of a T3 prime, which is equal to 2T1 or 2T2 minus [? 2T3 ?] minus T1 or T2. This is going to be side-centered. And the final choice where it is a T3 that's one half of T1 plus one half of T2 can once again be defined in terms of-- redefined in terms of a body-centered lattice. OK this is a coordinate system where all interaxial angles are 90 degrees. The translations, however, have arbitrary magnitude with respect to one another. All of these cells could be described as rhombuses. In this case all angles in the rhombus are 90 degrees, so this is called the orthorhombic system-- orthorhombic coordinate system. All the angles between the axes are 90 degrees. So we can get that out of any one of the rectangular groups that has twofold axes and symmetry planes in it. And conversely, take one of these lattices and let the symmetry elements that we found in the plane group extend through them and you've got a three dimensional space group. So without emphasizing the fact is we go along, we're picking up a healthy number of three dimensional space groups as we go along. AUDIENCE: [INAUDIBLE]? PROFESSOR: This is primitive. It's also orthorhombic -- same dimensionality, three translations, and they're all mutually orthogonal. That's true of these other cells except they're double cells in the case of the side-centered and double as well for the case of the body-centered. It's time we introduced some other conventions for notation in describing lattices. We've been using T1, T2, and T3 to designate the three translations that are used to define the cell. One has to have some sort of rules for picking a standard unit cell so that two people can do an x-ray diffraction experiment, let's say, on a particular material and end up defining the lattice in exactly the same way. So let me now list, so that we can use these labels as we go along further, conventions for selecting cell edges. OK one convention is that one should select the shortest translations in the lattice. And these shortest translations, if there's no symmetry, will be [? dignified ?] by calling them a, b and, c-- standing for the x, y, and z directions. One picks always, if there's nothing special about the translations in the triclinic system, the magnitude of b greater than a. And in a rational world-- every so often you come to a point like this and it feels almost the silliest talking about underground cells and jail cells-- in a logical, rational thinking world, you would do this. This is not what one does-- for reasons that have to do with the perverse nature of crystals. Take b greater than a, one does that always, but you pick c as the shortest translation. Why? It has something to do with the peculiar behavior of crystals. If you were working with a crystal that had the shape of a needle, in other words, greatly elongated in one direction so that the crystal looks something like this. How would you draw a picture of that? Would you put it straight up and down like this or would you put it on its side. Takes up a lot of room this way. You'd almost certainly instinctively put it this way. Now if you didn't know anything about the translations down inside the innards of that crystal-- and this was the position that the people who did crystallography solely on the basis of crystal shape and morphology-- and had no idea if there was even a lattice inside of the crystal that caused this regular shape. When you draw a coordinate system x, y, z, z always goes up. Who draws a Cartesian coordinate system with z going this way, x going this way, and y going this way? It's obscene. We always put z going straight up and down. You do the same if you don't know what the lattice translations are with the labels on the axes of the crystal. So this is a, this is b, and this is c. And there were tons of pages published with drawings of crystal morphology on them. And then along came x-rays. And when they begin to determine lattice constants of a the material that looked like this, almost invariably they found the translations inside of a needle-like crystal corresponded to a unit cell in the shape of the pancake. Crystals, when they have a very, very short lattice vector, almost always grow most rapidly in that direction. So this is a terrible pickle. Here are reams of books that published descriptions of crystals that had a needle-like-- an acicular shape, if you want to use a fancy term, which had c as a direction which corresponded to the shortest translation in the crystal. So what do you do? You don't want to have to redraw all these figures and redefine all these directions in the crystal. So what you do is cop out. And you make c the shortest direction in the crystal. You think this field is rigorous but people wimp out and they do something that's contrary to the logical thing to do. Interesting question is why crystals should grow most rapidly in the direction of the shortest lattice vector. And I think I know the answer to that-- I've never seen anybody express it in the writing. The clue comes from some early work that looked at the growth of the crystal from solution. And this was done first with-- I think it was some sort of iodide. I can look it up-- something like cadmium iodide. And this material in solution grew very, very slowly in the form of hexagonal plates. And continued to grow most rapidly in this direction. This was the direction of the shortest lattice translation. Then all a sudden something happened. And suddenly the rate of growth and the nature of the morphology changed. The crystal took off like a bat in these directions and finally ended up in a needle-like fashion like this. This was an observation that led to the dislocation theory of crystal growth. The interpretation was that in the early stages of growth, the crystal is growing by accretion of atoms on to these planar surfaces. And the probability of an atom sticking is less high when the atom is on the flat surface than when an atom is on the surface that might have a little crevice in it so it can bind to two planes at once. So the original postulate-- and now we all take this for granted-- is that if there was a flaw such that the crystal developed a dislocation on this surface, and an exposed ledge, atoms would accrete and stick more tenaciously to this exposed ledge of the dislocation. And this is the so-called dislocation-- screw dislocation mechanism-- for crystal growth. And after the crystal has grown for a while, you can see spiral steps on the surface of the crystal that correspond to this dislocation winding around. OK so why should crystals grow most rapidly along the direction of the shortest translation? It's going to be a lot easier to create a screw dislocation with a Burgers vector for a translation that is short, rather than the translation that's 10 or 20 angstroms that the Burgers vector would have to involve-- many, many layers of atoms. If the translation is very small, just two or three atoms, it doesn't cost you much work to make a screw dislocation. So I think that is the reason why crystals with very short lattice translations grow most rapidly in that direction, because it's easier to make the screw dislocation with a Burgers vector that's parallel to that translation. So here's the first flagrant bit of logic that flies in our face. Let me give a few more definitions and then we'll return to this. If we have a crystal in which two translations are equivalent by symmetry-- and we would have such a situation if we picked a translation that was perpendicular to a square net. And would we call the base of the cell something defined by translations a and b? If this is a fourfold axis, what goes on in this direction is identical to what goes on in this direction by symmetry. And if these two directions are identical, why call one of them a and one of them b? So if two directions are equivalent, and this is true of the square net, what you do is call one a1 and you call one a2-- because they're the same thing. So let's call them both a. And the third direction then is c. And in such a situation here, c is always defined as the unique direction. What do I mean by a unique direction? This is the direction of high symmetry in crystals that are based on square bases and hexagonal bases. Yes sir? AUDIENCE: You pretty much contradict yourself if c is the longest [INAUDIBLE]. PROFESSOR: Not always, not always. AUDIENCE: Well, I mean in that picture. PROFESSOR: Well I drew it that way. You know why? Because if I draw it like this, the top of my cell doesn't get in the way of the bottom of my cell. And if I want to draw a small c translation, then I get into geometrical complications that doesn't look nearly as nice. AUDIENCE: Well, then I'm just saying two [INAUDIBLE]. PROFESSOR: No, no, no. It's different for different systems. This is what is true for a triclinic system. And in the case of a brick-shaped unit cell-- orthorhombic. So this is done for triclinic. It's also done for orthorhombic. Not for monoclinic. For monoclinic, there is a special direction because there is a direction of a twofold axis or maybe a twofold axis perpendicular to a mirror plane. In that case, if we were rational, we'd call that 1c. OK I'll close with an opinion poll. I will invariably, and I'm sure most people do, draw a crystal with a square base, with the c axis the longest. And if we have someone drawing a sketch of a cell in which the base of the cell is a hexagonal net, we'd again call one a1, one a2. And perpendicular to that would be the same direction. Say look, there I did it again. I made c longer than a1 and a2. All right. Let's look at hexagonal crystals. Let me ask for a poll. What percent of all crystals with threefold and sixfold axes do in fact have the length of c greater than the length of a? 50/50? How many would say equally probable to have it the longest or the shortest? No opinion. Let me tell you the astounding fact. Seventy percent of all hexagonal crystals have c greater than a. Astounding. Let's look at the next-- haven't really defined these terms-- tetragonal crystals. These are crystals that have a fourfold axis in them. What percentage have c greater than a? This is almost a first cousin of the cubic crystal or all three translations are equal. How many would say more have c greater than a, fewer have c greater than a? One vote. More? That's a good answer, but it's only about 60 some percent. So let's go down to orthorhombic crystals. And orthorhombic crystals are the ones that have a brick-shaped unit cell. How many have c greater than a? It's declining. So if you extrapolate 50/50? AUDIENCE: Zero? PROFESSOR: Who said that? Are you a wise guy? You weren't supposed to give the right answer. It's 0 percent. It's 0 percent because there's nothing that makes one direction more special than any other. So you decide on the labelling of axes on the basis of their relative lengths. And you, by definition, make c the smallest axis. Congratulations. To the rest of you, gotcha. So that's a good note on which to quit. And a few remaining mysteries of coordinate systems and lattices will be expounded in the second half of our lecture.
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https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/8.06-spring-2018.zip
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PROFESSOR: Today, we're going to continue with the adiabatic subject. And our main topic is going to be Berry's Phase. It's interesting part of the phase that goes in adiabatic process. And we want to understand what it is and why people care about it. And then, we'll turn to another subject in which the adiabatic approximation is of interest. And it's a subject of molecules. So I don't think I'll manage to get through all of that today, but will we'll make an effort. So let me remind you of what we had so far. So we imagine we have a Hamiltonian that depends on time and maybe had no dependents before time equals 0 turns on. And it has no further variation after some time t. So the Hamiltonian changes like that. And the adiabatic theorem states that if you have a state at time equals 0, which is a particular instantaneous eigenstate, that is the instantaneous eigenstate, then that's the full wave function at time equals 0 and it coincides. Then, at time at any time in this process, if it's slow, the process, the state of the system, the full wave function, psi of t, will tend to remain in that instantaneous eigenstate. And the way it's stated precisely is that psi of t minus-- I'll write it like this-- psi prime n of t, the norm of this state is of order 1/T for any t in between 0 and T. So I'm trying to state the adiabatic theorem in a way that is mathematically precise. And let me remind you the norm of a wave function is you integrate the wave function square and take the square root. It's sort of the usual definition of the norm of a vector is the inner product of the vector with itself square root. So that's the norm for wave function. And here, what this means is that with some suitable choice of phase, the instantaneous eigenstate is very close to the true state. And the error is a Fourier 1/T. So if the process is slow, it means that the change occurs over long t, this is a small number. And there is some instantaneous eigenstate with some peculiar phase-- that's why I put the prime-- for which this difference is very small. And we calculated this phase, and we found the state, psi of t is roughly equal to e to theta n of t, e to the I gamma n of t, psi n of t. And in this statement, this is what I would call the psi n prime of t. And that's why the real state is just approximately equal to that one. And we have these phases in which theta of t is minus 1 over h bar integral from 0 to t E n of t prime dt prime. That is kind of a familiar phase. If you had a normal energy eigenstate, time independent 1, this would be 1 to the-- well, would be minus e times t over h bar with an I would be the familiar phase that you put to an energy eigenstate. Then it comes the gamma n of t, which is an integral from 0 to t of some new n of t prime dt prime. And this new n of t is I psi n of t psi n dot of t. I think I have it right. So the second part of the phase is the integral of this new function. And this new function is real, because this part we showed before is imaginary, where with an I, this is real. And this second part, this gamma n, is called the geometric phase. This is the phase that has to do with Berry's phase. And it's a phase that we want to understand. And it's geometrical because of one reason that we're going to show that makes it quite surprising and quite different from the phase theta. The phase theta is a little like a clock, because it runs with time. The more time you wait on an energy eigenstate, the more this phase changes. What will happen with this geometric phase is that somehow properly viewed is independent of the time it takes the adiabatic process to occur. So whether it takes us small time or a long time to produce this change of the system, the geometric phase will be essentially the same. That's very, very unusual. So that's the main thing we want to understand about this geometric phase, that it depends only on the evolution of the state in that configuration space-- we'll make that clear, what it means-- and not the time it takes this evolution to occur. It's a little more subtle, this phase, than the other phase. So I want to introduce this idea of a configuration space. So basically, we have that-- let me forget about time dependence for one second and think of the Hamiltonian as a function of a set of coordinates, or parameters. So the Rs are some coordinates. R1, R2, maybe up to R capital N are some coordinates inside some vector space RN. So its N components. And what does that mean? It means maybe that your Hamiltonian has capital N parameters. And those are these things. So you buy this Hamiltonian. It comes with some parameters. You buy another one. It comes with another set of parameters. Those parameters can be changed. Or you construct them in the lab, your Hamiltonians with different parameters. Those are the parameters of the Hamiltonian. And suppose you have learned to solve this Hamiltonian for all values of the parameters. That is whatever the Rs are you know how to find the energy eigenstates. So H of R times-- there are some eigenstates, psi n of R with energies En of R, psi n of R. And n maybe is 1, 2, 3. And these are orthonormal states, those energy eigenstates. So this equation says that you have been able to solve this Hamiltonian whatever the values of the parameters are. And you have found all the states of the system, n equal 1, 2 3, 4, 5, 6. All of them are in now. So this is a general situation. And now, we imagine that for some reason, these parameters start to begin to depend on time. So they become time dependent parameters-- can become time dependent. So that you now have R of t vector. These are 1 of t up to our Rn of t. So how do we represent this? Well, this is a Cartesian space of parameters. This is not our normal space. This is a space where one axis could be the magnetic field. Another axis could be the electric field. Another axis could be the spring constant. Those are abstract axis of configuration space. Or this could be R1, R2, the axis, R3. And those are your axes. And now, how do you represent in this configuration space the evolution of the system? What is the evolution of the system in this configuration space? How does it look? Is it a point? A line? A surface? What is it? Sorry? STUDENT: A path. PROFESSOR: It's a path. It's a line. Indeed, you look at your clock. And at time equals 0, well, it takes some values. And you're fine, OK, here it at time equals 0. At time equal 1, the values change. There's one parameter, which is time. So this traces a path. As time goes by, the core in this changing in time and this is a line parameterized by time-- so a path gamma parameterized by time. And that represents the evolution of your system. At time equals 0, this point could be R at t equals 0. And maybe this point is R a t equal T final. And this system is going like that. You should imagine the system as traveling in that configuration space. That's what it does. That's why we put the configuration space. And we now have a set-- not the set-- a time dependent Hamiltonian, because while H was a function of R from the beginning, now R is a function of time. So this is your new Hamiltonian. And this is time dependent-- dependent Hamiltonian. But now, the interesting thing is that the work you did before in finding the energy eigenstates for any position in this configuration space is giving you the instantaneous energy eigenstates, because if this equation here holds for any value of R, it certainly holds for the values of R corresponding to some particular time. So psi n of R of t is equal to En of R of t psi n of R of t. So it's an interesting interplay in which the act that you know your energy eigenstates everywhere in your configuration space allows you to find the time evolved states, the time dependent energy eigenstates, the instantaneous energy eigenstates are found here. So what we want to do now is evaluate in this language the geometric phase, this phase. I want to understand what this phase is in this geometric language.
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https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/8.334-spring-2014.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK. Let's start. So last lecture we started on the topic of doing renormalization in position space. And the idea, let's say, was to look at something like the Ising model, whose partition function is obtained, if you have insides, by summing over all the 2 to the n configuration of a weight that tends to align variables, binary variables that are next to each other. And this next to each other is indicated by this nearest neighbor symbol, sigma i, sigma j. Potentially, we may want to add the magnetic field like term that [INAUDIBLE]. The idea of the renormalization group is to obtain a similar Hamiltonian that describes interactions among spins that are further apart. We saw that we could do this easily in the case of the one-dimensional system. Well, let's say you have a line, and you have sites on the line. Each one of them wants to make their neighbor parallel to itself. And what we saw was that I could easily get rid of every other spin and keep one set of spins. If I do it that, I get a partition function that operates between the remaining spins. Was very easy to sum over the two variables that spin in between these two could have and conclude that after this step, which corresponds to removing half of the degrees of freedom, I get a new interaction, k prime, which was 1/2 log hyperbolic cosine of 2K, if h was zero. And we saw that that, which is also a prototype of other systems in one dimension, basically is incapable of giving you long-range order or phase transition at finite temperature that corresponds to finite k. So the only place where you could have potentially ordering over large landscape is when k becomes very large at zero temperature. We saw how the correlation length behaves and diverges as you approach zero temperature in this type of model. Now, the next step would be to look at something that is two-dimensional. And in this context, I describe how it would be ideal if, let's say, we start with a square lattice. We have interactions k between neighbors. And we could potentially do the same thing. Let's say remove one sublattice of spins, getting interactions among the other sublattice of spins that would again correspond to removing half of the spins in the system. But in terms of length scale change, it corresponds to square root of 2. This length compared to the old length is different by a factor of square root of 2. But the thing that I also indicated was that this spin is now coupled to all four of them. And once I remove the spin, I can generate new interactions operating between these spins. In fact, you will generate also a four-spin interaction, so your space of parameters is not closed under this procedure. The same applies to all higher dimensional systems, and so they are not really solvable by this approach unless you start making some approximations. So the particular approximation that I introduced here was done and applied shortly after Cavenaugh brought forth this idea of removing degrees of freedom and renormalization by-- I'll write this once, and hopefully I will not make a mistake. Niemeijer and van Leeuwen And it's a kind of cumulant expansion, as I will describe shortly. It's an approximation. And rather than doing the square lattice, it is applied to the triangular lattice. And that's going to be the hardest part of this class for me, to draw a triangular lattice. Not doing a good job. So basically, we put our spins, sigma i plus minus 1, on the sides of this. And we put an interaction k that operates between neighbors. And we want to do a renormalization in which we reduce the number of degrees of freedom. What Niemeijer and van Leeuwen suggested was the following. You can group the sublattices of the triangular lattice into three. I have indicated them by 1, 2, 3. Basically, there is going to be some selection of sublattice sites on this lattice. What they suggested was to basically define cells-- so this would be one cell. This would be another cell. This would be another cell. This would be another cell over here-- such that every side of the original lattice belongs to one and only one site of these cells. OK. So basically, I guess the next one would come over here. All right. So let's call the site by label i. And let's give the cells label that I will indicate by Greek. So sites, I will indicate i, j, cells, by Greek letters, alphabet, et cetera. So that, for example, we can regard this as cell alpha, this one, this triangle, as forming cell beta. Now, the idea of Niemeijer and van Leeuwen was that to each cell, we are going to assign a new spin that is reflective of the configuration of the site spins. So for this, they propose the majority rule. Basically, they said that we call the spin for site alpha to be the majority of the site spins. So basically, if all three spins are plus or all three of them are minus, you basically go to plus or minus. If two of them are plus and one of them is minus, you would choose the one that is a majority. So you can see that this also has only two possibilities, plus 1 and minus 1, which would not have taken place if I had tried to do a majority of two sites. It would have worked if I had chosen a majority of three sites on a one-dimensional lattice clearly. So that's the rule. So you can see that for every configuration that I have originally, I can do these kinds of averaging and define a configuration that exists for the cells. And the idea is that if I weigh the initial configurations according to the weight where the nearest neighbor coupling is k, what is the weight that governs these new configurations for the averaged or majority cell spins? Now, to do this problem exactly is subject to the same difficulty that I mentioned before. That is, if I somehow do an averaging over the spins in here to get the majority, then I will generate interactions that run over further neighboring, as we will see shortly. So to remove that, they introduced a kind of uncontrolled break-up of the Hamiltonian that governed the system. That is, they wrote beta H minus beta H, which is the sum over all neighboring site spins, as the sum of a part that then cancels perturbatively and a part of that will treat as a correction to the part that we can solve exactly. The beta H zero is this sum over all cells alpha. What you do is basically you just include the interactions among the cell spins. So I have K sigma alpha 1 sigma alpha 2 sigma alpha 2 sigma alpha 3 sigma alpha 3 sigma alpha 1. So basically, these are the interactions within a cell. What have I left out? I have left out the interactions that operate between cells, so all of these things, which, of course, are the same strength but, for lack of better things to do, they said OK. We are going to sum over all, you can see now, neighboring cells. So the things that I left out, let's say, interactions between this cell alpha and beta, evolve-- let's say the spin number 1 in this labeling of beta times spin number 2 of alpha and spin number 3 of alpha. Now, of course, what I call 1, 2, or 3 will depend on the relative orientation of the neighboring cells. But the idea is the same, that basically, for each pair of neighboring cells, there will be two of these interactions. So again, these no a priori reason to regard this as a perturbation. Both of them clearly carry the same strength for the bond, this parameter K. OK? The justification is only solvability. So the partition function that I have to calculate, which is the sum of all spin configuration with the original weight, I can write as a sum over all spin configurations of e to the minus beta H 0, and then minus u, and the idea of perturbation is, of course, to write the part that depends on u as a perturbation. It's expanding the exponential. Now, solvability relies on the fact that the term that just multiplies 1 is the sum of triplets that are only interacting between themselves. They don't see anybody else. So that's clearly very easily solvable, and we can calculate the partition functions 0 that describes that. And then we can start to evaluate all of those other terms, once I have pulled out e to the minus beta H zero sum over all configurations in Z 0. The series that I will generate are averages of this interaction calculated with the 0 [INAUDIBLE] Hamiltonian. So my log of Z-- OK. That's how if I was to calculate the problem perturbatively. Now I'm going to do something that is slightly different. So what I will do is, rather than do the sum that I have indicated above, I will do a slight variation. I will sum only over configurations. Maybe I should write it in this fashion. I will sum only over configurations that under averaging give me a particular configuration of site spins. So, basically, let's say I pick a configuration in which this is-- the cell spin is plus, the cell spin is minus, whatever, some configuration of cell spins. Now, depending on this cell spin being plus, there are many configurations-- not many-- but there are four configurations of the site spins that would correspond to this being plus. There are four configurations that would correspond to this. So basically, I specify what configuration of cell spins I want, and I do this sum. So the answer there is some kind of a weight that depends on my choice of [INAUDIBLE] sigma alpha [INAUDIBLE]. OK. So then I have the same thing over here. And then, in principle, all of these quantities will become a function of the choice of my configuration. OK. So this is a weight for cell configurations once I average out over all site configurations that are compatible with that. So if I take the log of this, I can think of that as an effective Hamiltonian that operates on these variables. And this is what we have usually been indicating by the prime interactions. And so, if I take the log of that expression, I will get log of z 0 that is compatible with the choice of cell spins. And then the log of this series-- we've seen this many times-- it starts with minus U 0 as a function of the specified interactions. And then I have the variance, again compatible with the interactions. OK? So now it comes to basically solving this problem. And I pick some particular cell. And I look at what configurations I could have that are compactible for a particular sign of the cell spin. As far as the site spins are concerned. And I will also indicate what the weight is that I have to put for that cell coming from minus beta H0. one thing that I can certainly do is to have the cell spin plus. And I indicated that that I can get either by having all three spins to be plus or just the majority, which means that one of the three can become minus. So there are these configurations that are consistent, all of them, with the sigma alpha prime, the majority that is plus. And there are four configurations that correspond to minuses we shall obtain by essentially flipping everything. And the weights are easy to figure out. Basically I have a triplet of spins that are coupled by interaction K. In this case, all three are positive. So I have three into the K factor, so I will get e to the 3K. Whereas if one of them becomes minus and two remain plus, you can see that there are two unhappy misaligned configurations. So I will get minus K, minus K, plus K. I will get e to the minus K. It doesn't matter which one of these three it is. If all three are minuses, then again, the sites are aligned. So I will get e to the 3K. If two minuses and 1 plus, 2 bonds are unhappy. One is happy. I will get e to the minus K, e to the minus K, e to the minus K. So once I have specified what my cell spin is, the contribution to the partition function is obtained by summing over contributions things that are compatible with that. So what is it if I specify that my cell spin is plus? The contribution to the partition function is e to the 3K plus e to the 3 to the minus K. It's actually exactly the same thing if I had specified that it is minus. So we see that this factor, irrespective of whether the choice here is that the cell spin is plus or minus, is log of e to the 3K plus 3e to the minus K. Pair any one of the cells and how many cells I have, 1/3 of the number of sites with the number of sites I had indicated by N, this would be N over 3. OK. Now, let's see what this U average is. So U-- I made one sign error. I put both of them as minuses, which means that in the notation I had-- no. That's fine. Minus U0. So I put the minus sign here -- is plus K sum over all pairs that are neighboring each other, for example, like the pair alpha beta I have indicated, but any other pair of neighboring cells. I have to write an expression such as this. So I have the K. I have sigma. I have beta 1 sigma alpha 2 plus sigma beta 1 sigma alpha 3. So this is the expression that I have for you. I have to take the average of this quantity, basically the average of the sum. I will have two of these averages. Now in my zero to order weight, there is no coupling between this cell and any other cell. So what the spin on each cell is on average cares nothing about what the spin is on any other cell, which means that these averages are independent of each other. I can write it in this fashion. So all I need to do is to calculate the average of one of these columns, given that I have specified what the cell is. So let's pick, let's say sigma alpha 1 average in this zero to order weight. Now I can see immediately that I will have two possibilities, the top four or the bottom four. The top four corresponds to sigma cell being plus. The bottom four correspond to the sigma cell being minus. So the top four is essentially I have to look at the average on this column. I have-- it is either plus, and then I get a weight e to the 3K. Or it is minus. I get weight e to the minus K plus e to the minus k plus e to the minus k. So once I add 2 plus e to the minus K and subtract 1e to the minus K, I really get this. Now, of course, I have to normalize by the weights that I have per cell. And the weights are really these factors but divided by e to the 3K plus 3 to the minus K. Whereas if I had specified that the cell spin is minus, and I wanted to calculate the average here, I would be dealing with these numbers. You can see that I will have a minus e to the 3K. I will have one plus and two minuses e to the minus K, so I will get minus e to the minus K e to the 3K plus 3e to the minus K, which is the normalizing weight. So it's just minus the other one. And I can put these two together and write it as e to the 3K plus e to the minus K e to the 3K plus 3e to the minus K sigma alpha prime. So the average of any one of these three site spins is simply proportional to what you said was the cell spin. The constant of proportionality depends on K according to this. So now if I substitute this over here, what do I find? I will find that minus U at the lowest level, each one of these factors will give me the same thing. So this K becomes 2K. I have a sum over alpha and betas that are neighboring. Each one of these sigmas I will replace by the corresponding average here. Add the cost of multiplying by one of these factors. And there are two such factors. So I basically get this. So add this order in the series that I have written, if I forget about all of those things, what has happened? I see that the weight that governs the cell spins is again something that only couples the nearest neighbor cells with a new interaction that I can call K prime. And this new interaction, K prime, is simply 2K into the 3K plus e to the minus K e to the 3K plus 3 e to the minus K squared. So presumably, again, if I think of the axes' possible values of K running all the way from no coupling at 0 to very strongly coupling at infinity, this tells me under rescaling where the parameters go. If I start here, where do I go back and forth? So let's follow the path that we followed in one dimension. We expect something to correspond to essentially no coupling at all. So we look at the limit where K goes to 0. Then you can see that what is happening here is that K prime is 2K. And then from here, when K goes to 0, all of these factors become 1. So numerator is 2, denominator is 4. The whole thing is squared. So basically in that limit, the interaction gets halved. So if I have a very [INAUDIBLE] coupling of 1/8, then it becomes 1/16 and then it becomes 1 over 32. I get pulled towards this. So presumably anything that is here will at long distance look disordered, just like one dimension. But now let's look at the other limit. What happens when K is very large, K goes to infinity? Then K prime is-- well, there's the 2K out front, but that's it. Because e to the 3K is going to dominate over e to the minus K when K is large, and this ratio goes to 1. So we see that if I start with a K of 1,000, then I go to 2,000 to 4,000, and basically I get pulled towards a behavior of infinity. So this is different from one dimension. In one dimension, you were always going to 0. Now we can see that in this two-dimensional model, recoupling disappears, goes to no coupling. Strong enough coupling goes to everybody's following in line and doing the same thing at large scale. So we can very well guess that there should be some point in between that separates these two types of flows. And that is going to be the point where I would have KC, or let's call it-- I guess I call it K star in the notes. So let's call it K star. So K star is 2K star e to the 3K star plus e to the minus K star e to the 3K star plus e to the minus-- 3e to the minus K star squared. So we can drop out the K star. You can see that what you have to solve is e to the 3K star plus e to the minus K star divided by e to the 3K star plus 3e to the minus K star. This ratio is 1 over square root of 2. I can multiply everything by e to the plus K star so that this becomes 1. And I have an algebraic equation to solve for e to the 4K star. So I will get root 2e to the 4K star plus root 2 is e to the 4K star plus 1. And I get the value of K star, which is 1/4 log of-- whoops, this was a 3-- 3 minus root 2 divided by root 2 minus 1. You put it in the calculator, and it becomes something that is of the order of 0.233. No, 27. So yes? AUDIENCE: So, first thing, do we want to name what is the length factor by which we change the characteristic length? PROFESSOR: Absolutely. Yes. So the next-- yeah. AUDIENCE: But we never kind of bothered to do it so far. PROFESSOR: We will need to do immediately. So just hold on a second. The next thing, I need this B factor. But it's obvious, I have reduced the number of degrees of freedom by 3, so the length scale must have in two dimensions increased by square root of 3. And you can do also the algebra analogous to this to convince you that the distance, let's say, from the center of this triangle to the center of that triangle is exactly [INAUDIBLE]. AUDIENCE: Also, when you're writing the cumulant expansion-- PROFESSOR: Yes. AUDIENCE: In all of our previous occasions when we did perturbations, the convergence of the series was kind of reassured because every perturbation was proportional to some scalar number that we claimed to be small, and thus series would hopefully converge. PROFESSOR: Right. AUDIENCE: But In this case, how can you be sure that for modified interaction and renormalized version, you don't need [INAUDIBLE]? PROFESSOR: Well, let me first slightly correct what you said before I think you meant correctly, which is that previously we had parameters that we were ensuring were small. That did not guarantee the convergence of the series or the lattice. In this case, we don't have even a parameter that we can make small. So the only thing that we can do, and I will briefly mention that, is to basically see what happens if we include more and more terms in that series and we compare results and whether there is some convergence or not. Yes? AUDIENCE: Can you explain again how we got the K prime equation? PROFESSOR: OK. So I said that I have some configuration of the cell spins. Let's say the configuration is plus plus minus plus. Whatever, some configuration. Now there are many configurations of site spins that correspond to that. So the weight of this configuration is obtained by summing over the weights of all configurations of site spins that are compatible with that. And that was a series that we had over here. And K prime, or the interaction, typically we put in the exponent, so I have to take a log of this to see what the interactions are. The log has this series that starts with the average of this interaction. OK? So this was the formula for U. It's over here. And then here, it says I have to take an average of it. Average, given that I have specified what the cell spins are. And I see that that average is really product of averages of site spins. And I was able to evaluate the average of a site spin, and I found that up to some proportionality constant, it was the cell spin. So if the cell spin is specified to be plus, the average of each one of the site spins tends to be plus. If the cell spin is specified to be minus, since I'm looking at this subset of configuration, the average is likely to be minus. And that proportionality factor is here. I put that proportionality factor here, and I see that this average is a product of neighboring cell spins, which are weighted by this factor, which is like the original weights that you write, except with a new K. Yes? AUDIENCE: So after renormalization, we get some new kind of lattice, which is not random. It's completely new. Because what you did here is you take out certain cells-- PROFESSOR: Yeah. AUDIENCE: And call them [INAUDIBLE]. PROFESSOR: Right. But what is this new lattice? This new lattice is a triangular lattice that this is rotated with respect to the original one. So it's exactly the same lattice as before. It's not a random lattice. AUDIENCE: Yes. But on the initial lattice, you specified that these cells would contribute to-- PROFESSOR: Yes. I separated K and K prime. Yes. K and U, yes. AUDIENCE: OK. So if you want to do a renormalization group again, we'll need to-- PROFESSOR: Yeah. Do this. AUDIENCE: Again [INAUDIBLE]. PROFESSOR: Exactly. Yeah. But we do it once, and we have the recursion relation. And then we stop. AUDIENCE: Yeah. PROFESSOR: OK. Yes? AUDIENCE: Is this possible for other odd number lattices? Will you still preserve the parameter? PROFESSOR: Yes. It's even possible for square lattices with some modification, and that's what you'll have in one of the problems. OK? Fine. But the point is-- OK. So I stopped here. So K star was 0.27, which is the coupling that separates places where you go to uncorrelated spins, places you go to everything ordered together. It turns out that the triangular lattice is something that one can solve exactly. It's one of the few things. And you'll have the pleasure of serving that also in a problem set. And you will show that KC, the correct value of the coupling, is something like 0.33. So that gives you an idea of how good or bad this approximation is. But the point in any case is that the location of the coupling is not that important. We have discussed that it is non-universal. The thing that maybe we should be more interested in is what happens if I'm in the vicinity of this, how rapidly do I move away? And actually I have to show that we are moving away. But because of topology, it's more or less obvious that it should be that way. So what I need to do is evaluate this at K star. OK. Now you can see that K prime is a function of K. So what you need to do is to take derivatives. So thers' some algebra involved here. And then, once you have taken the derivative, you have to put the value of K star. And here, some calculator is necessary. And at the end of the day, the number that you get, I believe, is something like 1.62. Yes. And so that says since being it's larger than 1, that you will be pushed away. But these things have been important to us as indicators of these exponents. In particular, I'm on the subspace that has symmetry, so I should be calculating yt here. As was pointed out, important to this step is knowing what the value of B is, which we can either look at by the ratio of the lattice constants or by the fact that I have removed one third of the spins. It has to be root 3 to the power yt. So my yt is log of 1.62 divided by log of root 3. So again you go and look at your calculator, and the answer comes out to be 0.88. Now the exact value of yt for all two-dimensionalizing models is 1. So again, this is an indicator of how good or bad you have done at this ordering perturbation tier. OK. Now, answering the question that you had before, suppose I were to go to order of U squared? Now, order of U squared, I have to take this kind of interaction, which is bilinear. Let's say pair of spins here, multiply two of them, so I will get a pair of spins here and pair of spins there. As long as they are distinct locations when I subtract the average squared, they will subtract out. So the only place where I will get something non-trivial is if I pick one here and one here. And by that kind of reasoning, you can convince yourself that what happens at next order is that in addition to interactions between neighbors, you will degenerate interactions between things that are two apart and things that are, well, three apart, so basically next nearest neighbors and next next nearest neighbors. So what you will have, even if you start with a form such as this, you will generate next nearest neighbor and next next nearest neighbor interactions. Let's call them K, L, M. So to be consistent, you have to go back to the original model and put the three interactions and construct recursion relations from the three parameters, K, L, M, to the new three parameters. More or less following this procedure, it's several pages of algebra. So I won't do it. Niemeijer and van Leeuwen did it, and they calculated the yt at next order by finding the fixed point in this three-dimensional space. It has one relevant direction, and that one relevant direction gave them an eigenvalue that was extremely close to 1. So I don't believe anybody has taken this to next order. You've got good enough, might as well stop. I think it's not going to improve and get better because this is an uncontrolled approximation. So it's likely to be one of those cases, that you asymptotically approach the good result and then move away. Now once I have yt, I can naturally calculate exponents such as alpha. First of all U, which is 1 over yt. 1 over 0.88 is something like 1.13. And the exact result would be the inverse of 1 which is 1. And I can calculate alpha, which is 2 minus d, which is 2 mu. With that value of U, I will get minus 0.26. Again, the correct result would be corresponding to a logarithmic divergence. So this zeroed order, OK. Those things, let's say, for the exponents to 10%, 20%. You would say that, OK, what about other exponents, such as beta, gamma, and so forth. Clearly, to get those exponents, I also need to have yh. OK. So to get yh, I will add, as an additional perturbation, a term which is h sum over i sigma i, which is, of course, the same thing as sum over alpha, sigma alpha of 1 plus sigma alpha 2 plus sigma alpha 3. And if I regard this as a perturbation, you can see that in the perturbative scheme, this would go under the transformation to the average of this quantity, and that the average of this quantity will give me 3 for each cell. So I will get 3h. And for each cell, I will get the average of a site spin, which is related to the cell spin through this factor that we calculated, e to the 3K e to the minus k e to the 3k plus 3e to the minus k sigma alpha prime. So we can see that by generating h prime, which is 3h times e to the 3K plus e to the minus K e to the 3K plus 3e to the minus K. And I can evaluate d to the yh as dh prime by dh evaluated at the fixed point. So I will get essentially 3 times this factor evaluated at the fixed point. But we can see that at the fixed point, this factor is 1 over root 2. So the answer is 3 over root 2. And my yh would be the log of 3 over root 2 divided by log of b that we said is square root of 3. Put it in the calculator, you get a number that is of the order of 1.4. And exact yh is 1.875. So again, once you have yh, you can go and calculate through the exponent scaling relations all the other exponents that you have like beta. So not bad, considering that if you wanted to go through epsilon expansion, how much difficulty you would have. And in any case, we are at two dimensions, which is far away from four. And getting results at 2 is worse than trying to get results at three dimensions. Now we want to do the procedure as an approximation that is even simpler than this. And for that-- so that was the Niemeijer-van Leeuwen procedure. The next one is due to Kadanoff again and Migdal. And it's called bond-moving. And again, we have to do an approximation. You can't do things exact. So let's demonstrate that by a square lattice, which is much easier to draw than the triangular lattice. And let's kind of follow the procedure that we had for the one-dimensional case. Let's say we want to do rescaling by a factor of 2. And I want to keep this spin, this spin, this spin, this spin, this spin and get rid of all of the other spins that I have. Not the circular round, much as I did for the one-dimensional case. And the problem is that if I'm summing over this spin over here, there are paths that connect that spin to other spins. So by necessity, once I sum over all of these spins, I will generate all kinds of interactions. So the problem is all of these paths that connect things. So maybe-- and this is called bond-moving-- maybe I can remove all of these things that are going to cause problem for-- So if I do that, then the only connection between this spin and this spin comes from that side, between this spin and that spin comes from that side. And if the original interaction was K and I sum over this, I will get K prime, which is what I have over there, 1/2 log cos 2K, because the only thing that I did was to connect this site to two neighbors, and then effectively, it's the same thing as I was doing for one dimension. So clearly this is a very bad approximation, because I have reproduced the same result as one dimension for the two-dimensional case. And the reason is that I weakened the lattice so drastically, I removed most of the bonds. So there isn't that much weight for the lattice to order. Kadanoff and Migdal suggested was, OK. Let's not to remove these bonds. Just move them to some place that they don't cause any harm. So I take this bond and I strengthen this bond. I take this bond, strengthen this one. This one goes to this one. Essentially what happens is you can see that each one of the bonds has been strengthened by 2. So I have this because of the strength. So as simple as you can get to construct a potential recursion relation for this square lattice. So this is a way that the parameter K changes, going from 0 to infinity. And we can do the same thing that we did over there. So we can check that for K going zero, if I look at K prime, it is approximately 1/2 log of hyperbolic cosine of something that is close to 0. So that becomes 1 plus the square of this quantity, which is 4K squared over 2. And taking the log of that, it becomes 4K squared. The factor of 4 does not really matter. If K is very small, like 1 over 100, K squared would be 10 to the minus 4. So basically, you certainly have the expected behavior of becoming disordered if you have a [INAUDIBLE] interaction. If you have a strong enough coupling, however, are we different from what we did for the one-dimensional case? Well, the answer is that in this case, K prime is 1/2 log hyperbolic cosine of 4K, starts as e to the 4K plus e to the minus 4K divided by 2. e to the minus 4K I can ignore. So you can see that in this case, I will have 2K. I can even ignore the minus log 2, which previously was so important because previously we had 1 here and now it became 2, which means that if I start to be 10,000, it will become 20,000, 40,000 and now you're going this direction. So again, by necessity almost, I must have a fixed point at some value in between. So I essentially have to solve for K star as K star is 1/2 log cos of 4K star. You can recast this as some algebraic equation in terms of e to the 4k e to the K star and manipulate it. And after you do your algebra, you will eventually come up with a value of K star, which I believe is 0.3. You can ask, well, the square lattice we will solve in class. I said the triangular lattice I will leave for you to solve. KC for the square lattice is something like 0.44. So you are off by about 25%. Of course, again, the quantity that you're interested in is b to t yt. b is 2 in this case. The length scale has changed by a factor of 2, which is dk prime by dk evaluated as K star, again, a combination of doing the algebra of derivatives, evaluating at K star, and then ultimately taking the log to convert it to a yt. And you come up with a value of yt that is around 0.75. And, as I said, the exact yt, which doesn't depend on whether you are dealing with a square lattice or a triangular lattice-- it's only a function of symmetry and dimensionality-- is 1. So you can see that gradually we are simplifying the complexity. Now we could, within this approximation, solve everything within one panel. Now this kind of approximation, again, is not particularly very good. But it's a quick and dirty way of getting results. And the advantage of it is that you can do this not only in two dimensions, but in higher dimensions as well. So let's say that you had a cubic lattice and you were doing rescaling by a factor of 2, which means that originally you had spins along the various diagonals and so forth-- around the various partitions of a square of size 2 x 2, and you want to keep interactions among these and get rid of the interactions among all of the places that you are not interested in. And the way that you do that is precisely as before. You move these interactions and strengthen the things that you have over here. Now whereas the number of bonds that you had to move for the square lattice was 2-- the enhancement factor was 2-- turns out that the enhancement factor in three dimensions would be 4. You essentially have to take one from here, one from there, one from there. So 1 plus 3 becomes 4. And you can convince yourself that if I had done this in d-dimensional hypercubic lattice, what I would have gotten is again the one-dimensional recursion relation, except for this enhancement factor, which is 2 to the power of d minus 1 in d dimensions. Actually, I could even do that for rescaling rather by a factor 2, by a factor of b. And hopefully, you can convince yourself that it will become b to the d minus 1 times 2K. And essentially that factor is a cross section that you have to remove. So the cross-sectional area that you encounter grows as the size to the power of d minus 1. And a kind of obvious consequence of that is that if I go to the limit of K going to infinity, you can see that K prime b would go like b to the d minus 1K. Essentially, if you were to have some kind of a system and you make pluses on one side, minuses on the one side, to break it, then, the number of bonds that you would have to break would grow like the cross-sectional area. So that's where that comes from. It turns out that, again, this approach is exact as we've seen for one dimension. As we go to higher dimensions, it becomes worse and worse. So I showed you how bad it was in two dimensions. If I calculate the fixed point and the exponents in three dimensions compared to our best numerical result, it is off by, I don't know, 40%, 50%, whereas 25% over there. So it gradually gets worse and worse. And so one approach that people have tried to do, which, again, doesn't seem to be very rigorous is to convert this into an expansion on dimensionality of 1. So it's roughly correct that it's going to be close to 1-- correct close to one dimension. But as opposed to the previous epsilon expansion, there doesn't seem to be a controlled way to do this. I showed you how to do this for Ising models. Actually, you can do this for any spin model. So let's imagine that we have some kind of a model that's in one dimension. At each site, we have some variable that as i that I will not specify what it is, how many variables-- how many values it takes. But it interacts only with its neighboring sites. And so presumably, there is some interaction that depends on the two sites. There may be multiple couplings implicit in this if I try to write this in terms of the dot product of spins or things like this. So if I were to calculate the partition function in one dimension-- I already mentioned this last time-- I have to do a sum over what this spin is of e to the K of si si plus 1 and a product over subsequent sites. And if I regard this as a matrix, which is generally called a transfer matrix, you can see that this multiplication involving the sum over all of the spins is equivalent to matrix multiplication. And, in particular, if I have periodic boundary conditions in which the last spin couples to the first spin, I would have trace of T to the power of N, where t is essentially this, e to the K of si si plus 1. Now, clearly I can write this as trace of T squared to the power of N over 2. Right. And this I can regard as the partition function of a system that has half as many spins. So I have performed the renormalization group like what we are doing in one dimension. I have T prime is T squared. And in general, you can see that I can write this as N to the b N over b. So the result of renormalization by a factor of b is simply one dimension to take the matrix that you have and raise it to b power. OK. And so I could parameterize my T by a set of interactions K, like we do for the Ising model. Raise it to the power of b, and I would generate the matrix that I could then parametrize by K prime, and I would have the relationship between K prime and K in one dimension. So this is d equals 1. And the way to generalize this Migdal Kadanoff, RG, to a very general system is simply to enhance the couplings. So basically, what I would write down is T prime, which, after rescaling by a factor of b, is a function of a set of parameters that I will K prime, is obtained by taking the matrix that I have for one set of couplings and raise it to the power of b. This is the exact one-dimensional result. And if I want to construct this approximation in d dimensions, I will just do this. So for a while, before people had sufficiently powerful computers to maybe simulate things as easily, this was a good way to estimate locations and exponents of phase diagrams, critical exponents, et cetera for essentially complicated problems that could have a set of parameters even here. Nowadays, as I said, you probably can do things much more easily by computer simulations. So I guess I still have another 10 minutes. I probably don't want to start on the topic of the next lecture. But maybe what I'll do is I'll expand a little bit on something that I mentioned very rapidly last lecture, which is that in this one-dimensional model where I solve the problem by transfer matrix, what I have is that the partition function is trace of some matrix raised to the N power. And if I diagonalize the matrix, what I will get is the sum over all eigenvalues raised to the N power. Now note that we expect phase transitions to occur, not for any finite system, but only in the limit where there are many degrees of freedom. And, actually, if I have such a sum as this in the limit of very large number of degrees of freedom, this becomes lambda max to the power of N. Now in order to get any one of these series of eigenvalues, what should I do? I should take a matrix, which is this e to the T-- e to the strength of the interactions. What did I write? e to the K of S and S prime. So there is a matrix. All of its elements are Boltzmann weights. There are all positive. And find the eigenvalues of this matrix. Now for the case of the Ising model without a magnetic field, the matrix is 2 x 2. It's e to the K, corresponding to the diagonal terms where the spins are parallel, e to the minus K when the spins are antiparallel. And clearly you can see that the eigenvalues corresponding to 1, 1 or 1, minus 1 as eigenvectors, are hyperbolic cosine of K-- strike this. e to the K plus e to the minus K e to the K minus e to the minus K. You can see that to get this, all I had to do was to diagonalize a matrix that corresponded to one bond, if you like. And just as in here, there is no reason to expect that these eigenvalues, which depend on this set of parameters, should non-analytical functions. There is no reason for non-analyticity as long as you are dealing with a single bond. We expect non-analyticities at their limit of large N. So if each one of these is analytic function of K, the only way spanning the K axis that I could encounter in non-analyticity is if two of these eigenvalues cross, because we have seen a potential mechanism for phase transitions. We discussed this in A333. That if we have sum of contributions, that each one of them is exponentially large in N, and two of these contributions cross, then your partition function will jump from one hill to another hill. And you will have a discontinuity, let's say, in derivatives or whatever. So the potential mechanism that I can have is that if I have as a function of changing one of my parameters or a bunch of these parameters, the ordering of these eigenvalues-- let's say lambda 0 is the largest one, lambda 1 is the next one, lambda 2, et cetera. So each one of them is going its own way. If the largest one suddenly gets crossed by something else, then you willl basically abandon one eigenvalue for another, and you will have a mechanism for a phase transition. So what I told you was that there is a theorem that for a matrix where all of the eigenvalues are positive, this will never happen. The largest eigenvalue will remain non-degenerate. And there is some analog of this, probably you've seen in quantum mechanics, that if you have a potential, the ground state is non-degenerate. The ground state is always a function that is positive everywhere. And the next excitation would have to have a node go from plus to minus. And somehow you cannot have the eigenvalues cross. So in a similar sense, it turns out the largest eigenvalue, the largest eigenvector for these matrices, will have all of the elements positive, and it cannot become degenerate. And so you are guaranteed that this will not happen. Now the second part of this story that I briefly mentioned was you can repeat this for two dimensions, for three dimensions, higher dimensional things. So one thing that you could do is, rather than sorting the Ising model on a line, you can solve it on a ladder, or a ladder that has two rungs. So solving the Ising model on this structure is not very difficult because you can say that there are eight possible values that this can take. And so I can construct a matrix that is 8 x 8 that tells me how I go from the choice of eight possibilities here to the eight possibilities there. And I will have an 8 x 8 matrix that has these properties and will satisfy this. It will be true if I go to a 4 strip. It will be a 16 x 16. No problem. I can keep going. And it would say, well, the two-dimensional-- and also you could do three-dimensional or higher dimensional models-- should not have a phase transition. Well, it turns out that all of this relies on having a finite matrix. And what Onsager showed was that, indeed, for any finite strip, you would have a situation such as this-- actually more accurately if I were to draw a situation such as this, where two eigenvalues would approach but will never cross. And one can show that the gap between them will scale as something like 1 over this length. And so in the limit where you go to a large enough system, you have the possibility of ascension to some singularity when two eigenvalues touch each other. So this scenario is very well known and studied in two dimensions. In higher dimensions, we actually don't really know what happens. OK. Any questions? AUDIENCE: So it appears that there are or there are not phase transitions [INAUDIBLE]? PROFESSOR: Well, we showed-- we were discussing phase transitions for the triangular lattice, for the square lattice. I even told you what the critical coupling is. AUDIENCE: But it seems to me that the conclusion of what you're-- of this part is that there aren't. PROFESSOR: As long as you have a finite strip, no. But if you have a 2-- an infinite strip, you do. So what I've shown you here is the following. If I have an L x N system in which you keep L finite and set N going to infinity, you won't see a singularity. But if I have an N x N system, and I said N goes to infinity, I will encounter a singularity in the limit of N going to infinity. Again, very roughly, one can also should develop a physical picture of what's going on. So let's imagine that you have a system that is a very large number but finite in one direction. And this other direction can be two, can be three, whatever. You have a finite size. But in this other direction, you basically can go as large as you like. Now presumably this two-dimensional model has a phase transition if it was infinite x infinite. And on approaching that phase transition, there would be a correlation length that would diverge with this exponent nu. So let's say I am sufficiently far away from the phase transition that the correlation length is something like this. So this patch of spins knows about each other. If I go closer to the transition, it will grow bigger and bigger. At some point, it will fit the size of the system, and then it cannot grow any further. So beyond that, what you will see is essentially there is one patch here, one patch here, one patch here. And you are back to a one-dimensional system. So what happens is that your correlation length starts to grow as if you were in two dimensions or three dimensions. Once it hits the size of the system, then it has to saturate. It cannot grow any bigger. And then this block becomes independent of the next block. So essentially, you would say that you would have effectively a one-dimensional system where the number of blocks that you have is of the order of N over L. So what we are going to do starting from next lecture is to develop again a more systematic approach, which is a series expansion about either low temperatures or more usefully about high temperature. And then we will take that high temperature expansion and gradually go in the direction to solve these two-dimensionalizing models exactly. And so we will see why I told you some of these results about exact value of KC, exact value yT. Where do they come from?
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https://ocw.mit.edu/courses/3-60-symmetry-structure-and-tensor-properties-of-materials-fall-2005/3.60-fall-2005.zip
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PROFESSOR: To mention, most everyone did extremely well on the quiz. But I sense that there's still some of you who have not yet come to terms with crystallographic directions and planes, and you feel a little bit awkward in distinguishing brackets around the HKL and parentheses around HKL. And there are some people who generally get that straightened out, but when I said point group, suddenly pictures of lattices with fourfold axes and twofold axes adorning them came in, and that isn't involved in a point group at also. Again, a point group the symmetry about point. A space group is symmetry spread out through all of space and infinite numbers. So let me say a little bit about resources. I don't know whether you've been following what we've been doing in the notes from Buerger's book that I passed out. That was hard to do is because we did the plane groups, and he doesn't touch them at all. So now we're back following once again Buerger's treatment quite closely. So read the book. And if you like, I can tell you with the end of each lecture, this stuff is on pages 57 through 62. The other thing. As you'll notice, this nonintrusive gentleman in the back is making videotapes of all the lectures. These are eventually going to go up on the website as OpenCourseWare. We were just speaking about that, and it takes a while before they get up, but I have a disk of every lecture. So if there's something you didn't follow or a place where I chewed my lines and you want to go back over it again-- not that that happens very often-- you are more than welcome to ask me to borrow and borrow the disk if you want to review it. So that's another resource. And then I will regularly throughout the term give hard-copy handouts of some of the things that we're doing, particularly when it involves geometry. And when we move on to three-dimensional geometry, unless of the graphics is really tight and precise, it's hard to follow what's going on. So in that vein, one of the first things I wanted to pass out-- the only other one for today-- is a demonstration that in fact in the Group 23 all you need is a single twofold axis oriented along the normal to a face and a single threefold axis coming out of one body diagonal. And that gives you all of the axes you are going to get into 23. So when this comes around, there are number of steps that you can perform letting the axes work on each other. And if you start with just the single twofold axis and the single threefold axis-- which the symbol suggests is all you need, there's only one kind of each-- if you look at a cube along its body diagonal, the twofold axis that's coming out of one face gets rotated into directions normal to all the other faces if you rotate by 120 degrees. So the little diagram in the upper right hand corner of the sheet hopefully convinces you of this. Now we've got three mutually orthogonal twofold axes and one threefold axis coming out of a body diagonal. So the vertical twofold axis swings that around by 180 degrees to give you another threefold axis along a body diagonal. And I labeled that one 3 prime in the middle diagram at right. And then if you take the axis 3 prime and repeat it by the second of the twofold axes that we have along face normals, that in the middle diagram on the right hand edge takes 3 prime and repeats it to 3 double prime. Then finally, the third twofold axis that we generated repeats the threefold prime axis to the remaining threefold axis along the fourth body diagonal. So the results start with one twofold axis oriented along the normal to a cube and one threefold axis along the diagonal of the cube, you get axes coming out of all faces and all diagonal. And there's a staple on that sheet for reasons that I don't understand, but it there was, probably on the surface of the Xerox machine when I went over there. So let me now return to our next step, and that is to add what in the language of group theory is called an extender, a new symmetry operation that can be added to a preexisting group that will generate new operations. And let's see what sort of theorems we need to describe what we should look for and in which particular orientation. We've got these 11 axial combinations, and these are frameworks that we can hang mirror planes on. So let's look at a first simple combination. Suppose we have a twofold axis, and the only nontrivial operation there is a rotation through 180 degrees. So this an axis A pi. And again, the ground rules are that if we're to add a mirror plane to this axis, which along with identity constitutes a group, we can't create any new axes. So there are two ways we can do this. One is the three-dimensional analog of something that we have already done, namely to pass a mirror plane through the twofold axis. And this is the group that we found as a two-dimensional Point Group, and we called it 2mm. And to do that, we used the theorem that says that if you take a rotation operation A alpha and combine it with a reflection operation that goes through it, you get a new reflection plane, sigma prime, that's at an angle alpha over 2 to the first. So what gave us the second mirror line in two dimensions, in three dimensions that would give us another mirror plane that's at right angles to the first. And this will give us a three-dimensional symmetry, which is also called 2mm. So it's nothing more than taking the two- dimensional Point Group and letting it come out at the board at you-- Man, that'll give you nightmares when these things are coming out of the paper at you-- and that is a valid group 2mm. There's another way that we can add a mirror plane to a rotation axis though which will not create any new axes, and that's to add the mirror plane-- the reflection operation sigma I should say since it is a combinations of operations-- exactly perpendicular to the locus of the rotation operation. And that reflection operation sigma then just flips the rotation axis end to end. And the rotation operation just swirls the locus of the reflection plane around within its own locus and doesn't create any new reflection plane. So in the particular combination A pi, if we add a mirror plane perpendicular to that as the operation sigma, this would be then all of the operations of a twofold axis over all of the operations of a mirror plane. So we've got now a combination of a twofold axis with a mirror plane. We've got the same thing here. To distinguish these two combinations, we'll write this as a fraction 2/m. And that literally in words is the way we've added the twofold axis. It is sitting over the mirror plane and gets reflected down into its other end when the mirror plane act on. So it's just language. It's nice though, as we said some weeks ago, if our language have some descriptive content to so when we look at it we can remind ourselves of what it means. So two 2mm means the mirror planes are parallel to the axis that contain it. 2/m means the twofold axis is over the mirror plane. It goes through and pierces it. What I would like to ask though is have we got a group yet? And let's take a first object-- let's call it right handed-- rotate it by 180 degrees. You get a second one, which will stay right handed. And then repeat it by a reflection operation in the mirror plane, and we'll get a third operation. Reflection changes chirality, so the third one is left handed. So we are performing the sequence of operations A pi followed by sigma. And let me append just so we make no mistake and not confuse it with 2mm that the reflection operation is normal to the mirror plane. So question, what is the net effect? And lo and behold, we have stumbled over-- if we had not been clever enough to invented it and suggested it early on-- the only way you can get from 1 to 3 in one shot is to turn it inside out and change its chirality by projecting it through a point which is the location where the twofold rotation pierces the mirror plane. And what this is going to do is to change the sense of all three coordinates. This is going to take the coordinates XYZ of object number 1 and change them into minus x, minus y, minus c. And what we're doing is inverting the motif, turning it inside out as it were, into an enantiomorph. And we have discovered as soon as we combine a pi with a perpendicular reflection plane, a new operation which we'll call in words inversion. And the symbol that used to describe this is a one with a bar over it, and we'll see why later on. But this is a onefold axis with an inversion center sitting on it, and that's what an inversion center by itself is. The implication is that they're going to be other axes that we can abbreviate such as 3-bar, 4-bar, and so on. So this is analogous to a situation that will come later where we really need a new notation. So we've fell headlong over a new type of symmetry operation, and we should consider taking inversion and adding that to the rotation axes by themselves as an extender. Obviously, if we take inversion and add it to a mirror plane, we're going to get A pi. If we take a twofold rotation, add it perpendicular to a mirror plane, we get inversion. If we put inversion and put it on a 180-degree rotation, we'll get the mirror plane back. So these things all permute one to another. You may even remember some time ago we asked in general terms when do two operations permute without changing anything. And the answer is if these operations to leave the locus of the other one alone. And the mirror plane obviously leaves the locus of the rotation operation unchanged. The rotation operation spins the mirror plane around in its own plane and doesn't create a new axis. And the inversion center leaves the mirror plane alone and takes the twofold axis top to bottom. So those three operations, inversion A pi, sigma are the three nontrivial operations that exist in the space. The fourth one is the identity operation. So here is the set of elements in the group that we will call 2/m. 2/m implies three operations inversion, a 180-degree rotation, reflection in a plane perpendicular to the axis, and the identity operation. And I'll leave it to yourself for you to convince yourself that I can rotate and then reflect or I can reflect and then invert. And all of these operations do not create any new motifs in the set. The group multiplication table, in other words, contains just the four elements, 1, 1-bar, A pi, and sigma. So the full pattern that corresponds to 2/m consists of four objects. It'll be a fourth one down here. The twofold axis tells you have this fellow is related to this one. Inversion tells you how this one is related to this one. And the mirror plane tells you how this one, number 1, is related to number 4. And the identity operation tells you how 1 is related to itself. So, again, as we've seen in the set of operations that constitute a group, there's a one-to-one correspondence between the transformations that are elements of the group and the number of objects in the pattern. So we've made one combination, and what we found from this is a new transformation inversion that involves changing the sign of all the coordinates in a space through a point which is called the inversion center. Questions? If not, let me quickly rattle off other Point Groups in this family. We could take the operation A pi/2 in 90-degree rotation and add this perpendicular to mirror plane. Let me now say something that I've said again many times before. The pattern of objects that will result is the pattern of objects that's produced by the initial group-- let's say a fourfold axis-- repeated by the extender. And the operation sigma perpendicular is the extender. So the pattern, without making any big deal about it, is going to look like this square of objects reflected down below the mirror plane. So it'll be one going down like this. One like this. One like this. The operation A pi sits perpendicular to the operation of reflection, so there will be an inversion center that arises at the point of intersection. And indeed the square above can be inverted through this point-- little hasty repairs there-- and every one up above gets inverted down to an enantiomorph below. So all of these guys up on top are right handed. All these guys down below are left handed. So this is another group. This is 4/m in international notation, a fourfold axis perpendicular to a mirror plane. There is also, unfortunately, a Schoenflies notation. The international notation tells you what you've got, a twofold axis or a fourfold axis perpendicular to a mirror plane. The Schoenflies notation tells you how you derived it. And the symbol for a twofold axis is C2, so this is a twofold axis. And what we did was to add an extender consisting of a horizontal mirror plane. Schoenflies calls this one C2h; Group C2, which is a twofold axis; a horizontal m is the extender. So Schoenflies tells you how you make it. The international notation tells you what you get as a result. Schoenflies notation here would be C4, that's the symbol for a fourfold axis, and the extender is an h. And then without making any big fuss about it, if I do the same with a sixfold axis, I will have six objects related by a sixfold rotation axis. If I take that sixfold axis and put a mirror plane perpendicular to it, these will be reflected down to a hexagon of enantiomorph equidistant below the mirror plane. That's not terribly good, but it's not terribly bad either. So this would be called 6/m, Schoenflies notation C6h. And the operation A pi exists in a sixfold axis, so there is an inversion center but also arises as a new symmetry element at the point of intersection. So with reckless abandon, you can continue on here and derive noncrystallographic groups for all the even-fold axes and derive an 8/m and a 12/m and 16/m. Lovely symmetries. I wouldn't want to draw them, but they're still symmetries. They all have an inversion center in them, but they're noncrystallographic. So we don't have to worry about them for prism purposes. I left one out because it introduces a complication that is kind of curious and interesting. Any questions on what we've done here? Yes? AUDIENCE: The 2mm, it's just the same-- Schoenflies notation is just C2-- PROFESSOR: Schoenflies notation for three dimensions is exactly the same as in two. So the three-dimensional version where this extends in a direction that is normal to the two-dimensional space of our two dimensions. Two dimensions it was this. Now just imagine them coming out of the blackboard at you. The symbol for this one was 2mm. The symbol for this one is also 2mm, the same thing. The mirror plane is a vertical mirror plane. So the Schoenflies notation is exactly the same as what we used per two dimensions. It's called C2v. We've added a vertical mirror plane. And again, horizontal and vertical. Horizontal is horizontal with respect to the axis of higher symmetry. Vertical is vertical with respect to the two-dimensional space of the two-dimensional Point Group, parallel to the axis in three dimensions. So that's the vertical indication. So let's, though, tuck that away for future reference. We've got two kinds of extenders. We've got a horizontal mirror plane, and we've got a vertical mirror plane, and these are extenders that we should consider adding. So we've taken care of 2/m, 4/m and 6/m. 2mm, 3m, 4mm, and 6mm are just the extensions into a third dimension of what we've seen and come to love in the two-dimensional space. Let me now turn to the threefold axis. And this is a curious one. Threefold axes require fewer symbols to indicate the vertical mirror planes because there's only one independent one. But let's see what would happen. And now I'm not going to attempt to draw these in three dimensions anymore. I'm going to use a stereographic projection. And what I'll do is use a solid dot for a point that's up above the equatorial plane and an open dot for one that's down below the equatorial plane. So my stereographic projection of 4mm would look like. This is the fourfold axis. This is the mirror plane. And I've got one up that gets reproduced by the axis to give me a set of four. All of these are, let's say, right handed. And then directly below them is another set of four repeated by reflection, and these are all left handed. And then there's an inversion center at the point of intersection. And I'll indicate that by the little open circle sitting right on the fourfold axis. So there is a projection of what 4/m looks like. So let me now do the same thing for a threefold axis. And I'll add to the triangle of points that a threefold axis would generate. So these guys are all of one chirality. Let's say right handed. Then I'll reflect them down, and I'll get three objects that are down. And that's what 3/m looks like. Is there an inversion center here? No. No, because the operation of A pi is missing. And it was the horizontal mirror plane combined with A pi that gave us the inversion center with all of the even rotation axis. So one of the things we have to say here is that there is no 1-bar that's present, which means in this instance, unlike the other ones, we have another option. So we can use the operation of inversion as an extender too. So we're going to get another group out of the threefold axis besides this one. And this one we will name 3/m or C3h in Schoenflies notation. They're six objects here, so there should be six operations in the group. So let me number these guys up on top as number 1, number 2, number 3. And 1 is related to itself by inversion. There's an operation A 2 pi/3. And that tells me how the one that's up is related to the second one that's up and how the left-handed one that's down is related to the one that's directly below number 2. There is an operation A 4 pi/3. And that's the same as saying 8 minus 2 pi/3. And that tells us how the things that are separated by 240 degrees are related, both up and down. I know how 3 up is related to 3 down and how 1 up is related to 1 down and 2 up is related to 2 down. This is all with the horizontal mirror plane, which I'll call sigma h. That is a total of one, two, three, four-- whoops-- one, two, three four operations. I need six. Let's ask how is this one that's up related to this one that's down? I just got rotation operations and reflection. The only way I can get from this one number 1 to this one number 3 left that's down is to take two steps to do it. I can't get from this one up here to this one down here unless I rotate 60 degrees and then invert. If I move over to 3-bar. This is A 2 pi/3 with 1-bar as an extender. The pattern would, again, look like what are threefold axis does. But then if I repeat this set of three by inversion, the two triangles above and below are skewed. The ones down below are enantiomorphs. The three that are up are of opposite chirality. And this is a new type of pattern. And in international notation, what do we call this? It's a threefold axis. But how do we indicate a symbol for three with inversion sitting on it? Let's ask if we know how each of these objects is related to each of the other. So here's 1, 2, and 3; 1 goes to 2 by A 2 pi/3; 1 goes to 3 by A minus 2 pi/3. Let's put some numbers on here for the ones down below. Let's call them 4, 5, and 6; 1 goes to 6 by an version. How do I get from 1 up to 4 that's down? I can do that only by taking two steps. Rotate 1 from here to number 2. Don't yet put it down. First invert it. So 1 to 4 involves the operation A 2 pi/3 followed immediately by inversion. And I go from 1 up to 5 down by doing the operation A minus 2 pi/3 followed immediately by inversion. And then 1 goes to 1 and itself by the identity operation. So I have six objects, one, two, three, four, five, six operations. Yes? AUDIENCE: In the cases where you're rotating and inverting, does it matter which way to the other? PROFESSOR: No, it shouldn't because they leave each other alone. So I can rotate from here to here and invert. Or I can invert from here to here and then rotate. It's the same transformation. Again, they permute if the two loci of the two operations leave the other locus alone. Maybe the enormity of what we've shown here has not sunk in. This is a new two-step operation. We can't describe it any simpler than saying, rotate and not put it down yet, follow up by inversion. Yes, sir? AUDIENCE: Couldn't we express that in another way by sort of extending the three-directional glide plane by saying invert, then transform by some vector that's parallel to the glide plane? PROFESSOR: Maybe they do in space group, but as soon as we introduce a glide plane, you've got an operation that's half a lattice translation. And that means you've got to have a lattice translation and double the lattice translation, so-- AUDIENCE: Oh, we don't have to worry about that. PROFESSOR: --when we're in a space group, yeah. That could be present. But not for a point group because the ground rules are at least one point has to remain immutably fixed in space. So this is a two-step operation, and what we're going to call it is rotoinversion. It consists of as a first step an operation by rotating alpha from 0.1 to a virtual point number 2. But before you put it down, you will invert it to a new object number 2 which is of opposite chirality. So here then are the operation of the group that results when you combine a threefold rotation axis and add to it an version center as an extended; A 2 pi/3; A minus 2 pi/3; a rotoinversion operation through 2 pi/3 and then inverting; a rotoinversion operation of A minus 2 pi/3 followed by inversion. And the symbol that is used to represent that new two-step operation is putting a bar over the top of the symbol for the axis. And then, finally, we have inversion by itself. So that's a group rank 6. The Schoenflies notation is called C3i because we got this group by adding an inversion to C3, the threefold axis. The international notation picks up on putting a bar over an axis to indicate a rotoinversion operation. So this is called 3-bar in the international notation. So there is a new group, and it is an oddball. It sort of stands alone from the other groups of the form C3h. This we derived by using the rotation of the threefold axis and adding 1-bar as an extender. So there's no mirror plane in this. AUDIENCE: That's not the same as-- PROFESSOR: That's the same as 3/m, no. 3/m is C3h. 3-bar is C3i, different extender added to the same subgroup 3C. AUDIENCE: What was the definition of 3/m? PROFESSOR: Oh, we never really finished that. That if we need the six operations that control the group, we'll have a sixfold rotoinversion axis. But this pattern looks just like the triangle produced by 3, and we add an reflection operation as an inversion, and the 3 go down. So if we ask how every one of the top is related to one underneath, that's by this horizontal mirror plane. If I want to know how I get from this one that's up to this one that's down, then I've got to rotate through 60 degrees and invert. So that would be a rotoinversion operation. Let us to extend this idea of a rotoinversion operation. And we would find this eventually in adding different extenders and falling headlong over this rotoinversion operation as we did here with 3-bar. But let me in this case start by defining a 4-bar operation. And this would contain the operation A pi/2 followed immediately by inversion. And we'll call this step A pi/2-bar. So let's try to do that and see what we get. Start with a first point, number 1. And that's up, so it's a solid dot. And let's say it's right handed. If we combine that with a rotation of 90 degrees. Not yet put it down. That's a virtual motif. Before putting it down, we inverted it. We would get one that's down, and it would be left handed. Do the operation again. I'll put the little tadpole inside. Do the operation again. Rotate 90 degrees and invert. We're back up again. So this was 1. This is 2. This was 3. And that's up. Do the operation again. Rotate and invert. And here is number 4, and it's down. Do it a fifth time, and we're back to where we started. So this is a crazy pattern. It's a pair of objects that's up and a pair of objects that's down. So there's a twofold axis in there. That twofold axis A pi leave the pattern invariant. But there is no way of specifying the relation between the two that are up and the two that are down other than doing this two-step process of rotating 90 degrees and then inverting. So there is actually in this pattern a new type of operation analogous to 3-bar, and it's called a 4-bar axis. And it's indicated geometrically by drawing a square because there's a 90-degree angular symmetry to this. But a twofold axis inscribed inside of it because this is a pattern that has a twofold symmetry. So something that has this symmetry is the symmetry of a tetrahedron. And if we draw a line from the upper edge to the lower edge, this is the locus of a 4-bar axis. International notation this is called 4-bar. That's how we generated the pattern. The Schoenflies notation is an S, little bit of exotica. This geometric solid is something that's called a sphenoid. And sphenoid. Is the Greek word for axe. And you can imagine a handle put onto this thing, and it does look kind of like an axe. You could splits firewood with a thing like that. It looks like a tetrahedron, but in a tetrahedron, it's either elongated along the 4-bar axis are squished. It doesn't have to be regular. So this is called S4, and the S stands for sphenoid. AUDIENCE: Is it part of a regular tetrahedron? PROFESSOR: No, no. A regular tetrahedron would be something where all of the edges had equal length. And what we're doing is taking one edge and the edge that's opposite it and either stretching it or squishing it. So there are two edges. This one, and this one, which are the same length. And then these four inclined edges have a different length. It could be either elongated or squished. But it's not a regular tetrahedron. If those three distances were equal, then geometrically it would be a tetrahedron. But strictly speaking, a tetrahedron is not a tetrahedron, just as a square prism with eight sides approximately equal can't claim to be a cube unless there's symmetry present that demands that this be true. In this case, the 4-bar requires that these four edges inclined to the 4-bar axis be of one length. And this have to have the same length. But there's nothing that constrains all six to the edges to be of identical length. So it's not a tetrahedron, so squished or deformed tetrahedron. So there is another two-step symmetry element that we would not have been clever enough to think of had we not discovered this sort of rotoinversion operation when we added inversion to a threefold axis. A 3-bar axis is a step that's present when you add inversion to a threefold axis. So 3-bar, what we call it for short, is identical to a threefold axis plus inversion sitting on it. A 4-bar is not equal to a fourfold axis with inversion added to it. A 4-bar is something that you cannot describe any more simply than saying there is a two-step operation in there, and it's a group of rank 4. Let me finish by setting up the task of going through this systematically. We have 11 axial combinations 1, 2, 3, 4, 6, 222, 32, 422, 622, 23, and 432. So there are 11 of those. We want to examine as extenders a vertical mirror plane that would be one extender. We should add that to each of these symmetries. We already done a lot of these. We've done pretty much up here. We could add a horizontal mirror plane. Or we've encountered inversion when we added a mirror plane perpendicular to an even-fold axis. We could add inversion as an extender. And to be complete, we should add to our list of axes in quotation marks, the 4-bar axis, having discovered it. And does that do it? Is there anything else we could do to these axes that would leave them invariant? . AUDIENCE: What about 2-bar? PROFESSOR: 2-bar; 2-bar would be rotate 180 degrees and invert. So 2-bar is identical to a horizontal mirror plane. So that's nothing new. We're already running a little over time so-- Yeah? AUDIENCE: 3-bar? PROFESSOR: 3-bar is 3 plus 1, and we call at 3i, so this one down here. This is 3-bar. We describe it for short as that, but it really is a threefold axis with an inversion center sitting on it. This thing is distinct because it's not a fourfold axis with inversion sitting on it. A 4-bar is a 4-bar is a 4-bar. You can't decompose it as a twofold axis is a subgroup. That's only half the story. I don't want to keep you anxious, not anxious to find out what the answer is but anxious to get on your way and go home. So let me submit that when we have more than one rotation axis, such as 222 or as in 32, if we put the mirror plane in normal to the principal axis, we'll call that a horizontal mirror plane. If we add the mirror plane through the principal axis, we could pass it through the threefold axis and the twofold axis, pass it through the vertical twofold axis and the horizontal twofold axis. We will call this a vertical mirror plane. And that's all we could say for a single axis, the mirror plane was perpendicular to the axis or passed through it. But when there's more than one axis, another thing we could do would be to snake the mirror plane in between the twofold axis. In that case, this twofold axis gets reflected into this one. But I haven't created any new axes. So that is going to leave the results of Euler's construction unchanged. I can't similarly put a vertical mirror plane through this first twofold axis but in between the other two. In that case, this is no longer 222 because these two mirror planes are equivalent by reflection, so I want to drop that at very least. So in any case, without belaboring the point, it's late. I could do for each of the groups that involved more than one axis I could add a diagonal mirror plane, or I should try to add a diagonal mirror planes. And this means interleaved between axes that are present in these combinations, added such that no new axis is created. But the addition clearly is going to be a new disposition of symmetry elements arranged in any different fashion and space. So the game's afoot. This is what remains to be done next. What I'll do for next time is prepare a chart that looks like this that has the results of all of the unique combinations shown and then hand out pictures of stereographic projections of all the results. I think once you know how to play the game to go through and do every single one in detail is probably not necessary. If you know how to do some of them and you know all the tricks for adding extenders, you could do it if you had to. All right. So, again, sorry we started late and sorry that we last long as well.
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https://ocw.mit.edu/courses/3-091sc-introduction-to-solid-state-chemistry-fall-2010/3.091sc-fall-2010.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So a couple of announcements. Weekly quiz Tuesday. And also on Tuesday, I want to draw your attention to the event in the slide. We have a poster here for a lecture that will occur right in this room, at 4:00 on Tuesday. And it's entitled the Wolf Lecture. The Wolf Lecture was established here about 30 years ago in honor of Professor John Wolf. John Wolf was an antecedent of mine. He was the person that invented 3091. And in 1961, he dared, as a metallurgist, to launch a variant of freshman chemistry. And you're the beneficiaries of John Wolf's initiative. And he also was a spectacular teacher, quite a showman. And in his honor, they instituted in the Department of Material Science and Engineering, the Wolf Lecture, which, as the poster says, is the entire community is invited to attend, but it's geared towards freshman. So this is a lecture that the rules are. Talk about whatever you want in terms of material science and engineering, but make it entertaining, and make it engaging and accessible to freshmen. You should go to lectures. If you can't go to this one, then by all means, go to other seminars. If all you do when you come to MIT is go to class, you're missing out on some of the richness here. We have all sorts of people coming through here every day. And you can go to these seminars, you can learn a lot. You might say, well, I'm just a freshman, I'm not going to understand everything. I go to these things. I don't understand everything. But I learn something, because the first few minutes of the lecture, if the speaker is any good, he or she is going to set up the topic for you. If you follow the first five minutes, you'll have something that, you know, gets you oriented to the topic. And then the other thing, the other reason to go-- you see this on the poster-- room 10-250, reception immediately following. That means there will be refreshments! Food, for that, you know, pick-me-up in the late afternoon. So you go to the seminar, you know, sometimes they have refreshments before the speaker. I'm not going to tell you have to go into the seminar, having eaten the food and partaken of the refreshments. You didn't hear me say that. But anyways, there's always food around here. Go to this one. I think you'll enjoy it. And the speaker is Professor Michael Rubner. He's a faculty member in course 3. An excellent teacher. As you see, MacVicar Faculty Fellow, which means he's been acknowledged as a good teacher. I think you'll learn a lot. And he's talking about nature-inspired, so biomimetics, how we see things in nature and then mimic that design in advanced material science. OK. Enough said. Let's get on with the lesson. Last day we started looking at solutions, and we recognized that bonding is the key to understanding, and it's encapsulated in that catch-all phrase, like dissolves like. And towards the end, we talked about Ksp as a metric that helps us understand common ion effect. Common ion effect, again, if I tell you you have, say, 100 units of sodium chloride that can go into solution, you would say, that's the solubility limit. But then if I tell you, I've got a solution that already contains 25 units of sodium chloride, that's a no-brainer. You've only got another 75 units to go. Now I say I've got potassium chloride in there. So now you stop and think, wait a minute. It's not sodium chloride, but it is a chloride. How do I think about this? Common ion effect through Ksp helps you reason through it. It's when you have more than one solute, and one of the constituents of the solute is already present. So today I want to talk about a subset of solutions, and in particular, I want to talk about acids and bases. And this is important, not only in materials processing, but we're going to have to understand this if we're going to go forward later and talk about biochemistry. So, you know, they're around us everywhere. You know, you probably got up this morning, you washed your hair with a pH-balanced shampoo, maybe had orange juice or grapefruit juice with citric and ascorbic acid. Your radio is powered by zinc alkaline batteries. I started my car, it's got the lead acid battery in it, and electricity for all my appliances came from coal-fired power plants, spewing out SO2, turning out acid rain. So we're off to a good start! It's Friday. So now we want to go back and understand this from the beginning. So we're going to start with a history lesson. And the history lesson starts in ancient times. Acids were known all the way back in early times for processing of food and materials. So who among us hasn't eaten something that has been pickled? Pickle means to have been processed in acid solution. In fact, the word acid comes from the word acidus in Latin, and acidus means either sour or tart. But the modern chemistry of acids and bases starts with Lavoisier in France in 1779. And I'm going to be naming many scientists from different countries today, so I'm going to use the international symbol. You know, these ovals that you put on the back of the car, so now if you see a car with this on it, you know it's a French registry. So he's probably got a Peugeot, and he's driving around with his Peugeot. And what Lavoisier said, in trying to understand acids and bases, is-- it's a really interesting story. He said that oxygen is present in all acids. And why? Because he spent most of his career studying combustion, so he was very concerned about oxygen. And the interesting thing is that oxygen, the name of the gas, oxygen, actually comes from the greek oxy, which means sharp, and, you know, the particle gen, as in to generate or to be born. OK? So it's interesting that this gas is named incorrectly. It's named for an attribute that is ascribed to acid, and it's wrong. And it's in other languages, too. The word oxygen translates into other languages as meaning something sharp, and it has nothing to do with it. But while we're on the topic of Lavoisier, Lavoisier did study oxygen. And there was an intense rivalry between Lavoisier in France, Joseph Priestley in Britain, and Scheele in Sweden. And all three of them were working simultaneously to study combustion and understand oxygen. And here I've got a very nice image. This is a portrait of Lavoisier with his wife. And it's hard to see. If the lights were a little bit dimmer, you'd see-- notice here, you see all of this chemical apparatus. There is bell jars, and various glass apparatus. His wife was his partner in the laboratory. She was unique among French women in that she read English. And in his rivalry with Priestley, he relied on his wife in order to read Priestley's writings. So they worked together. She even helped him in the laboratory. They married at the time he was 27 and she was 13, and they were-- what are you so shocked about? It's France, and it's 1779! Get over it! Anyways, she helped him a lot, and we'll say a little bit more about that later. OK. So now let's go on to the next part of the history lesson. So so far, we've got one explanation, and it's wrong. So let's go to the next one, and we'll go to Britain. Sir Humphry Davy in London in 1810. So we'll put GB here, and he's driving around in his little Jaguar, I suspect. And he said something that was correct. He said that it's hydrogen present in all acids. And that was pretty much it. He was a great scientist, did some very good work, but really didn't do anything quantitative here. So for the quantitative stuff, we have to wait for almost the end of the century. And we go up to Sweden again, to Arrhenius. And Arrhenius, in 1887-- so I'll give him a Volvo. And he said that the acid is a substance that disassociates to produce protons. So the acid is defined as a solution that dissociates to give protons. I'm going to say H plus, meaning the hydrogen ion, or P plus, and we're going to keep using the term proton. I'm not going to say hydrogen ion, I'm going to say proton. So proton in solution. So this is what he defines. And he further defines the base. He defines the base as the complement to the acid. And he says that the base is something that dissociates to give us hydroxyl. The base dissociates to give OH minus the hydroxyl. And so now we've got this whole concept of electrolytic disassociation, which is what wins him the Nobel prize in 1903 for this thing. And then we saw last day how the addition of ions to water gives charge carriers. And so the presence of protons and hydroxyl is, in fact, the way we have any charge carried through water. So up until now, when I said electrical conductivity, we were pretty much referring to electronic conductivity. But there's a second kind of conductivity, and we can have ionic conductivity. Ionic conductivity, as the name implies, is not by electrons, but by ions. And this is typically ions in solution. Ionic conductivity in solution, and a solution that is an ionic conductor is called an electrolyte. So we have electrolyte in our bodies, saline, about 5% chloride. We have electrolyte in batteries that are ionically conductive. And the term simply means, we have conduction by ions. And so it was the theory of electrolytic dissociation that won the Nobel prize. So how does this work? So we can start with something like HCl gas. I'm going to dissolve it in water, and this will give me proton, H plus. And I'm going to write aq, meaning that it's dissolved in water, and the chloride ion, also dissolved in water. And so this gives me an acid. This is a proton donor. And now let's look at something like sodium hydroxide. Sodium hydroxide at room temperature is an ionic compound, so it is a solid, but it is soluble in a polar, hydrogen-bonded liquid, so H20, to give hydroxyl aqueous plus sodium ion aqueous. So again, we see the dissociation. And then we can have, from here, a neutralization reaction. And the neutralization reaction is simply reconstitution of the solvent. So neutralization, another way to think about it, because we're not going to be confined to water by the end of the lecture, neutralization is simply a reaction that results in reconstitution of the solvents. So the solvent, in this case, is water. So how would we reconstitute water? We combine proton with hydroxyl. So let's do that and see the result of that reaction. So if I take, and run it in the vertical direction, proton plus hydroxyl, we'll give water again-- let's give H20 liquid-- and now you can see Na plus Cl will give me NaCl, aqueous dissolved. So this is what you probably learned in your high school chemistry, that acid plus base ca give you salt plus water. So you see all of this resulting from just the simple Arrhenius definition. So this is good. We've got off to a decent start here with Arrhenius. But then the theory has its limitations, as they all do. And how do we discover the limitations? With some new data. So let's look at some data. So it had been known for a long time that ammonia, when it's dissolved in water, can act as a solution capable of neutralizing an acid. So if ammonia dissolved in water neutralizes an acid, then this must be a base. But look, there's no hydroxyl here. So the Arrhenius definition of base is inadequate to account for this. So let's just get that down. Ammonia, which you know is a gas at room temperature, will neutralize acids. But no hydroxyl present. So something's going on here that we can't account for by the simple Arrhenius definition. So to get us out of this conundrum, we had to wait until 1923. And two scientists simultaneously enunciated the same ideas. So I'm going to put them both down. 1923. Bronsted, who was in Denmark, and Lowry, who was working in the UK. Also 1923. And I'll give him a Jaguar as well. And so the two of them proposed a broader definition to account for what's going on here. And what did they say? They said that acid-- they'll keep the same definition as Arrhenius. So an acid is going to be a proton donor. So that's good. Let's even put here, same as Arrhenius. But now here's the difference. The base is no longer confined to hydroxyl chemistry. They call it a proton acceptor. And that's different. So hydroxyl-free. Now, let's be careful here. If I propose this new definition, I can't throw out hydroxyl. So you've got to watch to make sure that by broadening the definition, we don't exclude hydroxyl. So the theory has to encompass what we already know, and then continue to encompass things that tend to contradict what we already know. So let's take a look at what this gives us. So I'm going to write a broader equation here. I'm going to write this as the acid. So what do I have? I have every acid has to have a proton in it, plus some residual. So I can rewrite the acid in this form. I'm going to react it with a base. So according to this definition, this has to be a proton donor. And this, you're going to watch me on this one. This is going to be a proton acceptor. And we'll put some identities here. Eventually we're going to write this with ammonia in it, but let's just do the broad definition. So what happens is, if this is a proton acceptor, on this side of the equation, it takes the proton from here. This is the proton donor. It gives it away, and we end up with a BH plus. So this proton acceptor has accepted the proton, leaving behind the deprotonated A minus. That's a nice little reaction. But now, if you'd nodded off for the last 90 seconds, and then opened your eyes and said, well, he said acid is a proton donor, this thing can give up a proton. So OK, I'm going to call this thing a proton donor. And this is most certainly a proton acceptor. Look, it has accepted the proton. So this is going to be a proton acceptor. So now I've got in this equation two proton donors and two proton acceptors. And they're linked. This proton donor, BH plus, is linked to B. So I'm going to put a little yolk over this one. And this HA, the original acid, is linked to this base, A minus. So I'm going to put a yoke over that. And the Latin word for yoke, to yoke, is jugere, from which we get the word conjugate. So I'm going to call these things conjugate acid-base pairs. And every equation has two. Because you've got a proton donor proton acceptor. So now this is, follow the bouncing ball. If you want to understand acid-base chemistry, just follow the proton. Here's proton here, goes over here. Here's something deprotonated, now it's protonated. That's the rhythm here. So the whole thing is, acid-base reactions can be defined as a proton transfer reactions. Remember we talked about ionicity and electron transfer in order to achieve octet stability? So we saw the whole life and times of electron transfer reactions. Now life and times of proton transfer reactions. OK. And last thing I want to do, is to just make sure that we see the definitions. You know, all queens are female, but not all females are queens. So this one here, what's this? This is definitely, it's a proton donor. So it's an acid. So it's definitely an Arrhenius acid, because it's got a proton, and it's also a Bronsted-Lowry acid. Now, what about this one? This is a base. It may not have hydroxyl in it. May or may not. So this is definitely a Bronsted-Lowry base, but I can't say for sure that it's an Arrhenius base, because Arrhenius has to be O minus. This one here, what about this? Unless it's OH minus, this one here is, what? It's Bronsted-Lowry base. And this one here, this can be both Bronsted-Lowry acid and it can be an Arrhenius acid. So this is how we can differentiate them. OK, good. Now let's go to the ammonia, see if we've got ourselves out of the conundrum so we can write now NH3. And I'm going to say that this is ammonia that's already dissolved. So ammonia is already in aqueous solution, and it reacts with water, H20 liquid, to do what? This is supposed to be a-- if this is a base, remember, this is the thing that we're trying to demonstrate. If it's a base, according to this, it's a proton acceptor. So I'm going to stick a proton on this, and I'm going to get NH4 plus the ammonium ion. And how did I get the ammonium ion? I took a proton from water, leaving behind OH. So can you see that how this thing works? By gobbling up protons, by becoming a proton vacuum cleaner, it takes protons out of water, leaving an excess of hydroxyl. And in effect, now we've got something that's tantamount to an Arrhenius base. But it all started off with this. So now we can link these two. I'm going to yoke ammonium and ammonia as conjugate acid-base, and there is water and hydroxyl as acid-base pairs. So again, let's get the colors going. Here we've got Bronsted-Lowry-- the bases are always going to be in blue. Blue and base both begin with the letter B. So this is definitely Brosted-Lowry base, but it's not an Arrhenius base. This is a Bronsted-Lowry base, and it's also an Arrhenius base. This one here, we can call this one-- in this case, it's a proton donor, so it's both an Arrhenius acid and a Bronsted-Lowry acid. But look at this. You'd say, OK, I know what he's doing. Arrheius, Bronsted-Lowry, Arrhenius, Bronsted-Lowry. This is Bronsted-Lowry only, this must be Bronsted-Lowry only. But look! This is a proton donor. It dissociates to give protons. So this is both Bronsted-Lowry and Arrhenius. Most people will miss that. We'll get to see, maybe, on a subsequent celebration, how many of you miss that. OK. So. So now, let's go into the chemistry here. Because I said up here at the beginning of the lecture-- what is it-- bonding is the key to understanding. So what is it about bonding here that defines the Bronsted-Lowry base? So I'm going to write this reaction in this way, now. I'm going to put-- here's the proton, which is coming from water. And I'm going to write that with NH3. I'm going to use the Lewis structure. So, you know, Lewis structure looks like this, Nitrate has got five valence electrons, three of them are bound, and now I've got this long pair. So what do we know about the electronic structure of proton? What does it look like? What's the Lewis structure of proton? How many electrons on proton? None. This is nothing. This is very needy. It's like the neediest friend you have. You know, the one that's always taking stuff from you, taking your emotional energy, giving back nothing? That's proton. That's the human equivalent of proton. So if proton wants to make a bond to form NH4, proton is going to come over here and exploit both electrons. So if this is going to be a proton acceptor, the only way it can be a proton acceptor is to have two unused electrons available. This is going to be a dative bond, agreed? Dative, because both electrons from the bond come from the nitrogen. The proton doesn't contribute anything. So that's the hallmark. The proton acceptor axiomatically must have an available non-bonding pair. So wherever you see non-bonding pairs, wherever you see a compound that has this capability, it could serve as a Bronsted-Lowry base. We know that water, because it has hydroxyl in it, has to satisfy this. So let's take a look. H20 we can do similarly. Oxygen. It's got 6 electrons, 2 are in the, bonds and we have 2 non-bonding pairs. So then when we attach proton here, we make something called hydronium. We make hydronium ion. So it's a little fancier than just saying, I've got this naked proton with no electrons swimming around in water. In fact, it's more coordinated like this. H30 plus, and while we're in the neighborhood, we might as well show. This is sp3 hybridized. We've got one, two, three, four orbitals. Hydrogen's on three. It's got a net charge of plus 1, and lone pair on the fourth side. So there's the structure of hydronium. And we can now write the equation. We can say H20 liquid plus H20 liquid gives me H3O plus, plus OH minus. So this is now the self-ionization or self-dissociation reaction. So now I can pair these as well. I can yoke these. I have acid-base pair. So this is acid here, and the conjugate base is water, H20. Hydroxyl, we know, has to be a base, and its conjugate acid is water. So you see, in the same equation, same place, same time, water is acting as both acid and base relative to these two species. And so we call such a such a compound that acts as both acid and base, we call it amphipathic. Has two moods. You know, amphi, like in amphitheater. Have you ever seen a Roman theater? It looks like this from the top down. It's got all this, you know, like this, and all the people are here, and there's the stage up here. But if I make a theater in the round, that is amphitheater. Or as it's commonly mispronounced, ampitheater. 3091ers do not say ampitheater, they say amphitheater. So this reaction here is self-dissociation. Right? This reaction is self-dissociation of water. Or because we're forming ions, it's also called self-ionization. Self-dissociation or self-ionization of water. Now, turns out that the chemical's power of H3O plus and OH minus is very high. So this reaction doesn't go very far to the right. The amount of the dissociation is tiny, and to be specific, at 25 degrees C, the amount of, if you take deionized, deaerated water, absolutely pure and pristine. The amount of native H30 plus and OH minus is on the order of one part in 10 million. So H3O plus concentration would be 10 to the minus 7 molar. And obviously, from the stoichiometry of the equation, you get the identical amount of OH minus. Well, that's in the self-dissociation. And so you have very, very few charge carriers, which is why high-purity water is a very poor conductor of electricity, even as an electrolyte. And that's, I've told you in the past, that could be one indicator of certain toxins. If you measure the conductivity of water and you discover it's abnormally high, and it's putatively supposed to be pure water, that's an indication that it's not pure water. There's some other charge carriers present. So we can write a Ksp we analogy. And it's called the water ionization constant, which is the product H3O plus and OH minus. And you can do the math. That's 10 to the minus 14 is the product. And you can use the common ion effect here. If I introduce some other acid, look at how this equation helps you the same way that Ksp did. If I have acid, in other words, protons donated from some other source, that means this number is going to be higher than 10 to the minus 7. If this number is higher than 10 to the minus 7, and the product must be 10 to the minus 14, this must be lower. So when I have, H3O plus goes up, then OH minus must go down, and under these circumstances, we have something that is called proton-rich. And we call this acid, simply because the proton concentration exceeds the hydroxyl. And likewise for the other. And so we can make a plot of this, and that's shown here. And all we've got is OH versus H3O plus. And you can see that it's a right hyperbola. But it goes back to that comment that I made in the past about how you graph data. I don't know what to do with that, because it's curved. I can't tell whether the data are good or bad. So instead of looking at something like this, y versus x, I want to transform so that can get a straight line. Because then I can make a judgment about goodness of fit and so on. So I need some f of x versus g of y to, quote unquote, straighten this out. And we have that, thanks to another Dane, by the name of Sorensen. Sorensen was the one that gave us another way to think about it. Sorensen, who in 1901, and he's got the DK on his-- he's probably got a Volvo, too. Or maybe he's driving a Saab. So he was a biochemist at the Carlsberg brewery. Yes, they had biochemists, because they wanted understand the chemistry of beer production. What a concept. Having people that understand the technology of the business. So he was working at the Carlsberg brewery, and he decided to take this acid-base business and turn it into something that's much more readily recognizable. And so he defined a concept called the chemical potential, which, if you like, is the reactivity, the chemical potential of hydronium. And he gave it the symbol. Lowercase p for potential, and uppercase H as, obviously, for the hydrogen ion. And he said, I'm going to make this a logarithm. So it's logarithm, and in those days, everybody was using slide rules, so it's log base 10 of the concentration of hydronium. And being a good engineer, he recognized since this starts off at 10 to the minus 7, the log of a number less than one is going to be negative. And who wants to deal with negative numbers? So you put a minus sign here. That way, pH normally is a positive number. And now you can see the results of Sorensen's straightening things out for us. So now you go over a wider range, and you can see, you have nice straight line relationship, and then you can get data on there, and so on. And this is taken from your book, and it shows the pH values of some common substances. If you start here at neutral pH 7, you have milk, you have human blood, normal things that you would expect, fluids and so on, are hovering around 7. If you take some coffee, you're going to go into the acidic region. You see tomatoes down here at about 4. Wine, you'll see in the reviews of wines, they'll say it has balanced acidity and so on. Yeah, it's down around pH 3. And carbonated soft drinks, some of them get down to 2. Vinegar is wine that has spoiled. Vin aigre, which means eager. The Middle French eager meant impetuous, or tart, or something. So this is spoiled wine. And the pH changes. So by measuring the pH of wine, you can tell if it's changing or not. There's lemon juice. Gastric juices in the stomach can get down to 1, pH of 1. But you want that happening only at one point in the digestive cycle. If you run around for long periods of time at pH 1, you will ulcerate, in other words, you'll puncture the walls. And if that's about to happen, you need to neutralize. So how do you neutralize? You go up with something that has a high pH. So you can start with something like Alka-Seltzer, sodium bicarbonate at about 8.4, or milk of magnesia, 10.5. Why is it milk of magnesia? From last day, because it's not a solution, it is a suspension. The magnesia is in suspension. The magnesium hydroxide. And that's one of those mandatory shake well before using, because all the magnesia is on the bottom, and you've got this almost clear, colorless liquid on the top. So you're just drinking water. Not going to help you if you've got stomach pain, but gastric juice is down here. If you're really desperate, don't reach for ammonia. You have to be patient. They have to work with milk of magnesia. OK. So that's the range. So what else do we have here? Let's take a look. So far we've been assuming that when we add acid to the system, we get 1 to 1 dissociation. So the next message that I want to give you here, is that not all acids are of equal strength. See, I could look up here and say, well, maybe the reason that the lemon juice is down it at 2 and the gastric juices are at 1 is simply a concentration effect. How much acid was introduced. No, there's another explanation for it. And that is that certain acids don't fully disassociate. So for example, if we look at 1 molar HCl, hydrochloric acid, you look at 1 molar hydrochloric acid, that goes 100% into solution and gives us 1 molar H30 plus. So 1 to 1 correspondence between how much hydrochloric acid we introduce, and how much hydroxyl that we generate. So this is total dissociation of the HCl. And as a result, we call this a strong acid. In contrast, let me show you a weak acid. So a weak acid is going to dissolve, but it doesn't dissociate. So there's really two steps here. First you've got to get the stuff in solution, and then once you get it in solution, it has to break apart. It's possible to go into solution and not break apart. In which case, you don't have the protons, but you've got the solution. That's not an acid. So a weak acid would be, let's, in contrast, look at a weak acid. A weak acid would be something like acetic acid, which is CH3COOH. And this is for historical reasons. Normally, the proton that's going to dissociate is at the front of the formula, but this was written long ago, before people understood this theory. And that's the way you'll see acetic acid written. So really, this is the proton, and the CH3COO is really the acetate anion, Ac, acetate anion. So we're going to put that into solution, like this. I'm going to put it into water, H20 liquid. So this is the salt out of solution, and all we're going to do is make it into the aqueous solution of same COOH, and I'm just going to write aq. Now, if this were strong acid with impunity, I would write whatever the concentration of acetic acid is, I break it apart, and I get the full amount of the proton. What happens is that this, now plus H20, gives me H3O plus plus the remaining acetate, which is CH3COO minus. And now, here's where the weakness is manifested. The degree to which the acetic acid grabs protons from here, and then ends up being protonated, is very, very small. It turns out that in the case of 1 molar-- I'm going to write acetic acid this way, now. HAc. So this thing here is the CH3COO minus, all right? It turns out that in 1 molar acetic acid, we get only 0.4% reaction. 0.4% reacts and dissociates. So to give protons, it hangs onto most of its protons. So we can then represent this in the form of an acid dissociation constant. And that it looks like this. K sub a for acid. So on the right side, what we're going to do is make a mass balance here. We're going to take the product of the proton, the acetate divided by the undisassociated acetic acid. So we'll put H3O plus concentration times the acetate concentration divided by the undissociated HAc aqueous. So I'm trying to represent what the ratio is here. And instead of being one to one, it's 10 to the minus 5. Very, very weak. You get a little bit, but not much. Now, you could say, well, 10 to the minus 5, this is, just for reference, it's 100 times 10 to the minus 7, which is the Kw. So the point here is, if you put in acetic acid, you'll end up with 100 times the proton population that you would have had by just self-dissociation. So it isn't acid. It is donating protons. But it's not doing it a lot. So this is called a weak acid, because it's a poor proton donor. And so we can then look at a comparison to, say, over here, HCl. Let's go back here, and we can write the acid dissociation constant for HCl, for the strong acid. And in that case, Ka over here, which is going to be the H3O plus Cl minus, H3O plus. In the case of HCl, we're going to get proton and chloride over undissociated HCl dissolved in water. And that's 10 to the plus 6, which for all intents and purposes is infinity. You put in 1 molar HCl, you get 1 molar H plus. And look at the ratio. 10 to the sixth, 10 to the minus 5. There's a 10 to the eleventh relationship between the weak acid and the strong acid. And here's the cartoon that shows this strong acid, put in this amount, 100% dissociation, weak acid gives you small amounts of proton, and a very weak acid is imperceptible amount of disassociation. And here's a table that quantifies it. So the strong acids up here with the Ka. This value of Ka has to be greater than 1, because it's saying that we're getting a lot of dissociation, the reactions moving to the right. And you have these numbers here. 10 to the ninth, 10 to the eighth, 10 to the sixth. Those are differences without distinction. if I say the ratio is a million to one, or a billion to one, for all intents and purposes, you have dissociation. And then here are the week acids. Look. 10 to the minus 3. There's phosphoric, hydrofluoric, and so on, all the way down. There's carbonic acid, 10 to the minus 7. So there's the list. And just another cute demonstration of the relationship between bonding and acidity, the acid strength increases from left to right, and the bond strength increases from right to left. You know, HF is very strong. It's a hydrogen-bonded. Very tight HF bond. But what determines if it's going to be a strong acid or not? It's how willing HF is to give up its H. But its H is tightly bound. So oddly enough, in HF, you don't give up much of the H. HF is over here. It's less than one. It's a weak acid. HCl is a pretty strong acid. If you want to go heavy-duty, HBr and HI are even stronger acids, because they have a weaker hold on the hydrogen within them, so they are much, much more generous proton donors. So you can see that from there. OK. So what's the takeaway message here? The takeaway message is that equal acid concentration does not mean equal acid strength. You have to be mediated by the Ka. Equal acid concentration, which means, how much did you dissolve? Does not equal acid strength. And what's the reason? Because equal acid concentration, this is a function of solubility, whereas this is a function of dissociation. So many dissolve, few dissociate. How about that? There's a nice tagline you can use. Use that at a party this weekend. Many dissolve, but few dissociate. All right. Now I want to go very, very high. Big concept, all right? Last definition of acid-base comes from the United States. G. N. Lewis. Same G. N. Lewis who gave us the Lewis cross and dot structures, covalent bonding, didn't stop inventing. So we're going to put U.S.A., with a Chevrolet out in California. And what did G. N. Lewis tell us? He said, I want to extend the Bronsted-Lowry concept. But you know, let me give you an analogy. Have you ever stood on a bridge and watched the traffic go? You see it starts from a stop sign or a traffic light. Light turns green, the first car pulls away, the second car pulls away, the third car pulls away. The other way you can look at it is, when the first car pulls away, there's a car vacancy. And instead of watching the cars go from right to left, watch the car vacancy move from left to right. And the two are linked. And point of fact, if you're the fourth car from the traffic light, light turns green and you hit the horn, why? You can't drive. You have to drive only into a car vacancy, and it takes time for the car vacancy to get to you. When the rate of vacancy flux is not matched by the rate of car flux, then we have a collision, and then we exchange papers. So it's a mass transport problem. So now what G. N. Lewis said was, instead of looking at this reaction as a proton transfer reaction, and looking at proton attachment, or proton acquisition, he said, why don't we look at this from the other side of the relationship? And so he said, let's look at something like, for example, we've got NH3 here. So let's put the N with the 2 electrons. And we've been talking about this from the perspective of the bond being formed here through the acquisition of the proton. Lewis said, I can view this from the other perspective, and say, I'm going to view it from the perspective of the nitrogen. From the perspective of the nitrogen, this reaction represents the donation of the electron pair. A very high concept. So this base is not about proton attachment. It's about donating an electronic pair. So let's get that down, because that's really good. So I'm going to say something that's a base looks like this. This is what a base is. It's a lone pair. This is electron pair, very high level base. And what's the proton give us? What we're doing, is we're talking about electrons and donation. So what's this other thing in that same category? An electron pair can match up with an empty orbital. That's the only place electron pairs can go. They go to the empty orbitals. So electron pair is a base, and this is just a circle around nothing. OK? This is nothing. But it's a special kind of nothing. This is called a vacant orbital. And I'm going to call a vacant orbital an acid, in the most general sense. So now I'm going to take an electron pair plus vacant orbital, and what did they react to give me? Covalent bond. You'd expect that from G. N. Lewis, because G. N. Lewis enunciated covalency. That's why you'd expect it from G. N. Lewis. So this is, vacant orbital plus electron pair gives covalent bond. Very high level. Very high level. So what is a base? A base is now not just a proton acceptor. It's an electron pair donor, and the acid is an electron pair acceptor. So now there's no chemistry here. No chemical identities. Which means, I can even go from-- I don't even have to be in solution. I don't have to be in a liquid. This could be a gas and a solid. Could be a gas plus a gas. So this is, like, Darwinian. You know, we've come out of the primordial ooze, and now we can fly, because we can talk about any chemical reaction we want. Any state of matter. It's fantastic. Absolutely fantastic. So here, let me do one last color drawing. We want to do this one here. This is Lewis and the biggest concept. So we're going to go like this. Base like this. Base with its lone pair hanging out plus acid gives us-- three colors. This is great. We've got all this colored chalk, it's Friday, what could be better. All right. So now we're going to end up with A, B, there it is. This is the Lewis concept. Lewis acid-base concept. In the broadest thing. Now I'm going to show you an example of how you can use this. I'm not going to stop here. OK, let's go. All right. So we're going to talk about acid rain from burning of coal. You know, about half of the electric power in this country comes from burning coal, and coal contains about 1% sulfur. You can purify it of sulfur, but it takes money to do so. A ton of coal will give you about 25 million British thermal units, and that's the number in SI units and joules. So 3 tons of coal will give you 1 megawatt per day. By the way, it takes about one gram of uranium to do the same thing. So someday, when you're sitting there as a policymaker, you've got a choice of 3 tons of coal or one gram of uranium, keep this fact in mind. A 10 megawatt plant burns 30 tons of coal per day, containing a third of a ton of sulphur, which makes 2/3 of a ton of SO2. And SO2 is a precursor to acid rain. Now we can reduce that SO2 emissions by reacting SO2 with a lime, CaO, according to this reaction, to make calcium sulfite. And now I'm going to bring in this. Lewis. Oh, here's from your textbook, there's the scrubbers with the calcium oxide, and water missed, and so on, and eventually they trap the SO2. They do nothing to the CO2. So all the CO2 is going up. So here's calcium oxide, which is a solid. It's lime. You know, stuff you sprinkle on football fields. And this is SO2, and it's got this structure with the resident bond double single. The oxygen here acts the way it did in the glasses, as a modifier. It goes in and breaks this double bond, and now we have three single bonds. So we can look at this from Lewis acid-base. You know, calcium oxides, electron pair donor. If it's an electron pair donor, it's a base. So it's a base. And you know SO2 better be an acid, or your theory is nuts, because SO2 is a precursor to acid rain. So it has to be an acid. It's an electronic pair acceptor, which is what we said. It's an electron pair acceptor. So this is a Lewis acid-base reaction that is a gas-solid reaction. So we started with aqueous solutions, and we end up generalizing all of these types of reactions. So that's pretty good. Here's a plot of acid concentration up in the sky, here. You're Here. Yeah. Last thing I'll do, I talked a lot about Lavoisier and Scheele and Priestley. This is a play, if you've got the little bit of time this weekend. You might try reading it. This is Carl Djerassi and Roald Hoffman, both Nobel Prize winners in Chemistry. Not the same year. They won independent Nobel Prizes, and actually speak to each other, and they collaborated on this play. Here's the setup, the premise. The premise of the play is that the Nobel committee decides to give Nobel Prizes retrograde. Before 1901. You see, you can't get a Nobel Prize posthumously. You've got to do something great, and you've got to keep on living until you get the prize. So they decided they're going to go backwards, you know, maybe give prizes to people like, I don't know, Maxwell or Faraday or whatever. And they decide, it should be simple in the old days. Because we didn't have all this dog-eat-dog competition in science, and big grants, and corporate interests. Should be simple. Until they hit the Nobel Prize for the discovery of oxygen. And then they hit the wall with these three different competitors. And so the story goes that the three competitors and their wives end up in Stockholm, and the interactions the take place. And it's really fascinating, because you have Lavoisier, who was the political conservative but the chemical radical, Priestley, who was the chemical radical but the political conservative-- by the way, Lavoisier. You know how he died? He was guillotined during the French revolution. And people think it's because he was a tax collector. Not true. He was a tax collector. The real reason is Marat, who was one of the chief justices of the Jacobin terror, was also a scientist. And around 1740, Marat had submitted an article for publication, and it was rejected. And Lavoisier was on the editorial board. And so sometimes when I get an article to review, and I have to give it a negative review, I think about this story. And sometimes, I will just say, you know, I'm really busy teaching 3091 and I won't get to this on a timely basis. OK, people. Have a nice weekend.
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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I would now like to show you how to calculate the moment of inertia of a typical continuous body. Let's consider a rigid rod, very thin. And what we want to do is calculate the moment of inertia of this body about the center of mass. Let's say the body is of length L, and it has total mass M. Now, recall that the moment of inertia about the center of mass we defined as an integral of dm r-squared integrated over the body. Now, our challenge today is to understand exactly what all these terms are in this expression. What is dm? What is r? And what do we mean by an integral over the body? So now let's do a stepwise interpretation of each term. The crucial thing is to introduce coordinates systems. So let's choose a coordinate system. And let's put the origin at the center of mass. Now, the most important thing is that we're going to do an integral. So we need to introduce the integration variable. And that's the hardest part of setting up intervals. So what we want to do is arbitrarily choose an element dm. So there's our elements dm. And here's our integration variable, it's a distance x from the origin. So that's the integration variable. Now, there's two things when we set up the integral. r is equal to that integration variable. r is abstract in this expression, but in this concrete realization it is the integration variable. The second place the integration variables shows up is in dm. dm is a mass in this small element. But if we want to express that in terms of our integration variable, we have to express it in terms of the differential length dx. So dm mass is equal to the total mass per unit length. We're assuming the rod is uniform times the length dx of our small piece. And now we've set up the two pieces that are crucial and all we have to think about now is what does an integral mean. Well, an integral means that we're dividing up the piece into a bunch of small elements and we're adding the contribution of each small element. So in particular, when we write Icm equals-- now, we can write it as m over Ldx. That was our dm. And the distance of dm from the point we're computing the axis is x-squared. Now, the question is what is our integration variable doing? Well, x is going from minus L over 2-- that's at this end-- to x equals plus L over 2 on the other end. So we have x minus L over 2x equals plus L over 2. And now we've set up the integral for the moment of inertia, and the rest is just doing an integral. Recall that the integral of dx x-squared is x-cubed over 3. And so this integral then simply becomes Icm equals m over L x-cubed over 3-- evaluating from the limits minus L over 2 to x equals plus L over 2. Again when you evaluate the limits, what we get is m over L we have to put in L over 2 cubed divided by 3 minus L over 2 cubed divided by 3, and that's 1 over 2 cubed is an eighth-- divide by third, that's the 24. 24 minus 24 is a 12. And so what we get for Icm is m over L, 12 L-cubed, or 1/12 m L-squared is the moment of inertia about the center of mass of our rigid rod. And this is a measure of how the mass is distributed about this axis.
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https://ocw.mit.edu/courses/3-320-atomistic-computer-modeling-of-materials-sma-5107-spring-2005/3.320-spring-2005.zip
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PROFESSOR: OK. Let me also remind you that we slightly changed the schedule for next week. On one of the older schedules, Tuesday appears as a lab date, but it's actually Thursday. So Tuesday will be a regular lecture here and Thursday will be the lab. And for the lab, we meet in 1115. That's on the handout for the first day. Also, if you're not registered for class, you don't have automatic access to Stellar, the website. So we can manually add you, but you'll have to let us know. Just send an email to Professor [? Marzari ?] and myself. And even better is if you send an email to both of us. And then we'll manually add that. And we'll put links there to things like articles and things that may have restricted access, copies of the lecture notes, things like that will appear there. OK. So what I want to go do is go back to their potential models. And what we're going to do today, is talk a little more in detail about the sort of formal or conceptual failures of pair potential models, and then how to get addressed with other empirical models. But before I do that, I wanted to-- let's see if I can get this into place, guess not-- go through a few practical issues, since the first lab is actually on using pair potentials to model things so you get a bit of an idea. So in a potential model, your energy is written pairwise. And so essentially what you need to do between every pair of atoms-- every couple iJ-- essentially to evaluate the distance between them and then put that in some kind of energy function. So this is clearly an N squared operation where N is the number of atoms, since you have to look for the distance between every atom and every other atom. So if systems get extremely big-- and that's often the reason why you want to use the pair potentials rather than, say, a quantum mechanical method-- N squared will get very large. And people do kind of tricks. If you have millions and millions of atoms in your simulation, if you know ahead of time that some of them are very far from the atom on which J you want to calculate-- energies or forces-- then you really never want to look further distance. So people do things like keeping neighbor lists. So literally, for every atom you would know, well, these other items or possibly in its neighbor environment. So then you really only have to look for distances with those atoms. So if you have a solid, that's great. Because if you have a solid, then the topological relation between atoms will not change much during the simulation. Of course, if you're simulating a gas or a liquid, they will. And then what you have to do is essentially update the neighbor lists regularly. And so there's overhead associated with that. But the benefit you get is that you essentially have an order N system rather than an N squared system. I'll show you in a second some large scale simulations. Very often, you will want to relax a system-- so get it to its lowest energy. And very often, that's done by calculating the forces on atoms and literally stepping down the force. If you want to do dynamics, Professor [? Marzari ?] will later teach you molecular dynamics. You also need the force because the force will tell you how much atoms accelerate or decelerate. You will need to calculate the force, which is a potential derivative. So you're essentially calculating the derivative of the total energy with respect to a given position. Same deal. That's an N squared operation. You have to sum over pairs of atoms. We have to sum over 1 all atoms for the force on a given atom. But since they're N atoms, it's an N squared operation. In a lot of codes, the minimization is done by very standard schemes-- things like conjugate gradient, Newton-Raphson, sometimes often trivial line minimizations where you literally just calculate the force and assume some kind of elasticity along the force path. And so if you know the force, you assume you know what the curvature is of the energy in that direction. You kind of can make an approximation to its minimum when you do that iteratively. So you can do that with very large simulations. And I wanted to show you sort of one of the leading edge ones. I got to get out of PowerPoint for that if I can make that work. OK, here we go. So what I'm going to show you is a order million atom simulation. This is essentially a cube of 1,000 by 1,000 by 1,000 atoms. That would make more than a million. I forgot to write down what it was. Yeah, I wrote down 1,000 by 1,000 by 1,000 atoms. That would be a billion atoms. And essentially what's being done-- and what you see is a notch on top and on bottom because there's so many atoms you don't see the discrete resolution anymore [? of ?] [? the ?] [? atoms, ?] [? so ?] this thing starts to look like a continuum. And it's being pulled from the side. So you'll see the cracked grow and dislocation spread. This is a simulation from Farid Abraham when he was at IBM. And of course, if you have big computers, you can do big simulations. So let me show it to you. So what they actually did was they showed just the dislocations by counting under-coordinated or over-coordinated atoms. So what you're actually seeing is the dislocation lines. I mean, if we plotted all the atoms, you wouldn't see anything. You'd just kind of see green everywhere. And that's sort of zoom up. So this is a molecular dynamic simulation. And you learn later more about that, where you essentially just do Newtonian mechanics on atoms. So this is one of these very large scale things that you can actually do with potentials. Now, you could ask yourself, what did I learn from this? This is the kind of work that my personal opinion is scientifically not necessarily all that relevant. But it's pioneering work, you know. This is like Lewis and Clark who went to the West. You could always say, what did we learn from them? Maybe not a lot, except that you could get the Pacific. But it's kind of pioneering work. And when people do these kind of simulations, it's not as much for the science as it is probably for pushing the envelope and seeing what can you do with enormous computational resources. You really have to fine tune your algorithms, your parallelization scheme. So it's kind of like flying to the moon which may have not brought us much. But there was a lot of technology fine tuned for that. So that's one of the reasons I wanted to show it. OK, let's go back. Another very practical issue that I'm sure a lot of you are familiar with. Even when you have a billion atoms, it's actually still very small. I actually think if you do the calculation, a million atoms is, for a lot of materials, a cube of just about 200 Angstrom on the side so you're still, if these were finite systems, would be very small systems. So to get rid of boundary effects, you almost always use what's called periodic boundary conditions in simulations in as many directions as you can. And of course, what a periodic boundary condition means is that if you have some unit, if you have some unit, you essentially repeat it next to it. So this end here is the same as that end, and this end here is the same as that end. So people always ask you how many atoms do you have in your simulation, and the answer is infinite. It's just that you only have a finite number of degrees of freedom because it's these atoms here do exactly the same as these here. But the answer is infinite. So we don't really work with small systems. Or, rarely. I mean, sometimes we work with finite systems. So what's the repeat unit you use? Well, if you're studying a periodic material, then it's easy. If you study, say, a perfect crystalline material, here's a perovskite unit, so that's what you repeat. Because the material already has periodicity implied in it. So at that point, you're actually not making any approximation. You're not forcing anything on the material-- not when you do a static calculation that it already doesn't have. Of course, when you do dynamics, you're enforcing something because you're saying that in dynamics, the translational symmetry is temporarily broken because this atom can vibrate independent of that one. And so, then, when you improve impulse periodicity, you are imposing that these vibrate together. So in some sense, you're imposing a cutoff on the frequencies of vibrations that can occur. If systems don't have periodicity, you typically tend to enforce it. If I took this perovskite and I made one defect-- say I wanted to study oxygen vacancy so I take out the oxygen vacancy, I lose all translational periodicity now. But what you do is you tend to impose it back by repeating the defect. It's going to be slow. So let's say I've taken out an oxygen here. I repeat the defect so that, in essence, I just have a bigger unit cell. But it's one with a defect in it. This is now my unit cell. And that's called a supercell approximation. So you make a cell that's big enough so that you hope that these defects-- I don't know how you call it-- don't interact because then, if you have N cells, your calculation just has the energy of N defects. OK? And the critical assumption is, of course, that they don't interact. Because if they interact, then you're not modeling an isolated defect. You're modeling a defect that's interacting with other ones. So how do they interact typically? It's kind of important to know. This is one of these things that you're going to do in the lab. You should always check conversions if you can. The problem is that often you already work at your computational limits for size of supercells, so it's easier said than done. In an ideal world, you say, well, I calculate bigger and bigger, and if the answer doesn't change, I take it. But usually, you work so much at your computational edge already that you can't make them bigger and bigger. A lot of people forget that [? to ?] [? check ?] [? conversions, ?] you also can make it smaller. And if it doesn't change much, you were probably already converged. But it's useful to know how these things usually interact with each other. So that gives you some sense of how far you have to go. There's often the direct sort of interaction. If you work with a pair potential, then you can actually see the range of the pair potential. That's usually not the problem. Rarely do you have potentials that are relevant, large, and over large distances that you need to make really large supercells. But there can be other energetic effects. If you have electrostatics in your system, that will give you very long range interaction. If that's a charge defect-- which it is in a perovskite-- you will have a very strong one. Actually, if we truly did a charge defect in that supercell, what would happen? If that supercell became charged-- so let's say I took out an oxygen 2 minus, which means the supercell now has a net positive charge-- what would happen? You'd have an infinite energy because you would essentially have a positive charge interacting with only other net positive charges. And at any distance, if you do an infinite system, that gives you an infinite energy. So that's something that you would have to explicitly take out. Now you say, well, I'll compensate with a negative charge because that's what happens in reality. If I take out an oxygen, those two electrons of that or O2 minus are going to go somewhere, so it's going to create a net positive charge. So then you have a dipole. So let's say I have sort of plus, minus, and I have a dipole now. So now I have dipoles interacting. Dipoles will not give you infinite energy, but they will give you a fairly long range energy tail. And if you're smart about it-- so the brute force way to convert supercell calculation is always to push him as hard as you can and hope and sort of see the energy converge of the defect. But if you're smart about it, you can often do much smaller cell calculations and explicitly take out the term that's the defect-defect interaction. Which is, if you know that this is a dipole, you know what the form is of it, and you can take it out. Other things that often cause havoc is relaxation. If I were to put a really big ion in the center of that perovskite unit cell, I would build up a strain field and the different defects would interact through their strain field. So, rarely do you actually interact through the direct energetics. You often interact through secondary effects. Another way they often interact which is very indirect. If you put in your defect and you relax the volume of the supercell, then that will interfere with the relaxation of the supercell next to it. So you get a strain field just through the homogeneous. You get an extra energy term just through the homogeneous strain interaction. So these are all things to consider when you do supercell calculations. We're trying something new where we save the annotations we make on the slides, but it seems to be really slow. So I may not be able to keep that up. OK, so this I went over-- the kind of convergence. You know, here's a typical example. I don't know if this is typical. This is the vacancy formation energy in aluminum. Now, this is done with quantum mechanics, versus the number of atoms in the supercell. And you know, this is one of these that you really don't know if you're converged. I mean, it depends on how accurate you want it. If you really only care that it's between 0.7 and 0.75, you're probably OK. But if you did this point and that point, you'd probably think you're converged. But then, if you look at the next one, you're actually going back up a little bit again. So it's sometimes very hard to figure out whether you're actually converged. OK. So that was a few practical issues to sort of start gearing you up for the lab. What I want to do now is talk explicitly about some of the formal failures of pair potentials and go on to correct them. And the first one is the vacancy formation energy which is, if you look at historically when people started doing simulations late 1950s, 1960s, into the '70s, people thought that whenever there was something wrong with the outcome that they just needed better parameters in the potential. And it's actually rather recent-- sort of 1980s-- that people figured out that there were really formal problems and that there were certain things you could not get right-- never-- with a potential because the problem of the form rather than of the parameters. And a critical number in that work was actually the vacancy formation energy which nobody could ever get right. And I'll show in a second why. Here's an FCC crystal. So let's calculate the pair approximation to the vacancy formation energy. What do I do when I make a vacancy? What I do is that I take this central atom and I move it somewhere else in the bulk. So in some sense, I make my system one side bigger. So let's calculate the energy balance of that. So the vacancy formation energy. OK. What do I destroy? I destroy 12 bonds because this atom is coordinated 12-fold in FCC. So I destroy 12 bonds. I lose 12 times the bond energy. But then, of course, I put it back somewhere in the bulk. And so I gained the cohesive energy of one atom. OK? And what's the cohesive energy per atom in FCC? It's 6 times the bond energy because there are 12 bonds around an atom, but every bond is shared between two. OK? So there's six per atom. So the vacancy formation is really minus 12 times the bond energy plus 6 times the bond energy. So what it is is minus 6 times the bond the energy. So essentially, it's minus the cohesive energy. The cohesive energy per atom. And that's essentially what you will always find repair potentials. Now, I did a static approximation. I assumed that all of these bonds were kind of unchanged as I took the atom out. In reality, if I take that central atom out, the atoms around it will relax somewhat and change their bond energy. But you'll see this when you do this in the lab. That will make it from one time the cohesive energy to something like 0.9 or 0.95 times the cohesive energy. Essentially, you find in pair potentials that the vacancy formation energy is about the cohesive energy. If you look, on the other hand, at experimental results for the cohesive energy-- [INAUDIBLE]---- here's some experimental data. So this is for FCC metals, and this is for Leonard-Jones, and for noble gases. The ratio of the vacancy formation energy to the cohesive energy. And what you see is that, in the noble gases-- which you think of as well-described by pair potentials-- you're reasonably close. Because remember, noble gases are just inert shells interacting through Van der Waals interaction, and that's how we came up with the Leonard-Jones potential. But if you look at the metals, the vacancy formation energy is actually only a small fraction of the cohesive energy. And we can never get that low with pair potentials. Never. OK? Unless you maybe really have the most odd form of a pair potential. But the reason is because there are certain physics missing from pair potentials. So actually, let me go back for a second to the pair potentials. What's the physics here of why the vacancy formation energy is so low in real materials? Remember what I said before, that bonding in metals, the bonding energy goes not linear in the coordination number. It's more something like a rough approximation would be the square root of the coordination number. But essentially, when you go from 1 to 2 bonds, your energy goes down a lot more than when you go from 11 to 12 bonds. So why is the vacancy formation energy so low? Well, if you think about what you do when you take out an atom, you make a lot of the atoms around it go from 12 coordination to 11. So they lose a lot less than 1/12 of the cohesive energy because the cohesive energy you could think of as the average of all bonds. But around the vacancy, you are just reducing the coordination from 12 to 11. You can actually make a simple model if you said that the cohesive energy was some constant times the square root of the coordination. So rather than linear in the coordination, which is what we would get in pair of potentials, if you calculate the vacancy formation energy in this model-- let's do it for FCC. In FCC, Z is 12. The vacancy formation energy, what would it be? Well, if we put the atom somewhere back in, we would gain Z times the square root of 12. And then, as we take it out, what do we lose? For 12 atoms, we go from coordination 12 to 11. So this is the atom in the bulk. This is actually the change around where we made the hole. And if you actually calculate that-- the vacancy formation energy over the cohesive energy-- you find that that's 1 plus 12 times square root of 11 over square root of 12 minus 1. And it's about 0.49. So in a square root model, you're already much closer to the experiment where the vacancy formation energy is a much smaller fraction of the cohesive energy. Another critical piece of information was surface relaxation, which people had been seeing with [? LEEDs. ?] If you look at a metal, typically if you create a surface, the difference between the first and the second layer becomes smaller. So in a metal, the first layer tends to relax inward. And can somebody explain to me why that happens? And typically, you don't see this. In pair potentials, you very often have about the same distance. So the first to second neighbor distance is often bulk-like and in some cases actually relaxes outwards because of competing forces. But in almost all real systems, it tends to relax inward. Why do you think that is? Yes, [INAUDIBLE]? AUDIENCE: [INAUDIBLE] PROFESSOR: True. But I'm not sure that that necessarily makes it relax inward. The reason it relaxes inward is because the bonds between the first and the second layer strengthen. That's really what is a sign of them. And why do they strengthen? Because the surface is under-coordinated. Remember, if you go to lower coordination, you're effectively-- although you're losing energy and that's why there's surface energy-- but the energy per bond is actually becoming stronger. Remember that I showed you the potentials for copper? Remember that for the copper II molecule-- so where you just had one bond-- had the strongest potential. The equation of state gave you where your were 12-fold coordinate gave you much weaker potential. So you see that in surfaces as well. If you cut away the bonds here, the remaining ones strengthen and they pull the layer in. The other problem that's common in pair potential which I won't say much about is what's called the Cauchy problem. If you write down the relation between the stress and the strain tensor and so this is the matrix of elastic constant that for pair potential C12 and C44 are always the same. And this has something to do with there actually being a zero-energy strain mold. Remember that I showed you this? It's related to the same issue that in a cubic or in a square model, you can actually kind of slide it without any energy change. And that actually puts a constraint on the elastic constant, which essentially translates to C12 being C44. And the last one-- if you're a practicing material scientist-- may be one of the important ones. But you can essentially not predict crystal structure in anything that's covalent. And metals are covalent. I know they teach you there's metallic, covalent, then ionic bonding. But metallic is just delocalized covalent. And the reason is-- I think you can already see-- first of all, potentials tend to go to close packing always because they just they want to have as many atoms around them as possible because that's how you bring the cohesive energy down. And they don't count that cost, that if you bring more around you, your bound strength weakens. And they have no directional dependence, so they miss those two important components to get crystal structure right. You can actually do some theory about this, which was done already in the '70s that you actually need what's called at least the fourth moment of the density of states to get crystal structure right. And that turns out to be at least a four-body effect. So if you actually thought of potentials as an expansion-- so I'm going to write the energy as a constant plus an expansion in two-body terms, three-body terms, four-body terms, et cetera, which we don't even know would converge-- you would have to go at least up to the fourth order to get crystal structure differences right. OK, so how do you fix the problem? There are essentially, if you look at the literature of empirical models, two avenues to go. You can either make what's called cluster potentials. And so, if you don't like a two-body potential, you can make a three-body potential. And the advantage of that, once you have a three-body potential, you can see angular dependence. Because if I have three atoms and they don't interact just pairwise, but there's a [? two ?] three-body, I can include this bonding angle. A bonding angle is truly a three-body concept. So you can start building an angular dependence. The other direction to go is go to pair functionals. So you still want to write the [? energies ?] pairwise-- some pairwise summation. You just don't want the cohesive energy per atom to go linear with what you sum pairwise or what you sum as its environment pairwise. So we're going to deal with both in class. Today, I'm going to mainly do pair functionals. Pair functionals tends to be most applicable to metals. Where I would say the square root depends, that effect is strongest, whereas cluster potentials or many body potentials in general tend to be more favored in the organic literature-- both organic chemistry and polymer chemistry. And so, going to that. So there's a whole class of methods which are called effective medium theories. The version of it that I'm going to discuss is what's called the embedded atom method, which is probably the most popular one. This was developed by [? Mike ?] [? Baskis ?] [? and ?] [? Murray ?] [? Dahl ?] I think going back to the '80s. Mid '80s, or something like that. And it's probably now the most popular of the sort of empirical energy schemes. If you look back at those papers, those papers have now over 1,000 citations. And the idea is extremely simple. It's, if you understand the source of the problem-- that our energy should not be linear with coordination-- that's what you're going to fix. So in these models, what you do is you write the energy per atom as a function of the number of bonds around it. It's just that F should not be linear. And so that's why they call it energy functionals rather than a pair potential. The question is, how do you measure number of bonds? And if you sort think about it, if you think yourself back in the '80s, how do you want to measure number of bonds? You don't really want to do it discreetly. Because you could say, well, maybe I'm just going to count atoms within a two-Angstrom radius, or some radius. The problem is, as soon as you measure it discretely, you're going to have to deal with discontinuities. As atoms move in and out of that radius, suddenly you've got 14 things around you instead of 15. So you would always have super large forces because of those discontinuities around your [? cutoff ?] interface. So what you need is a smooth count of coordination. And the one they came up with in the embedded atom method is literally, you measure the electron density being projected on you from your neighbors. It's ironic. I'll show you later the formal justification for the embedded atom method which has a lot more to do with quantum mechanics. But that was actually a justification after the fact. The original idea came up with sort of, you really want to have a continuous measure of bumbling around you. And so the idea is that you're going to do something like, I have a central atom. I have other atoms around me. How do I measure how much these contribute to my neighborhood? Well, these have some electron density functions around them. Their atomic electron density I'm sort of drawing. And they all project onto that central atom. You count up how much electron density there and that's somehow a measurement of your neighborhood. And you see, that's going to roughly have the right behavior. It's continuous. As atoms move farther away, they still contribute, but they contribute less. So it is a continuous measure. And the more density you project, either the closer by the atoms are or the more of them there are. OK. So this is the idea of the electron density. You're going to actually calculate the electron density on a site as a measurement in some sense of the coordination as a sum of electron densities coming from its neighboring atoms, J. OK? What electron densities you use is kind of less relevant. In the original embedded atom and in most embedded atom codes still, they use essentially the atomic densities. So people have calculated the atomic electron densities with [INAUDIBLE] methods. And they're actually tabulated in these Clementi and Roetti tables. And you know, I'm just sort of showing you that they're basically parameterized by a bunch of Gaussians, but you can get those. So you can get those functions. We skipped through that. OK. So remember that if you have-- yes, sir. AUDIENCE: [INAUDIBLE]? PROFESSOR: Correct. Correct. Well, it's a distance dependent function. AUDIENCE: [INAUDIBLE] PROFESSOR: Exactly. Yeah. So it's sort of what people in quantum mechanics think of as sort of non self-consistent methods. You literally treat the electron density of the crystal as the overlap of the atomic densities of the atoms. The reason that saves you here is that you're actually not going to do quantum mechanics on that density, and you're just using that density as some measure of coordination. So [INAUDIBLE] I skipped some slides. Somehow, I lost the slide in here. Sorry about that. OK. You know what? I put it in the wrong place. Here we go. So the way you write the energy in the embedded atom method is you write it as the embedding energy, which is you sum over every atom. And you're looking for it's cohesive energy, which is some function F of the neighborhood measurements, which is the electron density at the site of i. And that's your summing from these functions F which are the projected densities from all the other atoms J. So this is the nonlinear part of the energy. Typically, that is supplemented with a pair potential. So the standard embedded atom method is a pair potential which is often just used for having a repulsive part and a nonlinear embedding energy. So that means we have to solve three problems, really-- the one we've solved. You need to know what the electron densities are. Take the Clementi and Roetti tables. You need to know what the embedding function is, and you need to know what the pair potential is. The pair potential, to be honest, you can use almost anything. It's mainly used for the hardcore repulsion part. You don't want the atoms to fly into each other. Typically, they use some screened electrostatic form which is essentially the product of charge distribution functions on atoms. But you know, nothing would stop you if you really wanted to from using a Leonard-Jones. I'll show you in a second that there is ambiguity about the division of energy between the embedding part and the pair potential is not unique. So that means, every time you change your pair potential, you'll have to change your embedding function. I'll show that in a second. What do you use for the embedding function? There are two ways that it can be done. People have written down analytical forms with some theoretical justification that I'm not going to go into. There's actually an excellent review article on the embedded atom methods which we post just for internal use on the website. It's a 1993 Materials Science and Engineering report. It's a really long paper, but it's really an excellent paper. So there are analytical forms. And here's a few of them given. And people have come up with other ones. More and more now, what people simply do is tabulate the embedding function. And really what they do is they tabulate it so that you exactly reproduce the equation of state. What is the equation of state? The equation of state is energy versus volume or lattice parameter. So say at every level parameter, you could try to get that energy with a better method-- say, quantum mechanics-- and then fit your embedding function so that you exactly reproduce that. And so, often, the embedding function is tabulated. But commonly it has this form. It's decaying, it's sort of decreasing convex, decreasing in energy and convex, and it needs both those properties. Well, why does it need to be decreasing in energy? So this is the embedding density and this is the value of the embedding function. Of course, as you put more density on a site, that comes from more neighbors. So your cohesive energy should be going down. That's why the energy of the embedding function goes down. But it needs to be convex because convexity gives you sort of the correct behavior. Convexity really tells you that adding a bond going from here to here-- let's say a bond is adding 0.01. So if we added a bond here, we'd get this change in embedding energy. If we added it here, we'd only get that change. So the function has to be convex in some sense. Decrease has to be per unit, and electron density has to be less for higher densities than for lower. And that's the definition of convex. The slope of it has to be increasing of this function. And you see, that's going to by construction now get some of our problems right because this shape of the function, the convexity is what tells you that the low coordination bonds are stronger than the high coordination bonds. So you can already guess from this that we're going to get a lot of things right. I'll show you some examples in a second. What's the physical concept that sort of justifies the embedded atom method? There are things you can do formally in ab initio theory showing that in fairly simple tight binding models. So these are things that have the structure of a proper quantum mechanical Hamiltonian, but they are parameterized forms of a Hamiltonian. But still, what they all tend to show is that the cohesive energy in highly delocalized systems to first order goals like the square root of the coordination. So you can get some justification for this nonlinear behavior. A somewhat intuitive concept of where the bonding energy comes from is that actually, if you put an atom in a metallic solid, where does it get a cohesive energy from? For the most part, it gets its cohesive energy from the delocalization of its electrons. And that's largely a kinetic energy effect. Remember that the kinetic energy in quantum mechanics is the curvature of the wave function. It's d squared phi, dx squared if you it in one dimension. So highly curved wave functions have high kinetic energy. So if you can actually delocalize, your wave function becomes, in some sense, much smoother with much less curvature, and you have lower kinetic energy. So if you believe that a lot of bonding cohesive energy and delocalized source comes from that, then the measure of cohesive energy is essentially how well you can delocalize it. So what do you measure with the embedding density? If you measure the electron density coming from atoms, the reason this works is that that's probably a very good measure over the number of states you can delocalize over. Things with higher electron density tend to have a higher number of states-- higher densities of states. And so you can delocalize over more states. And that's the somewhat intuitive explanation for why the embedded atom works. I mean, people have given somewhat other explanations. They could say, well, the electron density that I see from a neighboring atom measures the number of electrons I can bond with. But that's not totally true because if those electrons come from a filled shell, there's nothing I can do with them. So somehow, I tend to believe more it's really a measure of the delocalization you see. The embedded atom method is-- yes, sir? AUDIENCE: Did you say [INAUDIBLE]?? PROFESSOR: Correct. AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah. Yeah. So in an atom, you actually have a much higher kinetic energy because your state is much more localized. The embedded atom method is probably the most popular one of all these effective medium theories. But you'll see other ones with other names that are in spirit very much the same-- things like the glue model. Finnis-Sinclair potentials, which slightly predate the embedded atom method, which are also non-linear potentials. So they're really particle potential pair functionals. And the equivalent crystal models Smith and Banerjee. These are all exactly in the same spirit. The brilliance of the embedded atom method was essentially that it was something for nothing. The computational cost is essentially the same as for doing pair potentials because you sum things only pairwise. If you think of what you need to do an embedded atom calculation, you have your pair potential. That's a pairwise sum. But your embedding energy is also a pairwise sum. Because what you have to do per atom, you just have to sum the electron density coming from a bunch of other atoms around you. So that's a pairwise sum. And then you just have to stick it in F, which is an extra evaluation. But embedded atom fundamentally scales only like N squared. So that was the brilliant part of it. Or in the end, that's what made it successful. You could essentially do much better things with almost the same effort as pair potentials. And that's why it sort of took off as a method. And pretty much anybody in metals now would use embedded atom if they want to get any physics right. If you want to do demonstration projects or things like want to do billions of atoms, there can be reasons to go through much simpler forces still to evaluate. But in metals, there's really no reason not to use EM over simple pair potentials. What do people fit to? People fit to anything, like I told you before. The original EM potentials were fit to a standard set of data, which was lattice parameter, sublimation energy, elastic constants, and vacancy formation energies. Now you'll regularly see papers in the literature claiming, I have a better potential than you for aluminum, and I have a better potential than you for nickel. And some of that is true. So people are not in the game of kind of super optimizing these potentials for what it's worth. But the original database was a set of standard potentials-- largely unknowable metal alloys. People have spent a lot of time finding better potentials on technologically important alloys like nickel aluminum. You'll find at least five potentials who all claim to be better than somebody else's potential on the nickel aluminum system. So what I want to do now is just show you some results. And we're not going to analyze these in great deal. But I want to show you so you get some idea of what people do with this stuff. Here's linear thermal expansion for a bunch of late transition metals calculated in EM and calculated in experiment. And these agree pretty well. That's maybe not a big surprise in close packed solids. The thermal expansion is largely a measure of the asymmetry of your equation of state. I mean, in complex materials, that's not true. But in simple, close packed metals. Remember, your equation of state is slightly asymmetric. It curves less towards the high lattice parameters and the low lattice parameter. So if you draw energy versus volume, it's asymmetric around the minimum. And that's why you have thermal expansion. Think about it. It's slightly easier to fluctuate to the higher volume than to the lower volume. Yes, sir? AUDIENCE: [INAUDIBLE] PROFESSOR: Oh, good point. There is, of course, no temperature dependence in the potential itself. So to do something like thermal expansion, you either have to do a dynamic simulation like molecular dynamics-- and we'll come to that-- or you have to explicitly calculate the entropic factors that give you thermal expansion, which is essentially the dependence of the phonon frequencies on volume. So you can indirectly calculate the thermal expansion. But we sort of haven't got into-- we will go into micro dynamics. But of course, in the potential, there's no temperature dependence. Here's activation barriers for self-diffusion in metals. These are almost too good to believe it. You actually don't get them this good with quantum mechanics. So it's almost surprising to me that they're this good, if you compare EM to external. So this is the activation barrier for-- I forget. But I thought it was interstitials self-diffusion, but I'm not for sure whether it was interstitial or vacancy self-diffusion. You know, here's a bunch of surface energies. These are not quite as good as the experiment. If you, for example, look at copper, here's the experimental number in ergs per centimeter squared. And here is the calculation on different facets. Because of course, often experimentally you measure some average of different facets. Unless you do careful single crystal work, you have some average. On the calculations you have to calculate the energy on a specific surface. And that's actually one of the things you're going to do in the lab. So these are a little lower. And that seems to be a rather consistent problem that is, for example, also in platinum. These are quite a bit lower than the experiment. So that may tell you that either there are more complicated quantum mechanical effects going on that you don't capture with these simple energy models. But sometimes, it also means the experiment is wrong. I think in this case, platinum is a fairly easy metal to deal with clean. But surface energies of materials is sort of notoriously difficult to get experimental. Here's a phonon dispersion curve for copper. Again, don't worry about the details. I just wanted to show you the kind of things people do. Calculations are the lines and the points are the measurements. So, reasonably good at some of these. High frequencies there seems to be some problems with. Melting point. Again, this would have been done with molecular dynamics and then free energy integration. And again, for most materials, these come out remarkably well. So, look at this. Except again for the very late transition metals like palladium and platinum, there just seems to be some amount of problems. I can tell you in a second about the limitations of these methods. And they may be related to the unusual electronic structure of palladium and platinum. You know, here's a pair correlation function in a liquid. [? People, ?] [? we're ?] [? good? ?] A grain boundary. This is something you simply could not do with a pair potential in metals because when you calculate a grain boundary, you really are dealing with different coordinated environments. And the painful thing about a grain boundary is that you don't know ahead of time what kind of coordinations you're going to see. And you're going to see multiple coordinations. See, but people with potentials always get, if I know I'm going to deal with a surface, I can make a pair potential that's pretty good for those coordinations at the surface. If I deal with the bulk, I make a pair potential that's pretty good with bulk coordinations. It's when you work with grain boundaries that you kind of end up with all kinds of coordinations because the atoms in the grain boundary, some are high coordinated, some are low coordinated, depending on how the grains contact each other. So it's a great method, but it still has its limitations. One is that the bonding is spherical. You literally sum all the electron density coming from around you, and you really just sum it up. You don't keep track of its orientation dependence. So having one atom there next to me and one atom on that side is the same as having both straight in front of me because I just sum up the electron density coming from them So it's purely spherical. That can be fixed. About in the '90s, [? Mike ?] [? Baskis ?] developed what's called a MEAM, and that just stands for modified embedded atom method, which basically has embedding functions that keep track of the angular dependence of the electron density around you, essentially by mapping the problem on to spherical harmonics. MEAM is not nearly as popular as the standard EM because it adds a level of complexity to the method. And it's not totally clear that you're making things work because the physics is better or you just have more fitting parameters. [? Mike ?] [? Baskis ?] originally developed the MEAM to work on silicon where angular dependence clearly seems to play an important role. Yes, sir? AUDIENCE: [INAUDIBLE] PROFESSOR: Well, your force-- AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah. Your force is directional dependent because you're taking the derivative of energy with respect to a vector, which is some distance. But-- AUDIENCE: [INAUDIBLE] PROFESSOR: But again, the force will be directional dependent. Let me show you an example. Let's say I have a central atom here and I got four atoms around it. The force will be directional dependent because-- what is the force? The force is how the energy of an atom changes as I move it into a particular direction. So the force along this direction is the energy change as I move the atom along this direction because it is the derivative of the total energy with respect to some unit vector in that direction. See, if I change in this direction, I'm going to significantly increase the overlap with that atom. But if I change it, say, in this direction, that's not necessarily the same derivative than in this direction. So the force will be directional dependent just because of the crystallography, but the energy is not. So what I mean with that is that-- let's call this some unit length L, and they're all at L. If I do this, and here's my central atom, and these are all L, that gives me exactly the same embedding energy as in the first problem. Of course, the total energy will be different because these guys interact very closely. But the bonding energy of the central atom is exactly the same. And that's what I mean with no angular dependence. It doesn't in the end see what orbitals that electron density comes from. And when you think about angular dependence, in the end, what it is, it is polarization of orbitals. Think of sp3 bonding. What is that really telling you? If I do an sp3 hybrid along this direction, I'm really starting to polarize the orbitals along-- what is it-- 109 degrees in other directions. So I really need to try to get my bonds in those directions. So that's essentially what angular dependence comes from. Does that help? AUDIENCE: [? I ?] [? think ?] [? so. ?] PROFESSOR: OK. You think so. Well, let me know if it doesn't. OK. The other problem is-- which is more a technical problem-- is that the potential is not unique. And this is something you just have to be careful with if you start messing around with potential files yourself. If you're going to use it as a black box, you're probably OK. Let me get you some notes on that. The way you can see that the potential is not unique. If I have an embedding function F of rho, if I define another embedding function-- we call it G of rho-- they're always the electron density. And I say, that's F of rho plus some linear term in rho. You could make the total energy function exactly the same with either of the two embedding functions just by changing the pair potential. And let me show you that. If I write the energy as the sum over atoms, the embedding function on that atom, plus some air potential part. Let me substitute this in. So I get that this is sum i. Fi rho plus 1/2 sum iJ psi iJ, RiJ, plus-- let me-- sum i k sum J [? not equal ?] i of f rj where rho i was sum j f. OK. Essentially, if you want to look through the math, don't worry too much about it. If you do a linear transformation on the embedding function, you see this term is linear in things coming from the atoms around it. It's linear in these electron density functions. So anything that's linear in functions that I sum from the other atoms is just like a pair potential because that's what a pair potential is. I sum contributions, OK? I add up contributions from the things around me. So I'm linear in that contribution. So any linear part of the embedding function just looks like a potential. This is a pair potential. This is the sum over [? 2n. ?] This is i an J, and I add something up. So this I could just add the psi. So if I add psi to this and use the embedding function F, I have exactly the same as using the embedding function G and using just the potential psi. So the linear part you can move arbitrarily between the embedding function and the pair potential. So that tells you something else that's kind of important. When is a pair potential alone good in metals? It's when the embedding part doesn't change. If my atoms stay roughly at the same density, then all the changes come from the pair potential part. OK? So what would you do-- [INAUDIBLE]---- if you're having an embedding function as a function of rho. If you're working, say, at this density? Let's sort of say the density of the bulk, and you're studying bulk problems. You could take the slope of this, take the linear term, make a linear approximation to that and add that to the pair potential, and now you wouldn't have to worry about the embedding function. So at constant density, pair potentials are actually pretty good. It's when the density around you-- the electron density around you changes which is, in essence, a measure of the coordination change that you run into trouble. And so, people have gone into that game of making density dependent potentials. And so you would have a different potential when you're near a surface than when you're near the bulk. So my summary on effective medium theories which I showed by embedded atom methods is, in essence, that there's no reason not to use them. If you've decided you're going to use an empirical model for metals, there's really no reason at all to use pair potentials. Embedded atom basically does better at very little cost. Saying that it does better doesn't say that the does everything. And I remember talking to people like [? Mike ?] [? Baskis ?] who invented this. They're embarrassed for kind of things people now do with the embedded atom method. Like with any successful method, people push it way too hard and try to calculate all kinds of subtle energy effects that it just never was made for. And you see people write papers on, if I change my constant by 1,000 here, I can get this phase correct. But in the end, it gives you these gross coordination effects right, but it does not give you very subtle hybridization effects right. Why, for example, were platinum and palladium not so good? Because platinum and palladium have almost filled or filled D bands. And as a function of the environment around platinum and palladium, there is very subtle electron transfer between the D band and the S band. OK? And when you're just measuring electron density, that all looks the same, but those states behave very differently. So there will be always subtle electronic change that you simply won't get. And if your system is really influenced strongly by those, you won't get them right. For example, people try to get complex crystal structure differences in metals with EM. You will just never get there, period. Let's say you make an alloy-- palladium ruthenium, whatever-- the crystal structure energy difference in that [? material, ?] like in most metals over the order of 5 to 10 millielectron [? volt. ?] You cannot get things right on that scale with these kind of simple models, so you shouldn't try too hard. But on the other hand, if you're going to study, say, I don't know, fracture in a material in a simple element like one of these big movies I showed you, that's to a large extent a topological event. You will get a lot of essential pieces of that right. Because what is that? That's a coordination change event, largely. You will get a lot of that right by doing EM over potentials. This is a great website I only added that this year to the lecture I found. There's a group in Japan that called themselves the [? EAMers. ?] They even have a blog now. And so there's a lot of cool stuff you can download. They have potential files. You can get EM code from the Sandia site. It's a little hard to get to, because Sandia being a national lab, they're totally paranoid about downloading anything. But there are sources where you can get the EM code. It is technically freely available. It's just not necessarily trivial to get to it. So we talked about pair functionals, which was this direction. So the other direction is obviously to go to cluster potentials. Well, you know, I don't like my expansion in just two-body terms. I'll start adding three-body terms. It's the obvious extension. And if you don't like that, you can go to four-body terms. That's definitely done. And for reasons I'll go into in the next lecture on Tuesday, it's extremely popular in the world of organics, where it's actually really, really powerful. And we'll go into why that is the case. But a typical material in condensed matter where obviously this was done was silicon. Silicon is a low coordinated solid-- only fourfold coordinated in the diamond cubic structure. It's extremely hard to get that with a pair potential because again, your systems always try to form FCC or HGP, essentially. So what you can do is build a three-body potential that favors these 190-degree tetrahedral angles. And as soon as you do that, if you actually make that force strong enough, you can do almost nothing else than form diamond cubic. Because diamond cubic is basically the way to pack atoms and put them under all these 109-- what is it-- .4 or something degree angles. So if you make a strong enough three-body force, you will by definition end up in the diamond cubic structure. If you work with three coordinates, you can also transform that and instead work with two distances and an angle. So if I have three atoms, the coordinates of them, I can also always say that to have the relation between them, I just need to have these two distances and this angle that fully defines the three-body problem. And that's usually how it's done. People have pair potentials that take care of the distance dependence of that. And then they have an angular potential which takes care of the angle between these. And often, the prefactor of the angular potential will have some distance depends in it. Because of course, maybe you want to enforce this 190-degree angle when your two other atoms are close. But when they're very far, you don't really care anymore. What's the possible choices for angular potentials? The simplest one is simply a quadratic-- a harmonic function in the angle. If you want to impose an angle theta naught, you make a potential that's literally a quadratic around theta naught. So, extremely simple to evaluate. So that function's actually often used. You can also use-- the disadvantage of this one is that it's not periodic. Angle measurements should be periodic. So you can kind of get into trouble with this if you go too far from equilibrium. So another thing you can use is a cosine function which has the proper periodicity. If you take-- I don't know-- 109 degrees or whatever angle you want, or you take its complement-- 360 minus that-- you should get the same answer. And so, cosine functions will do that right. And this one, cosine theta plus 1/3, this part is minimal when theta is 109 with tetrahedral angle. [CLEARS THROAT] Excuse me. Losing my voice. So it works working with three-body potentials. It's, of course, more work. It's not [? order ?] N cubed. Because, for every atom, you now have to look at distance and angles with pairs of atoms. OK so it becomes an N cubed operation, so it's more work. And calculating the forces as the derivatives of the energy is also quite a bit more work. The most famous three-body potential is probably the Stillinger Webber potential which, as you can see, has this cosine theta form for the-- this is the angular part. This is the distance dependence of the angular part. So how much should you penalize angular deviations [INAUDIBLE].. And then this is the two-body part. Just sort of a more [? slight ?] potential. Originally, Stillinger Webber, when they actually wrote down the function, they didn't even fit it very carefully. I mean, it was a great form, but they just kind of did a rough fit. And so, the original Stillinger Webber parameters and the ones people now use sometimes more recently are not the same anymore. The potential has the same form. But some people have done recently better fits. So when somebody says that Stillinger Webber potential, there are variations of it for the constants. So what do you get right with a [? sale ?] something like a Stillinger Webber potential? Well, you can kind of guess why people did this. Because silicon is so important, they wanted to be able to model silicon right. So it's important to understand which pieces you get right. One of the critical issues was surface reconstruction. If you take silicon 111-- sorry, 100. If you truncate it, it actually undergoes a 2 times 1 reconstruction. What essentially happens if you take this top layer atoms, they're only bonded to two atoms below them, and so they dimerize on the surface. So basically, these move together in sort of double rows. I need a better color. And then these will move together. And so they double the periodicity in one direction. And so you have these dimers that form. And because of that, the surface atom is bonded to three other atoms-- two below it, one all the way in the surface. And so they're better bonded. And that's the 2 by 1 reconstruction. The 2 by 1 reconstruction you get right with the Stillinger Webber potential. You'd actually get it right with almost anything. It's essentially a topological reconstruction. You know, atoms like to have more atoms around them to bond, period. And so that's what drives a 2 by 1 reconstruction. At the top layer, atoms now are at least threefold coordinated and they're actually almost fourfold coordinated instead of twofold coordinated at just an unreconstructed terminated surface. But here's the reconstruction on the 111 surface. The reconstruction on the 111 surface is 7 by 7. So it forms a much larger unit cell on the surface. And that is not reproduced by the Stillinger Webber potential. So clearly, that's a more subtle effect. Silicon, being such an important material, has a multitude of potentials. Everybody sort of has made one and claims theirs is better than yours. And I just show this graph not for the details, but to show you that this isn't even a tenth of the potentials made for silicon, but some of the famous ones. SW-- Stillinger Webber-- [? bismuth, ?] [? Harman, ?] and the Tersoff potentials, which are quite famous for silicon, are not even on here. But if you see, this is the angular part of these potentials. It's very different. They all give the same answer, unlike static properties like the elastic properties of silicon, the cohesive energy, the lattice parameter. But it's sort of divided differently between pair parts and angular parts. This one clearly wouldn't give you the same stiffness of bond bending around the typical dihedral angle. And that shows up. There's what I thought was a really nice paper. Oh, and I'm sorry, I-- oh, no. The reference [? isn't ?] here. This is the [? Norman ?] [? and ?] [? paper ?] from Physical Review B. They did what I thought was brilliant. They did exactly the same simulation with three different potentials. Left is the Stillinger Webber potential, and then middle column is Tersoff [? 2. ?] This is the Tersoff potential, and then Tersoff [? 3. ?] So this is exactly the same Monte Carlo simulation with the three potentials. And this is the [? 001 ?] or [? 100 ?] surface of silicon. So remember, that's the one I showed you that dimerizes. And so the top line is at low temperature. Tersoff [? 3 ?] doesn't dimerize. Tersoff 2 does sort of dimerize. And so does Stillinger Webber, but they look slightly different. But these are in reasonable agreement. If you go higher in temperature, then Tersoff [? 3 ?] dimerizes. So does the Stillinger Webber. But Tersoff [? 2 ?] starts kind of looking quite different already. So these are potentials that, if you looked at how they reproduce static properties, they all do extremely well. But it's extremely hard to fit to dynamical properties because you don't have that as quantitative information very often. Like, first of all, we don't know what the right answer is here. But even if we did, let's say, somebody does brilliant STM and we know the right answer, it's very hard to feed that back into a potential. And so the reason I'm telling you this is not to dissuade you from using [INAUDIBLE] when you need them. It's just to make sure you're cautious. There's a big reason that people do ab initio calculations these days. It's not because they necessarily like it, but it's because the results tend to be more reliable, or at least predictable. The errors are more predictable, let's put it that way, which is a useful thing to have. Having error is one thing. Having unpredictable error is a lot worse. But still, there will always be problems where they're just too complex, or you want to do them fast where you want to use pair potentials. Just don't always believe everything you see. Try to fit as close to where you're going to explore the physics that's essentially [? meaning. ?] OK, here's a bunch of references. I'll actually put these on the Stellar website. We'll finish the empirical potential lecture. We'll go into organics and a little bit of polymers on Tuesday. And then Professor [? Marzari ?] will start giving you introduction to basic quantum mechanics. And then we'll start sort of ab initio energy methods, and Professor [? Marzari ?] will do that. So I'll be still lecturing on Tuesday. So have a good weekend, and I'll see you on Tuesday.
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https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/8.04-spring-2013.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: All right. So today's task is going to be to outline some of the basic experimental facts that we will both have to deal with and that our aim should be to understand and model through the rest of the course. Physics doesn't tell you some abstract truth about why the universe is the way it is. Physics gives you models to understand how things work and predict what will happen next. And what we will be aiming to do is develop models that give us an intuition for the phenomena and allow us to make predictions. And these are going to be the experimental facts I would like to both explain, develop an intuition for, and be able to predict consequences of. So we'll start off with-- so let me just outline them. So, first fact, atoms exist. I'll go over some of the arguments for that. Randomness, definitely present in the world. Atomic spectre are discrete and structured. We have a photoelectric effect, which I'll describe in some detail. Electrons do some funny things. In particular electron diffraction. And sixth and finally, Bell's Inequality. Something that we will come back to at the very end of the class, which I like to think of as a sort of a frame for the entirety of 8.04. So... we'll stick with this for the moment. So everyone in here knows that atoms are made of electrons and nuclei. In particular, you know that electrons exist because you've seen a cathode ray tube. I used to be able to say you've seen a TV, but you all have flat panel TVs, so this is useless. So a cathode ray tube is a gun that shoots electrons at a phosphorescent screen. And every time the electron hits the screen it induces a little phosphorescence, a little glow. And that's how you see on a CRT. And so as was pithily stated long ago by a very famous physicist, if you can spray them, they exist. Pretty good argument. There's a better argument for the existence of electrons, which is that we can actually see them individually. And this is one of the most famous images in high-energy physics. It's from an experiment called Gargamelle, which was a 30-cubic meter tank of liquid freon pulsing just at its vapor pressure 60 times a second. And what this image is is, apart from all the schmut, you're watching a trail of bubbles in this de-pressurizing freon that wants to create bubbles but you have to nucleate bubbles. What you're seeing there in that central line that goes up and then curls around is a single electron that was nailed by a neutrino incident from a beam at CERN where currently the LHC is running. And this experiment revealed two things. First, to us it will reveal that you can see individual electrons and by studying the images of them moving through fluids and leaving a disturbing wake of bubbles behind them. We can study their properties in some considerable detail. The second thing it taught us is something new-- we're not going to talk about it in detail-- is that it's possible for a neutrino to hit an electron. And that process is called a weak neutral current for sort of stupid historical reasons. It's actually a really good name. And that was awesome and surprising and so this picture is both a monument to the technology of the experiment, but also to the physics of weak neutral currents and electrons. They exist if you can discover neutrinos by watching them. OK. Secondly, nuclei. We know that nuclei exist because you can shoot alpha particles, which come from radioactive decay, at atoms. And you have your atom which is some sort of vague thing, and I'm gonna make the-- I'm gonna find the atom by making a sheet of atoms. Maybe a foil. A very thin foil of stuff. And then I'm gonna shoot very high-energy alpha particles incident of this. Probably everyone has heard of this experiment, it was done by Rutherford and Geiger and Marsden, in particular his students at the time or post-docs. I don't recall-- and you shoot these alpha particles in. And if you think of these guys as some sort of jelly-ish lump then maybe they'll deflect a little bit, but if you shoot a bullet through Jello it just sort of maybe gets deflected a little bit. But Jello, I mean, come on. And I think what was shocking is that you should these alpha particles in and every once in a while, they bounce back at, you know, 180, 160 degrees. Rutherford likened this to rolling a bowling ball against a piece of paper and having it bounce back. Kind of surprising. And the explanation here that people eventually came up upon is that atoms are mostly zero density. Except they have very, very high density cores, which are many times smaller than the size of the atom but where most of the mass is concentrated. And as a consequence, most of the inertia. And so we know that atoms have substructure, and the picture we have is that well if you scrape this pile of metal, you can pull off the electrons, leaving behind nuclei which have positive charge because you've scraped off the electrons that have negative charge. So we have a picture from these experiments that there are electrons and there are nuclei-- which, I'll just write N and plus-- which are the constituents of atoms. Now this leads to a very natural picture of what an atom is. If you're a 19th-century physicist, or even an early 20th-century physicist, it's very natural to say, aha, well if I know if I have a positive charge and I have a negative charge, then they attract each other with a 1 over r minus q1 q2-- sorry, q1 q2 over r potential. This is just like gravity, right. The earth and the sun are attracted with an inverse-r potential. This leads to Keplerian orbits. And so maybe an atom is just some sort of orbiting classical combination of an electron and a nucleus, positively charged nucleus. The problem with this picture, as you explore in detail in your first problem on the problem set, is that it doesn't work. What happens when you accelerate a charge? It radiates. Exactly. So if it's radiating, it's gotta lose energy. It's dumping energy into this-- out of the system. So it's gotta fall lower into the potential. Well it falls lower, it speeds up. It radiates more. Because it's accelerating more to stay in a circular orbit. All right, it radiates more, it has to fall further down. So on the problem set you're going to calculate how long that takes. And it's not very long. And so the fact that we persist for more than a few picoseconds tells you that it's not that-- this is not a correct picture of an atom. OK. So in classical mechanics, atoms could not exist. And yet, atoms exist. So we have to explain that. That's gonna be our first challenge. Now interestingly Geiger who is this collaborator of Rutherford, a young junior collaborator of Rutherford, went on to develop a really neat instrument. So suppose you want to see radiation. We do this all the time. I'm looking at you and I'm seeing radiation, seeing light. But I'm not seeing ultra high energy radiation, I'm seeing energy radiation in the electromagnetic waves in the optical spectrum. Meanwhile I'm also not seeing alpha particles. So what Geiger wanted was a way to detect without using your eyes radiation that's hard to see. So the way he did this is he took a capacitor and he filled the-- surrounded the capacitor with some noble gas. It doesn't interact. There's no-- it's very hard to ionize. And if you crank up the potential across this capacitor plate high enough, what do you get? A spark. You all know this, if you crank up a capacitor it eventually breaks down because the dielectric in between breaks down, you get a spontaneous sparking. So what do you figure it would look if I take a capacitor plate and I charge it up, but not quite to breakdown. Just a good potential. And another charged particle comes flying through, like an alpha particle, which carries a charge of plus 2, that positive charge will disturb things and will add extra field effectively. And lead to the nucleation of a spark. So the presence of a spark when this potential is not strong enough to induce a spark spontaneously indicates the passage of a charged particle. Geiger worked later with-- Marsden? Muller. Heck. I don't even remember. And developed this into a device now known as the Geiger counter. And so you've probably all seen or heard Geiger counters going off in movies, right. They go ping ping ping ping ping ping ping ping ping, right. They bounce off randomly. This is an extremely important lesson, which is tantamount to the lesson of our second experiment yesterday. The 50-50, when we didn't expect it. The white electrons into the harness box then into a color box again, would come out 50-50, not 100 percent. And they come out in a way that's unpredictable. We have no ability to our knowledge-- and more than our knowledge, we'll come back to that with Bell's Inequality-- but we have no ability to predict which electron will come out of that third box, white or black, right. Similarly with a Geiger counter you hear that atoms decay, but they decay randomly. The radiation comes out of a pile of radioactive material totally at random. We know the probabilistic description of that. We're going to develop that, but we don't know exactly when. And that's a really powerful example-- both of those experiments are powerful examples of randomness. And so we're going to have to incorporate that into our laws of physics into our model of quantum phenomena as well. Questions? I usually have a Geiger counter at this point, which is totally awesome, so I'll try to produce the Geiger counter demo later. But the person with the Geiger counter turns out to have left the continent, so made it a little challenging. OK. Just sort of since we're at MIT, an interesting side note. This strategy of so-called hard scattering, of taking some object and sending it at very high velocity at some other object and looking for the rare events when they bounce off at some large angle, so-called hard scattering. Which is used to detect dense cores of objects. It didn't stop with Rutherford. People didn't just give up at that point. Similar experience in the '60s and '70s which are conducted at Slack, were involved not alpha particles incident on atoms but individual electrons incident on protons. So not shooting into the nucleus, but shooting and looking for the effect of hitting individual protons inside the nucleus. And through this process it was discovered that in fact-- so this was done in the '60s and '70s, that in fact the proton itself is also not a fundamental particle. The proton is itself composite. And in particular, it's made out of-- eventually people understood that it's made out of, morally speaking, and I'm gonna put this in quotation marks-- ask me about it in office hours-- three quarks, which are some particles. And the reason we-- all this tells you is that it's some object and we've given it the name quark. But indeed there are three point-like particles that in some sense make up a proton. It's actually much more complicated than that, but these quarks, among other things, have very strange properties. Like they have fractional charge. And this was discovered by a large group of people, in particular led by Kendall and Friedman and also Richard Taylor. Kendall and Friedman were at MIT, Richard Taylor was at Stanford. And in 1990 they shared the Nobel Prize for the discovery of the partonic structure out of the nucleons. So these sorts of techniques that people have been using for a very long time continue to be useful and awesome. And in particular the experiment, the experimental version of this that's currently going on, that I particularly love is something called the relativistic heavy ion collider, which is going on at Brookhaven. So here what you're doing is you take two protons and you blow them into each other at ultra high energy. Two protons, collide them and see what happens. And that's what happens. You get massive shrapnel coming flying out. So instead of having a simple thing where one of the protons just bounces because there's some hard quark, instead what happens is just shrapnel everywhere, right. So you might think, well, how do we interpret that at all. How do you make sense out of 14,000 particles coming out of two protons bouncing into each other. How does that make any sense? And the answer turns out to be kind of awesome. And so this touches on my research. So I want to make a quick comment on it just for color. The answer turns out to be really interesting. First off, the interior constituents of protons interact very strongly with each other. But at the brief moment when protons collide with each other, what you actually form is not a point-like quirk and another point-like quark. In fact, protons aren't made out of point-like quarks at all. Protons are big bags with quarks and gluons and all sorts of particles fluctuating in and out of existence in a complicated fashion. And what you actually get is, amazingly, a liquid. For a brief, brief moment of time the parts of those protons that overlap-- think of them as two spheres and they overlap in some sort of almond-shaped region. The parts of those protons that overlap form a liquid at ultra high temperature and at ultra high density. It's called the RHIC fireball or the quark-gluon plasma, although it's not actually a plasma. But it's a liquid like water. And what I mean by saying it's a liquid like water, if you push it, it spreads in waves. And like water, it's dissipative. Those waves dissipate. But it's a really funny bit of liquid. Imagine you take your cup of coffee. You drink it, you're drinking your coffee as I am wont to do, and it cools down over time. This is very frustrating. So you pour in a little bit of hot coffee and when you pour in that hot coffee, the system is out of equilibrium. It hasn't thermalized. So what you want is you want to wait for all of the system to wait until it's come to equilibrium so you don't get a swig of hot or swig of cold. You want some sort of Goldilocks-ean in between. So you can ask how long does it take for this coffee to come to thermal equilibrium. Well it takes a while. You know, a few seconds, a few minutes, depending on exactly how you mess with it. But let me ask you a quick question. How does that time scale compare to the time it takes for light to cross your mug? Much, much, much slower, right? By orders of magnitude. For this liquid that's formed in the ultra high energy collision of two protons, the time it takes for the system-- which starts out crazy out of equilibrium with all sorts of quarks here and gluons there and stuff flying about-- the time it takes for it to come to thermal equilibrium is of order the time it takes for light to cross the little puddle of liquid. This is a crazy liquid, it's called a quantum liquid. And it has all sorts of wonderful properties. And the best thing about it to my mind is that it's very well modeled by black holes. Which is totally separate issue, but it's a fun example. So from these sorts of collisions, we know a great deal about the existence of atoms and randomness, as you can see. That's a fairly random sorting. OK so moving on to more 8.04 things. Back to atoms. So let's look at specifics of that. I'm not kidding, they really are related to black holes. I get paid for this. So here's a nice fact, so let's get to atomic spectra. So to study atomic spectra, here's the experiment I want to run. The experiment I want to run starts out with some sort of power plant. And out of the power plant come two wires. And I'm going to run these wires across a spark gap, you know, a piece of metal here, a piece of metal here, and put them inside a container, which has some gas. Like H2 or neon or whatever you want. But some simple gas inside here. So we've got an electric potential established across it. Again, we don't want so much potential that it sparks, but we do want to excite the H2. So we can even make it spark, it doesn't really matter too much. The important thing is that we're going to excite the hydrogen, and in exciting the hydrogen the excited hydrogen is going to send out light. And then I'm going to take this light-- we take the light, and I'm gonna shine this on a prism, something I was taught to do by Newton. And-- metaphorically speaking-- and look at the image of this light having passed through the prism. And what you find is you find a very distinct set of patterns. You do not get a continuous band. In fact what you get-- I'm going to have a hard time drawing this so let me draw down here. I'm now going to draw the intensity of the light incident on the screen on this piece of paper-- people really used to use pieces of paper for this, which is kind of awesome-- as a function of the wavelength, and I'll measure it in angstroms. And what you discover is-- here's around 1,000 angstroms-- you get a bunch of lines. Get these spikes. And they start to spread out, and then there aren't so many. And then at around 3,000, you get another set. And then at around 10,000, you get another set. This is around 10,000. And here's the interesting thing about these. So the discovery of these lines-- these are named after a guy named Lyman, these are-- these are named after a guy named-- Ballmer. Thank you. Steve Ballmer. And these are passion, like passion fruit. So. Everyone needs a mnemonic, OK. And so these people identified these lines and explained various things about them. But here's an interesting fact. If you replace this nuclear power plant with a coal plant, it makes no difference. If you replace this prism by a different prism, it makes no difference to where the lines are. If you change this mechanism of exciting the hydrogen, it makes no difference. As long as it's hydrogen-- as long as it's hydrogen in here you get the same lines, mainly with different intensities depending upon how exactly you do the experiment. But you get the same position of the lines. And that's a really striking thing. Now if you use a different chemical, a different gas in here, like neon, you get a very different set of lines. And a very different effective color now when you eyeball this thing. So Ballmer, incidentally-- and I think this is actually why he got blamed for that particular series, although I don't know the history-- Ballmer noticed by being-- depending on which biography you read-- very clever or very obsessed that these guys, this particular set, could be-- they're wavelengths. If you wrote their wavelengths and labeled them by an integer n, where n ran from 3 to any positive integer above 3, could be written as 36. So this is pure numerology. 36, 46 angstroms times the function n squared over n squared minus 4, where N is equal to 3, 4, dot dot dot-- an integer. And it turns out if you just plug in these integers, you get a pretty good approximation to this series of lines. This is a hallowed tradition, a phenomenological fit to some data. Where did it come from? It came from his creative or obsessed mind. So this was Ballmer. And this is specifically for hydrogen gas, H2. So Rydberg and Ritz, R and R, said, well actually we can do one better. Now that they realized that this is true, they looked at the whole sequence. And they found a really neat little expression, which is that 1 over the wavelength is equal to a single constant parameter. Not just for all these, but for all of them. One single numerical coefficient times 1 over m squared minus 1 over n squared-- n is an integer greater than zero and greater in particular than m. And if you plug in any value of n and any value of m, for sufficiently reasonable-- I mean, if you put in 10 million integers you're not going to see it because it's way out there, but if you put in or-- rather, in here-- if you put any value of n and m, you will get one of these lines. So again, why? You know, as it's said, who ordered that. So this is experimental result three that we're going to have to deal with. When you look at atoms and you look at the specter of light coming off of them, their spectra are discrete. But they're not just stupidly discrete, they're discrete with real structure. Something that begs for an explanation. This is obviously more than numerology, because it explains with one tunable coefficient a tremendous number of spectral lines. And there's a difference-- and crucially, these both work specifically for hydrogen. For different atoms you need a totally different formula. But again, there's always some formula that nails those spectral lines. Why? Questions? OK. So speaking of atomic spectra-- whoops, I went one too far-- here's a different experiment. So people notice the following thing. People notice that if you take a piece of metal and you shine a light at it, by taking the sun or better yet, you know, these days we'd use a laser, but you shine light on this piece of metal. Something that is done all the time in condensed matter labs, it's a very useful technique. We really do use lasers not the sun, but still it continues to be useful in fact to this day. You shine light on a piece of metal and every once in a while what happens is electrons come flying off. And the more light and the stronger the light you shine, you see changes in the way that electrons bounce off. So we'd like to measure that. I'd like to make that precise. And this was done in a really lovely experiment. Here's the experiment. The basic idea of the experiment is I want to check to see, as I change the features of the light, the intensity, the frequency, whatever, I want to see how that changes the properties of the electrons that bounce off. Now one obvious way-- one obvious feature of an electron that flew off a piece of metal is how fast is it going, how much energy does it have. What's its kinetic energy. So I'd like to build an experiment that measures the kinetic energy of an electron that's been excited through this photoelectric effect. Through emission after shining light on a piece of metal. Cool? So I want to build that experiment. So here's how that experiment goes. Well if this electron comes flying off with some kinetic energy and I want to measure that kinetic energy, imagine the following circuit. OK first off imagine I just take a second piece of metal over here, and I'm going to put a little current meter here, an ammeter. And here's what this circuit does. When you shine light on this piece of metal-- we'll put a screen to protect the other piece of metal-- the electrons come flying off, they get over here. And now I've got a bunch of extra electrons over here and I'm missing electrons over here. So this is negative, this is positive. And the electrons will not flow along this wire back here to neutralize the system. The more light I shine, the more electrons will go through this circuit. And as a consequence, there will be a current running through this current meter. That cool with everyone? OK. So we haven't yet measured the kinetic energy, though. How do we measure the kinetic energy? I want to know how much energy, with how much energy, were these electrons ejected. Well I can do that by the following clever trick. I'm going to put now a voltage source here, which I can tune the voltage of, with the voltage V. And what that's going to do is set up a potential difference across these and the energy in that is the charge times the potential difference. So I know that the potential difference it takes, so the amount of energy it takes to overcome this potential difference, is q times V. That cool? So now imagine I send in an electron-- I send in light and it leads an electron to jump across, and it has kinetic energy, kE. Well if the kinetic energy is less than this, will it get across? Not so much. It'll just fall back. But if the kinetic energy is greater than the energy it takes to cross, it'll cross and induce a current. So the upshot is that, as a function of the voltage, what I should see is that there is some critical minimum voltage. And depending on how you set up the sign, the sign could be the other way, but there's some critical minimal voltage where, for less voltage, the electron doesn't get across. And for any greater voltage-- or, sorry, for any closer to zero voltage, the electron has enough kinetic energy to get across. And so the current should increase. So there's a critical voltage, V-critical, where the current running through the system runs to zero. You make it harder for the electrons by making the voltage in magnitude even larger. You make it harder for the electrons to get across. None will get across. Make it a little easier, more and more will get across. And the current will go up. So what you want to do to measure this kinetic energy is you want to measure the critical voltage at which the current goes to zero. So now the question is what do we expect to see. And remember that things we can tune in this experiment are the intensity of the light, which is like e squared plus b squared. And we can tune the frequency of the light. We can vary that. Now does the total energy, does that frequency show up in the total energy of a classical electromagnetic wave? No. If it's an electromagnetic wave, it cancels out. You just get the total intensity, which is a square of the fields. So this is just like a harmonic oscillator. The energy is in the amplitude. The frequency of the oscillator doesn't matter. You push the swing harder, it gets more kinetic energy. It's got more energy. OK. So what do we expect to see as we vary, for example, the intensity? So here's a natural gas. If you take-- so you can think about the light here as getting a person literally, like get the person next to you to take a bat and hit a piece of metal. If they hit it really lightly they're probably not going to excite electrons with a lot of energy. If they just whack the heck out of it, then it wouldn't be too surprising if you get much more energy in the particles that come flying off. Hit it hard enough, things are just gonna shrapnel and disintegrate. The expectation here is the following. That if you have a more intense beam, then you should get more-- the electrons coming off should be more energetic. Because you're hitting them harder. And remember that the potential, which I will call V0, the stopping voltage. So therefore V0 should be greater in magnitude. So this anticipates that the way this curve should look as we vary the current as a function of v, if we have a low voltage-- sorry, if we have a low-intensity beam-- it shouldn't take too much potential just to impede the motion. But if we have a-- so this is a low intensity. But if we have a high-intensity beam, it should take a really large voltage to impede the electric flow, the electric current, because high-intensity beam you're just whacking those electrons really hard and they're coming off with a lot of kinetic energy. So this is high intensity. Everyone down with that intuition? This is what you get from Maxwell's electrodynamics. This is what you'd expect. And in particular, as we vary-- so this is our predictions-- in particular as we vary-- so this is 1, 2, with greater intensity. And the second prediction is that V-naught should be independent of frequency. Because the energy density and electromagnetic wave is independent of the frequency. It just depends on the amplitude. And I will use nu to denote the frequency. So those are the predictions that come from 8.02 and 8.03. But this is 8.04. And here's what the experimental results actually look like. So here's the intensity, here's the potential. And if we look at high potential, it turns out that-- if we look, sorry, if we look at intermediate potentials, it's true that the high intensity leads to a larger current and the low intensity leads to a lower current. But here's the funny thing that happens. As you go down to the critical voltage, their critical voltages are the same. What that tells you is that the kinetic energy kicked out-- or the kinetic energy of an electron kicked out of this piece of metal by the light is independent of how intense that beam is. No matter how intense that beam is, no matter how strong the light you shine on the material, the electrons all come out with the same energy. This would be like taking a baseball and hitting it with a really powerful swing or a really weak swing and seeing that the electron dribbles away with the same amount of energy. This is very counter-intuitive. But more surprisingly, V-naught is actually independent of intensity. But here's the real shocker. V-naught varies linearly in the frequency. What does change V-naught is changing the frequency of the light in this incident. That means that if you take an incredibly diffuse light-- incredibly diffuse light, you can barely see it-- of a very high frequency, then it takes a lot of energy to impede the electrons that come popping off. The electrons that come popping off have a large energy. But if you take a low-frequency light with extremely high intensity, then those electrons are really easy to stop. Powerful beam but low frequency, it's easy to stop those electrons. Weak little tiny beam at high frequency, very hard to stop the electrons that do come off. So this is very counter-intuitive and it doesn't fit at all with the Maxwellian picture. Questions about that? So this led Einstein to make a prediction. This was his 1905 result. One of his many totally breathtaking papers of that year. And he didn't really propose a model or a detailed theoretical understanding of this, but he proposed a very simple idea. And he said, look, if you want to fit this-- if you want to fit this experiment with some simple equations, here's the way to explain it. I claim-- I here means Einstein, not me-- I claim that light comes in packets or chunks with definite energy. And the energy is linearly proportional to the frequency. And our energy is equal to something times nu, and we'll call the coefficient h. The intensity of light, or the amplitude squared, the intensity is like the number of packets. So if you have a more intense beam at the same frequency, the energy of each individual chunk of light is the same. There are just a lot more chunks flying around. And so to explain the photoelectric effect, Einstein observed the following. Look, he said, the electrons are stuck under the metal. And it takes some work to pull them off. So now what's the kinetic energy of an electron that comes flying off-- whoops, k3. Bart might have a laugh about that one. Kinetic, kE, not 3. So the kinetic energy of electron that comes flying off, well, it's the energy deposited by the photon, the chunk of light, h-nu well we have to subtract off the work it took. Minus the work to extract the electron from the material. And you can think of this as how much energy does it take to suck it off the surface. And the consequence of this is that the kinetic energy of an electron should be-- look, if h-nu is too small, if the frequency is too low, then the kinetic energy would be negative. But that doesn't make any sense. You can't have negative kinetic energy. It's a strictly positive quantity. So it just doesn't work until you have a critical value where the frequency times h-- this coefficient-- is equal to the work it takes to extract. And after that, the kinetic energy rises with the frequency with a slope equal to h. And that fits the data like a champ. So no matter-- let's think about what this is saying again. No matter what you do, if your light is very low-frequency and you pick some definite piece of metal that has a very definite work function, very definite amount of energy it takes to extract electrons from the surface. No matter how intense your beam, if the frequency is insufficiently high, no electrons come off. None. So it turns out none is maybe a little overstatement because what you can have is two photon processes, where two chunks hit one electron at the right, just at the same time. Roughly speaking the same time. And they have twice the energy, but you can imagine that the probability of two photon hitting one electron at the same time of pretty low. So the intensity has to be preposterously high. And you see those sorts of multi-photon effects. But as long as we're not talking about insanely high intensities, this is an absolutely fantastic probe of the physics. Now there's a whole long subsequent story in the development of quantum mechanics about this particular effect. And it turns out that the photoelectric effect is a little more complicated than this. But the story line is a very useful one for organizing your understanding of the photoelectric effect. And in particular, this relation that Einstein proposed out of the blue, with no other basis. No one else had ever seen this sort of statement that the electrons, or that the energy of a beam of light should be made up of some number of chunks, each of which has a definite minimum amount of energy. So you can take what you've learned from 8.02 and 8.03 and extract a little bit more information out of this. So here's something you learned from 8.02. In 8.02 you learned that the energy of an electromagnetic wave is equal to c times the momentum carried by that wave-- whoops, over two. And in 8.03 you should have learned that the wavelength of an electromagnetic wave times the frequency is equal to the speed of light, C. And we just had Einstein tell us-- or declare, without further evidence, just saying, look this fits-- that the energy of a chunk of light should be h times the frequency. So if you combine these together, you get another nice relation that's similar to this one, which says that the momentum of a chunk of light is equal to h over lambda. So these are two enormously influential expressions which come out of this argument from the photoelectric effect from Einstein. And they're going to be-- their legacy will be with us throughout the rest of the semester. Now this coefficient has a name, and it was named after Planck. It's called Planck's Constant. And the reason that it's called Planck's Constant has nothing to do with the photoelectric effect. It was first this idea that an electromagnetic wave, that light, has an energy which is linearly proportional not to its intensity squared, none of that, but just linearly proportional to the frequency. First came up an analysis of black body radiation by Planck. And you'll understand, you'll go through this in some detail in 8.044 later in the semester. So I'm not going to dwell on it now, but I do want to give you a little bit of perspective on it. So Planck ran across this idea that E is equal to h/nu. Through the process of trying to fit an experimental curve. There was a theory of how much energy should be emitted by an object that's hot and glowing as a function of frequency. And that theory turned out to be in total disagreement with experiment. Spectacular disagreement. The curve for the theory went up, the curve for the experiment went down. They were totally different. So Planck set about writing down a function that described the data. Literally curve-fitting, that's all he was doing. And this is the depths of desperation to which he was led, was curve-fitting. He's an adult. He shouldn't be doing this, but he was curve-fitting. And so he fits the curve, and in order to get it to fit the only thing that he can get to work even vaguely well is if he puts in this calculation of h/nu. He says, well, maybe when I sum over all the possible energies I should restrict the energies which were proportional to the frequency. And it was forced on him because it fit from the function. Just functional analysis. Hated it. Hated it, he completely hated it. He was really frustrated by this. It fit perfectly, he became very famous. He was already famous, but he became ridiculously famous. Just totally loathed this idea. OK. So it's now become a cornerstone of quantum mechanics. But he wasn't so happy about it. And to give you a sense for how bold and punchy this paper by Einstein was that said, look, seriously. Seriously guys. e equals h/nu. Here's what Planck had to say when he wrote a letter of recommendation to get Einstein into the Prussian Academy of Sciences in 1917, or 1913. So he said, there is hardly one among the great problems in physics to which Einstein has not made an important contribution. That he may sometimes have missed the target in his speculations as in his hypothesis of photons cannot really be held too much against him. It's not possible to introduce new ideas without occasionally taking a risk. Einstein who subsequently went on to develop special relativity and general relativity and prove the existence of atoms and the best measurement of Avogadro's Constant, subsequently got the Nobel Prize. Not for Avogadro's Constant, not for proving the existence of atoms, not for relativity, but for photons. Because of guys like Planck, right. This is crazy. So this was a pretty bold idea. And here, to get a sense for why-- we're gonna leave that up because it's just sort of fun to see these guys scowling and smiling-- there is, incidentally there's a great book about Einstein's years in Berlin by Tom Levenson, who's a professor here. A great writer and a sort of historian of science. You should take a class from him, which is really great. But I encourage you to read this book. It talks about why Planck is not looking so pleased right there, among many other things. It's a great story. So let's step back for a second. Why was Planck so upset by this, and why was in fact everyone so flustered by this idea that it led to the best prize you can give a physicist. Apart from a happy home and, you know. I've got that one. That's the one that matters to me. So why is this so surprising? And the answer is really simple. We know that it's false. We know empirically, we've known for two hundred and some years that light is a wave. Empirically. This isn't like people are like, oh I think it'd be nice if it was a wave. It's a wave. So how do we know that? So this goes back to the double-slit experiment from Young. Young's performance of this was in 1803. Intimations of it come much earlier. But this is really where it hits nails to the wall. And here's the experiment. So how many people in here have not seen a double-slit experiment described? Yeah, exactly. OK. So I'm just going to quickly remind you of how this goes. So we have a source for waves. We let the waves get big until they're basically plane waves. And then we take a barrier. And we poke two slits in it. And these plane waves induce-- they act like sources at the slits and we get nu. And you get crests and troughs. And you look at some distant screen and you look at the pattern, and the pattern you get has a maximum. But then it falls off, and it has these wiggles, these interference fringes. These interference fringes are, of course, extremely important. And what's going on here is that the waves sometimes add in-- so the amplitude of the wave, the height of the wave, sometimes adds constructively and sometimes destructively. So that sometimes you get twice the height and sometimes you get nothing. So just because it's fun to see this, here's Young's actual diagram from his original note on the double-slit experiment. So a and b are the slits, and c, d and f are the [INAUDIBLE] on the screen, the distant screen. He drew it by hand. It's pretty good. So we've known for a very long time that light, because of the double-slit experiment, light is clearly wavy, it behaves like a wave. And what are the senses in which it behaves like a wave? There are two important senses here. The first is answered by the question, where did the wave hit the screen? So when we send in a wave, you know, I drop a stone, one big pulsive wave comes out. It splits into-- it leads to new waves being instigated here and over here. Where did that wave hit the screen? Anyone? AUDIENCE: Everywhere. PROFESSOR: Yeah, exactly. It didn't hit this wave-- the screen in any one spot. But some amplitude shows up everywhere. The wave is a distributed object, it does not exist at one spot, and it's by virtue of the fact that it is not a localized object-- it is not a point-like object-- that it can interfere with itself. The wave is a big large phenomena in a liquid, in some thing. So it's sort of essential that it's not a localized object. So not localized. The answer is not localized. And let's contrast this with what happens if you take this double-slit experiment and you do it with, you know, I don't know, take-- who. Hmm. Tim Wakefield. Let's give some love to that guy. So, baseball player. And have him throw baseballs at a screen with two slits in it. OK? Now he's got pretty good-- well, he's got terrible accuracy, actually. So every once in a while he'll make it through the slits. So let's imagine first blocking off-- what, he's a knuckle-baller, right-- so every once in a while it goes, the baseball will go through the slit. And let's think about what happens, so let's cover one slit. And what we expect to happen is, well, it'll go through more or less straight, but sometimes it'll scrape the edge, it'll go off to the side, and sometimes it'll come over here. But if you take a whole bunch of baseballs, and-- so any one baseball, where does it hit? Some spot. Right? One spot. Not distributed. One spot. And as a consequence, you know, one goes here, one goes there, one goes there. And now, there's nothing like interference effects, but what happens is as it sort of doesn't-- you get some distribution if you look at where they all hit. Yeah? Everyone cool with that? And if we had covered over this slot, or slit, and let the baseballs go through this one, same thing would have happened. Now if we leave them both open, what happens is sometimes it goes here, sometimes it goes here. So now it's pretty useful that we've got a knuckle-baller. And what you actually get is the total distribution looks like this. It's the sum of the two. But at any given time, any one baseball, you say, aha, the baseball either went through the top slit, and more or less goes up here. Or it went through the bottom slit and more or less goes down here. So for chunks-- so this is for waves-- for chunks or localized particles, they are localized. And as a consequence, we get no interference. So for waves, they are not localized, and we do get interference. Yes, interference. OK. So on your problem set, you're going to deal with some calculations involving these interference effects. And I'm going to brush over them. Anyway the point of the double-slit experiment is that whatever else you want to say about baseballs or anything else, light, as we've learned since 1803 in Young's double-slit experiment, light behaves like a wave. It is not localized, it hits the screen over its entire extent. And as a consequence, we get interference. The amplitudes add. The intensity is the square of the amplitude. If the intensities add-- so sorry, if the amplitudes add-- amplitude total is equal to a1 plus a2, the intensity, which is the square of a1 plus a2 squared, has interference terms, the cross terms, from this square. So light, from this point of view, is an electromagnetic wave. It interferes with itself. It's made of chunks. And I can't help but think about it this way, this is literally the metaphor I use in my head-- light is creamy and smooth like a wave. Chunks are very different. But here's the funny thing. Light is both smooth like a wave, it is also chunky. It is super chunky, as we have learned from the photoelectric effect. So light is both at once. So it's the best of both worlds. Everyone will be satisfied, unless you're not from the US, in which case this is deeply disturbing. So of course the original Superchunk is a band. So we've learned now from Young that light is a wave. We've learned from the photoelectric effect that light is a bunch of chunks. OK. Most experimental results are true. So how does that work? Well, we're gonna have to deal with that. But enough about light. If this is true of light, if light, depending on what experiment you do and how you do the experiment, sometimes it seems like it's a wave, sometimes it seems like it's a chunk or particle, which is true? Which is the better description? So it's actually worthwhile to not think about light all the time. Let's think about something more general. Let's stick to electrons. So as we saw from yesterday's lecture, you probably want to be a little bit wary when thinking about individual electrons. Things could be a little bit different than your classical intuition. But here's a crucial thing. Before doing anything else, we can just think, which one of these two is more likely to describe electrons well. Well electrons are localized. When you throw an electron at a CRT, it does not hit the whole CRT with a wavy distribution. When you take a single electron and you throw it at a CRT, it goes ping and there's a little glowing spot. Electrons are localized. And we know that localized things don't lead to interference. Some guys at Hitachi, really good scientists and engineers, developed some really awesome technology a couple of decades ago. They were trying to figure out a good way to demonstrate their technology. And they decided that you know what would be really awesome, this thought experiment that people have always talked about that's never been done really well, of sending an electron through a two-slitted experiment. In this case it was like ten slits effectively, it was a grading. Send an electron, a bunch of electrons, one at a time, throw the electron, wait. Throw the electron, wait. Like our French guy with the boat. So do this experiment with our technology and let's see what happens. And this really is one of my favorite-- let's see, how we close these screens-- aha. OK. This is going to take a little bit of-- and it's broken. No, no. Oh that's so sad. AUDIENCE: [LAUGHTER] PROFESSOR: Come on. I'm just gonna let-- let's see if we can, we'll get part of the way. I don't want to destroy it. So what they actually did is they said, look, let's-- we want to see what happens. We want to actually do this experiment because we're so awesome at Hitachi Research Labs, so let's do it. So here's what they did. And I'm going to turn off the light. And I set this to some music because I like it. OK here's what's happening. One at a time, individual photons. [MUSIC PLAYING] PROFESSOR: So they look pretty localized. There's not a whole lot of structure. Now they're going to start speeding it up. It's 100 times the actual speed. [MUSIC PLAYING] PROFESSOR: Eh? Yeah. AUDIENCE: [APPLAUSE] PROFESSOR: So those guys know what they're doing. Let's-- there were go. So I think I don't know of a more vivid example of electron interference than that one. It's totally obvious. You see individual electrons. They run through the apparatus. You wait, they run through the apparatus. You wait. One at a time, single electron, like a baseball being pitched through two slits, and what you see is an interference effect. But you don't see the interference effect like you do from light, from waves on the sea. You see the interference effect by looking at the cumulative stacking up of all the electrons as they hit. Look at where all the electrons hit one at a time. So is an electron behaving like a wave in a pond? No. Does a wave in a pond at a spot? No. It's a distributed beast. OK yes, it interferes, but it's not localized. Well is it behaving like a baseball? Well it's localized. But on-- when I look at a whole bunch of electrons, they do that. They seem to interfere, but there's only one electron going through at a time. So in some sense it's interfering with itself. How does that work? Is an electron a wave? AUDIENCE: Yes. PROFESSOR: Does an electron hit at many spots at once? AUDIENCE: No. PROFESSOR: No. So is an electron a wave. No. Is an electron a baseball? No. It's an electron. So this is something you're going to have to deal with, that every once in awhile we have these wonderful moments where it's useful to think about an electron as behaving in a wave-like sense. Sometimes it's useful to think about it as behaving in a particle-like sense. But it is not a particle like you normally conceive of a baseball. And it is not a wave like you normally conceive of a wave on the surface of a pond. It's an electron. I like to think about this like an elephant. If you're closing your eyes and you walk up to an elephant, you might think like I've got a snake and I've got a tree trunk and, you know, there's a fan over here. And you wouldn't know, like, maybe it's a wave, maybe it's a particle, I can't really tell. But if you could just see the thing the way it is, not through the preconceived sort of notions you have, you'd see it's an elephant. Yes, that is the Stata Center. So-- look, everything has to happen sometime, right? AUDIENCE: [LAUGHTER] PROFESSOR: So Heisenberg-- it's often, people often give the false impression in popular books on physics, so I want to subvert this, that in the early days of quantum mechanics, the early people like Born and Oppenheimer and Heisenberg who invented quantum mechanics, they were really tortured about, you know, is it an electron, is it a wave. It's a wave-particle duality. It's both. And this is one of the best subversions of that sort of silliness that I know of. And so what Heisenberg says, the two mental pictures which experiments lead us to form, the one of particles the other waves, are both incomplete and have the validity of analogies, which are accurate only in limited cases. The apparent duality rises in the limitation of our language. And then he goes on to say, look, you developed your intuition by throwing rocks and, you know, swimming. And, duh, that's not going to be very good for atoms. So this will be posted, it's really wonderful. His whole lecture is really-- the lectures are really quite lovely. And by the way, that's him in the middle there, Pauley all the way on the right. I guess they were pleased. OK so that's the Hitachi thing. So now let's pick up on this, though. Let's pick up on this and think about what happens. I want to think in a little more detail about this Hitachi experiment. And I want to think about it in the context of a simple two-slit experiment. So here's our source of electrons. It's literally a gun, an electron gun. And it's firing off electrons. And here's our barrier, and it has two slits in it. And we know that any individual electron hits its own spot. But when we take many of them, we get an interference effect. We get interference fringes. And so the number that hit a given spot fill up, construct this distribution. So then here's the question I want to ask. When I take a single electron, I shoot one electron at a time through this experiment, one electron. It could go through the top slit, it could go through the bottom slit. While it's inside the apparatus, which path does it take? AUDIENCE: Superposition. PROFESSOR: Good. So did it take the top path? AUDIENCE: No. PROFESSOR: How do you know? [INTERPOSING VOICES] PROFESSOR: Good, let's block the bottom, OK, to force it to go through the top slit. So we'll block the bottom slit. Now the only electrons that make it through go through the top slit. Half of them don't make it through. But those that do make it through give you this distribution. No interference. But I didn't tell you these are hundreds of thousands of kilometers apart, the person who threw in the electron didn't know whether there was a barrier here. The electron, how could it possibly know whether there was a barrier here if you went through the top. This is exactly like our boxes. It's exactly like our box. Did it go through-- an electron, when the slits are both open and we know that ensemble average it will give us an interference effect, did the electron inside the apparatus go through the top path? No. Did it go through the bottom path? Did it go through both? Because we only see one electron. Did it go through neither? It is in a-- AUDIENCE: Superposition. PROFESSOR: --of having gone through the top and the bottom. Of being along the top half and being along the bottom path. This is a classic example of the two-box experiment. OK. So you want to tie that together. So let's nuance this just a little bit, though, because it's going to have an interesting implication for gravity. So here's the nuance I want to pull on this one. Let's cheat. OK. Suppose I want to measure which slit the electron actually did go through. How might I do that? Well I could do the course thing I've been doing which is I could block it and just catch the-- catch electrons that go through in that spot. But that's a little heavy-handed. Probably I can do something a little more delicate. And so here's the more delicate thing I'm going to do. I want to build a detector that uses very, very, very weak light, extremely weak light, to detect whether the particle went through here or here. And the way I can do that is I can sort of shine light through and-- I'm gonna, you know, bounce-- so here's my source of light. And I'll be able to tell whether the electron went through this slit or it went through this slit. Cool? So imagine I did that. So obviously I don't want to use some giant, huge, ultra high-energy laser because it would just blast the thing out of the way. It would destroy the experiment. So I wanna something very diffuse, very low energy, very low intensity electromagnetic wave. And the idea here is that, OK, it's true that when I bounce this light off an electron, let's say it bounces off an electron here, it's true it's going impart some momentum and the electron's gonna change its course. But if it's really, really weak, low energy light, then it's-- it's gonna deflect only a little tiny bit. So it will change the pattern I get over here. But it will change it in some relatively minor way because I've just thrown in very, very low energy light. Yeah? That make sense? So this is the experiment I want to do. This experiment doesn't work. Why. AUDIENCE: You know which slit it went through. PROFESSOR: No. It's true that it turns out that those are correlated facts, but here's the problem. I can run this experiment without anyone actually knowing what happens until long afterwards. So knowing doesn't seem to play any role in it. It's very tempting often to say, no, but it turns out that it's really not about what you know. It's really just about the experiment you're doing. So what principle that we've already run into today makes it impossible to make this work? If I want to shine really low-energy, really diffuse light through, and have it scatter weakly. Yeah. AUDIENCE: Um, light is chunky. PROFESSOR: Yeah, exactly. That's exactly right. So when I say really low-energy light, I don't-- I really can't mean, because we've already done this experiment, I cannot possibly mean low intensity. Because intensity doesn't control the energy imparted by the light. The thing that controls the energy imparted by a collision of the light with the electron is the frequency. The energy in a chunk of light is proportional to the frequency. So now if I want to make the effect the energy or the momentum, similarly-- the momentum, where did it go-- remember the momentum goes like h over lambda. If I want to make the energy really low, I need to make the frequency really low. Or if I want to make the momentum really low, I need to make the wavelength what? Really big. Right? So in order to make the momentum imparted by this photon really low, I need to make the wavelength really long. But now here's the problem. If I make the wavelength really long, so if I use a really long-wavelengthed wave, like this long of a wavelength, are you ever going to be able to tell which slit it went through? No, because the particle could have been anywhere. It could have scattered this light if it was here, if it was here, if it was here, right? In order to measure where the electron is to some reasonable precision-- so, for example, to this sort of wavelength, I need to be able to send in light with a wavelength that's comparable to the scale that I want to measure. And it turns out that if you run through and just do the calculation, suppose I send in-- and this is done in the books, in I think all four, but this is done in the books on the reading list-- if you send in a wave with a short enough wavelength to be able to distinguish between these two slits, which slit did it go through, the momentum that it imparts precisely watches-- washes out is just enough to wash out the interference effect, and break up these fringes so you don't see interference effects. It's not about what you know. It's about the particulate nature of light and the fact that the momentum of a chunk of light goes like h over lambda. OK? But this tells you something really interesting. Did I have to use light to do this measurement? I could have sent in anything, right? I didn't have to bounce light off these things. I could have bounced off gravitational waves. So if I had a gravitational wave detector, so-- Matt works on gravitational wave detectors, and so, I didn't tell you this but Matt gave me a pretty killer gravitational wave detector. It's, you know, here it is. There's my awesome gravitational wave detector. And I'm now going to build supernova. OK. And they are creeping under this black hole, and it's going to create giant gravitational waves. And we're gonna use those gravitational waves and detect them with the super advanced LIGO. And I'm gonna detect which slit it went through. But gravitational waves, those aren't photons. So I really can make a low-intensity gravitational wave, and then I can tell which slit it went through without destroying the interference effect. That would be awesome. What does that tell you about gravitational waves? They must come in chunks. In order for this all to fit together logically, you need all the interactions that you could scatter off this to satisfy these quantization properties. But the energy is proportional to the frequency. The line I just gave you is a heuristic. And making it precise is one of the great challenges of modern contemporary high-energy physics, of dealing with the quantum mechanics and gravity together. But this gives you a strong picture of why we need to treat all forces in all interactions quantum-mechanically in order for the world to be consistent. OK. Good. OK, questions at this point? OK. So-- oh, I forgot about this one-- so there are actually two more. So I want to just quickly show you-- well, OK. So, this is a gorgeous experiment. So remember I told you the story of the guy with the boat and the opaque wall and it turns out that's a cheat. It turns out that this opaque screen doesn't actually give you quantum mechanically isolated photons. They're still, in a very important way, classical. So this experiment was done truly with a source that gives you quantum mechanically isolated single photons, one at a time. So this is the analogue of the Hitachi experiment. And it was done by this pretty awesome Japanese group some number of years ago. And I just want to emphasize that it gives you exactly the same effects. We see that photons-- this should look essentially identical to what we saw at the end of the Hitachi video. And that's because it's exactly the same physics. It's a grating with something like 10 slits and individual particles going through one at a time and hitting the screen and going, bing. So what you see is the light going, bing, on a CCD. It's a pretty spectacular experience. So let's get back to electrons. I want another probe of whether electrons are really waves or not. So this other experiment-- again, you're going to study this on your problem set-- this other experiment was done by a couple of characters named Davisson and Germer. And in this experiment, what they did is they took a crystal, and a crystal is just a lattice of regularly-located ions, like diamond or something. Yeah? AUDIENCE: Before you go on I guess, I wanted to ask if the probability of a photon or an electron going through the 10 slits is about the same? PROFESSOR: Is what, sorry? AUDIENCE: Is exactly the same. PROFESSOR: You mean for different electrons? AUDIENCE: Yeah. PROFESSOR: Well they can be different if the initial conditions are different. But they could be-- if the initial conditions are the same, then the probabilities are identical. So every electron behaves identically to every other electron in that sense. Is that what you were asking? AUDIENCE: It is actually like through any [INAUDIBLE] the probability of it going like [INAUDIBLE]? PROFESSOR: Sure, absolutely. So the issue there is just a technological one of trying to build a beam that's perfectly columnated. And that's just not doable. So there's always some dispersion in your beam. So in practice it's very hard to make them identical, but in principle they could be if you were infinitely powerful as an experimentalist, which-- again, I was banned from the lab, so not me. So here's our crystal. You could think of this as diamond or nickel or whatever. I think they actually use nickel but I don't remember exactly. And they sent in a beam of electrons. So they send in a beam of electrons, and what they discover is that if you send in these electrons and watch how they scatter at various different angles-- I'm going to call the angle here of scattering theta-- what they discover is that the intensity of the reflected beam, as a function of theta, shows interference effects. And in particular they gave a whole calculation for this, which I'm not going to go through right now because it's not terribly germane for us-- you're going to go through it on your problem set, so that'll be good and it's a perfect thing for your recitation instructors to go through. But the important thing is the upshot. So if the distance between these crystal planes is L-- or, sorry, d-- let me call it d. If the distance between the crystal planes is d, what they discover is that the interference effects that they observed, these maxima and minima, are consistent with the wavelength of light. Or, sorry, with the electrons behaving as if they were waves with a definite wavelength, with a wavelength lambda being equal to some integer, n, over 2d sine theta. So this is the data-- these are the data they actually saw, data are plural. And these are the data they actually saw. And they infer from this that the electrons are behaving as if they were wave-like with this wavelength. And what they actually see are individual electrons hitting one by one. Although in their experiment, they couldn't resolve individual electrons. But that is what they see. And so in particular, plugging all of this back into the experiment, you send in the electrons with some energy, which corresponds to some definite momentum. This leads us back to the same expression as before, that the momentum is equal to h over lambda, with this lambda associated. So it turns out that this is correct. So the electrons diffract off the crystal as if they have a momentum which comes with a definite wavelength corresponding to its momentum. So that's experimental result-- oh, I forgot to check off four-- that's experimental result five, that electrons diffract. We already saw the electron diffraction. So something to emphasize is that-- so these experiments as we've described them were done with photons and with electrons, but you can imagine doing the experiments with soccer balls. This is of course hard. Quantum effects for macroscopic objects are usually insignificantly small. However, this experiment was done with Buckyballs, which are the same shape as soccer balls in some sense. But they're huge, they're gigantic objects. So here's the experiment in which this was actually done. So these guys are just totally amazing. So this is Zellinger's lab. And it doesn't look like all-- I mean it looks kind of, you know. It's hideous, right? I mean to a theorist it's like, come on, you've got to be kidding that that's-- But here's what a theorist is happy about. You know, because it looks simple. We really love lying to ourselves about that. So here's an over. We're going to cook up some Buckyballs and emit them with some definite known thermal energy. Known to some accuracy. We're going to columnate them by sending them through a single slit, and then we're going to send them through a diffraction grating which, again, is just a whole bunch of slits. And then we're going to image them using photo ionization and see where they pop through. So here is the horizontal position of this wave along the grating, and this is the number that come through. This is literally one by one counts because they're going bing, bing, bing, as a c60 molecule goes through. So without the grating, you just get a peek. But with the grating, you get the side bands. You get interference fringes. So these guys, again, they're going through one by one. A single Buckyball, 60 carbons, going through one by one is interfering with itself. This is a gigantic object by any sort of comparison to single electrons. And we're seeing these interference fringes. So this is a pretty tour de force experiment, but I just want to emphasize that if you could do this with your neighbor, it would work. You'd just have to isolate the system well enough. And that's a technological challenge but not an in-principle one. OK. So we have one last experimental facts to deal with. And this is Bell's Inequality, and this is my favorite one. So Bell's Inequality for many years languished in obscurity until someone realized that it could actually be done beautifully in an experiment that led to a very concrete experiment that they could actually do and that they wanted to do. And we now think of it as an enormously influential idea which nails the coffin closed for classical mechanics. And it starts with a very simple question. I claim that the following inequality is true: the number of undergraduate-- of the number of people in the room who are undergraduates, which I'll denote as U-- and not blonde, which I will denote as bar B-- so undergraduates who are not blonde-- actually let me write this out in English. It's gonna be easier. Number who are undergrads and not blonde plus the number of people in the room who are blonde but not from Massachusetts is strictly greater than or equal to the number of people in the room who are undergraduates and not from Massachusetts. I claim that this is true. I haven't checked in this room. But I claim that this is true. So let's check. How many people are undergraduates who are not blonde? OK this is going to-- jeez. OK that's-- so, lots. OK. How many people are blonde but not from Massachusetts? OK. A smattering. Oh God, this is actually going to be terrible. AUDIENCE: [LAUGHTER] PROFESSOR: Shoot. This is a really large class. OK. Small. And how many people are undergraduates who are not from Massachusetts? Yeah, this-- oh God. This counting is going to be-- so let's-- I'm going to do this just so I can do the counting with the first two rows here. OK. My life is going to be easier this way. So how many people in the first two rows, in the center section, are undergraduates but not blonde? One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen. We could dispute some of those, but we'll take it for the moment. So, fourteen. You're probably all undergraduates. So blonde and not from Massachusetts. One. Awesome. Undergraduates not from Massachusetts. One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, fifteen. Equality. AUDIENCE: [LAUGHTER] PROFESSOR: OK. So that-- you might say well, look, you should have been nervous there. You know, and admittedly sometimes there's experimental error. But I want to convince you that I should never, ever ever be nervous about this moment in 8.04. And the reason is the following. I want to prove this for you. And the way I'm gonna prove it is slightly more general, in more generality. And I want to prove to you that the number-- if I have a set, or, sorry, if the number of people who are undergraduates and not blonde which, all right, is b bar plus the number who are blonde but not from Massachusetts is greater than or equal to the number that are undergraduates and not from Massachusetts. So how do I prove this? Well if you're an undergraduate and not blonde, you may or you may not be from Massachusetts. So this is equal to the number of undergraduates who are not blonde and are from Massachusetts plus the number of undergraduates who are not blonde and are not from Massachusetts. It could hardly be otherwise. You either are or you are not from Massachusetts. Not the sort of thing that you normally see in physics. So this is the number of people who are blonde and not from Massachusetts, number of people who are blonde, who are-- so if you're blonde and not from Massachusetts, you may or may not be an undergraduate. So this is the number of people who are undergraduates, blonde, and not from Massachusetts plus the number of people who are not undergraduates, are blonde and are not from Massachusetts. And on the right hand side-- so, adding these two together gives us plus and plus. On the right hand side, the number of people that are undergraduates and not from Massachusetts, well each one could be either blonde or not blonde. So this is equal to the number that are undergraduates, blonde, and not from Massachusetts, plus-- remember that our undergraduates not blonde and not from Massachusetts. Agreed? I am now going to use the awesome power of-- and so this is what we want to prove, and I'm going to use the awesome power of subtraction. And note that U, B, M bar, these guys cancel. And U, B bar, M bar, these guys cancel. And we're left with the following proposition: the number of undergraduates who are not blonde but are from Massachusetts plus the number of undergrad-- of non-undergraduates who are blonde but not from Massachusetts must be greater than or equal to zero. Can you have a number of people in a room satisfying some condition be less than zero? Can minus 3 of you be blonde undergraduates not from Massachusetts? Not so much. This is a strictly positive number, because it's a numerative. It's a counting problem. How many are undergraduates not blonde and from Massachusetts. Yeah? Everyone cool with that? So it could hardly have been otherwise. It had to work out like this. And here's the more general statement. The more general statement is that the number of people, or the number of elements of any set where each element in that set has binary properties a b and c-- a or not a, b or not b, c or not c. Satisfies the following inequality. The number who are a but not b plus the number who are b but not c is greater than or equal to the number who are a but not c. And this is exactly the same argument. And this inequality which is a tautology, really, is called Bell's Inequality. And it's obviously true. What did I use to derive this? Logic and integers, right? I mean, that's bedrock stuff. Here's the problem. I didn't mention this last time, but in fact electrons have a third property in addition to-- electrons have a third property in addition to hardness and color. The third property is called whimsy, and you can either be whimsical or not whimsical. And every electron, when measured, is either whimsical or not whimsical. You never have a boring electron. You never have an ambiguous electron. Always whimsical or not whimsical. So we have hardness, we have color, we have whimsy. OK. And I can perform the following experiment. From a set of electrons, I can measure the number that are hard and not black, plus the number that are black but not whimsical. And I can measure the number that are hard and not whimsical. OK? And I want to just open up the case a little bit and tell you that the hardness here really is the angular momentum of the electron along the x-axis. Color is the angular momentum of the electron along the y-axis. And whimsy is the angular momentum of the electron along the z-axis. These are things I can measure because I can measure angular momentum. So I can perform this experiment with electrons and it needn't be satisfied. In particular, we will show that the number of electrons, just to be very precise, the number of electrons in a given set, which have positive angular momentum along the x-axis and down along the y-axis, plus up along the y-axis and down along the z-axis, is less than the number that are up. Actually let me do this in a very particular way. Up... zero down at theta. Up at theta, down at-- two theta is greater than the number that are up at zero and down at theta. Now here's the thing-- two theta. You can't at the moment understand what this equation means. But if I just tell you that these are three binary properties of the electron, OK, and that it violates this inequality, there is something deeply troubling about this result. Bell's Inequality, which we proved-- trivially, using integers, using logic-- is false in quantum mechanics. And it's not just false in quantum mechanics. We will at the end of the course derive the quantum mechanical prediction for this result and show that at least to a predicted violation of Bell's Inequality. This experiment has been done, and the real world violates Bell's Inequality. Logic and integers and adding probabilities, as we have done, is misguided. And our job, which we will begin with the next lecture, is to find a better way to add probabilities than classically. And that will be quantum mechanics See you on Tuesday. AUDIENCE: [APPLAUSE]
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https://ocw.mit.edu/courses/7-01sc-fundamentals-of-biology-fall-2011/7.01sc-fall-2011.zip
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PROFESSOR: Recombinant DNA, often referred to also as genetic engineering. This is a series of techniques, series of methods that allow us to manipulate DNA for a variety of reasons. Now, we take it for granted. It's very much part of our everyday life in the laboratory. It's made a huge impact on the biotechnology and pharmaceutical industries as well. It wasn't always so uncontroversial. In fact, in the 1960s and '70s, when this technology was first being developed, it was great concern about scientists manipulating DNA, manipulating genetic material. In the city of Cambridge, in fact, had a moratorium for a while on the practice of genetic engineering or recombinant DNA technology, which fortunately, ultimately was overcome with good practices and I think good education about what the real limits of risk and benefit were so that now it's used very widely and, I would also say, very safely. The bottom line for what we're going to talk about today and for this section of the class is the ability to isolate and to amplify specific DNA sequences. They might be particular genes of interest to us. They might be whole genomes. They might be other regions of DNA, but we need to be able to isolate them from the cells of the organisms of interest, amplify them up into large quantities in order to be able to study them in detail, and to make modifications in them. This is done in a variety of organisms for a variety of purposes. One of them, and this is by no means the only one, but one that sort of strikes home is the ability to make therapeutic proteins, to be able to manufacture in the laboratory or in a company proteins they could have benefit for patients who are, for example, lacking the function of a particular protein or enzyme leading to a disease state. One can use these methods to produce that protein in the laboratory, and treat the individual. This can be done in bacteria such as E. coli. E. coli, which is the common gut bacterium that we all carry, a very useful organism. We use it in the lab. We're going to use in our demonstrations today. This is the standard vehicle in which we'd grow up, recombinant DNA, amplify recombinant DNA for further uses, but we can also make therapeutic proteins using the manufacturing capabilities of the bacterium. Plants, likewise, can be a source of therapeutic proteins. We can modify the genomes of plants so they will produce in large quantities therapeutic proteins of interest, which then can be ingested by the individual, which can reduce production costs significantly, and animals, likewise. We can manipulate the genes of animals so that they will express a protein of interest and secrete that protein, for example, into the milk. So there are so-called transgenic cows and transgenic goats that produce therapeutic proteins in the mammary gland and secrete those therapeutic proteins into the milk. So the individual just needs to drink the milk and receive the relevant dose. So there are lots of ways, that this technology can be helpful including in the context of medicine. We can also engineer organisms. I've already given you examples of that, but that was really for the purposes of using those organisms as a factory to make something of interest to us. But we can manipulate the genes of plants, for example, to make them resistant to various pests to make them more robust, to give them a longer shelf life. We can manipulate them for the better production of things that are valuable to us. Again, this is a bit of a controversial area, genetically modified foods, not always well-accepted by everyone because again, the thought is, this might be disrupting the food chain in important ways, and this might be ultimately not so beneficial. Personally don't agree with that, but lots of people do feel that way. We're going to teach you the methods that we use to allow us to do this. And again, in work that I've alluded to from my own lab, we use these methods to manipulate the genes of animals, in our case it's mice, but there are lots of animal species that one can use in order to create, for example, disease models. We talked to you about, Professor Brown talked to you about genetic diseases. They have specific alterations in genes. We can use these methods to make similar alterations in the genes of mice and other animal species and then study the disease process in those animals, develop new treatments in those animals in ways that are hard to do in people. So there's lots and lots, and these are just a few examples, lots and lots of important uses for genetic engineering and recombinant DNA technology. One that we'll emphasize in just a few lectures and is becoming extremely common and popular nowadays is DNA sequencing. You need to isolate the DNA from the organism of interest and then have it in such a fashion that you can sequence it's nucleotides. This is a very popular activity now. And specifically with respect to human health, DNA sequencing, But some of the other methods that we're going to talk about as well, allow us to characterize disease-causing mutations at molecular detail. So we know what causes x disease or y disease. We can understand the consequences for the individual-encoded proteins for the pathways that they regulate, and we can come up with better medicines. I think about this with respect to cancer, but it's really changing the treatment of all diseases as we understand more the molecular basis, the genetic basis of those diseases. And these techniques have been essential to allow us to do that. I want to cover a little bit of history so that you know from whence this came. Things really began to pick up in this field in the late 1960s. And the critical advance at this stage, was the discovery of a method to cleave DNA into defined fragments, to start with a genome or chromosome and to be able to cut it particular places reproducibly so that one could isolate fragments of DNA away from the mass of DNA, isolate a particular region of the DNA away from everything else using this method. It wasn't enough just to cut the DNA up. You had to amplify it up. In order to amplify it up, you needed a vessel in which to do the amplification, and the vessel of choice as I mentioned was bacteria. And this relied on a method that actually was known for decades before but actually was not used for this purpose until the early 1970s, and it's called bacterial transformation. You can transform bacteria by adding a new DNA sequence, and the bacteria will take up that new DNA sequence and begin to express the genes that are present on that DNA sequence as if it was one of their own. So the transfer of DNA a was critical, this process of bacterial transformation. And then the final thing which also occurred in the 1970s was the identification of methods to amplify DNA sequences once they got inside of bacteria. It wasn't enough to actually get them in there. You needed some special way to cause the bacterium to amplify, that is, to replicate the DNA that was present within them. And these three events, which all came together in about a five to ten year period, really initiated, launched what we now call the recombinant DNA revolution and initiated the biotechnology industry, which started in the mid 1970s. And an individual above all others who is credited with launching the biotechnology industry was Robert Swanson who in 1966 or so was sitting where you are as an MIT freshman. Bob Swanson was class of '69, actually. And at the age of 28, in 1976, founded the company Genentech with a scientist, Herb Boyer, who was very instrumental in some of those breakthroughs that I just mentioned to you. And I mention Bob to you now because of your connections through MIT. But he was a wonderful guy and sadly passed away from glioblastoma at the age of 52. So he didn't really get to fully realize the benefits of what he had started, but he did, in fact, start a great deal. And this is a famous statue at Genentech which shows Herb Boyer and Bob Swanson talking about the idea for the first time and you might notice over a beer, very important part of science, discussions over a beer. In addition to this connection to MIT and really in honor of Bob's connections to MIT, and a great pride that Bob had actually in MIT, credited MIT greatly with him learning about science, the importance of science, and also business. He got a degree in chemistry as well as a degree from the Sloan School. We decided when we launched the Koch Institute to name a very large space in the Koch Institute for Bob Swanson. It's called the Swanson Biotechnology Center. You can visit it. So it's a series of core facilities that support all of our researchers. And indeed, researchers across the MIT campus, and this is a nice quote from Bob's widow, Judy Swanson, who has been very supportive of this effort. So Bob Swanson and MIT, in many ways, actually, have a lot to do with the technology and the revolution that we're going to talk about now. All right. So as I mentioned, we are going to do a demo. We're going to teach you a real-life example now of how this work is done. And our goal is to clone a gene. You can actually wait right there, Anna. Thanks. We're going to clone a gene, and we're going to clone a particular gene. It's a toxin gene. It's a gene from a pathogenic bacterium. And you might ask the question, a reasonable question, why the heck would you do that? Why would you clone a toxin gene? So what do you think? What's the purpose in the laboratory of cloning out a toxin-encoding gene from a pathogenic bacterium? So several people have suggested the obvious was that you want to engage in global terror, which we actually don't support here at MIT, so we're going to take that down. But maybe somebody else is going to do that. So perhaps it would be good if we got one step ahead of the game, isolated the gene, manufactured the protein, and then made a vaccine against that toxin so that we could prevent the bad consequences of exposure to the toxin. Or maybe that thing is actually very interesting independent of its bad uses. We could learn stuff, which might be helpful ultimately in related activities. So for the general purpose of biomedical research, we often study how these organisms work because it can teach us things, sometimes surprising things that are useful down the road. The organism in question is Streptococcus pyogenes. Streptococcus pyogenes, which causes in certain cases, in certain individuals a disease called necrotizing fasciitis. Necrotizing fasciitis, which is otherwise more commonly called the flesh-eating disease. And you might think I'm joking, but I'm not. This is a true thing. This is a true thing, and some of you might be squeamish, and if you are, and I'm being serious here. If you're squeamish looking at ugly, nasty, disgusting pictures, close your eyes for a second. I'll tell you when you can open them. But this is an individual who was exposed to this bacterium and developed necrotizing fasciitis. That's a real-world case, so it is really pretty bad. You can open your eyes now if you closed them. I hesitate to show that slide because in the past, I've had a few boys throw up when I showed that slide. So what are we going to do? Well, we're going to isolate this gene. And in order to isolate this gene, we need to be able to separate it, this gene, from the chromosome in which it is contained. And the chromosome from which it is contained is the chromosome of S. pyogenes. So this is the S. pyogenes chromosome. It's about four million base pairs and it has about 1,000 genes. So spread throughout this circular chromosome, there are lots of genes. We're interested in one of them. Now, this chromosome also has another thing on it, which I hope you all know about from the material that Professor Sive just covered for you. What is the one piece of DNA material that all chromosomes need in order to replicate? AUDIENCE: Origin of replication. PROFESSOR: An origin of replication, very good. An origin of replication. It has an origin of replication, and then it has a bunch of genes. It has gene A. I'm just making this up. It has gene B, and it has a whole bunch of other genes as well. And then it has -- and imagine that this orange chalk was red because it's more effective if it's red. It has the T gene, and that's the toxin gene. So our goal is to transfer the T gene and not the rest of this stuff, because we don't actually care about the rest of the stuff. We just care about the T gene -- into the E. coli cells for the reasons that I mentioned up there. And in order to do that, we have to grow large amounts of the organism that's going to in a sense donate the DNA. We then isolate the chromosomal DNA, and we'll show you how. We're then going to use this method to fragment the DNA not randomly, but in specific places. And then we're going to transfer the DNA of interest to E. coli using this method of transformation. We're going to take this fragment and move it through the membrane of the E. coli so that it becomes resident inside the E. coli cell. So now on to our demonstration and my lab assistant, Anna Deconinck, will help me here. So what we've done is to, in the laboratory, isolate S. pyogenes as well as E. coli, grow them up in large quantities. You've got your gloves, right? So I'll take the buffers. So we have various solutions and buffers that will allow us to sort of wash the stuff we don't want away from the bacterial cells, lice the bacterial membrane, isolate the nucleic acid away from all the other stuff that's inside the cells, and then we'll purify the chromosomal DNA. So Anna has grown up E. coli and S. pyogenes, taken that suspension of cells, and used a centrifuge to spin those cells down to the bottom of these tubes here-- you can show them, Anna-- these tubes here. And the first thing we need to do is get rid of the supernate, the broth that the cells grew in. So first, we're going to decant. Here, you can decant the pyogenes, but be careful with it. So we're going to decant this in order to grow up the amounts of bacteria that we need. Now we're doing this in a very small quantities. In fact, in industrial scale, you do it in huge-- ANNA: Sorry. PROFESSOR: Ah, that's a problem. It's actually a bit more of a problem. ANNA: I just have a buffer to wash. PROFESSOR: I don't think that's going to do it, Anna. Dude, we may need to actually skip, we may need to cancel. This is a little more serious. Wait a minute. Well, I don't know. Maybe. Let's just see if it's safe or not. I think it'll be okay. All right. That was a joke. She did very well though, don't you think? She did very well. That was outstanding. Anybody want some apple juice? You're welcome to it. ANNA: It needs ice. PROFESSOR: Of course we would never bring pathogenic bacterium to class.
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https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/8.04-spring-2016.zip
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Three-dimensional case. Now, in the future homework, you will be doing the equivalent of this calculation here with the Laplacians-- it's not complicated-- so that you will derive with the current is. And the current must be a very similar formula as this one. And indeed, I'll just write it here. The current is h bar over m, the imaginary part of psi star. And instead of ddx, you expect the gradient of psi. That is the current for the probability in three dimensions. And the analog of this equation, d rho dt plus dj dx equals 0, is d rho dt plus divergence of j is equal to 0. That is current conservation. Perhaps you do remember that from your study of electromagnetism. That's how Maxwell discovered the displacement current when he tried to figure out how everything was compatible with current conservation. Anyway, that argument I'll do in a second so that it will become clearer. So one last thing here-- it's something also-- you can check the units here of j is 1 over l squared times 1 over t, so probability per unit area and unit time. So what did we have? We were doing the integral of the derivative of the integral given by n. It was over here, dn dt. We worked hard on it. And dn dt was the integral of d rho dt. So it was the integral of d rho dt dx. But we showed now that d rho dt is minus dj dx. So here you have integral from minus infinity to infinity dx of dj dx. And therefore, this is-- I should have a minus sign, because it was minus dj dx. This is minus the current of x equals infinity times p minus the current at x equals minus infinity nt. And as we more or less hinted before, since the current is equal to h over 2im psi star [INAUDIBLE] psi dx minus psi [INAUDIBLE] psi star dx, as you go to plus infinity or minus infinity, these things go to 0 given the boundary conditions that we put. Because psi or psi star go to 0 to infinity, and the derivates are bound at the infinity. So this is 0, dn dt 0. All is good. And two things happened. In the way of doing this, we realized that the computation we have done pretty much established that this is equal to that, because dn dt is the difference of these two integrals, and we showed it's 0. So this is true. And therefore, we suspect h is a Hermitian operator. And the thing that we should do in order to make sure it is is put two different functions here, not two equal functions. This worked for two equal function, but for two different functions, and check it as well. And we'll leave it as an exercise. It's a good exercise. So this shows the consistency. But we discover two important ideas-- one, the existence of a current for probability, and two, h is a Hermitian operator. So last thing is to explain the analogy with current conservation. I think this should help as well. So the interpretation that we'll have is the same as we have in electromagnetism. And there's a complete analogy for everything here. So not for the wave function, but for all these charge densities and current densities. So we have electromagnetism and quantum mechanics. We have rho. Here is the charge density. And here is the probability density. If you have a total charge q in a volume, here is the probability to find the particle in a volume. There is a j in Maxwell's equations as well, and that's a current density. Amber's law has that current. It generates the curl of b. And here is a probability current density. So that's the table. So what I want to make sure is that you understand why these equations, like this or that, are more powerful than just showing that dn dt is 0. They imply a local conservation of probability. You see, there has to be physics of this thing. So the total probability must be 1. But suppose you have the probability distributed over space. There must be some relation between the way the probabilities are varying at one point and varying in other points so that everything is consistent. And those are these differential relations that say that whenever you see a probability density change anywhere, it's because there is some current. And that makes sense. If you see the charge density in some point in space changing, it's because there must be a current. So thanks to the current, you can learn how to interpret the probability much more physically. Because if you ask what is the probability that the particle is from this distance to that distance, you can look at what the currents are at the edges and see whether that probability is increasing or decreasing. So let's see that. Suppose you have a volume, and define the charge inside the volume. Then you say OK, does this charge change in time? Sure, it could. So dq dt is equal to integral d rho dt d cube x over the volume. But d rho dt, by the current conservation equation-- that's the equation we're trying to make sure your intuition is clear about-- this is equal to minus the integral of j-- no, of divergence of j d cube x over the volume. OK. But then Gauss's law. Gauss's Lot tells you that you can relate this divergence to a surface integral. dq dt is therefore minus the surface integral, the area of the current times that. So I'll write it as minus jda, the flux of the current, over the surface that bounds-- this is the volume, and there's the surface bounding it. So by the divergence theorem, it becomes this. And this is how you understand current conservation. You say, look, charge is never created or destroyed. So if you see the charge inside the volume changing, it's because there's some current escaping through the surface. So that's the physical interpretation of that differential equation, of that d rho dt plus divergence of j is equal to 0. This is current conservation. And many people look at this equation and say, what? Current conservation? I don't see anything. But when you look at this equation, you say, oh, yes. The charge changes only because it escapes the volume, not created nor destroyed. So the same thing happens for the probability. Now, let me close up with this statement in one dimension, which is the one you care, at this moment. And on the line, you would have points a and b. And you would say the probability to be within a and b is the integral from a to b dx of rho. That's your probability. That's the integral of psi squared from a to b. Now, what is the time derivative of it? dp ab dt would be integral from a to b dx of d rho dt. But again, for that case, d rho dt is minus dj dx. So this is minus dx dj dx between b and a. And what is that? Well, you get the j at the boundary. So this is minus j at x equals bt minus j at x equal a, t. So simplifying it, you get dp ab dt is equal to minus j at x equals bt plus j of a, t. Let's see if that makes sense. You have been looking for the particle and decided to look at this range from a to b. That's the probability to find it there. We learned already that the total probability to find it anywhere is going to be 1, and that's going to be conserved, and it's going to be no problems. But now let's just ask given what happens to this probability in time. Well, it could change, because the wave function could go up and down. Maybe the wave function was big here and a little later is small so there's less probability to find it here. But now you have another physical variable to help you understand it, and that's the current. That formula we found there for j of x and t in the upper blackboard box formula is a current that can be computed. And here you see if the probability to find the particle in this region changes, it's because some current must be escaping from the edges. And let's see if the formula gives it right. Well, we're assuming quantities are positive if they have plus components in the direction of x. So this current is the current component in the x direction. And it should not be lost-- maybe I didn't quite say it-- that if you are dealing with a divergence of j, this is dj x dx plus dj y dy plus dj z dz. And in the case of one dimension, you will have those, and you get this equation. So it's certainly the reduction. But here you see indeed, if the currents are positive-- if the current at b is positive, there is a current going out. So that tends to reduce the probability. That's why the sign came out with a minus. On the other hand, if there is a current in a, that tends to send in probability, and that's why it increases it here. So the difference between these two currents determines whether the probability here increases or decreases.
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https://ocw.mit.edu/courses/7-016-introductory-biology-fall-2018/7.016-fall-2018.zip
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PROFESSOR: So what are we going to do today? So, today we're going to continue with amino acids, peptides, and proteins. And I want to talk about a different protein variant that is the causative, the cause of sickle cell anemia. And it's a very interesting structural issue. But let me very briefly recap what we did last time and then talk to you a little bit about a process known as denaturation. So last time, we discussed how the primary sequence of a polypeptide chain defines its folded structure. The folded structure is put in place with secondary and tertiary interactions, non-covalent interactions. Secondary just amongst backbone and its tertiary sort of everything else, even including backbone amides, but either with water, or a side chain, and so on. And then there are some proteins that dissociate into quaternary structure. So these monomer subunits, as they would be called-- and I'm going to depict this as a closed circle or an open circle-- may form dimers of some kind. The dimers may be heterodimers. Or they may be homodimers. Or you could form trimers, tetramers, and so on. And when we talk about hemoglobin, which is the protein that gets, that has a problem-- that is the cause of sickle cell anemia, you'll see that that is a heterotetrameric protein. So in this sort of rendition, you would kind of draw it like this where there are four subunits. Two are of one flavor and two are of the other. And that's the quaternary structure of hemoglobin. Now proteins fold. There are weak forces that are holding them together. But there's a lot of weak forces. But if you subject a protein to various treatments that may break up those weak forces, the protein will undergo a process of denaturation. So can anyone think of what kinds of things would cause protein DNA denaturation? Yes. AUDIENCE: Some heat. PROFESSOR: Heat is a bad one, is a serious one, obviously. And heat-- yes, I'll write them all down. What's yours? AUDIENCE: pH. PROFESSOR: pH. So pH. Acidity. Basicity. And we'll talk about why those things cause changes. Any other thoughts? Yes? AUDIENCE: [INAUDIBLE] PROFESSOR: Oh. Yeah. So for example, salt. Organic solvents. And a process that a lot of people don't necessarily think about, but as engineers some of you will, is shear forces. So if you're shooting a protein through a very narrow tubing and there's high shear forces, those who will also denature nature proteins. So with heat, it's very clear. You're going to break those weak bonds. And then they can either reform. Or if you go to too high heat, the unfolded protein starts to form aggregates. And anyone who has ever scrambled an egg knows that that is an irreversible process. You don't get to cram the egg back into the shell. It's not the same anymore. Because what you're doing when you're scrambling eggs is denaturing proteins through heat treatment. So that's what heat does. It breaks the forces. The proteins stretch out into their denatured state. And instead of refolding to a compact structure, they just start aggregating with each other. And that's pretty much irreversible. pH is interesting. Why would pH break up at low temperature? Why would pH cause changes? Yeah. AUDIENCE: [INAUDIBLE] amino acids have a certain structure. So they're either protonated or deprotonated, then the pH, that would change. PROFESSOR: OK. So pH, perfect. So pH will change the charge states of many of your sight chains. And once you've changed it, you might have had a lovely electrostatic interaction. But then you go and protonate the carboxylic acid. And it can't form-- in fact, it wants the form, it wants to break apart as opposed to come together. So that is changing charged state, which causes denaturation. Salts and organics. For example, they may make interactions with parts of the protein. For example, organics, organic molecules may slip into a hydrophobic core and break them up. Just push them apart. They want to be there. And then too much of a high concentration of an organic solvent that is miserable with water. And we would say ethanol, acetonitrile, DMSO. But you don't need to worry about too much of which details. Well actually, once you get above 10% or so, we'll just start denaturing proteins, sometimes reversibly but often irreversibly. So this is very important to know that proteins are stable, but you've got to treat them nicely. There are some human diseases that are a result of misfolded or aggregated proteins. So for example, all the prion diseases are proteins gone bad, pretty much, where they are not in a folded structure anymore, but they are in aggregates that cause problems with cellular processes and toxicity. So Alzheimer's disease. Mad cow disease. A lot of those are neurologic disorders caused by poorly folded or very misfolded proteins, for example. So these are the things we talked about last time with respect to the flux from primary to secondary, to tertiary to quaternary. And that's a perfect time for me to introduce to you what we'll talk about today. So last time we talked about structural proteins. And I showed you how collagen, just with a simple defect, changing a glycine an alanine in one of its subunits, really alters the quaternary structure of the protein to make very weak collagen that's no longer supportive of bone strength. But what I'm going to talk to you about today is a defect in a transport protein that carries oxygen around the body. So we're going to talk about hemoglobin. These diseases are what are known as inborn errors of metabolism, or that's kind of a complex term. Or genetically linked diseases, because there is a single defect in a DNA strand that then gets transcribed into an RNA strand. So one base defect that then becomes an amino acid defect in your protein strand. So these are tiny changes in the protein that cause dramatic changes in the structure and function of the protein. And what you will see with hemoglobin is it causes a real problem with the quaternary structure and causes proteins to aggregate. So hemoglobin is the dominant protein in red blood cells. Or erythrocytes. And in fact, the differentiation of the red blood cell as it comes from progenitor cells goes through a process where the red blood cell dumps out its nucleus so it can't divide anymore. And basically, the content of the cell is extremely high in hemoglobin. You've packed the hemoglobin into the red blood cell at the cost of losing the nucleus. So that's terminally differentiated. Can't become a red blood cell. It can't divide anymore. And it has about a half-life, they have about a half-life of 100 days. So they turn over, and then that's it. And when red blood cells turn over, the hemoglobin has to be taken care of in order that it's not toxic. Red blood cells are red because of a particular molecule that's in the hemoglobin called the heme molecule, which is bound to iron, which provides the hemoglobin with the capacity to pick up oxygen in your lungs, travel it around the body, and then leave it where it's needed. And then replace the oxygen with CO2 and take the CO2 back to the lungs in order for you to respire it out. OK? So hemoglobin carries oxygen and CO2, from oxygen from the lungs, CO2 back to the lungs. And the reason why you need the iron is that the iron is coordinated to the oxygen. So the heme molecule-- I won't draw it. If you want to see it, it's a big, complex organic structure. Very interesting structure. But something for another day here. But I want to just stress to you that the iron heme complex is red. That's why your blood cells are red. Your blood cells don't have a nucleus so they can cram in lots more hemoglobin. So it's kind of a fascinating situation. So hemoglobin is an example of a homotetrameric protein. And it has four subunits. Two of one flavor and two of another. So we call this an alpha 2 beta 2 protein, differentiating the alpha subunits and the beta ones. Yes. AUDIENCE: Why isn't it homotetrameric? PROFESSOR: Why isn't it homotetrameric? AUDIENCE: [INAUDIBLE] PROFESSOR: You could ask why is it? I don't know. I mean, there will be interactions amongst the subunits that favor that particular packaging. The subunits are kind of similar in shape. They have what's called a globin fold. You can more or less pick out those tubes, remember, alpha helices. They could form tetramers that are all the same, but the energetically favored form is the two and two. Hemoglobin is a tetrameric protein, because that's really advantageous for picking up oxygen and dropping off oxygen in a very narrow oxygen range. So there are proteins called globins that just one of these that can bind oxygen. Hemoglobin is tetrameric because it has a cooperative oxygen binding. So in a very narrow range of oxygen, it fills all four sites in the tetrameric protein with an oxygen Molecule. So it's very advantageous from a physics perspective that it responds to very narrow changes in oxygen. Does that make sense to everyone? Yeah. AUDIENCE: [INAUDIBLE] PROFESSOR: OK. It means, anything that's cooperative means that one, let's say I've got a tetramer of hemoglobin. One oxygen binds to one of them. So I'm a binding oxygen here. And then binding to the next, the next, and the next gets easier and easier. So they sort of want to come in as a team. And that's handy for maximizing oxygen transport around the body in a narrow oxygen range, which we can only deal with what's out there in the atmosphere, so we have to make this work. Does that answer your question? OK. All right. So where was I? OK. So what we're going to do today. We're going to look at hemoglobin. It's the tetramer. Those discoid structures are the hemes that I just mentioned. I've drawn them as this sort of four-leafed clover here just for simplicity. And there is a single defect in the sequence of the single monomer subunits in hemoglobin. So each of these-- let's go here. So there are four proteins-- beta globin, two copies of beta globin, and two copies of alpha globin. They are all-- let me see. What's the size? Do, do, do, do. [INAUDIBLE] You know, I can never see things when I'm up at the screen. But they're about 150. 156. OK. So they're about 146 amino acids long in each of them. And a single defect in the beta globin where you have a change from glutamic acid residue 6 to valine at residues 6-- one change in beta globin, which means two changes in the whole structure, because there are two beta globins-- alters the properties of the hemoglobin and causes what's called sickling of your red blood cells. So let's take a look at what that would look like at the amino acid level. Glutamic acid is one of your charged amino acids. I'm just going to draw a little bit of it as it were in a peptide. And it's at position 6 in the sequence. So it's six residues from the amino terminus because we always write things in this direction. And the change takes place to put in place a valine. And there's a pretty big change in identity and personality of those residues. You've gone from polar charged, to neutral, big, fluffy, hydrophobic residue. And it's really amazing. So the beta globin is expressed on chromosome 11. It's 134 million base pairs. One base has changed. So what you have in the DNA, in the normal DNA that encodes the normal beta globin gene, there's a particular sequence of nucleic acids. This is what the double strand would look like. We're going to see way more about nucleic acids next week. When that gets converted to the messenger RNA, you get a particular code that in the genetic code codes for glutamate acids. Everything's normal. A single change, if we change the center nucleic acid within the DNA, it makes a different messenger RNA. And one base pair puts in valine instead of glutamic acid out of 134 million base pairs. So what happens in the normal hemoglobin, you have normal behavior. You had this tetrameric structure. It cooperatively carries oxygen. It moves around the blood no problem. Excuse me, it sits in the erythrocytes or red blood cells no problem. The minute you have that mutation, the hemoglobin molecules start to associate into clusters like fibrillar clusters, because each tetramer gets glued to another tetramer, and another one, and another one. So you have hemoglobin not behaving as this beautiful, independent quaternary structure, but rather sticking to, physically sticking to other Molecules And those tangles get, those molecules get so large that they start to form long and inflexible chains. And it's such a dramatic change that that discoid structure that you're familiar with for red blood cells suddenly becomes a sickle shape. So that would be the normal cell with normal hemoglobin. But sickle cell, they look like this. They're kind of curved, odd, a very odd shape. And the problem is red blood cells have evolved to move really smoothly through your capillaries. As soon as you get a different shape that's sort of not that discoid structure, they start clogging in the capillaries. And when you have the defect where all of your hemoglobin is messed up with this variation, it's incredibly painful, because think of all your capillaries going out to the farther reaches of your joints. Those very thin blood vessels are blocked up with the sickle red blood cells that are caused by the variation in hemoglobin. So that one little defect takes us all the way to a serious disease. All right? So what I want to do very briefly is show you the molecular basis for this. All right. And the defect actually appears on the two beta globin chains, but right on the outside of the protein, not in the middle of the protein. Because this is a defect that affects how proteins interact with other proteins, not the function of the protein on its own. Probably still carries oxygen just fine. But it's the mechanical change in the hemoglobin that causes the disease. OK. So sickle cell anemia, the hemoglobin is now called hemoglobin S with that mutation that I just described. And when people are heterozygous, it means they have one good copy of the gene that's normal and the copy of the gene that's the variant. And you'll learn much more about this in human genetics when we talk about that later on. So you have a mixture of the OK hemoglobin and the sickle cell hemoglobin. People who are homozygous for the defect, all of their hemoglobin is disrupted, and those are the people who really end up in hospital with a lot of transfusions, and so on. The heterozygous, actually, you can manage quite well with. And I'm going to show you in a minute that in some parts of the world, being heterozygous-- i.e., having some of your hemoglobin with a defect and some without it-- actually confers an advantage. It's a really cool story. So what I'm going to do is quickly show you the wire structure. OK, so this is the structure that elucidated the real reason for the interaction. What happens when you have this mutation. And it was a structure that was captured of a dimer of hemoglobin molecules where you could really see what was happening at the interface and the sorts of changes that had been put in place by that variation from the charged to the neutral structure. So for any of you who want to pop by, I can start to show you how to manipulate PyMOL. We can do that separately from class. But this is a dimer of tetramers. And if I just show you just some of the subunits, I can actually show you how there's two of each subunit in each structure. So if I, go I can pick some out. Every other one. And then I can color them a different color. You can see where the globins, where the beta globin are and where the alpha globins are. That's still looks like chicken wire. It's very unsatisfactory. So what I can do is I can show you everything as a cartoon and get rid of all those little lines. And then you can see perfectly the structure where you see two beta globins and two alpha globins in each structure. OK? So what we're going to do next is zoom in to see what's happening where we've done this mutation, what's going on with the placement of the valine in that structure. All right? And wherever I put a four-letter code-- so that one was 2HBS-- that's what's known as the protein data bank code, and it enables you to go fetch the coordinates of that protein. So if any of you for the late project want to do a protein structure and print it, come to me and I'll explain a lot more about that. Or the TAs can also do that. So let me now move you to looking in closely to the variations. So what I've done here is I've actually colored-- the beta globin is purple, and the alpha globin is cyan colored. You can see the hemes in each of the subunits. Those are those red wire things. And now we've zoomed into the place where the mutation is where you have a valine instead of carboxylic acid. And what you can see from this image which should stop is that the valine on one subunit in one homotetramer interacts with a sticky patch on another subunit that's made up of phenylalanine 85 in the adjacent protein and leucine 88 in the adjacent protein. So this sticky patch on one surface glues onto a sticky patch on the surface of another tetramer. If you had glutamic glutamate there, would that form? No. In fact, it would be quite deterred from forming because you don't want to cram that negatively charged element into those two hydrophobic residues. So what you've gone from is a situation where this really is fine on the surface. It's hydrated. It's not sticking to anything. To another situation where you have phenylalanine and leucine, which are both hydrophobic, providing a patch on the one tetramer where the valine from the other tetramer combined. And because the molecule's a tetramer, on each of the subunits, there is also another valine that will go off and do that elsewhere, and another valine. And there's one you can't see that's tucked behind. So that's why the hemoglobin forms these structures, because every hemoglobin molecule has two places to stick to another hemoglobin tetramer, and so on. So think of the repercussions from one nucleic acid change that's really quite remarkable. So what we've seen here is that that change occurs. And just a couple of moments for you to think about this, you can have variations at that site that won't cause a problem. Which ones of these do you think are least likely to cause a sickle cell type of phenomenon? So tyrosine, serine, aspartic acid, and lysine? So I'm going to change the glutamate to something else. Which one's going to have a perfectly normal hemoglobin? There's one that stands out. Yeah. AUDIENCE: [INAUDIBLE] PROFESSOR: Aspartic. That's fine. No problem. It just switched it for its younger brother. Well, which one of the others? And in many cases here, you could probably argue your way to all of them. But one would be pretty bad. Which one would be pretty bad? Tyrosine, exactly. It's another. Even though it's got that OH group, it's still pretty hydrophobic because of that ring system there. What about the other two, serine and lysine? What do you think? Which one would probably be, in fact, the least detrimental of those remaining two? And give me the reason as well. Yes. AUDIENCE: Lysine. PROFESSOR: Lysine. I think it would be lysine because lysine is now positively charged. It's equally unlikely to want to do this goofy interaction because it is also charged, just charged in the other direction. But one could also argue that serine would be OK because it's a little bit more polar, so it wouldn't cause as much problem. OK. Finally, this issue with sickle cell anemia, there's some fascinating data that shows in parts of the world-- for example, during a drug trial for plasmodium falciparum, one of the causative agents of malaria, they found that 1 out of 15 people with the sickle cell trait was infected with malaria, whereas then the people who were healthy, normal homozygotes for the right hemoglobin, 14 out of 15 were infected with plasmodium falciparum. Now why do you think that is? How can we relate the infectivity of a parasite with the shape of a cell? We've gone from these juicy-looking red blood cells, nice and round and probably quite open, to a cell that's sort of difficult to shape. So it turns out that the parasite doesn't want to infect the sickle cell red blood cells anywhere near as well. And there are, for example, other bloods tested which shows the same correlation. And here's a map of Africa where you see a massive overlap of the prevalence of the sickle cell trait and the presence of plasmodium falciparum. So there is an evolutionary advantage to having the heterozygous variant where you have some normal hemoglobin but some of the sickling hemoglobin, because it confers you some resistance to malaria. It's not good to have both of them, the variant that causes sickling, because that's painful and it really causes a lot of health disorders. It's just when you have one of each gene encoding both variants. OK? All right. Great. OK. So now we're going to talk about enzymes. And these are the proteins that catalyze reactions. Any questions about that? So while a lot of disease states actually might be bred out because someone would be at a disadvantage with a particular disease, in this case, that trait has been maintained because it offers a very different advantage with respect to disease. OK. Let's talk about enzymes for a moment. Or for the rest of the class, in fact. OK. So enzymes are the heavy lifters of the protein world because they catalyze all the reactions in metabolism, in biosynthesis, all kinds of transformations that make you want you are. Enzyme is a protein-based catalyst. You all know that. Terrible writing again. There were a couple of other times I just quickly want to give you. So an enzyme, there is also a term known as an isozyme. And an allozyme. You may see them. You'll see allozyme less commonly, but you'll see isozyme quite commonly. An isozyme of one enzyme is a variation on the enzyme that catalyzes the same reaction, but it's expressed on a different gene. An allozyme is the same enzyme, but with a variation in it. So it's encoded by an allele of one gene. So it's just a variation of the gene that might have happened through a mutation. Still catalyzes the reaction, but there's a slight change in the sequence. But they're coded by the same gene. Same gene, with a variation. And as I said, you will see the isozyme term more commonly than the allozyme term. Now why do we need enzymes? Well, the problem is there are physiologic reactions that we need to carry out that are just too hard to carry out at room temperature pH 7 in water. They just don't occur. So you need enzyme catalysis for all of your metabolic reactions. Let me just give you one trivial example. This bond you already know nicely now. Peptide or amide bond. If I want to hydrolyze that, if I want to break it open, pH 7, physiologic temperature, so 37c, in water, it would take me-- how many years is it? The half-life of that bond would be 600 years. OK? That's pretty untenable for digesting a Big Mac even that even under the best of circumstances. So we need enzymes to speed up breaking down proteins and carrying out reactions because otherwise, we just can't-- we can't do anything. So what I want to describe to you are some of the details of how enzymes work and then how we can control the function of enzymes. So typical enzymes take a substrate to a product. Some enzymes may take two substrates and make one product. Some enzymes maybe take one substrate and make two products. It just depends on the transformation that you're doing. Enzymes are classified into a bunch of different families. But the thing that will tell you that something you're reading about is an enzyme is the suffix ASE at the end of the name of the enzyme. So the enzyme that hydrolyzes the peptide bond or hydrolyzes proteins is called, no big surprise, a protease. And you'll see later on ribonuclease, DNAs, oxidoreductases, all kinds of reactions where if you see this term at the end of the name it's telling you quite loud and clear that it's an enzyme. Just a very sort of simple way of remembering that. Now enzymes promote reactions in order that we can have them carried out at room temperature. But we want to think about how they carry out these changes and transformations. What is it about the structure of the protein that enables these reactions? But the first thing we have to do is take a look at the thermodynamics and kinetics of a transformation. So before I go anywhere, what I want to do is describe to you how enzymes work by thinking about the physical parameters that we describe the energetics of a transformation. So in thermodynamics, you all know delta G is delta H minus T delta S. And we're really only going to worry about one of these terms. We're going to worry about delta G, and I'll explain why. So delta G is the Gibbs free energy. H is the enthalpy T is the temperature in Kelvin. And then S is entropy. So these are the two terms when you're looking at an energy diagram, we generally think about reactions where we describe the y-coordinate as the change in delta G, the change in the free energy, and the x-coordinate is your reaction coordinate. So in going from a substrate to a product, we generally have a situation where we have a substrate at a certain energy, and then maybe a product at a different energy. And we're going to talk about the details of that. So why do we deal with Gibbs free energy, not enthalpy? Does anyone know? OK. Enthalpy describes the energies of all the bonds in a molecule. But when you're doing an enzyme-catalyzed transformation, you're not busting open all of those bonds. You're not breaking something down to carbon, hydrogen, and oxygen. You're only dealing with parts of the energetics of the molecule. You're only dealing with what's known as the free energy changes. So looking at the enthalpy changes isn't going to get you very far. It's not going to describe the reaction because the enthalpy changes would be enormous breaking down that molecule. And that's not what you want to achieve. In a chemical transformation, we care about delta G. Now the next thing to think about is what are the energetics of the reaction, and how does an enzyme-catalyzed reaction manipulate those energetics? So the key thing here is we want to talk about Gibbs free energy. I shouldn't have written quite this much stuff here because I need the Blackboard. All right. So when you describe a reaction, you want to understand how far that reaction goes and how fast that reaction goes. So when you go through a reaction, we can describe how far the reaction goes by thinking about the free energy of the substrates and the products. So in this case, the substrate is at a higher energy than the products. So you will go a long way through the reaction to make quite a lot of products in a transformation. So that describes how far the reaction goes. So that is the difference between the energy of the substrate and the product. How fast the reaction goes is described in a different part of this diagram. Does anyone know what it is? Yes. AUDIENCE: Activation rate. PROFESSOR: Yes, exactly. How fast the reaction goes is literally how high the mountain is that you have to get over to carry out the transformation. And that height is described as the energy of activation. So that tells you how fast, and the difference here tells you how far. The energy of activation is a really important parameter because it's actually what gets manipulated when you're dealing with catalyzed reactions. So the energy of activation-- the higher that mountain is, the slower the reaction will be because it's a much harder transformation to go through. The reactions in our bodies can be of different flavors depending on the difference in energy of the substrate and the product. So shown there, substrate going to product where the product is at lower energy than the substrate, we would call this an exergonic reaction because we're releasing energy in the transformation. So S higher than P. Exergonic. And if we have a different reaction-- and I'll sketch this one in here-- where the product is higher energy-- and this is a reaction coordinate-- then that will be an endergonic reaction. Both reactions happen in enzyme-catalyzed systems. And we'll explain why you're able to catalyze even ones that require energy. So exergonic releases energy. And endergonic requires. OK. What else have I got on here? We also, in the situations where energy is produced, the exergonic reactions, we call these catabolic processes. And if you have trouble remembering catabolic and anabolic, just join me in that because I always forget which is which. But the ones that produce energy are catabolic. The ones that require energy are anabolic. And when we think about metabolism, the catabolic reactions are when we're breaking molecules down because we need energy. We need to use it to do something. The anabolic reactions are when we want to store things. Store fats, build proteins, because they're going to be endergonic. They're going to be requiring energy to take place. I just forgot one thing that I have shamefully done. Remember, this axis is kilocalories per mole most commonly when we're talking about delta G, or kilojoules per mole if you're in a different part of the world. But it's important to have units on these diagrams. So that tells us a little bit about enzyme-catalyzed reactions. We need the enzyme to do something about this energy of activation. Because if we didn't have a high energy of activation and I brought a Snickers bar to eat during class, I would just burst into flames, right? It needs a high energy of activation to keep it stable under regular conditions, but only break down the bonds at times when you require that breakdown. All right. So what did the catalyst do? OK. Now I'll show you the simple reaction. The enzymes are a very large structure. It binds to a substrate, chemistry happens, and it releases a product. But at the same time, you can't disobey the principles of thermodynamics. So there are certain criteria we have to think about when we consider an enzyme-catalyzed reaction. So first of all, do not disobey whichever law of thermodynamics it is. They do not change delta G. Delta G is a property of the two reactants. You're not going to change it with a catalyst. It's going to have a much more, a more important impact on a different parameter. Which parameter do enzymes change and help lower? Over there. AUDIENCE: [INAUDIBLE] PROFESSOR: Right. So catalysts do change and in fact lower energy of activation. And we'll talk about how they do that the end. And then the last rule about a catalyst is you can recover them unchanged after a reaction. It would be a lousy catalyst if it did its chemistry and then you've used up the catalyst. So enzyme catalysis are the ultimate green reagents. You can keep using them thousands and thousands of times to continuously turnover transformation. So you haven't changed a catalyst. So the things that we want to think about is how-- what are the processes that enzymes can manipulate? And I should probably just quickly run through these slides so we've talked about these entities. But I put them on the board because they're particularly important. So the energy of activation of a catalyzed reaction is lower than the uncatalyzed. And I'm not going to bore you with these questions because you can work this out quite readily. So delta G is the free energy that changes. And these are endergonic because the energy of the products is lower. So this is the slide I want to get to with respect to the enzyme-- to enzyme catalysis. So we always think, well, gosh, the enzyme is really large relative to the size of the product. That's because all the energy within the protein-folded structure is very useful for lowering the energy of activation of a transformation. So let's say I have a reaction that involves two substrates coming together to make a product. If I'm off the enzyme, these guys, it's going to take them a long time to bump into each other to do chemistry. The way enzymes catalyze those types of reactions is they have binding sites for both of those compounds. In fact, the enzyme acts as a stage. One substrate binds. The other substrate binds. They're binding close to each other on the enzyme. Chemistry can happen. It favors reactions that involve multiple molecules. What about another situation where you have a bond-- for example, the amide bond-- the proteases break? It's hard to think of how that-- how can we make that more easy? Well, amides are most stable when they are flat and planar through this arrangement of atoms. But what can happen on the enzyme is that they can twist bonds to make them less stable and then more easy to hydrolyze. So the structure of that enzyme basically holds onto the substrate and twists or distorts the bond that you're trying to do chemistry on to once again lower the energy of activation. Another way enzymes work is in a reaction where you're breaking this bond, you might make charged intermediates. The enzyme's there to hold those charged intermediates in order to stabilize them. Once again, to lower energy of activation. So it's funny when you get the question that's well, how do enzymes catalyze reactions? There is no one rule. You want to think about the reactions and then just think about the ways in which an enzyme could contribute to that. For example, orienting two substrates ready to do chemistry. Causing physical strain in a bond that you want to break. Or comforting electric charges that form during a reaction coordinate. So there are loads and loads of different principles, and it's a really important study that is carried out. So finally, I think I have a couple-- oh no, I have a couple of minutes. But I want to just describe this to you. It'll also be covered in the sections, because I'm going to rush it a bit because this last bit features a little bit on the P set. So finally, enzymes are very commonly the targets of drugs. We like to think that some drugs are important targets. If we deactivate the enzyme, we might mitigate the symptoms of a disease. Now you can't go in and heat the enzyme or denature the enzyme if you're trying to treat a person. So we do a lot of work to mitigate disease by inhibiting enzymes with small molecules. So in these slides, I describe to you the types of molecules that may alter the chemistry of a transformation. So if a substrate binds to an enzyme-active side-- we often do this Pac-Man rendition-- you could design a molecule that binds there instead and basically inhibits the substrate from getting there. This would be called a simple reversible inhibitor that's competitive with the active site. There are other inhibitors that will bind to the enzyme but do chemistry with it and stay blocked at the enzyme. And that would be called an irreversible competitive inhibitor. You can't get the inhibitor off. And there's differences in the way you can reverse this. Because for example, up here, if I add a lot more substrate and these are equilibria, I can get my reaction to happen any way. But here, I could add as much substrate as possible but it won't help. It won't reverse the transformation. OK? And there's a question here to restore the reaction. The answer really is, you just have to start with a new enzyme cause you covalently changed the protein structure. The last type of inhibitors that are important are the ones that bind at different sites on the enzymes. And they are called allosteric. Allo always means different. So if you have a compound that's an allosteric inhibitor, it might bind on another face of the enzyme, but it will alter the active side so it doesn't work. That's an allosteric inhibitor. And the final type of compound is an allosteric activator that may bind somewhere else on the enzyme but make it more active. So these are the way small molecules work. I'd like to encourage the TAs to just cover this in a little bit more detail because I've rushed It. And I'll also re-mention it at the beginning of the next class. But bear in mind, we should have everything covered now so the problem set 1. And if you have any questions, reach out to us. Covered them in section. And I'll reiterate a little bit of this in the next class. And finally, there's a little bit of reading. If you would like to prepare, we'll talk about carbohydrates next time, one of my favorite molecules. And there's also a fabulous set of videos on how enzymes work at the Protein Data Bank site. And you will see this little handout on the version of the slides that's posted.
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https://ocw.mit.edu/courses/8-286-the-early-universe-fall-2013/8.286-fall-2013.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK. Good morning, everybody. Let's get started. Let me just begin by asking if there are any questions, either about logistical issues or about physics issues? OK. Today we'll be finishing our discussion of black-body radiation by talking about the actual spectrum of the cosmic microwave background that we find in our universe. And then move on to talk about the rather exciting discovery in 1998 of the fact that our universe today appears to have a nonzero cosmological constant. So I want to begin by reviewing what we did last time. And one of the reasons why I do this is I think it's a good opportunity for you to ask questions that don't occur to you the first time we go through. And that, from my point of view, has been an extraordinary success. I think you've asked great questions. So we'll see what comes up today. We began the last lecture by recalling, I think from the previous lecture actually, the basic formulas for black-body radiation, which is just the radiation of massless particles at a given temperature. And we have formulas for the energy density, the pressure, the number density, and the entropy density, all of which are given in terms of two constants, little g and little g star, which is the only place where the actual nature of the matter comes in. G and g-star are both equal to 2 for photons, but these formulas allow us to talk about other kinds of black-body radiation as well, black-body radiation of other kinds of particles. As neutrinos are also effectively massless, so they contribute. And in addition, e plus e minus pairs, if the temperature gets hot enough so that the mass of the electrons becomes negligible compared to kt, also contribute to the cosmic background radiation. And if we want the higher temperatures, other particle will start to contribute. And at the highest temperatures all particles act like black-body radiation. The general formula for g and g star is that there is a factor out front that depends on whether the particle is a boson or a fermion, a particle which does or does not obey the Pauli exclusion principle. Fermions do not, bosons-- excuse me, I said that backwards. Fermions obey the Pauli exclusion principle, bosons do not. G and g star are both 1 for bosons. But for fermions there's a factor of 7/8 for g and 3/4 for g star. Yes? AUDIENCE: Would you mind quickly restating why the positron-electron pairs act like radiation above that temperature? PROFESSOR: OK. The question is, why do electron-positron pairs act like radiation at these high temperatures? And the answer is that radiation is just characterized by the fact that the particles are effectively massless. And the effective energy scale is kt, that's the average thermal energy in a thermal mix. So as long as m e c squared is small compared to kt, electrons and positrons think that they're massless and act like they're massless. And as I said, if you go to higher temperatures still, all particles will act like they're massless. Coming back to the story of g and g star, we have the factor out front which depends on whether they're bosons or fermions. And then that just multiplies the total number of particle types, whereby a particle type-- we made a complete specification of what kind of a thing it is. And that includes specifying what species of particle it is, whether it's a particle or an anti-particle if that distinction exists, and what the spin state is if the particle has spin. So we can try this out now on some examples. First example, will be neutrinos which play a very important role in the early universe, and even in the particle number balance of today's universe. Neutrinos actually have a small mass, as we talked about last time and as I'll review again this time. But nonetheless, as far as cosmology is concerned, they effectively act like massless particles although the story about why they act like massless particles is a little complicated. It's more than just saying that they're mass is small, for reasons that we'll see. But anyway, I'm nonetheless going to start by describing neutrinos as if they were massless, as was believed to be the case really until 15 years ago or so. The massless model of the neutrino was a particle which was always left-handed. And by left-handed what I mean is that for neutrinos, if you took the angular momentum of the neutrino in the direction of the momentum, p hat there means dotted with the unit vector in the direction of the spatial momentum, you'd always get minus 1/2 in units of h bar. And conversely, all new bars are right-handed which just means the same equation holds with the opposite sign. So neutrinos always have spins that oppose the direction of motion, and anti-neutrinos always have spins aligned with the direction of motion. Now, it's not obvious but, if neutrinos were massless this would be a Lorentz invariant statement. If neutrinos have a mass, that statement is obviously not learn Lorentz invariant. As you can see by imagining a neutrino coming by, and you can get into a rocket ship, chase it, and pass it, and then see it going the other way out your window because you're going faster than it. You would see the momentum in the opposite direction from the way it looked to begin with. But the spin would look like it was the same direction as it did to begin with, and therefore the spin would now be aligned with the momentum instead of opposite the momentum. So this could not possibly hold universally if the neutrino has a mass. But for the time being our neutrinos are massless. So we're going to take this as a given fact. And it certainly is a fact for all neutrinos that have ever been actually measured. Given this model of the neutrino, the g for neutrinos is 7/8 because they're fermions. Then there's a factor of 3, because there are three different species of neutrinos- electron neutrinos, muon neutrinos, tau neutrinos. Neutrinos come in particles or anti-particles which are distinct from each other, we think. So there's a factor of 2 associated with the particle anti-particle duality. And there's only one spin state. The spin that's anti-aligned with the momentum, or aligned for the anti-neutrinos. But only one spin state in either case. So just a factor of 1 from spin states, and multiplying that through we get 21/4 for g, and 9/2 for g star. Yes? AUDIENCE: If we found out that they were Majorana, that they were their own anti-particles, would that change what we expect the temperature [INAUDIBLE] to be? PROFESSOR: No, it would not. OK. The question was, if we find that they're Majorana particles-- which I'm going to be talking about in a minute-- where the particles would be their own anti-particles, which would mean that the right-handed anti-neutrino would really just be the anti-particle of the left-handed neutrino, it would not change these final numbers at all. What it would do is, instead of having the 2 for particle anti-particle, we would have a 2 for spin states. So there would still be two kinds of neutrinos, but instead of calling them the neutrino and the anti-neutrino, the right words would be right-handed neutrino and left-handed neutrino. But the product would still be the same. AUDIENCE: Wait, they have mass and they are Majorana? PROFESSOR: If they have mass and Majorana, what I just said applies. The fact that they have a mass would mean at the lowest possible temperatures they would not act like black-body radiation. Kt would have to be bigger than their mass times c squared. But that's only on the order of electron volts at most. So I'll talk later about why the true model neutrinos which have masses give the same result as this. OK. Then we can also, just as an exercise, calculate g and g star. It's more than an exercise. We like to know the results. We can calculate g and g star for e plus e minus pairs, which is relevant for when kt is large compared to the rest energy of an electron. And again, they are fermions so we get a factor of 7/8 appearing in the expression for g, and 3/4 appearing in the expression for g star. And then we just have to multiply that times the total number of types of electrons that exist. There's only one species called an electron, so we only get a factor of one in the species slot of the product. There are both electrons and anti-electrons where the anti-electrons are usually called positrons. So we get a factor of 2 in particle anti-particle. Two spin states because an electron can be spin up or spin down, and that gives us 7/2 and 3. Given that, we can go ahead and calculate what the energy density and radiation should be for the present universe given the temperature of the photons, the temperature of the cosmic microwave background. And in doing that there's an important catch which is something which is the subject of a homework problem that you'll be doing on problem set seven. When the electron-positron pairs disappear from the thermal equilibrium mix, if everything were still in thermal contact, its heat would be shared between the photons and the neutrinos in a way that would keep a common temperature. But in fact, when the e plus e minus pairs disappear, things are not in thermal contact anymore. And in particular, the neutrinos have decoupled. They're effectively not interacting with anything anymore. So the neutrinos keep their own entropy and do not absorb any entropy coming from the e plus e minus pairs. So all the entropy of the e plus e minus pairs gets transferred only to the photons. And that heats the photons relative to the neutrinos in a calculable amount, which you will calculate on the homework problem. And the answer is that the temperature of the neutrinos ends up being only 4/11 to the 1/3 power, times the temperature of the photons. And that's important for understanding what's been happening in the universe since this time. That ratio is maintained forever from that time onward. So if we want to write down the formula for energy density and radiation today it would have two terms. The 2 here is the g for the photons, and this, times that expression is the energy density in photons. The second term is the energy density in neutrinos. And it has the factor of 21/4 which was the g factor for neutrinos. But then there's also a correction factor for the temperature, because on the right hand side here I put t gamma to the fourth. So this factor corrects it to make it into t neutrino to the fourth, which is what we need there to give the right energy density for the neutrinos today. And this is just that ratio to the fourth power. And once you plug in numbers there it's 7.01 times 10 to the minus 14th joules per meter cubed, which is, from the beginning, what we said was the energy density in radiation of the universe today. OK. Finally I'd like to come back to this real story of neutrinos and their masses and why, even though they have small masses, the answers that we gave for the massless model of the neutrino are completely accurate for cosmology. We've never actually measured the mass of a neutrino. But what we have seen is that neutrinos of one species can oscillate into neutrinos of the other species. And it turns out, theoretically, that that requires them to have a mass. And by seeing how fast they oscillate you can actually measure the difference in the mass squareds between the two species. So it's still possible actually, in principle, that one of the species could have zero mass. But they can't all have zero mass because we know the differences in the squares of their masses. So in particular, delta m squared 2 1 times c to the fourth, meaning the mass expressed as an energy, is 7.5 times 10 to the minus 5 eV squared. And larger values obtained for 2 3, which is 2.3 times 10 to the minus 3 eV squared. We're still talking about fractions of an eV. The other of the three possible combinations here are just not known yet. Now, if neutrinos have a mass, that does actually change things rather dramatically because of what we said about-- the statement that the neutrinos always align their spins with their motion just cannot be true if neutrinos have a mass. And more generally, for any particle with a mass of arbitrary spin j, the statement is that, the component of j along any particular axis-- and we'll call it the z-axis-- always takes on the possible values in terms of h bar going from minus j up to j with no emissions. It's different for massless particles. For massless particles every one of these elements on the right hand side is independent and, by itself, a Lorentz invariant possibility. But, coming back to neutrinos-- if the neutrinos have a mass, in addition to the left-handed neutrinos there has to be a right-handed partner. And the question then is, what's the story behind that? And it turns out we don't know the story. But we know two possible stories. And one of the possible stories is that could be what's called a Dirac mass. And for Dirac mass what it means is that, the right-handed neutrino is simply a new type of particle which just happens to be a particle that we've never seen, but a particle which would have a perfectly real existence. It would however, to fit into theory and observation, be an extremely weakly interacting particle. The interactions of the right-handed one do not have to be the same as the interactions of the left-handed one. That is, the interactions can depend explicitly on p hat dot j. So depending its value, you could affect what the interactions are, again, in a Lorentz invariant way. And in practice, the right-handed neutrinos would indirect so weakly that we would not expect to see them in the laboratory. And we would not expect even in the early universe that they would have been produced in any significant number. So even though it would be a particle that, in principle, exists, we would not expect to see it. And we would not expected it to affect the early universe. Alternatively-- and in some ways a more subtle idea-- is that the mass of the neutrino could be what's called a Majorana mass. Where Majorana, like Dirac, is the name of a person-- perhaps less well known than Dirac, but made important contributions in this context nonetheless. In this case, it can only occur if lepton number is not conserved. And if lepton number is not conserved then there are really no quantum numbers that separate the particle that we call a neutrino from the particle that we call an anti-neutrino. And if that's the case, the particle that we call the anti-neutrino could, in fact, just be the right-handed partner of the neutrino. So for the Majorana mass case we don't need to introduce any new things that we haven't already seen. We just have to rename the thing that we've been calling the anti-neutrino the neutrino with helicity plus 1 instead of minus 1-- it's with j hat dot p with j dot p hat, equal to plus 1/2 instead of minus 1/2. So these would just be the two spin states of the neutrino instead of the neutrino and the anti-neutrino. And that's a possibility. And this would also change nothing as far as the counting that we did. It would just change where the factors go instead of having a factor of 2 for particle anti-particle and this type counting. We'd have a factor of 2 in the spin state factor, and a factor of 1 in the particle anti-particle [INAUDIBLE]. The particle and the anti-particle would be the same thing. OK. Any questions about that? OK. Finally-- and I think this is my last slide of the summary. At the end of the lecture we just pointed out a number of tidbits of information. We can calculate the temperature of the early universe at any time from the formulas that we already had on the slide. We know how to calculate the energy density at any time. And by knowing about black-body radiation we can convert that into a temperature. And for an important interval of time, which is when kt is small enough so that you don't make muon anti-muon pairs, but large enough so that electron-positron pairs act like they are massless, and this very large range kt is equal to 0.860 m e v, divided by the square root of time where time is measured in seconds. So in particular, at 1 second kt is 0.86 m e v. And it does apply at 1 second. Because 0.86 m e v is in this range. We also then talked about the implications of the conservation of entropy. If total entropy is conserved, the entropy density has to just fall off like 1 over the cube of the volume. Total entropy is conserved for almost all processes in the early universe. So the entropy falls off like 1 over a cubed. And that means that, as long as we're talking about a period of time during which little g does not change-- and little g only changes when particles freeze out, like when the electron-positron pairs disappear-- but as long as little g doesn't change, s [INAUDIBLE] 1 over a cubed means simply that the temperature falls like 1 over a. And when little g changes you can even calculate corrections to this as, effectively you're doing when you calculate this relationship between the neutrino temperature and the photon temperature. And finally, we talked about the behavior of the atoms in the universe as the universe cools. For temperatures above about 4,000 degrees the universe, which is mainly hydrogen, is mainly a hydrogen plasma. Isolated protons and electrons zipping through space independently. At about 4,000 Kelvin-- and this is a stat [? mac ?] calculation, which we're not doing-- but using the answer. At about 4,000 Kelvin-- which is a number which depends on the density of hydrogen in the universe, it's not a universal property of hydrogen-- but for the density of hydrogen in the universe, at about 4,000 Kelvin hydrogen recombines. It becomes neutral atoms. And slightly colder, at about 3,000, the degree of ionization becomes small enough so that the photons become effectively free. The photons decouple. In between 4,000 and 3,000 the hydrogen is mostly neutral, but they're still enough ionized so that the photons are still interacting. So the most important temperatures-- the 3,000 Kelvin, when the photons are released, when the photons are no longer trapped with the matter of the universe. And last time we estimated the time at which that happens. That should be a small t, sorry. The time of decoupling is about 380,000 years. And that number is actually very accurate, even though we didn't calculate it very accurately. And that's the end of my summary. Any questions about the summary? OK. In that case, let's go on to talk first about the spectrum of the cosmic background radiation. And then we'll move on to talk about the cosmological constant. CMB is cosmic microwave background. And that's a very, very standard abbreviation these days. So when the cosmic microwave background was first discovered by Penzias and Wilson in 1965-- which, I might point out, is going to have its 50th anniversary in the coming year-- they only measured it at one frequency. It was a real tour de force to measure it at the one frequency and to convince themselves that the buzz that they were hearing in their detector was not just some kind of random electrical noise, but really was some signal coming from outer space. And the main clue that it was some signal coming from outer space was that they were able to compare it with a cold load, a liquid helium-cooled source, and find that that comparison worked the way they expected. And the main reason for believing it was cosmological rather than local is that they got the same reading no matter what direction they pointed their antenna. This just took a lot of radio technique skill to convince themselves that it wasn't just some radio tube that was malfunctioning or something. They even worried that it may have been caused by pigeon droppings in their antenna, I actually read about in Weinberg's book. But they finally convinced themselves that it was real. They were still not convinced really that it was a sign for the big bang and-- you may recall, again, from reading Weinberg that there were two papers published back-to-back. The experimental paper by Penzias and Wilson, which really just described the experiment, mentioning that a possible explanation was in this other paper by Dickie, Peebles, Roll, and Wilkinson which described the theory that this was radiation that originated with the big bang. But it's all based on one point at one frequency. Shortly afterwards, I guess within the same year, Roll and Wilkinson were able to measure it at a slightly different frequency. And when I wrote my popular-level book I tabulated all of the data that was known in 1975. And this mess is the graph. This shows sort of the full range of interesting frequencies. The solid line here is the expected theoretical curve corresponding to a modern measurement of the temperature 2.726 degrees Kelvin. All of the interesting historical points are in this tiny little corner on the left, which is magnified above. The original Penzias and Wilson point is way down here at very low frequencies by the standards of radiation at 2.726 degrees Kelvin. The Roll and Wilkinson point is there. These blobs indicate error bars. The [? cyanogen ?] points that you read about in Weinberg are shown there and there. The first measurement that showed that, it didn't only go up but started to go down like black-body radiation should, was a balloon flight-- this 1971 balloon flight which produce that blob and that bound. This was an experiment by MIT's own Ray Weiss. And it was very important in the history because it was the first evidence that we weren't just seeing some straight line, but we were seeing something which did indeed rise and fall the way black-body radiation should. A later balloon flight in 1974 produced error bars that are shown by this gray area. Incredibly broad. So the bottom line that this graph was intended by me to illustrate is that, in 1975 you could believe that this was black-body radiation if you so wished. But there was not really a lot of evidence that it was black-body radiation. The situation did not get better quickly. The next significant measurement came in 1987 which was a rocket flight, which was a collaboration between a group at Berkeley and a group at Nagoya, Japan. I believe it was the Japanese group that supplied the rocket and the American group that supplied the instrumentation. And they measured the radiation at three points. I can give you the number that goes with those graphical points. I guess what I have tabulated here is the effect of temperature that those points correspond to. As you can see from the graph, those points are all well above the black-body curve. Significantly more radiation than what was expected by people who thought it should be black-body. And 0.2 up there would correspond to a temperature of 2.955 plus or minus 0.017 K. The size of the vertical bars there are the error bars that the experimenters found. And 0.3 was t equals 3.175 plus or minus 0.027 K. So these were higher temperatures then the 2.7 that fit the lower part of the spectrum. And very, very small error bars. So this data came out in 1987. And, in truth, nobody knew what to make out of it. The experimental group were well aware that this was not what people wanted them to find. And they certainly examined their data very carefully to figure at what could have conceivably gone wrong. And they were going around the country giving talks about this. And I heard one or two of them in which they described how surprised they were by the results, but emphasized that they analyzed the experiment very, very carefully and couldn't find anything wrong with it. And this was the situation for awhile. I should point out that I think this point number three is something like 16 standard deviations off of the theory. And usually when somebody makes a measurement that's three or four standard deviations off of your theory, you really start to worry. 16 standard deviations is certainly a bit extreme. Nonetheless, nobody had any good explanation for this. So, well, different people had different attitudes. There were some people who tried to construct theories that would account for this. And there were others who waited for it to go away. I'm pretty sure I was among those who waited for it to go away, and we were right. So the next important piece of data came from the first satellite dedicated to measuring the cosmic background radiation. The famous COBE Satellite-- Cosmic Background Explorer-- I guess I didn't write down the name here. Oh, it's in the title. Preliminary measurement of the cosmic microwave background spectrum by the Cosmic Background Explorer, COBE Satellite. So COBE was the first satellite dedicated to measuring the cosmic background radiation. It was launched in 1989, I guess, and released its first data in January of 1990. Back in those days there was no internet or archive. So you may or may not know that the way physics results were first announced to the world were in the form of what were called pre-prints, which were essentially xeroxed copies of the paper that were sent out to a mailing list. Typically, I think, institutions had mailing lists of maybe 100 other institutions. And every physics department had a pre-print library that people can go to and find these pre-prints. So this is the COBE pre-print. 90-01, the first pre-print from 1990. And this is the data. So it is kind of breathtaking, I think. It suddenly changed the entire field, and in some sense really change cosmology for the field. Where we only had approximate ideas of the way things worked, to suddenly having a really precise science in which precise measurements could be made, and cleared up the issue of the radiation. It wasn't just a mess like this, or a terrible fit like that, but a fantastically good fit. Really nailing the radiation as having a thermal spectrum. So the history is that John Mather presented this data at the January 1990 American Physical Society meeting, and was given a standing ovation. And he later won the Nobel Prize for this work. He was the head of the team that brought this data. He won the Nobel Prize in 2006 along with George Smoot, who was responsible for one of the other experiments on the COBE satellite. Yes? AUDIENCE: So do we know what happened with the other measurements? PROFESSOR: To tell you the truth, I don't think the other measurements ever-- the other people ever really published what they think happened. But the widespread rumor, which I imagine is true, is that they were seeing their own rocket exhaust. And there were, I think, some arguments going on between the Americans and the Japanese, with the Americans more or less accusing the Japanese of not really telling them how the rocket was set up. Yes? AUDIENCE: Are the error bars plugged on those points, or is it just that good? PROFESSOR: Those are the error bars? AUDIENCE: OK. PROFESSOR: And even more spectacular, a couple years later, I guess it was-- this was actually just based on nine minutes of data or something like that. But a couple years later they published their full data set, where the size of the error bars were reduced by a factor of 10. And still a perfect fit. They didn't even know how to plot it, so I think they plotted the same graph, and said the error bars are a factor of 10 smalled than what's shown. It was gorgeous. So I think I forgot to tell you what the spectrum is supposed to look like exactly. And this is just a formula that I want you to understand the meaning of, but not the derivation of. We-- as with the other stat mech results that we're relying on, we're going to relegate their derivation to the stat mech course that you either have taken or will take. But the spectrum is completely determined because the principle of thermal equilibrium is sort of absolute in statistical mechanics. And in order for a black-body radiating object to be in thermal equilibrium with an environment at that temperature, it has to have not only the right emission rate but also the right spectrum. If the spectrum weren't right you could imagine putting in filters that would trap in some frequencies and let out others. And then you would move away from thermal equilibrium if the spectrum were right or wrong because you'd be trapping in more radiation-- you could arrange for the filters to trap in more radiation than they are letting out. So the spectrum is calculable. And in terms of-- I guess this is energy density. I have to admit, I usually call energy density u and in these notes here it's called rho. We'll figure out the units after I write it down and make sure that it is energy density. Rho sub nu of nu d nu, means-- with this product it means the total energy density, energy per volume, per frequency interval, d nu-- well, it's times d nu, so if you multiply by times nu, this is the total energy for frequencies between nu and nu plus d nu. And the formula is 16 pi squared h bar nu cubed, divided by c cubed times 1 over e to the 2 pi h bar nu over kt, minus 1 d nu. OK. And actually, the unit's not that transparent. I believe this is energy density and not mass density. But maybe I'll make sure of that and let you know next time. And this is what produces that curve that you saw on the slides. I've included the subscript nu here to indicate that it's the number which, when you multiply it by d nu, gives you the energy density between nu and nu plus d nu. If instead you wanted to know the energy density between lambda and lambda plus d lambda, there'd be a kinematic factor that you'd have to put in here-- the factor that relates d lambda to d nu. And you could imagine working that out. I might add that, in Weinberg's book, he actually plots both sub lambda of nu. So his curve looks somewhat different than the curves that I showed you. This is not exactly the same thing. Now, what this extremely accurately black-body curve proves is that the early universe really was very accurately in thermal equilibrium. And that can only happen if the early universe was very dense. And of course, our model of the universe goes back to infinite density. So the model predicts that it should be in thermal equilibrium. But in particular, the numbers that we have here, if you ask how much could you change the model and still expect these curves the answer is roughly that, all of the important energy-releasing processes have to have happened before about one year after the Big Bang. Anything that happened after one year would still show up as some glitch in the black-body spectrum. So the big bag model really is confirmed back to about one year on the basis of this precise measurement of the spectrum of the cosmic background radiation. And the COBE measurement is still, by the way, the best measurement of the spectrum. We've had other very important experiments, that we'll talk about later, which measure the non-uniformity of the black-body radiation. Which is very small, but nonetheless very, very important [? effect. ?] So we've had WMAP and now Planck which have been dedicated to measuring the anisotropies of the radiation. COBE also made initial measurements of the anisotropies. And we'll be talking about anisotropies later in the course. Yes? AUDIENCE: [INAUDIBLE]. PROFESSOR: Sorry? AUDIENCE: The units of the right-hand side are energy density. PROFESSOR: Energy density. OK. Thanks. OK. Good. So my words were right. I should have called it u, I think, to be consistent with my usual notation. Thanks. OK. Any other questions about the CMB? Because if not, we're going to change gears completely and start talking about one of the other crucially important observational discoveries in cosmology in the last 20 years. OK. So what I want to talk about next is the very important discovery originally made in 1998-- also resulting in a Nobel Prize-- that the universe is accelerating. And this was a discovery that involved two experimental groups, and a total of something like 52 astronomers between the two groups. Which actually meant that-- I'm exaggerating slightly, I suppose. But it really involved the majority of the astronomers of the world, and therefore there weren't a lot of astronomers to argue with them about whether or not the result was right. But there still was some argument. The announcement was initially made at a AAS meeting in January of 1998 by-- which group was first? I think that was the High-Z Supernova-- where are they? Yeah. That was the High-Z Supernova Search Team. And then there was also a group largely based at Berkeley. The High-Z Supernova Search Team was actually fairly diffused, although based to some extent at Harvard. And the Supernova Cosmology Project was based rather squarely in Berkeley, headed by Saul Perlmutter. And they both agreed. And what they found was, by looking at distant supernovae of a particular type-- type 1a-- they were able to use these supernovae as standard candles. And because supernovae are brighter than other standard candles that had been studied earlier in history, they were able to go out to much greater distances. And that means to look much further back in time than previous studies. And what they discovered was that the expansion rate of the universe today was actually faster, and not slower, than the expansion rate about five billion years ago. And that was a big shock because until then everybody expected that gravity would be slowing down the expansion of the universe. And when these guys started to make these measurements they were just simply trying to figure out how fast the universe was slowing down. And they were shocked to discover that it was not slowing down, but instead speeding up. Initially there was some controversy about it. People did try to invent other explanations for this data. But the data has, in fact, held up for the period from 1998 to the present. And in fact, it has been strongly supported by evidence from these anisotropies in the cosmic microwave background radiation, which we'll be talking about later. But it turns out, you can get a lot of information from these anisotropies in the cosmic background radiation. So the picture now is really quite secure, that the acceleration-- the expansion of the universe is actually accelerating, and not decelerating. And the simplest explanation for that, which is the one that-- well, certainly because it's the most plausible, and the one that most of us take seriously, and it's the only one that fits the data extraordinarily well. So we've not seen any reason not to use this explanation. The simplest explanation is that there's a nonzero energy density to the vacuum, which is also what Einstein called the cosmological constant. So we should begin by writing down the equations that describe this issue. So we've learned how to write down the second order Friedmann equation, which describes how the scale factor of the universe accelerates. And on the right-hand side, once we included materials with nonzero pressures, we discover that we need on the right-hand side, rho plus 3 p, over c squared-- excuse me-- times a. Now when the cosmological constant was born, was when Einstein first turned his theory of general relativity to cosmology. Einstein invented the theory of general relativity in 1916. And just one year later, in 1917, he was applying it to the universe as a whole to see if he could get a cosmological model consistent with general relativity. Einstein at that point was under the misconception that the universe was static, as Newton had also thought, and as far as we know, as everybody between Newton and Einstein thought. If you look up at the stars, the universe looks pretty static. And people took this very seriously. In hindsight, it's a little hard to know why they took it so seriously, but they did. So when Einstein discovered this equation he was assuming that the universe consisted of basically non-relativistic stuff. Stars are essentially non-relativistic hunks of matter. So he thought that rho would be positive, the effective pressure would be zero. And he immediately noticed that this equation would imply that the scale factor would have a negative acceleration. So that if you tried to set up a static universe it would instantly collapse. And as we talked about earlier, Newton had talked himself out of that conclusion. And I think the real difference, as I think we also talked about earlier, was that Newton was thinking of the law of gravity as an action at a distance, where you determine the total force on something by integrating the forces caused by all other masses. And then things get complicated and divergent, actually, for an infinite, static universe. And Newton managed to convince himself that you could have a static universe of that type, a statement that we now consider to be incorrect even in the context of Newtonian mechanics. But this fact that it's incorrect even in the context of Newtonian mechanics was really not discovered until Einstein wrote down this equation. And then Einstein himself also gave a Newtonian argument showing that, at least with a modern interpretation of Newtonian mechanics. It doesn't work in Newtonian gravity either to have a static universe. But Einstein was still convinced that the universe was static. And he realized that he could modify his field equations-- the equations that we have not written down in this course, the equations that describe how matter create gravitational fields-- by adding a new term with a new coefficient in front of it which he called lambda. And this extra term, lambda, could produce a kind of a universal gravitational repulsion. And he realised he had to adjust the constant to be just right to balance the amount of matter in the universe. But he didn't let that bother him. And if he adjusted it to be just right, and the universe was perfectly homogeneous, he could arrange for it to balance the standard force of gravity. We can understand what lambda does to the equations because it does, in fact, have a simple description in terms of things that we have discussed and do understand. That is, you could think of lambda as simply corresponding to a vacuum energy density. Einstein did not make that connection. And not being an historian of science, I can speculate as much as I want. So my speculation is that, the reason this did not occur to Einstein is that Einstein was a fully classical physicist who was not at this time or maybe never accepting the notions of quantum theory. And in any case, quantum field theory was still far in the future. So in classical physics the vacuum is just plain empty. And if the vacuum is just plain empty it shouldn't have any energy density. The quantum field theory picture of the vacuum, however, is vastly more complex. So to a modern quantum field theory-oriented theoretical physicist the vacuum has particle, anti-particle pairs appearing and disappearing all the time. We are now convinced that there's also this Higgs field that has even a nonzero mean value in the vacuum. So the vacuum is a very complicated state which, if anything characterizes it, it's simply the state of lowest possible energy density. But because of, basically, the uncertainty principles of quantum mechanics, the lowest possible energy density does not mean that all the fields are just zero and stay zero. They're constantly fluctuating as they must according to the uncertainty principle, which applies to fields as well particles. So we have no reason anymore to expect the energy density of the vacuum to be zero. So from a modern perspective it's very natural to simply equate the idea of the cosmological constant to the idea of a nonzero vacuum energy density. And there are some unit differences-- just constants related to the historical way that Einstein added this term his equations. So the energy density of the vacuum-- which is also the mass density of the vacuum times c squared-- is equal to Einstein's lambda times c to the fourth, over 8 pi G. And this is really just an historical accident that it's defined this way. But this is the way Einstein defined lambda. Now, if the vacuum has an energy density, as the universe expands the space is still filled with vacuum. At least, if it was filled with vacuum. If it was matter it would thin out. But we can imagine a region of space that was just vacuum, and as it expands it would have to just stay vacuum. What else could it become? And that means that we know that, for a vacuum, rho dot should equal zero. Now we've also learned earlier, by applying conservation of energy to the expanding universe, that rho dot in an expanding universe, is equal to minus 3 a dot over a. Or we could write this as h times rho plus p over c squared. This is basically a rewriting of d u equals minus p d v, applying it to the expanding universe. So I won't re-derive it. We already derived it. Actually, I think you derived it on the homework, was the way it actually worked. But in any case, this immediately tells us that if rho dot is going to be 0 for vacuum energy, this has to be 0. And therefore p vacuum has to be equal to minus rho vacuum times c squared. And if we know the energy density in the pressure of this stuff called vacuum, that's all we need to know to put it into the Friedmann equations and find out how things behave. Otherwise this vacuum energy behaves no differently from anything else. It just has a particular relationship between the pressure and the energy density, with a very peculiar feature- that the pressure is negative. And that's an important feature because we had commented earlier that a negative pressure can drive acceleration. And now we're in a good position to see exactly how that works. To sort of keep things straight I'm going to divide the mass density of the universe into a vacuum piece and a normal piece, where normal represents matter, or radiation, or anything else, if we ever discover something else. But in fact it will just be matter or radiation for anything that we'll be doing in this course, or anything that's really done in current cosmology. And similarly, I'm going to write pressure as p vac plus p normal. "N" is for normal. But p vac I don't really need to use, because p vac I can rewrite in terms of rho vac. So in the end I can express everything just in terms of rho vac. And I can write down the second order Friedmann equation. And it's just a matter of substituting in that rho and that p into the Friedmann equation that we've already written. And we get minus 4 pi over 3 G, times rho normal plus 3 p normal, over c squared. And the vacuum pieces-- have two pieces because there's a vacuum piece there and a vacuum piece there. it can all be expressed in terms of rho vac and collected. And what you get is minus 2 rho vac times a. Showing just what we were talking about. That because of that minus sign, multiplies that minus sign, vacuum energy drives acceleration, not deceleration. And that's why vacuum energy can explain these famous results of 1998. And we'll see later that, for the same reason vacuum energy or things like vacuum energy can actually drive the expansion of the universe in the first place in what we call inflation. Yes? AUDIENCE: So for the equation without the cosmological constant it's, let's say, rho and p are about the constant, then wouldn't that be the equation for a simple harmonic function [INAUDIBLE] or the oscillation of a [INAUDIBLE] is some negative constant times a? PROFESSOR: That's right, except that you would probably not believe the equations with the bounds. AUDIENCE: OK. PROFESSOR: And when a went negative you wouldn't really have a cosmological interpretation anymore, I don't think. But it is, in fact, true that if rho and p were constants-- I'm not sure of any model that actually does that-- this would give you sinusoidal behavior during the expanding and contracting phase. Yes? AUDIENCE: [INAUDIBLE] the vacuum energy is constant over time, is it also makes sense [INAUDIBLE]? AUDIENCE: Are you asking, does it make sense for maybe the vacuum energy to change with time? I think, if it changed with time, you wouldn't call it vacuum energy. Because the vacuum is more or less defined as the lowest possible energy state allowed by the laws of physics. And the laws of physics, as far as we know, do not change with time. It's certainly true that, in a completely different context, you might imagine the laws of physics might change with time. And then thing would get more complicated. But that would really take you somewhat outside the sphere of physics as we know it. You could always explore things like that, and it may turn out to be right. But at least within the context of physics as we currently envision it, vacuum energies are constant, pretty much by definition. Now I should maybe qualify that within the context of what we understand, there may, in fact, be multiple vacua. For example, if you have a field theory one can have a potential energy function for one or more fields. And that potential energy function could have more than one local minimum. And then any one of those local minima is effectively a vacuum. And that could very likely be the situation that describes the real world. And then you could tunnel from one vacuum to another, changing the vacuum energy. But that would not be a smooth evolution. That would be a sudden tunneling. OK. So this is what happens to the second order Friedmann equation. It is also very useful to look at the first order Friedmann equation, which is a dot over a squared, 8 pi over 3 G. And in its native way of being written we would just have 8 pi over 3 G rho, minus k over-- kc squared over a squared. And all I want to do now is replace rho by rho vac plus rho n. And this is a first order Friedmann equation. And we can expand rho n if we want more details, as rho matter plus rho radiation. And rho matter, we know, varies with time proportional to 1 over a cubed. Rho radiation behaves with time as 1 over a to the fourth. So all of the terms here, except for rho vac, fall off as a grows. And that implies that if you're not somehow turned around firsts, which you can be-- you could have a closed universe that collapses before vacuum energy can take over. But as the universe gets larger, if it doesn't turn around, eventually rho vac will win. It will become larger than anything else because everything else is just getting smaller and smaller. And once that starts to happen everything else will get smaller and smaller, faster and faster, because a will start to grow exponentially. If rho vac dominates-- which it will, as I said, unless the universe re-collapses first-- so for a large class of solutions rho vac will dominate-- then you can solve that equation. And you have h, which is a dot over a, approaches, as a goes to infinity, the square root of 8 pi over 3 G rho vac. So h will approach a fixed value for a universe which is ultimately dominated by rho vacuum. And if a dot over a is a constant, that means that a grows exponentially. So we could maybe give this a name-- h vac. The value h has when it's completely dominated by the vacuum energy. And then we can write that a of t is ultimately going to be proportional to e to the h vac times t. Which is what you get when you solve the equation, a dot over a equals this constant. OK. Now one thing which you can see very quickly-- let's see how far I should plan to get today. OK. I'll probably make one qualitative argument and then start a calculation that won't get very far. I will continue next time. One qualitative point which you can see from just glancing at these equations is that the cosmological constant, when added to the other ingredients that we've already put into our model universe, will have the effect of increasing the age of the universe for a given value of h. And that's something that we said earlier in the course, we're looking forward to. Because the model of the universe that we're been constructing so far have always turned out to be too young for the measured value of h. That is, the oldest stars look like they're older than the universe. And that's not good. So we'd like to make universe look older. And one of the beauties of having this vacuum energy, as far as making things fit together, is that it does make the universe older. And the easiest way to see that-- at least a way to see that-- is to imagine drawing a graph of h versus t. Hubble expansion rate versus t. And if we look at the formula for h here we see that the rho vac piece just puts in a floor as h evolves with time, instead of going to 0 as it wood in most models-- at least, as it would in open-universe model. It stops at some floor. And certainly for the models that we've been dealing with, h just decreases to some-- this is supposed to represent the present time. So this is previous models. Now as you might say, that what I'm trying to describe here is not quite a theorem if you considered closed universes where this k piece could be causing a positive-- a negative contribution to h, which is then decreasing with time. Things can get complicated. But for the models that we've been considering which are nearly flat, that k piece is absent. And then we just have pieces that go like, 1 over a cubed, 1 over a to the fourth, and constant. All of which are positive. Then in the absence of vacuum energy we would have h falling. And with the presence of vacuum energy it would not fall as fast because we have this constant piece that would not be decreasing. So this is previous models. This is with rho vac. And I'm always talking about positive rho vac because that is what our universe has. So this would be the two different behaviors of h for the model without vacuum energy and the model with vacuum energy. And if we're trying to calculate the age of the universe we would basically be extrapolating this curve back to the point where h was infinite, at the big bang. And we could see that, since this curve is always below this curve, it will take longer before it turns up and becomes infinite. So the age will always increase by adding vacuum energy. With rho vac h equals infinity is further to the left. And notice that I'm comparing two different theories, both of which are the same age today. Because that's what we're interested in. We've measured the value of h. We're trying to infer the age of the universe. OK. Maybe I'll just say a couple words about the calculation that we'll be starting with next time. We want to be able to precisely calculate things like the age of the universe, including the effect of this vacuum energy. And we'll be able to do that in a very straightforward way by using this first order Friedmann equation. We know how each term in this Friedmann equation varies with a. And we can measure the amount of matter, and the amount of radiation, and in principal the amount of curvature-- it's negligibly small-- in our current universe. And once you have those parameters you can use that equation to extrapolate, to know what h was at any time in the past. And that tells you how the derivative-- it tells you the value of a dot at any time in the past. And if you know the value of a dot at any time in the past, it's a principle just a matter of integration to figure out when a was 0. And that's the calculation that we'll begin by doing next time. And we'll be able to get an integral expression for the age of the universe for an arbitrary value. We'll, at the end, express the matter density and the radiation density as fractions of omega, fractions of the critical density. And for any value of omega matter, omega radiation, and we'll even express the curvature as an omega curvature. The effective fraction of the critical density that this term represents. And in terms of those different omegas, we'll be able to write down an integral for the total age of the universe. And that really is going to be state of the art. That is what the Planck team uses when they're analyzing their data to try to understand what the age of universe is according to the measurements that they're making. So we will finally come up to the present as far as the actual understanding of cosmology by the experts. So that's all for today. see you all next Tuesday.
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https://ocw.mit.edu/courses/5-111-principles-of-chemical-science-fall-2008/5.111-fall-2008.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: All right. Let's get started. Why doesn't everyone go ahead and take ten more seconds on the clicker question. All right, and let's see how we did. Alright, excellent job, 86% of you, that's right. What we had just done a clicker question on is discussing light as a particle and the photoelectric effect, so we're going to finish up with a few points about the photoelectric effect today. And then we're going to try a demo to see if we can convince ourselves that the kind of calculations we make work out perfectly, and we'll do a test up here about half way through class. And we'll also talk about photon momentum as another example of light behaving up as a particle. After that, we'll move on to matter as a wave, and then the Schrodinger equation, which is actually a wave equation that describes the behavior of particles by taking into account the fact that matter also has these wave-like properties. So, starting back with the photoelectric effect -- yes. STUDENT: [INAUDIBLE] class last time [INAUDIBLE] notes. PROFESSOR: Oh, sure. Can one of the TAs maybe come up and hand around anyone that didn't get notes? We have not yet perfected the art of entering and exiting this classroom yet, we're still working on that. Raise your hand if you need notes and we'll make sure we get those to you. All right, so where we left off with the photoelectric effect was when we first introduced the effect, we were talking about it in terms of frequencies. So, for example, we were talking about a threshold frequency as in a minimum frequency of light that you need in order to eject an electron from a metal surface. What Einstein then clarified for us was that we could also be talking about energies, and he described the relationship between frequency and energy that they're proportional, if you want to know the energy, you just multiply the frequency by Planck's constant. So, now we can talk about it in different terms, for example, talking about e sub i, which is the incident energy or the energy of the light that comes in, or talking about work function here, and that's just another way to say threshold energy. So, the work function's the minimum amount of energy that's required in order to eject an electron, and most of you understand this relationship here, which is a little bit cut off, but it is all the way on in your notes, and that is what you saw the clicker question on -- how you can figure out, for example, the kinetic energy of the ejected electron by looking at the difference between how much energy you put in and how much energy is required to eject that electron in the first place. So, in this class we'll be talking about energy a lot, and it's often useful to draw some sort of energy diagram to visualize the differences in energy that we're discussing. So, we do this here for the photoelectric effect, and in terms of the photoelectric effect, what we know the important point is that the incoming photon has to be equal or greater in energy then the work function of the metal. So here we have energy increasing on the y-axis, and you see this straight line at the bottom here is lower down on the graph, and that's the energy of a bound electron, so that's going to be a low stable energy. But we see if we have a free electron, as we do in this dotted line here, that's going to be a higher energy that's less stable. So, if we want to go from that stable state to that less stable state, we need to put in a certain amount of energy to our system, and that's what we define as the work function here -- that difference between the free electron and the electron bound to the metal. So, the most basic case to understand, which is what we just saw is a case where we have the incident energy coming in, and that incident energy is greater than the work function, and in that case what we see is that we have an electron that is ejected. That makes sense and it also makes sense that this little extra bit here, that's the amount of energy that we have that goes into the kinetic energy of the electrons. So, that's how we could also graph figuring out the kinetic energy. So, in the second case what we have is what happens if we have the incident energy at some amount that's less than the work function, and in this case we're showing 1/2 of the work function. So in this case, we don't have enough energy to eject an electron, so, an electron is not ejected. And that's pretty clear, too, and the question I want to pose to you is instead the third case here. So in the third case what I'm showing is that we have -- now we're not just talking about 1 photon, we're talking about 3 photons -- let's say we shoot them all at the same time at our metal, each of them having some energy that's let's say 1/2 the work function. So, just to take a little bit of an informal survey, who thinks here that we will have an electron that is ejected in this case? So a couple hands, all right. And what about who thinks that we will not have enough energy here? All right. We've got a big majority, and both are logical ways of thinking, but it turns out that the majority is correct, which is not always the case, but the electron is not ejected in this case. And the reason for this, and this is a very important point about the photoelectric effect, and the point here is that the electrons here are acting as particles, you can't just add those energies together. One individual particle is being absorbed by the metal and exciting an electron. So, having other particles around that have the same energy that you could technically add up if you were adding them up like a wave, you can't do the same thing with particles, they're all separate. So, the take-home message is whether you have three photons or 3,000,000 photons that you're shooting at your metal, if you're not at that minimum frequency or that minimum energy that you need, nothing is going to happen. So, you might ask then well what is the significance of shooting different amounts of photons at a metal? Is there any significance at all, for example, in the number of photons that are hitting the metal or being absorbed by the metal. And there is a relationship here, and that is that the number of photons absorbed by the metal are related to the number of electrons ejected from the metal. So, in this figure here what I'm actually showing is these little sunshines, which let's say are each one individual photon. So we had six photons going in, so the maximum number of electrons that we're going to have coming out is also six because the maximum scenario that we could have that would maximize the number of electrons is that each one of those photons comes in, excites an electron, ejects it from the surface of the metal. It's important to note, of course though, it's not just the number, it's really important that the energy of each one of these individual photons is, of course, greater than the work function of the metal. So, that's, in fact, it's that number of photons that we're talking about when we refer to the intensity of light, and the intensity of light is proportional to that number, because when we talk about intensity, really we're talking about the amount of energy that a stream of particles, a stream of photons, has per second. So, if we have a high intensity, we're talking about having more photons per second, and it's important to know also what that does not mean. So it does not mean that we have more energy per photon. This is a really important difference. Intensity, if we increase the intensity, we're not increasing the energy in each photon, we're just increasing the number of photons that we're shooting out of our laser, whatever our light source is. And when we talk about intensity in terms of units, we usually talk about watts, so if you change your lightbulb, usually you see the intensity in terms of watts. But in terms of SI units, which become much more useful if you're actually trying to use intensity in a problem and cancel out your units, we're just talking about joules per second is what intensity is. So at this point, you should be able to have all the background you need on the photoelectric effect to solve any type of problem that we throw at you, and you see three on this problem set, and we'll probably give you one more on your next problem set, and the reason we ask you so many questions about the photoelectric effect is because it actually is very similar to ionization energy that we'll talk about later, also problems dealing with photoelectron spectroscopy. So, we want to make sure that this is something the entire class is 100% solid on. Sometimes the questions are worded quite differently, so I just want to sum up here the different ways they could be worded. For example, if we talk about photons, of course, we also just mean light, sometimes we refer to this as electromagnetic radiation, and there's several ways that you might be asked this in a problem or that you might be asked to answer. Sometimes we might just directly tell you the energy of the photon -- that's probably the easiest scenario, because when we think about work functions those are usually reported in energy. So since that's the easiest scenario, you can probably be sure it's not going to be too frequently that you're just given the energy, right, that might be too easy. So really what we'll probably do is instead either give you the wavelength or the frequency and you'll go ahead and calculate the energy from there. In terms of talking about the electrons, I wanted to point out that in the book and other places you might see electrons referred to as photoelectrons. That's sometimes confusing for people, because it seems like okay, is it a photon or is it an electron. I just want to clarify that it is an electron. It's called this just because it's an electron that results when an electron absorbs a photon's worth of energy, so thus it's a photoelectron. And if we talk about electrons or photoelectrons, again we can describe it in terms of energy, we can talk about velocity, and from there, of course, you can figure out the energy from 1/2 m v squared, and actually we can also describe the electron in terms of wavelength. So you don't actually know this yet from this class, you'll know it by the end of the class that electrons can, in fact, have a wavelength. So once we cover it, it will then be fair game to ask these photoelectron spectroscopy or these photoelectric effect questions using the wavelength of the electron. Also to point out, a lot of times you'll see electron volts instead of joules, this is the conversion factor here just so you all have it in your notes. All right. So let's test what we, in fact, know about the photoelectric effect, and before we do that actually, we're going to calculate what we would predict, so when we do the demo it will be meaningful and we can tell whether we're successful or not. So hopefully we will be successful. And as I point this out, we now know how to do any kind of photoelectric effect problem, also this means you should be able to go back to Monday's notes where we filled in all those graphs, which were what different scientists were observing when they were measuring either the frequency or the intensity of light that was irradiating different types of metals, and also the number of electrons ejected, and the kinetic energy of those electrons ejected. You should be able to maybe print out a blank copy of those notes from the website and fill in all those graphs -- not for memorizing them, but now just understanding how the photoelectric effect works. All right. So let's do an in-class problem, and this will be done with zinc. We have a zinc plate up here, and we're going to -- in a minute I'll describe how we can probe if electrons are coming off of it. But we're going to irradiate it with two different light sources. We have a UV lamp right here, which is centered at a wavelength of 254 nanometers. And then since we have my red laser pointer, we will also try with the red laser pointer, which is centered at wavelength of 700 nanometers. So, there are a few questions that we need to answer first. So we want to see, do we expect to eject electrons off of this metal surface, or do we expect that we don't have enough energy? So that means we're going to need to figure out what is the energy per photon that's emitted by that UV light. Also, what's the energy per photon of this red laser pointer, and then it's also worth trying a calculation dealing with intensity. So let's also try calculating the numbers of photons that would be emitted by this laser pointer, if, for example, we were to use it for 60 seconds and this were a one milliwatt laser. So, let's do some of these calculations starting first with what is the energy per photon, and let's start with the UV lamp. So we know that energy is equal to Planck's constant times nu, but what we know about the lamp is its wavelength, or the light that's emitted. We know that nu is equal to c over wavelength. So we can figure out the energy of each photon emitted by our UV lamp by saying e is equal to h c over wavelength. So let's just plug in these numbers here. That means our energy is equal to 6.626 times 10 to the -34 joules times seconds. And then we have c, the speed of light, 2.998 times 10 to the 8 meters per second. And we want to divide all of that by our wavelength, and to keep our units the same we'll do meters. So that's 254 times 10 to the -9 meters. So hopefully if some of you have your calculators with you, you can confirm the answer that I got, which is that the energy is 7.82 times 10 to the -19 joules. So, remember what we're talking about here is the amount of energy that's in each photon. So if we think about the work function for zinc, and the work function for zinc is 6.9 times 10 to the -19 joules, do we expect that when we shine our UV light on the zinc, we'll be able to eject electrons? What do you think? Yes. Good. OK, anyone disagree? No, OK and that's correct, because each photon of light actually has more energy than is needed to eject an electron. So, we would expect to see electrons ejected with the UV light source. So let's now think about using instead the amount of energy per photon in that red laser pointer. So again, we know that energy is equal to h c divided by wavelength, and energy is equal to -- you have written out in your notes what the actual value for h c is, but now our wavelength is 700 times 10 to the -9 meters. And what we end up with for the energy then is 2.84 times 10 to the -19 joules. All right. So please raise your hand now if you think there'll be sufficient energy to eject electrons from the metal surface? And raise your hands if you think there won't be. OK. Good hand raising technique. Yes. In fact, there is not enough energy in a single photon to go ahead and eject an electron from this zinc surface. So our last question we ask is what's the total number of photons emitted if we give this given intensity for 60 seconds? So, keep in mind that one milliwatt is just the same as saying 1 times 10 to the -3 joules per second. So we have 1 times 10 to the -3 joules per second, and we want to multiply that by -- or cancel out how much energy we have per photon, first of all, so how much energy do we have per photon if we're talking about the red laser pointer? Right. So this value right here. So for every photon we have 2.84 times 10 to the -19 joules. We're saying let's do this for 60 seconds. So what we end up with for the number of photons in this laser beam of light is 2.1 times 10 to the 17 photons. So this gives you a little bit of an idea of just how many individual photons there are in a laser beam of light. This is a huge number of photons. So the question is does this matter? How about if we shoot this many photons? Does it make any difference at all in terms of whether we can eject an electron? No, it actually doesn't. It is an impressive number, it is very, very large, but it doesn't make a difference. So we see that we do not eject electrons in the case of the laser pointer, even if we have this intensity, even for 60 seconds -- it is still not related to the energy of an individual photon, so we won't see an effect. All right. So let's hope that we can confirm our predictions here by actually doing it, and Professor Drennen well help me out by loading up our device with electrons, and I'll explain exactly what our set up here is as she does that. So basically what we have is this zinc plate here. So that's what we want to load up with electrons, and then see if we can remove some. But that's a little bit hard, we aren't all that good at seeing electrons with our eyes, so we need to think of a way to do this. So what she's going to do is start loading up the electrons, and you see this wand here move slowly, and it takes a while to do it, start become perpendicular. The reason for that is because all of this is connected, so we're moving electrons everywhere in the system. And since we have two bars that are together like this, once they're both loaded up with electrons there's going to be negative charges that repel, so the electrons will want to get as far away as possible, and they're on their slow way to doing that, to getting as far away from each other as possible. And if we do, in fact, hit it with light to get the electrons off, it will go back to the straight up in position, or if it gets knocked hard enough it does that, too. Sometimes it's easier actually not touch it to the metal, I should have-- TA: It's hard to see if it's moving or not. PROFESSOR: So, our technology TA is also our paper TA. Darcy will hold up the yellow paper. Right, there we go, now we're making a little progress. TA: It was moving before, you just couldn't see it. PROFESSOR: So, does anyone have any questions about the set up here, does it make sense what we're looking for the bar to go back once we make some progress. This demo works wonderfully in the winter months in Boston when we will all be full of static at all times. We're still close enough to the summer that the air is not just filling us up with extra static electricity, so it's a little more challenging here. We'll try to make this happen only once. I think that's probably, if we can get one more. So, it works, I think it's just getting too much [INAUDIBLE]. Sometimes it helps to not actually hit the metal, just put it next to the -- there we go. I wonder if there's some UV light out of this new lighting set up in our classroom here. That would be a little tricky. All right. I think this -- if this sticks. Yeah, it's the pressure of the paper. I think that's good enough, we'll be able to see. If you can keep showing that, though, Darcy, we'll try different scenarios and I'll try not to put laser in your eye. Actually you can look down as well as an added precaution. OK, let's try it with that. That's enough then. So, the first thing we're going to try is with the red laser pointer, because that we are expecting not to have an effect, and that will prevent Professor Drennen from having to charge up our apparatus again. So, Darcy will look down at this moment and we will hit this with the laser pointer, and what we see is nothing is happening at all. OK, good. Control one working. So now very carefully take our UV light source -- Darcy again will divert her eyes and her skin. Let me make sure this is actually on. OK, so we've got UV light here, and let's see what we can see, and we lose electrons, if that's what's happening. And it often doesn't go all the way, because actually this device gets stuck right there. So let's charge it up again and see if we can check again. But did you see movement? Are you buying our story here? This is actually very representative of when you do research in the laboratory, you will find often things do not work quite exactly as they worked 20 minutes ago when you just checked it in your office, for example. And sometimes it's a matter of factors that you need to figure out what it is, and maybe it's that there's extra light in the room we don't know about. It might just be -- so, we did get it back to the starting position. Next time maybe we'll charge it up before class. All right. So we kind of saw what was happening here, you saw it move a little bit. They'll keep trying to get it going, but maybe we should move on with our lives here while this is happening, and we'll click it back at the end, and if we have a nice set up at any point, I'll just stop and we'll go back and we'll look at it again. Since, I think that's just not going to happen right now. So let's switch, actually, back to our notes. So, ideally what we did see was, in fact, it does have enough energy with the UV lamp, it wasn't a dramatic shift you saw because we didn't start very high and then it went to that stuck point. But luckily we had the control of the red laser pointer where nothing moved at all. So hopefully you're convinced that your predictions worked well and you are able to predict what's going on when you're looking at the photoelectric effect. So, it turns out that the photoelectric effect is not the only evidence for the fact that light has these particle-like characteristics. And one thing that Einstein put forth is he figured if well, what we're saying is that light is, in fact, a stream of particles, each one of those particles or photons must, therefore, have a momentum. And that's really neat to think about, because photons, of course, are massless particles, they have no mass, so it's neat to think about something that has no mass, but that actually does have a momentum. And the relationship that he put forth is that the momentum is equal to Planck's constant times nu divided by the speed of light, or it's often more useful for us to think about it in terms of wavelength. So, since the speed of light equals lambda nu, we can say that momentum is equal to h divided by lambda. And there was experimental evidence that came along that supported this, and this is called the Compton scattering experiment, and this was done by Arthur Compton, and basically what he did was he took x-ray light, which had some frequency, which was a very high frequency because it was x-rays, and he shot it at a stationary electron. And what he was able to observe was that the electrons scattered and now had some momentum, and that both the frequency, and therefore, the momentum of the light that he shot in, went down once it was scattered. So what he's showing here is, first of all, that the light has some momentum and when it hits an electron it can actually transfer some of that momentum to the electron. So the transfer of momentum from a photon to an electron is what was being observed, and it was seen as completely separate evidence to the photoelectric effect that, yes, in fact, light is behaving in these particle-like ways. So up to this point, before it was really established that yes, light is like a particle sometimes, there was this very strong distinction between what is light and what is matter. And the distinction was when we're talking about light, light is a wave, and when we're talking about matter, well, matter is a particle. And these behave completely separate, they don't overlap at all in terms of behavior, but then, of course, with the photoelectric effect with Compton's scattering, what we see is that, oh actually, sometimes photons behave as if they're particles. So now this relationship's beginning to get a little bit fuzzy in terms of what is the difference between how we treat light and matter. And actually, this was taken a step further by Louis de Broglie who in his PhD thesis, as part of his work as a graduate student, put forth the idea that, OK, Einstein says, and everyone agrees that, in fact, light is particle-like at times, and light, in fact, of course has a wavelength, and if it has a wavelength we're saying that it can have momentum. And what de Broglie said is well, if it's true that light, which has a wavelength can have momentum, then it must also be true that matter, which has momentum, also has a wavelength. And you can look at this in two different ways. One is that he's just re-arranged an equation here and gotten both his PhD thesis and a Nobel Prize, but I think the more representative way to think about this is the real revolutionary idea that he put forth, which is that matter can actually behave as a wave. And in terms of equations that we use, it's sometimes easier to plug in the fact, since momentum is equal to mass times velocity. We can know the wavelength of any matter -- and he's not limiting this, for example, to electrons. What de Broglie is saying we can know the wavelength of any matter at all, as long as we know its mass and it's velocity. And Einstein credited de Broglie, which is a fair statement of lifting a corner of the great veil, because really there was this fundamental misunderstanding about what the difference was between matter and light, and the reality is that they can both be like-particles and they can both show characteristics of waves. So I mentioned, however, that in terms of de Broglie's work. This was Nobel Prize worthy, absolutely, but it was also his PhD thesis. So, we can think about what would happen if we're on his thesis defense, we're on his thesis committee, we would need to think of some pretty mean, hard, nasty questions to be asking de Broglie about this theory -- that's what happens when you defend your thesis. This is necessary, it's hard to find holes in a Nobel Prize worthy idea. But let's just try maybe one of the basic questions they could ask, and they can say, all right, de Broglie, so you say that all matter, absolutely all matter has wave-like behavior. Why is it that we're never observing this, for example, why is it the table doesn't defract as we bring it through the door? Why don't we see the influence of the wave-like behavior on every day matter? So it turns out that he could have picked anything to explain this, and hopefully done out the calculation, and we'll do this ourselves. And the example we'll pick is considering, for example, a Matsuzaka fastball. So, many of you are new to the Boston area, now I still realize, and I want to let you know it's not required that you be a Red Sox fan to be at MIT. We do encourage it, however, and in general, I find you don't have to give up that old team, you can keep your old team, even if it's teams I won't name, just keep them to the side. And you can join on to the Red Sox nation on top of that, and part of being a good Red Sox fan is knowing the statistics of your team. For example, if we're talking about a pitcher, like Matsuzaka, we might want to know the speed of his average fastball. We might want to know his ERA. If you're really into it and you're at MIT, maybe you want to know the wavelength of these average fastballs. So, let's go ahead and look at that. So, if we're trying to figure out the wavelength of a Matsuzaka fastball, we need to consider the velocity first, which is 42 miles per hour. We don't usually do our chemistry calculations in miles per hour, so let's switch that to 42 meters per second, so it's -- sorry, it's 94 miles per hour. And we can use the de Broglie relationship that wavelength should be equal to h over mass times volume. And we can put up here Planck's constant -- and I want to make note that instead of writing joules per second, I actually wrote out with a joule is. A joule is a kilogram meter squared per second squared. Occasionally, you'll find you need to cancel out units, because, of course, you're always doing unit analysis as you solve your problems, and sometimes you'll need to convert joules to kilogram meters square per second squared. We divide that by the mass, so 0.12 kilograms, that's the mass of a regulation baseball for the major leagues, and the velocity of the baseball is 42 meters per second. So, we can cross out our units doing our unit analysis. The seconds cross out, the kilograms cross out, one of the meters crosses out from the top, so we're left with an answer in meters. It's always good when we're looking for a wavelength that our answer is in a unit of length, that's a good sign already. And what we find out is the wavelength of a Matsuzaka fastball is 1.1 times 10 to the -31 meters. So, this is really small, this is undetectably small. And especially when we consider it, what tends to be important is the size of wavelength in relationship to its environment. So 1.1 times 10 to the -31 meters is not, in fact, a significant number when we're comparing it, for example, to the length of a ball, or the size of the baseball field. So that would probably be de Broglie's answer for why, in fact, we're not observing the wavelength behavior of material on a day-to-day life. So, that's for Matsuzaka, and even if you don't memorize all the wavelengths for all the pitchers. I would expect, whether you're a Red Sox fan or not, you to be able to look at a list of different pitchers and their average velocity for their fastball, and tell me who has the longest or the shortest wavelength. You should all be able to know that relationship. So why don't we go to a clicker question here, and see if you can tell us this. So we have 4 different pitchers we're showing here -- they all have different strengths. It's not always how fast you throw the fastball, sometimes it's your different styles or the different ways that you decide when to throw what. So, first we have Matsuzaka at 94 miles per hour. So, click one if you think that he's going to have the longest wavelength. Tim Wakefield on the DL right now throws a lot slower, because he has that tricky knuckle ball, he doesn't need to throw as fast. Then we have Beckett who can get up 96 just on a regular old day. And Timlin who is about 91 miles per hour, one of our relievers. So, why don't you take ten seconds to do that. If you can't decide, Timlin is my favorite ever, so that would be a good back up choice if you forgot the relationship between wavelength and the relationship between speed. It looks like, in fact, people did not forget that relationship, and only 1% of you humored me. So, let's see what the correct answer is, and it is, in fact, Wakefield, right, because there's an inverse relationship between how fast a particle is going and what its wavelength is. So, in terms of wavelength, Wakefield has the largest wavelength, but in terms of being significant, we're still not even close. It's still undetectably small. Yes. STUDENT: Why doesn't wavelength go to infinity as it stops, like a standing [INAUDIBLE]. PROFESSOR: As it stops. So, let's think. I would think that it would approach inifinity, and I would need to think about it and get back to you in terms of why we don't actually hit it and see something with an infinite wavelength. I'm sure there's some upper limit as there are to most things, like if we think of wavelengths and different types of light, there is so large that you can get, but you would be approaching that level. All right. So we can switch back actually to our notes here -- oh, do we have--? OK. We're going to just try this one more time just so you can see it. It'll still likely get stuck in that spot, but we'll just show you one more time the effects of the UV light, and actually we'll throw in an extra trick here, too. We know that UV light gets absorbed by glass, so it shouldn't be able to go through the glass. So first if Professor Drennen can try it through the glass, and we see nothing's happening. Let's move the glass away. All right. [APPLAUSE] PROFESSOR: All right. Good. So we can fully believe what our calculations were now, which is a nice thing to do. Let's go back to considering the wavelengths of different objects. We considered a baseball, but let's also think about now an electron. And an electron is something where, in fact, we might be able to, if we calculate it and see how that works out, actually observe some of its wave-like properties. So, if we do this calculation for an electron, saying it moves at 10 to the 5 meters per second, then what we end up with for a wavelength is 7 times 10 to the -9 meters. A lot of times we talk about these kind of distances either in nanometers or in angstroms so we can say this is 70 angstroms. So this is, first of all, even just on an absolute scale, this is way, way larger than the wavelengths we're talking about for baseball. In addition, if we compare this to the diameter of an atom, which is on the order of somewhere between one and ten angstroms, now we're seeing that, in fact, this wavelength is significantly larger than its environment. So certainly we would expect to see that it has an effect in terms of seeing its wave-like properties. And this was experimentally validated, hopefully, even more clearly than our experiment here. And at first this was done by Davidson and Germer, and they were American scientists who tried defracting electrons from a nickel crystal. They did this in Bell Laboratories, and they found that, in fact, the electronis did defract. And G.P. Thompson showed a similar thing. What he did was he defracted electrons through a very thin gold foil, and this is a picture -- oops, that is not. OK. It is a picture from your book here showing the defraction pattern of an electron going through that gold foil. So, you can see that, in fact, it's confirmed that an electron can have both wavelength and particle-like behavior. And it turns out that Davidson and Thompson shared a Nobel Prize for this discovery of seeing the wave-like behavior of electrons. So, this is actually kind of neat to point out, because we all remember J.J. Thomson from our second lecture, and J.J. Thomson got a Nobel Prize in 1906 for showing that electrons exist in that they are particles. And it turns out that G.P. Thompson, well, that's his son, so we can actually think of this -- and I'm sure this wasn't the case, but I like to think of it as a little bit of child rebelling against the father. So, the father gets a Nobel Prize for showing that an electron is a particle, and the son says, well, what can I do to top that? I'm going to show the exact opposite. I'm going to say that an electron's a wave no matter how much my father says differently, and I'm going to get a Nobel Prize for that, and he does. But the nice part of the story is, it turns out they're both right. An electron is a particle, but an electron's also a wave. So, father and son, happy ending, they both have their Noble prizes. So, what happens now that we, in fact, know that matter is a wave? Well, this allows us to try to go back and explain some phenomena that over the years, mounting evidence was building that couldn't be explained. So, for example, when people, and we'll talk about this next class, were looking at different characteristics spectra of different atoms, what they were seeing is that it appeared to be these very discreet lines that were allowed or not allowed for the different atoms to emit, but they had no way to explain this using classical physics. And it turns out that the Schrodinger equation is an equation of motion in which you're describing a particle by describing it as a wave. So you're basically having a wave equation for a particle, and for our purposes we're talking about a very particular particle. What we're interested in is the electron. So basically describing electrons by their wave-like properties. And this is Erwin Schrodinger, and this is the equation that he put forth where we have h hat psi being equal to e psi. So, let's explain what these are. So this symbol here is actually what we call a wave function. That doesn't mean a whole lot in itself, it will mean more in about two lectures from now. But right now, what I want you to be thinking of a wave function as is just some representation of an electron. So, it's some way of describing an electron. Specifically, we'll talk more about this, it's talking about different orbitals, it's the spatial part of an orbital. But before we get to that, in terms of thinking just think, OK, this is representing my particle, this is representing my electron that's what the wave function is. This e term here is the energy, or in our case when we talk about an electron in a hydrogen atom, for example, the binding energy of that electron to the nucleus. So, e is binding energy. And h with the carrot or the hat here, well, that carrot or hat tell us it must be an operator, and this is called the Hamiltonian operator. So when you operate on the wave function, what you end up with is getting the binding energy of the electron, and the wave function back out. When we need to describe the wave function term a little bit more specifically so we can describe, for example, the position of the electron, and I just want to mention that we do have two choices if we're trying to describe this, we could use cartesian coordinates, or we could use polar coordinates where we're either talking about x y z or r theta and phi. So, I just want to point out that when you look at wave functions, we are going to be using those spherical polar coordinates, and the reason is because a very important interaction here is the interaction between the electron and the nucleus, which we want to describe the distance of in terms of r. So, you can see, it's much easier to describe that as one term, r here, instead of using both y and z. Another reason I wanted to point this out in terms of the polar coordinates that we're using, is I think they're actually flipped from what you're used to seeing in physics. Sometimes different disciplines have different conventions, which can be very confusing because the whole point of what's happening now is there's so much interplay between different disciplines, but still I think this might be one remaining one where in our case theta is that distance from z, that angle there, where phi is this distance or angle from the x-axis. So just keep it in mind that it's flipped. It turns out we won't really using it, needing to identify it on the graph so much in chemistry. We'll be using the solutions, so you shouldn't have a problem, but I wanted to point it out so it does not look too strange to you. In terms of the Schrodinger equation, we now can write it in terms of our polar coordinates here. So we have the operation on the wave function in terms of r, theta, and phi and remember this e is just our binding energy for the electron, and we get back out this wave function. So, you might ask, this looks pretty simple up here, right, just with that h hat. It turns out, we can write it out fully. It's three different second derivatives in terms of the three different parameters. It's a little bit complicated. You won't have to solve it in this class, you can wait till you get to 18.03 to start solving these types of differential equations, and hopefully, you'll all want the pleasure of actually solving the Schrodinger equation at some point. So, just keep taking chemistry, you'll already have had 18.03 by that point and you'll have the opportunity to do that. What I want to point out also is that this h hat, the Hamiltonian operator written out for the simplest case we can even imagine, which is a hydrogen atom where we only have one electron that we're dealing with, and of course, one nucleus. So you can imagine it's just going to get more and more complicated as we get to other types of atoms, and of course, molecules from there. So, we just want to appreciate that what we'll be using in this class is, in fact, the solutions to the Schrodinger equation, and just so you can be fully thankful for not having to necessarily solve these as we jump into the solutions and just knowing that they're out there and you'll get to solve it at some point, hopefully, in your careers. So, we'll pick up with that, with some of the solutions and starting to talk about energies on Friday.
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https://ocw.mit.edu/courses/8-03sc-physics-iii-vibrations-and-waves-fall-2016/8.03sc-fall-2016.zip
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YEN-JIE LEE: So to make the demo done coherently with your lecture, a lot of preparation is actually needed in advance. So first of all, you need to make a list of the demo which you would like to include in your lecture. And secondly, you need to work with our technical instructors to set up those demos so that they work nicely during the class. Usually, I come to the class, like, half hour to one hour before the lecture to make sure that everything is working as I expect. And also there were a lot of communication between me and my technical instructors to tune the parameters to what we would like to demo during the class, and to make sure that the physical phenomena can be seen clearly from the demo of our choice. And finally, you need to decide when do you want to do those demos. And what you are going to get is that, OK, omega, omega D around omega 0, then you are going to get large type amplitude. OK? So now, what I am going to do is to continue-- As I mentioned before, I usually like to include the demo after the calculation is done. Sometimes maybe it's a good idea to use the demo as a teaser for the calculation and see that, OK, we can reproduce this result by physics intuition and mathematical calculation. So that's actually the kind of consideration when I choose which kind of demo to be included in the lecture and the kind of preparation before the class, and finally, the timing, which I decide where to insert those either experiments or analytical calculation using program during my lecture.
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https://ocw.mit.edu/courses/5-61-physical-chemistry-fall-2017/5.61-fall-2017.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. BOB FIELD: I'm Bob Field. This is my 44th year at MIT, and I've taught this course roughly half the time, so it is my favorite course. And I'm a spectroscopist, and that means that I use quantum mechanics almost like breathing. And I'm going to try to convey some of the beauty and utility of quantum mechanics. This is not a typical undergraduate quantum mechanics class. It's not about history. It's not about philosophy. It's about use-- use for understanding complicated stuff, use for insight. And so a lot of the material is not in the assigned text, which is a very good book, but it's mostly a safety net. The printed notes, all of which are posted except for a few lectures that I'm still working on-- and that is the text. Everything in the notes, you're responsible for. There are sections of the notes which I call non-lecture. That's explanation of what's in the lecture or a little bit going beyond it. You're responsible for that. If I don't finish all the material in the notes, you're responsible for the material I didn't finish. You can ask me questions about it. You can ask your TAs questions about it. But there's a supplementary text, which is a book that I wrote which is intended to take undergraduates like you to the frontier of research in spectroscopy. This is the book that I used to use to teach graduate quantum mechanics. So I've written a lot of stuff which is already available on the web, but I just want to make sure that you know the structure of the course. There are going to be nine problem sets, always in weeks there is not an exam-- when there is not an exam. And the problem sets are worth 100 points. There will be three 50-minute exams, to which you will be allotted 90 minutes. And they'll be evening exams and on Thursday nights. OK, there will be a final during final exam period. And so the points for the course add up to 600. For each exam, you will be entitled to bring one 8 1/2 by 11 page that you've prepared with as dense or as coarse writing as you want-- a different one page, only one page, for each exam and for the final. One page-- the exam-- one page for the exam. One page for the final. Not four pages for the final, not two pages for the second exam, et cetera-- one page. I believe that writing these pages of notes or pages of things you should remember provide structure that you actually learn by preparing this, and so I stick to this. My job here is to force you to suspend what you expect about the way matter and light behave. You expect things based on your experience with macroscopic objects, and you're very smart. You're here at MIT because you are able to integrate those expectations and to express them in mathematical analysis of your observations. So you're pretty good at that. But quantum mechanics forces you to step away from what you think you understand perfectly because it doesn't apply in microscopic systems. And so I'm going to destroy your expectations about how things work at the beginning, and by the end, I will give them back to you in the form of things called wave packets. Wave packets are quantum mechanical particle-like objects. They obey the rules, and they do a lot of the things you expect. But initially, you have to suspend that belief in particles. OK, so I'm going to start by throwing a piece of chalk. I have to throw it at my TAs. So here is a piece of chalk, and the chalk followed a trajectory. If you're a major league outfielder, you look at the initial part of a trajectory, and you can pretty much figure out where you have to go. This sort of thing would be the subject of 801, but the outfielders don't know physics. They just have instincts, and they know that if they look at the beginning of a trajectory, they can predict where it will end and when it will end, and this is really important. That's the macroscopic view. We talk about trajectories. In quantum mechanics, there are no trajectories. We have possibly an observation of what was the initial condition and what was the detection event, but we can't describe what's going on by observing, point by point, the position and momentum. We're only allowed to do what we call click-click experiments. We start something in some kind of a well-prepared state, and then we detect what has happened at the end. We might do something to the system in between, but we cannot observe everything as the system evolves. And here, I will attempt to a little bit justify that you need to suspend what you believe. I threw a piece of chalk. It could have been a baseball. And suppose now I said, let's decrease the mass of the thrower, the catcher, and the object by 100. You pretty much know exactly what will happen. Now let's decrease it by a factor of 10 to the 20, which is like going from a baseball to an electron. You think you might know, but I guarantee you don't know. And you're going to be surprised, and it's because in the microscopic world, observation modifies the state of the system, and any sort of interaction of the evolving system is going to affect what you observe. And so we have to create a formal structure, which is a kind of measurement theory. What measurements are we allowed to make, and what do they tell us? Because you cannot observe the time dependence. You can calculate what the time dependence is if you know enough, but you cannot observe the thing evolving except by destroying it. OK, so quantum mechanics is beautiful because it describes the microscopic world, and it doesn't tell you you're wrong about the macroscopic world. It matches everything the macroscopic world does. OK, some of the key ideas of quantum mechanics-- so you do a series of identical experiments. You don't get the same result. It's probabilistic, and that should bother you because if you do an experiment carefully, you're trained to think that you do it carefully, you'll get the same result. But quantum mechanics says, tough luck. You can't do an experiment that carefully. There's also this wave-particle duality. You know what particles are and what the properties of particles are. You learn that in 8.01. You know what waves are, and you probably also learned that in 8.01 or 8.02. But you know instinctively that particles and waves behave differently. But in quantum mechanics, everything is both particle-like and wave-like, and this is also something that should bother you. And the third thing is energy quantization. Now, I said I'm a spectroscopist, so I live and die by these spectra, which consist of transitions between energy levels. And these transitions between energy levels encode everything we want to know about the mechanics of an object-- its structure, what it does. And so quantum mechanics is an elaborate encoder of information, and it's usually encoded in the form of these quantized energy levels. And so you're going to want to be able to calculate how the energy levels or the spectrum that you would observe for an object is related to the thing that describes what it can do. And that's the Hamiltonian, and the Hamiltonian we'll look at in many useful ways. Now, I just want to warn you that there are two ways of presenting quantum mechanics-- the Schrodinger picture, which is differential equations and wave functions, and the Heisenberg picture, which is matrix mechanics, where we have matrices and we have eigenvectors. And I am an advocate. I'm passionate about the matrix picture because what it does is presents everything in a way that you can then organize your insights, whereas the Schrodinger picture mostly involves solving one complicated differential equation after another, and the focus is on the mathematics, and it's much harder to see the big picture. Now, that means you need to know a little bit of linear algebra. Now, most of you haven't taken a course in linear algebra, but that doesn't matter because the amount of linear algebra you need to know for quantum mechanics is extremely small, and I will probably present all of it in lectures. But you will definitely have the TAs as a resource, and it's possible that they will give some formal lectures on linear algebra or some handouts, but it's not complicated. It's beautiful. OK, so light. Light is electromagnetic radiation. You've heard that. And it is both wave-like and particle-like. Now, the particle-like aspect of light is going to bother you, but it's very easy to show that it's necessary. OK, so first of all, what are some wave characteristics? If you have a wave, what kind of measurements are you going to make on it? OK, Sasha. AUDIENCE: Intensity at a point in space. BOB FIELD: There can be intensity at a point in space, but there could also be particles that are impinging on that point, so you need something a little bit more that is definitely wave-like, and you have-- yes? AUDIENCE: It has a frequency and a wavelength and an amplitude. BOB FIELD: Yes. I'm angling for something more, but it does have a frequency and a wavelength, and wavelengths are how you understand interference effects. And when you put light through a lens, there's refraction. When you put light on a grating, there's diffraction-- or through a pinhole. And there's the two-slit experiment, where you send light on an object which has two slits, and you can't tell which slit the light went through. This is a very beautiful thing which I will talk about in a coming lecture-- in fact, lecture number three. So the key properties of waves are refraction, diffraction, two-slit experiment. And behind that is interference effects. So we can have just a comic book picture or a cartoon. So here is a wave, and here is another wave. And this other wave has exactly the same frequency and phase. And if we add these two guys together, we get something that looks like that. And in fact, the addition is a little bit nonlinear. So we have constructive interference, or we can have destructive interference and anything in between. If we add these two guys, what you get is nothing. So here, you get an intensified wave-- constructive. And here, we get cancellation. Now, it's a little bit stressful to think that if particles have wave characteristics, they can annihilate each other, so you'll have to be prepared for that. So quantum mechanics exploits constructive and destructive interference. That's at the core, and you have to get used to that, and you have to get used to seeing particles do that. OK, so waves have a frequency and a velocity and a wavelength. And now here, what is nu? Nu is c over lambda. If someone on the telephone or on Skype constantly asks you "what is new," you can put an end to that behavior by saying c over lambda. And if you do it often enough, it'll stop. It may be that the question will be rephrased, but it's useful to remember this as a repellent and also as a crucial organizing principle. OK, waves are electromagnetic, and so that means we have transverse electromagnetic waves. And so here is one part of the wave, and here is another part of the wave. And this is the electric part of the wave. And so we have zero, and here is E0. And this is the magnetic part of the wave, zero and B0. And this is in the xz plane. This is the z direction. This is the z direction. And this is the yz plane. Now, this corresponds to a wave which is linearly polarized along the x-axis. And so this is-- it's kind of hard to draw a transverse electromagnetic wave, but there's an electric part and a magnetic part. And for the most part, I forget about this. The magnetic resonance-- this is the only thing you care about, but they're both part of light. And OK. So we have these waves, and now, the intensity of the light is proportional to the E0 squared-- the electric field squared. And it's also measured in watts per square centimeter. And this is in-- so this is in volts per centimeter squared. Now, I'm an old fashioned guy. I use centimeters instead of meters. You'll just have to forgive me for that, not using MKS units. But the important thing is that the intensity is related to the square of an electric field. So suppose we have light impinging on a hunk of metal. So a metal is something where the electrons are free to move around, and so we like a metal because of that. We could ask, well, what happens if we put light on salt or on some organic molecule? And in some sense, there is going to be a relationship, but the easy thing is if we put a beam of light onto a metal and it's pushing the electrons around. That's what an electric field does. It moves the electrons. And so you would expect classically that at high enough intensity, you're going to start ripping electrons out of the metal, and that would be wrong. And this is the photoelectric effect. This is what we observe. 0-- the current divided by the charge of the electron. So that's the number of electrons per second, and this is the intensity. And if we have infrared radiation, nothing happens, no matter how strong it is. If we have ultraviolet radiation, we have something happening. As the intensity of the light increases, the number of electrons per second ejected from the metal increases linearly. So why is that? We expect that it would be intensity that determines it, but it's the color of the light that determines. We can also do something like this, where we again ask, what is the current divided by the charge of the electron versus frequency? And again, we have 0 here. And what we see is nothing up until some critical point, and then this is-- so this is some special frequency, which is different for every metal. And then, all of a sudden, we start getting electrons. And this increases linearly, or at least for a while increases linearly with the frequency. So there's something about the frequency of light that rips out the electrons. It has to be above a certain minimum, and then the electrons increase with the frequency, but in a little bit more complicated way than linearly. And so I have something in the notes which is not quite right, but this onset is important. So these observations suggest that the electron is bound to the metal by some energy which we call the work function, and it's called phi. And so this is the energy that it takes to rip an electron out of a metal. And so it's called the work function, and work functions for metals range from a little over 1 electron volt to around 5 electron volts. I'm going to use those units. Well, you can forget that. But all metals are within a relatively narrow range of work functions, and this is somehow related to nu 0. So this is an energy, and this is a frequency. And so we expect that there is some relationship between nu 0 and this energy, and there is some proportionality concept, which I can call anything I want, but I'm going to call it h because it's going to become Planck's constant. That's the proportionality concept. And so the onset of ejection of electrons is when the frequency of light is greater than the work function. As I said, every metal has a different work function. OK, so this is looking like somehow, the light does not act in an additive way on the metal. It acts in a singular way somehow. There are particles of light that have definite energy. This is what it looks like, and we're going to call those particles of light photons. And so now we're trying to think of, all right, electromagnetic radiation comes as particles. What are the properties of particles? Well, particles-- well, I'm getting ahead of myself, but as long as I said this, particles have kinetic energy and momentum. Kinetic energy is a scalar quantity. Momentum is a vector quantity. And right now, what we want to do is measure the kinetic energy of the electrons that are produced by the annihilation of photons. OK, and so we can imagine an apparatus like this. Here is our metal, and here is the light impinging on the metal. And we have grids. We have a ground, and we have another grid and a detector. So we have ground voltage is 0. Here we have a voltage less than 0. Electrons don't like that. But these two grids are at the same potential. So the electrons don't know which way to go. And then there's another grid where we have the potential is V plus V stop. And so what happens is if the electron is ejected with enough kinetic energy to go uphill and cross through this potential, this grid, then it will make it to the detector and we'll count it. So this is just a crude apparatus. If you were going to do this experiment, you would probably design it better. But the idea is what we want to do is to measure what is the voltage that we apply that causes the electrons to stop reaching the detector. And that's the way we measure the kinetic energy of the electrons. When all of the electrons stop hitting the detector, we know the voltage, the stopping voltage. And we can now plot the stopping voltage. V stop. Versus frequency. And here is nu 0. And what we see is this. We see that below nu 0, there are no electrons ejected. Once we're above nu 0, electrons start being ejected. And they have a kinetic energy, which is measured by V stop. And it increases linearly with frequency. Now, when we do this experiment on different metals, the slope is constant, the same. It's universal. So the kinetic energy of the ejected electrons is equal to some constant. It goes as minus nu 0. And this is the same constant that we saw before. And this is Planck's constant. And now what we've seen is that these particles of light have definite energy. They have definite kinetic energy. And we can stop and they transfer that energy. And so we can draw an energy level diagram. So here's 0 and here. So this is energy. And this is minus the work function. And so we have a photon, which has this energy. And we have the kinetic energy of the electrons. And this is h nu minus h nu 0. OK. This story is complete. The photoelectron effect is the easiest thing to understand at the beginning of quantum mechanics. And it says that light comes in particles, which we call photons. And the energy of a photon is h nu. h is a fundamental constant. And we know what nu is. OK. I'm going to talk about Compton scattering. But Compton scattering is a little bit harder to understand than the photoelectron effect. And I'm going to put equations on the board which I'm not going to derive. They're easily derived. But this is just to complete the picture of the particle-like characteristic of Planck. So for Compton scattering, we have a beam of x-rays. X-ray is a form of light. It's a very high energy form of light. We have a block of paraffin. And what we observe is there are electrons kicked out. And there is x-ray scattered. So the experiment is we're looking for the particle characteristics. Particles have kinetic energy and they have vector momentum. And you have been trained painfully and completely in conservation of energy and momentum, because it enables you to solve all sorts of useful problems. So suppose the x-ray comes in, hits the block. And so we have the incident momentum and the x-ray is backscattered. And the scattering angle theta is measured this way. So the difference in momentum for this scattering is large for backscattering and much less for forward scattering. Now, this momentum has to be transferred to the electron. And so we can draw conservation diagrams. Or for this case, the backscattering. We have pn. We have the electron, p electron. And we have the momentum of the x-ray. OK, so this diagram determines the momentum transferred to the electron. And if the x-ray photon transfers energy to the-- it transfers momentum to the electron, it also transfers energy. So what we're going to see-- and I'm just going to wave my hands, because the mathematical analysis is something you can do, but I don't want to go through it. I just want to show the structure of the argument. What you measure is the wavelength of the x-ray out minus the wavelength of the x-ray in. And so if the x-ray has transferred energy to the electron, it has less energy when it leaves, and it has longer wavelength. So this is measurable, the change in the wavelength of the x-ray, and that's going to be-- so we'll call that delta lambda. And that's going to be a function of the scattering angle. So Compton scattering basically says, OK, light is a particle. Particles have kinetic energy and momentum. Conservation of energy and momentum predicts a difference, a redshift of the photon, which depends on the scattering angle. The rest is all 8.01. You can do that. OK. And so what you end up finding is that if the scattering is 0, then delta lambda is 0. There is no momentum. The photon just goes through. If the scattering is pi, in other words, it's perfectly backscattered, then delta lambda you can calculate is 2h over the mass of the electron times the speed of light. Now, this quantity, h over the mass of the electron times the speed of light, has dimensions of length. And it's 0.02 or 3 angstroms. And this is called the Compton wavelength of the electron. Because the photon is scattering an electron. The change in wavelength of the photon is determined by this. This is the momentum transfer. This gives you the momentum transferred. And so this is a universal constant. It says electrons have, when they're scattered out of any material, have this behavior. Again, universal and strange. The actual experiment, since what you actually want to measure is the change in lambda over lambda. The change in lambda doesn't depend on lambda but this does. And so what you want to do is make the observable thing large. And so you go to a short wavelength to make this a large fractional change that's easily measured. OK. The equations are derived in the lecture notes and better in texts. And what I want to do now is just tell you where we're going. We've talked about the wave and particle nature of photons, of light. And the particle nature is a surprise, but it's perfectly understandable and observable. The next thing we want to do is determine the wave nature of things that we call particles. And so we're going to do other kinds of simple scattering experiments where we discover that the electron-- so we have some solid. And we have UV light or x-rays. And we go through this solid. And it's mostly transparent. It's mostly free space. And so this light is interacting with the electrons. And maybe with the other stuff that is involved in matter, the nuclei. But the important thing is that the electrons that get scattered are in a kind of a diffuse state. This is mostly nothing. This material doesn't scatter x-rays. It's mostly transparent. And so an explanation for that is the Rutherford planetary model, where we have a nucleus with a charge and electrons orbiting that nucleus. And that's all very nice. It's a way of saying, well, OK, the electrons are forced to choose distances from the nucleus. And there's mostly empty space. But then you look more closely, and the electrons are doing this. They're oscillating in space, and they're going to radiate energy. And this radiation of energy will cause the electrons to spiral in and combine with the charge in the middle. And that's a problem. And so we have an explanation for why matter is mostly empty space and a conundrum. The electron should recombine with the nucleus. So how can matter be relatively non-compressible? And the best explanation is this planetary idea. But then we have this thing we have to explain. Now, there's two hypotheses for this, which are really just ad hoc things. One is the Bohr model where it says, in order for the electron to obey some laws of physics as it orbits the nucleus, it conserves angular momentum. Well, if it were rotating combined with the nucleus, it wouldn't do that. And so that's sort of OK. And then there's de Broglie, who was a very smart person. And he said that in order for the electron not to annihilate itself as it goes around the orbit, it has to have an integer number of half wavelengths along that trajectory. Then it won't annihilate itself. And that is much better than Bohr's hypothesis. But both of these hypotheses lead to line spectra hydrogen where you can build a really simple model and you can predict to eight decimal places all the absorption transitions of hydrogen atom. So a very simple model leads to an incredibly powerful and mysterious result. There is another experiment that I will talk about next lecture. And that is suppose we have a thin sheet of metal. We have photons, x-rays, or UV photons, impinging on this metal. There's diffraction because there is regular distances between the atoms in the particle in the foil. And so what we measure is diffraction rings for the electron and for the photon. And the diffraction rings are identical when you choose the wavelength of the particle to be equal to the wavelength of the photon. And so this is another way of showing wave particle duality. We can show that they have wavelengths and we can calculate to make them the same. So this is all very exciting. And this is about as much philosophy and history as you're going to get from me. Once we understand that there is something that we have to do, which is called quantum mechanics, we're going to start solving problems and then being able to understand incredibly complicated effects beyond the simple problems. And that should be exciting, but it should also be a little bit disturbing, because I will use a technique called perturbation theory, which for some reason all of the textbooks for a course at this level either ignore or treat in the most superficial way. But perturbation theory is a way of taking problems that we understand and can solve exactly. And exactly means an infinite number of states, all of which are given to us by solving one equation. And we can then use them to understand problems we can't solve exactly. And there'll be several ways in which I deal with that. OK, so it's time to stop. And I hope you're excited about what lies ahead, because it is strange and wonderful. And it says we can look inside a molecule. We can make measurements. We can understand what this molecule is going to do. But we have to develop our new way of doing that. OK, thank you.
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https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/8.334-spring-2014.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. OK, let's start. So let's go back to our starting point for the past couple of weeks. The square lattice, let's say, where each side we assign a binary variable sigma, which is minus 1. And a weight that tends to make subsequent nearest neighbors [? things ?] to be parallel to each other. So into the plus k, they are parallel. Penalized into the minus k if they are anti-parallel. And of course, over all pairs of nearest neighbors. And the partition function that is obtained by summing over 2 to the n configurations. Let's make this up to give us a function of the rate of this coupling, which is some energy provided by temperature. And we expect this-- at least, in two and higher dimensions-- to capture a phase transition. And the way that we have been proceeding with this to derive this factor as the hyperbolic cosine of k. 1 plus r variable t, which is the hyperbolic [? sine ?] of k. Sigma i, sigma j. And then this becomes a cos k to the number of bonds, which is 2n on the square lattice. And then expanding these factors, we saw that we could get things that are either 1 from each bond or a factor that was something like t sigma sigma. And then summing over the two values of sigma would give us 0 unless we added another factor of sigma through another bond. And going forth, we had to draw these kinds of diagrams where at each site, I have an even number of 1's selected. Then summing over the sigmas would give me a factor of 2 to the n. And so then I have a sum over a whole bunch of configurations. [? There are ?] [? certainly ?] one. There are configurations that are composed of one way of drawing an object on the lattice such as this one, or objects that correspond to doing two of these loops, and so forth. So that's the expression for the partition function. And what we are really interested is the log of the partition function, which gives us the free energy in terms of thermodynamic quantities that potentially will tell us about phase transition. So here we will get a log 2 to the n. Well actually, you want to divide everything by n so that we get the intensive part. So here we get log 2 hyperbolic cosine squared of k. And then I have to take the log of this expression that includes things that are one loop, disjointed loop, et cetera. And we've seen that the particular loop I can slide all over the place-- so if you have a factor of n-- whereas things that are multiple loops have factors of n squared, et cetera, which are incompatible with this. So it was very tempting for us to do the usual thing and say that the log of a sum that includes these multiple occurrences of the loops is the same thing as sum over the configurations that involve a single loop. And then we have to sum over all shapes of these loops. And each loop will get a factor of t per the number of bonds that are occurring in that. Of course, what we said was that this equality does not hold because if I exponentiate this term, I will generate things where the different loops will coincide with each other, and therefore create types of terms that are not created in the original sum that we had over there. So this sum over phantom loops neglected the condition that these loops, in some sense, have some material to them and don't want to intersect with each other. Nonetheless, it was useful, and we followed up this calculation. So that's repeat what the result of this incorrect calculation is. So we have log of 2 hyperbolic cosine squared k, 1 over n. Then we said the particular way to organize the sum over the loops is to sum over there the length of the loop. So I sum over the length of the loop and count the number of loops that have length l. All of them will be giving me a contribution that is t to the l. So then I said, well, let's, for example, pick a particular point on the lattice. Let's call it r. And I count the number of ways that I can start at r, do a walk of l steps, and end at r again. We saw that for these phantom loops, this w had a very nice structure. It was simply what was telling me about one step raised to the l. This was the Markovian property. There was, of course, an important thing here which said that I could have set the origin of this loop at any point along the loop. So there is an over-counting by a factor of l because the same loop would have been constructed with different points indicated as the origin. And actually, I can go the loop clockwise or anti-clockwise, so there was a factor of 2 because of this degeneracy of going clockwise or anti-clockwise when I perform a walk. And then over here, there's also an implicit sum over this starting point and endpoint. If I always start and end at the origin, then I will get rid of the factor of n. But it is useful to explicitly include this sum over r because then you can explicitly see that sum over r of this object is the trace of that matrix. And I can actually [INAUDIBLE] the order, the trace, and the summation over l. And when that happens, I get log 2 cos squared k exactly as before. And then I have 1 over n. I have sum over r replaced by the trace operation. And then sum over l-- t, T raised to the l divided by l-- is the expansion for minus log of 1 minus t T. And there's the factor of 2 over there that I have to put over here. So note that this plus became minus because of the expansion for log of 1 minus x is minus x minus x squared over 2 minus x cubed over 3, et cetera. And the final step that we did was to note that the trace I can calculate in any basis. And in particular, this matrix t is diagonalized by going to Fourier representation. In the Fourier representation, the trace operation becomes sum over all q values. Sum over all q values, I go to the continuum and write as n integral over q, so the n's cancel. I will get 2 integration over q. These are essentially each one of them, qz and qz, in the interval. Let's say 0 to 2 pi or minus pi to 2 pi, doesn't matter. Interval of size of 2pi. So that's the trace operation. Log of 1 minus t. Then the matrix that represents walking along the lattice represented in Fourier. And so basically at the particular site, we can step to the right or to the left. So that's e to the i qx, e to the minus i qx, e to the i qy, e to the minus i qy. Adding all of those up, you get 2 cosine of qz plus cosine of qy. So that was our expression. And then we realized, interestingly, that whereas this final expression certainly was not the Ising partition function that we were after, that it was, in fact, the partition function of a Gaussian model where at each site I had a variable whose variance was 1, and then I had this kind of coupling rather than with the sigma variable, with these Gaussian variables that will go from minus to infinity. But then we said, OK, we can do better than that. And we said that log z over n actually does equal a very similar sum. It is log 2 hyperbolic cosine squared k, and then I have 1 over n. Sum over all kinds of loops where I have a similar diagram that I draw, but I put a star. And this star implied two things-- that just like before, I draw all kinds of individual loops, but I make sure that my loops never have a step that goes forward and backward. So there was no U-turn. And importantly, there was a factor of minus 1 to the number of times that the walk crossed itself. And we showed that when we incorporate both of these conditions, the can indeed exponentiate this expression and get exactly the same diagrams as we had on the first line, all coming with the correct weights, and none of the diagrams that had multiple occurrences of a bond would occur. So then the question was, how do you calculate this given that we have this dependence on the number of crossings, which offhand may look as if it is something that requires memory? And then we saw that, indeed, just like the previous case, we could write the result as a sum over walks that have a particular length l. Right here, we have the factor of t to the l. Those walks could start and end at the particular point r. But we also specified the direction mu along which you started. So previously I only specified the origin. Now I have to specify the starting point as well as the direction. I have to end at the same point and along the same direction to complete the loop. And these were accomplished by having these factors of walks that are length l. So to do that, we can certainly incorporate this condition of no U-turn in the description of the steps that I take. So for each step, I know where I came from. I just make sure I don't step back, so that's easy. And we found that this minus 1 to the power of nc can be incorporated through a factor of e to the i sum over the changes of the orientation of the walker-- as I step through the lattice-- provided that I included also an additional factor of minus. So that factor of minus I could actually put out front. It's an important factor. And then there's the over-counting, but as before, the walk can go in either one of two directions and can have l starting points. OK, so now we can proceed just as before. log 2 hyperbolic cosine squared k, and then we have a sum over l here, which we can, again, represent as the log of 1 minus t-- this 4 by 4 matrix t star. Going through the log operation will change this sign from minus to plus. I have the 1 over 2n as before. And I have to sum over r and mu, which are the elements that characterize this 4n by 4n matrix. So this amounts to doing the trace log operation. And then taking advantage of the fact that, just as before, Fourier transforms can at least partially block diagonalize this 4n by 4n matrix. I go to that basis, and the trace becomes an integral over q. And then I would have to do the trace of a log of a 4 by 4 matrix. And for that, I use the identity that a trace of log of any matrix is the log of the determinant of that same matrix. And so the thing that I have to integrate is the log of the determinant of a 4 by 4 matrix that captures these steps that I take on the square lattice. And we saw that, for example, going in the horizontal direction would give me a factor of t e to the minus i qx. Going in the vertical direction-- up-- y qy. Going in the horizontal direction-- down. These are the diagonal elements. And then there were off-diagonal elements, so the next turn here was to go and then bend upward. So that gave me in addition to e to the minus i qx, which is the same forward step here. A factor of, let's call it omega so I don't have to write it all over the place. Omega is e to the pi pi over 4. And the next one was a U-turn. The next one was minus t e minus i qx omega star. And we could fill out similarly all of the other places in this 4 by 4 matrix. And then the whole problem comes down to having to evaluate a 4 by 4 determinant, which you can do by hand with a couple of sheets of paper to do your algebra. And I wrote for you the final answer, is 1/2 2 integrals from minus pi 2 pi over qx and qy divided by 2 pi squared. And then the result of this, which is log of 1 plus t squared squared minus 2t 1 minus t squared cosine of qx plus cosine of qy. This was the expression for the partition function. OK, so this is where we ended up last time. Now the question is we have here on the board two expressions, the project one and the incorrect one-- the Gaussian model and the 2-dimensionalizing one. They look surprisingly similar, and that should start to worry us potentially because we expect that many when we have some functional form, that functional form carries within it certain singularities. And you say, well, these two functions, both of them are a double integral of log of something minus something cosine plus cosine. So after all of this work, did we end up with an expression that has the same singular behavior as the Gaussian model? OK, so let's go and look at things more carefully. So in both cases, what I need to do is to integrate A function a that appears inside the log. There is an A for the Gaussian, and then there is this object-- let's call it A star-- for the correct solution. So the thing that I have to integrate is, of course, q. So this is a function of the vector q, as well as the parameter that is a function of which I expect to have a phase transition, which is t. Now where could I potentially get some kind of a singularity? The only place that I can get a singularity is if the argument of the log goes to 0 because log of 0 is something that's derivatives of singularities, et cetera. So you may say, OK, that's where I should be looking at. So where is it this most likely to happen when I'm integrating over q? So basically, I'm integrating over qx and qy over a [INAUDIBLE] [? zone ?] that goes from minus pi to pi in both directions. And potentially somewhere in this I encounter a singularity. Let's come from the site of high temperatures where t is close to 0. Then I have log of 1, no problem. As I go to lower and lower temperatures, the t becomes larger. Then from the 1, I start to subtract more and more with these cosines. And clearly the place that I'm subtracting most is right at the center of q equals to 0. So let's expand this in the vicinity of q goes to 0 in the vicinity of this place. And there what I see is it is 1 minus. Cosines are approximately 1 minus q squared over 2. So I have 1 minus 4t, and then I have plus t qx squared plus qy squared, which is essentially the net q squared. And then I have order of higher power z, qx and qy. Fine. So this part is positive, no problem. I see that this part goes through 0 when I hit tc, which is 1/4. And this we had already seen, that basically this is the place where the exponentially increasing number of walks-- as 4 to the number of steps-- overcomes the exponentially decreasing fidelity of information carried through each walk, which was t to the l. So 4dt being 1, tc is of the order 1/4. We are interested in the singularities in the vicinity of this phase transition, so we additionally go and look at what happens when t approaches tc, but from this side above, because clearly if I go to t that is larger than 1/4, it doesn't make any sense. So t has to be less than 1/4. And so then what I have here is that I can write this as 4tc, so this is 4 times tc minus t. And this to the lowest order I can replace as tc-- q squared plus higher orders. This 4 I can write as 1 over tc. So the whole thing I can write as q squared plus delta t divided by tc, and there's a factor of 4. And delta t I have defined to be tc minus 10. How close I am to the location where this singularity takes places. So what I'm interested is not in the whole form of this function, but only the singularities that it expresses. So I focus on the singular part of this Gaussian expression. I don't have to worry about that term. So I have minus 1/2. I have double integral. The argument of the log, I expanded in the vicinity of the point that I see a singularity to take place. And I'll write the answer as q squared plus 4 delta t over tc. If I am sufficiently close in my integration to the origin so that the expansion in q is acceptable, there is an additional factor of tc. But if I take a log of tc, it's just a constant. I can integrate that out. It's going to not contribute to the singular part, just an additional regular component. Now if I am in the vicinity of q equals to 0 where all of the action is, at this order, the thing that I'm integrating has circular symmetry so that 2-dimensional integration. I can write whether or not it's d qx, d qy as 2 pi q dq divided by 2 pi squared, which was the density of state. And this approximation of a thing being isotropic and circular only holds when I'm sufficiently close to the origin, let's say up to some value that I will call lambda. So I will impose some cut-off here, lambda, which is certainly less than one of the order of pi-- let's say pi over 10. It doesn't matter what it is. As we will see, the singular part, it ultimately does not matter what I put for lambda. But the rest of the integration that I haven't exclusively written down from all of this-- again, in analogy to what we had seen before for the Landau-Ginzburg calculation will give me something that is perfectly analytic. So I have extracted from this expression the singular part. OK, now let's do this integral carefully. So what do I have? I have 1 over 2 pi with minus 1/2, so I have minus 1 over 4 pi. If I call this whole object here x-- so x is q squared plus this something-- then we can see that dx is 2 q dq. So what I have to do is the integral of dx log x, which is x log x minus x. So essentially, let's keep the 2 up there and make this 8 pi. And then what I have is x log x minus x itself, which I can write as log of e. And then this whole thing has to be evaluated between the two limits of integration, lambda and [INAUDIBLE]. Now you explicitly see that if I substitute in this expression for q the upper part of lambda, it will give me something like lambda squared plus delta t-- an expandable and analytical function. Log of a constant plus delta t that I can start analytically expanding. So anything that I get from the other cut-off of the integration is perfectly analytic. I don't have to worry about it. If I'm interested in the singular part, I basically need to evaluate this as it's lower cut-off. So I evaluate it at q equals to 0. Well first off all, I will get a sign change because I'm at the lower cut-off. I will get from here 4 delta t over tc. And from here, I will get log of a bunch of constants. It doesn't really matter. 4 over e delta t over tc. What is the leading singularity? Is delta t log t-- log delta t? And so again, the leading singularity is delta t log of delta t. There's an overall factor of 1 over pi. [? Doesn't match. ?] You take two derivatives. You find that the heat capacity, let's say that is proportion to two derivatives of log z by delta t squared. You take one derivative, it goes like the log. You take another derivative of the log, you find that the singularity is 1 over delta t. That corresponds to a heat capacity divergence with an exponent of unity, which is quite consistent with the generally Gaussian formula that we had in d dimensions, which was 2 minus t over 2. So that's the Gaussian. And of course, this whole theory breaks down for t that is greater than tc. Once I go beyond tc, my expressions just don't make sense. I can't integrate a lot of a negative number. And we understand why that is. That's because we are including all of these loops that can go over each other multiple times. The whole theory does not make sense. So we did this one to death. Will the exact result be any different? So let's carry out the corresponding procedure-- for A star that is a function of q and t. And again, singularities should come from the place where this is most likely to go to 0. You can see it's 1 something minus something. And clearly when the q's are close to 0 is when you subtract most, and you're likely to become negative. So let's expand it around 0. This as q goes to 0 is 1 plus t squared squared. And then I have minus. Each one of the cosines starts at unity, so I will have 4t 1 minus t squared. And then from the qx squared over 2 qy squared over 2, I will get a plus t 1 minus t squared q squared-- qx squared plus qy squared-- plus order of q to the fourth. So the way that we identified the location of the singular part before was to focus on exactly q equals to 0. And what we find is that A star at q equals to 0 is essentially this part. This part I'm going to rewrite the first end slightly. 1 plus t squared squared is the same thing as 1 minus t squared squared plus 4t squared. The difference between the expansion of each one of these two terms is that this has plus 2t squared, this has minus 2t squared, which I have added here. And then minus 4t 1 minus t squared. And the reason that I did that is that you can now see that this term is twice this times this when I take the square. So the whole thing is the same thing as 1 minus t squared minus 2t squared. So the first thing that gives us reassurance happens now, whereas previously for the Gaussian model 1 minus 4t could be both positive and negative, this you can see is always positive. So there is no problem with me not being able to go from one side of the phase transition to another side of the phase transition. This expression will encounter no difficulties. But there is a special point when this thing is 0. So there is a point when 1 minus 2t c plus tc squared is equal to 0. The whole thing goes to 0. And you can figure out where that is. tc is 1. It's a quadratic form. It has two solutions-- 1 minus plus square root of 2. A negative solution is not acceptable. Minus. I have to recast this slightly. Knowing the answer sometimes makes you go too fast. tc squared plus 2tc minus 1 equals to 0. tc is minus 1 plus or minus square root of 2. The minus solution is not acceptable. The plus solution would give me root 2 minus 1. Just to remind you, we calculated a value for the critical point based on duality. So let's just recap that duality argument. We saw that the series that we had calculated, which was an expansion in high temperatures times k, reproduced the expansion that we had for 0 temperature, including islands of minus in a sea of plus, where the contribution of each bond was going from e to the k to e to the minus k. So there was a correspondence between a dual coupling. And the actual coupling, that was like this. At a critical point, we said the two of them have to be the same. And what we had calculated based on that, since the hyperbolic tang I can write in terms of the exponentials was that the value of e to the minus e to the plus 2kc was, in fact, square root of 2 plus 1. And the inverse of this will be square root of 2 minus 1, and that's the same as the tang. That's the same thing as the things that we have already written. So the calculation that we had done before-- obtain this critical temperature based on this Kramers-Wannier duality-- gave us a critical point here, which is precisely the place that we can identify as the origin of the singularity in this expression. So what did we do next? The next thing that we did was having identified up there where the critical point was-- which was at 14-- was to expand our A, the integrand, inside the log in the vicinity of that point. So what I want to do is similarly expand a star of q for t that goes into vicinity of tc. And what do I have to do? So what I can do, let's write it as t is tc plus delta t and make an expansion for delta t small. So I have to make an expansion or delta t small, first of all, of this quantity. So if I make a small change in t, I'll have to take a derivative inside here. So I have minus 2t minus 2, evaluate it at tc, times the change in delta t. So that's the change of this expression if I go slightly away from the point where it is 0. Yes? AUDIENCE: I have a question. T here is [? tangent ?] of k where we calculated the [INAUDIBLE] to [? be as ?] temperature. PROFESSOR: Now all of these things are analytical functions of each other. So the k is, in fact, some unit of energy divided by kt. So Really, the temperature is here. And my t is tang of the above objects, so it's tang of j divided by kt. So the point is that whenever I look at the delta in temperature, I can translate that delta in temperature to a delta int times the value of the derivative at the location of this, which is some finite number. Of basically up to some constant, taking derivatives with respect to temperature, with respect to k, with respect to tang k, with respect to beta. Evaluate it at the finite temperature, which is the location of the critical point. They're all the same up to proportionality constants, and that's why I wrote "proportionality" there. One thing that you have to make sure-- and I spent actually half an hour this morning checking-- is that the signs work out fine. So I didn't want to write an expression for the heat capacity that was negative. So proportionalities aside, that's the one thing that I better have, is that the sign of the heat capacity, that is positive. So that's the expansion of the term that corresponds to q equals to 0. The term that was proportional to q squared, look at what we did in the above expression for the Gaussian model. Since it was lower order, it was already proportional to q squared. We evaluated it at exactly t equals to tc. So I will put here tc 1 minus tc squared q squared, and then I will have high orders. Now fortunately, we have a value for tc that we can substitute in a couple of places here. You can see that this is, in fact, twice tc plus 1, and tc plus 1 is root 2. So this is minus 2 root 2. I square that, and this whole thing becomes 8 delta t squared. This object here, 1 minus tc squared-- you can see if I put the 2 tc on the other side-- 1 minus tc squared' is the same thing as 2 tc. So I can write the whole thing here as 2 tc squared q squared. And the reason I do that is, like before, there's an overall factor that I can take out of this parentheses. And the answer will be q squared now plus 4 delta t over tc squared. It's very similar to what we had before, except that when we had q squared plus 4 delta t over tc, we have q squared plus 4 delta t over tc squared. Now this square was very important for allowing us to go for both positive and negative, but let's see its consequence on the singularity. So now log z of the correct form divided by n-- the singular part-- and we calculate it just as before. First of all, rather than minus 1/2 I have a plus 1/2. I have the same integral, which in the vicinity of the origin is symmetric, so I will write it as 2 pi qd q divided by 4 pi squared. And then I have log. I will forget about this factor for consideration of singularities 4 delta t over tc squared. And now again, it is exactly the same structure as x dx that I had before. So it's the same integral. And what you will find is that I evaluate it as 1/8 pi integral, essentially q squared plus 4 delta t over tc squared, log of q squared plus 4 delta t over tc squared over e evaluated between 0 and lambda. And the only singularity comes from the evaluation that we have at the origin. And so that I will get a factor of minus. So I will get 1/8 pi. Actually then, I substitute this factor. The 4 and the 8 will give me 2. And then I evaluate the log of delta t squared, so that's another factor of 2. So actually, only one factor of pi survives. I will have delta t over tc squared log of, let's say, absolute value of delta t over tc. So the only thing that changed was that whereas I had the linear term sitting in front of the log, now have a quadratic term. But now when I take two derivatives, and now we are sure that taking derivatives does not really matter whether I'm doing it with respect to temperature or delta t or any other variable. You can see that the leading behavior will come taking two derivatives out here and will be proportional to the log. So I will get minus 1 over pi log of delta t over tc. So that if I were to plug the heat capacity of the system, as a function of, let's say, this parameter t-- which is also something that stands for temperature-- but t goes between, say, 0 and 1. It's a hyperbolic tangent. There's a location which is this tc, which is root 2 minus 1. And the singular part of the heat capacity-- there will be some other part of the heat capacity that is regular-- but the singular part we see has a [INAUDIBLE] divergence. And furthermore, you can see that the amplitudes-- so essentially this goes approaching from different sides of the transition, A plus or A minus log of absolute value of delta a. And the ratio of the amplitudes, which we have also said is universal, is equal to 1. And you had anticipated that based on [? duality ?] [INAUDIBLE]. All right, so indeed, there is a different behavior between the two models. The exact solution allows us to go both above and below, has this logarithmic singularity. And this expression was first written down-- well, first published Onsager in 1944. Even couple of years before that, he had written the expression on boards of various conferences, saying that this is the answer, but he didn't publish the paper. The way that he did it is based on the transfer matrix network, as we said. Basically, we can imagine that we have a lattice that is, let's say, I parallel in one direction, l perpendicular in the other direction. And then for the problems that you had to do, the transform matrix for one dimensional model, we can easy to do it for a [? ladder ?]. It's a 4 by 4. For this, it becomes a 2 to the l by 2 to the l matrix. And of course, you are interested in the limit where l goes to infinity so that you can come 2-dimensional. And so he was able to sort of look at this structure of this matrix, recognize that the elements of this matrix could be represented in terms of other matrices that had some interesting algebra. And then he arguably could figure out what the diagonalization looked like in general for arbitrary l, and then calculate log z in terms of the log of the largest eigenvalue. I guess we have to multiply by l parallel. And showed that, indeed, it corresponds to this and has this phase transition. And before this solution, people were not even sure that when you sum an expression such as that for partition function if you ever get a singularity because, again, on the face of it, it's basically a sum of exponential functions. Each one of them is perfectly analytic, sums of analytical functions. It's supposed to be analytical. The whole key lies in the limit of taking, say, l to infinity and n to infinity, and then you'll be able to see these kinds of singularities. And then again, some people thought that the only type of singularities that you will be able to get are the kinds of things that we saw [INAUDIBLE] point. So to see a different type of singularity with a heat capacity that was actually divergent and could explicitly be shown through mathematics was quite an interesting revelation. So the kind of relative importance of that is that after the war-- you can see this is all around the time of World War II-- Casimir wrote to Pauli saying that I have been away from thinking about physics the past few years with the war and all of that. Anything interesting happening in theoretical physics? And Pauli responded, well, not much except that Onsager solved the 2-dimensionalizing model. Of course, the solution that he has is quite obscure. And I don't think many people understand that. Then before, the form that people refer to was presented is actually kind of interesting, because this paper has something about crystal statistics, and then there's the following paper by a different author-- Bruria Kaufman, 1949-- has this same title except it goes from number two or something. So they were clearly talking to each other, but what she was able to show, Bruria Kaufman, was that the structure of these matrices can be simplified much further and can be made to look like spinners that are familiar from other branches of physics. And so this kind of 50-page paper was kind of reduced to something like a 20-page paper. And that's the solution that is reproduced in Wong's book. Chapter 15 of Wong has essentially a reproduction of this. I was looking at this because there aren't really that many women mathematical physicists, so I was kind of looking at her history, and she's quite an unusual person. So it turns out that for a while, she was mathematical assistant to Albert Einstein. She was first married to one of the most well known linguists of the 20th century. And for a while, they were both in Israel in a kibbutz where this important linguist was acting as a chauffeur and driving people around. And then later in life, she married briefly Willis Lamb, of Lamb shift. He turned out the term, 1949. She had done some calculation that if Lamb had paid attention to, he would have also potentially won a Nobel Prize for Mossbauer effect, but at that time didn't pay attention to it so somebody else got there first. So very interesting person. So to my mind, a good project for somebody is to write a biography for this person. It doesn't seem to exist. OK, so then both of these are based on this transfer matrix method. The method that I have given you, which is the graphical solution, was first presented by Kac and Ward in 1952. And it is reproduced in Feynman's book. So Feynman apparently also had one of these crucial steps of the conjecture with his factor of minus 1 to the power of the number of crossings, giving you the correct factor to do the counting. Now it turns out that I also did not prove that statement, so there is a missing mathematical link to make my proof of this expression complete. And that was provided by a mathematician called Sherman, 1960, that essentially shows very rigorously that these factors of minus 1 to the number of crossings will work out and magically make everything happen. Now the question to ask is the following. We expect things to be exactly solvable when they are trivial in some sense. Gaussian model is exactly solvable because there is no interaction among the modes. So why is it that the 2-dimensionalizing model is solvable? And one of the keys to that is a realization of another way of looking at the problem that appeared by Lieb, Mattis, and Schultz in 1964. And so basically what they said is let's take a look at these pictures that I have been drawing for the graphs. And so I have graphs that basically on a kind of coarse level, they look something like this maybe. And what they said was that if we look at the transfer matrix-- the one that Onsager and Bruria Kaufman were looking at. And the reason it was solvable, it looked very much like you had a system of fermions. And then the insight is that if we look at these pictures, you can regard this as a 1-dimensional system of fermions that is evolving in time. And what you are looking at these borderline histories of two particles that are propagating. Here they annihilate each other. Here they annihilate each other. Another pair gets created, et cetera. But in one dimensions, fermions you can regard two ways-- either they cannot occupy the same site or you can say, well, let them occupy in same site, but then I introduce these factors of minus 1 for the exchange of fermions. So when two fermions cross each other in one dimension, their position has been exchanged so you have to put a minus 1 for crossing. And then when you sum over all histories for every crossing, there will be one that touches and goes away. And the sum total of the two of them is 0. So the point is that at the end of the day, this theory is a theory of three fermions. So we have not solved, in fact, an interacting complicated problem. It is. But in the right perspective, it looks like a bunch of fermions that completely non-interacting pass through each other as long as we're willing to put the minus 1 phase that you have for crossings. One last aspect to think about that you learn from this fermionic perspective. Look at this expression. Why did I say that this is a Gaussian model? We saw that we could get this, z Gaussian, by doing an integral essentially over weights that were continuous by i. And the weight I could click here as kij phi i phi j. Because in principle, if I only count an interaction once, let's say I count it twice. I could put a factor of 1/2 if I allow i and j to be some. Especially I said the weight that I have to put here is something like phi i squared over 2. So this essentially you can see z of the Gaussian ultimately always becomes something like z of the Gaussian is going to be 1 over the square root of a determinant of whatever the quadratic form is up there. And when you take the log of the partition function, the square root of the determinant becomes minus 1/2 of the log of the determinant, which is what we've been calculating. So that's obvious. You say this object over here that you write as an answer is also log of some determinant. So can I think of these as some kind of a rock that is prescribed according to these rules on a lattice, give weights to jump according to what I have, then do a kind of Gaussian integration and get the same answer? Well, the difficulty is precisely this 1/2 versus minus 1/2, because when you do the Gaussian integration, you get the determinant in the denominator. So is there a trick to get the determinant in the numerator? And the answer is that people who do have integral formulations for fermionic systems rely on coherent states that involve anti-commuting variables, called Grassmann variables. And the very interesting thing about Grassmann variables is that if I do the analog of the Gaussian integration, the answer-- rather than being in the determinant, in the denominator-- goes into the numerator. And so one can, in fact, rewrite this partition function, sort of working backward, in terms of an integration over Gaussian distributed Grassmannian variables on the lattice, which is also equivalent to another way of thinking about fermions. Let's see. What else is known about this model? So I said that the specific heat singularity is known, so we have this alpha which is 0 log. Given that the structure that we have involves inside the log something that is like this, you won't be surprised that if I think in terms of a correlation length-- so typically q's would be an inverse correlation length, some kind of a q at which I will be reaching appropriate saturation for a delta t-- I will arrive at a correlation range that diverge as delta t to the minus 1. Again, I can write it more precisely as b plus b minus to the minus mu minus. The ratio of the b's this is 1, and the mus are the same and equal to 1. So the correlation length diverges with an exponent 1, and one can [INAUDIBLE] this exactly. One can then also calculate actual correlations at criticality and show that that criticality correlations decay with separation between the points that you are looking at as 1 over r to the 1/4. So the exponent that we call mu is 1/4 in 2 dimensions. Once you have correlations, you certainly know that you can calculate the susceptibility as an integral of the correlation function. And so it's going to be an integral b 2r over r to the 1/4 that is cut off at the correlation length. So that's going to give me c to the power of 2 minus 1/4, which is 7/4. So that's going to diverge as delta t 2 to the minus gamma. Gamma is, again, 7/4. So these things-- correlations-- can be calculated like you saw already for the case of the Gaussian model with a appropriate modification. That is, I have to look at walks that don't come back and close on themselves, but walks that go from one point to another point. So the same types of techniques that is described here will allow you to get to all of these other results. AUDIENCE: Question? PROFESSOR: Yes? AUDIENCE: Do people try to experiment on the builds as things that should be [INAUDIBLE] to the Ising model? PROFESSOR: There are by now many experimental realizations of [INAUDIBLE] Ising model in which these exponents-- actually, the next one that I will tell you have been confirmed very nicely. So there are a number of 2-dimensional absorb, systems, a number of systems of mixtures in 2 dimensions that phase separate. So there's a huge number of experimental realizations. At that time, no, because we're talking about 70 years. So the last thing that I want to mention is, of course, when you go below tc, we expect that-- let's call it temperate-- when I go below temperature, there will be a magnetization. That has always been our signature of symmetry breaking. And so the question is, what is the magnetization? And then this is another interesting story, that around 1950s in a couple of conferences, at the end of somebody's talk, Onsager went to the board and said that he and Bruria Kaufman have found this expression for the magnetization of the system at low temperature as a function of temperature or the coupling constant. But they never wrote the solution down until in 1952, actually CN Yang published a paper that has derived this result. And since this goes to 1 at the critical point-- why duality [INAUDIBLE] 2k [INAUDIBLE] 2k-- [? dual ?] plus 1. This vanishes with an exponent beta, which is 1/8, which is the other exponent in this series. So as I said, there are many people who since the '50s and '60s devoted their life to looking at various generalizations, extensions of the Ising model. There are many people who try to solve it with a finite in 2 dimensions, a finite magnetic field. You can't do that. This magnetization is obtained only at the limit of [INAUDIBLE] going to 0. And clearly, people have thought a lot about doing things in 3 dimensions or higher dimensions without much success. So this is basically the end of the portion that I had to give with discrete models and lattices. And as of next lecture, we will change our perspective one more time. We'll go back to the continuum and we will look at n component models in the low temperature expansion and see what happens over there. Are there any questions? OK, I will give you a preview of what I will be doing in the next few minutes. So we have been looking at these lattice models in the high t limit where expansion was graphical, such as the one that I have over there. But the partition function turned out to be something of two constants, as 1 plus something that involves loops and things that involve multiple loops, et cetera. This was for the Ising model. It turns out that if I go from the Ising model to n component space. So at each side of the lattice, I put something that has n components subject to the condition that it's a unit vector. And I try to calculate the partition function by integrating over all of these unit vectors in n dimension of a weight that is just a generalization of the Ising, except that I would have the dot product of these things. And I start making appropriate high temperature expansions for these models. I will generate a very similar series, except that whenever I see a loop, I have to put a factor of n where n is the number of components. And this we already saw when we were doing Landau-Ginzburg expansions. We saw that the expansions that we had over here could be graphically interpreted as representations of the various terms that we had in the Landau-Ginzburg expansion. And essentially, this factor of n is the one difficulty. You can use the same methods for numerical expansions for all n models that are these n component models. You can't do anything exactly with them. Now the low temperature expansion for Ising like models we saw involved starting with some ground state-- for example, all up-- and then including excitations that were islands of minus in a sea of plus. And so then there higher order in this series. Now for other discrete models, such as the Potts model, you can use the same procedure. And again, you see that in some of the problems that you've had to solve. But this will not work when we come to these continuous spin markets, because for continuous spin models, the ground state would be when everybody is pointing in one direction, but the excitations on this ground state are not islands that are flipped over. They are these long wavelength Goldstone modes, which we described earlier in class. So if we want to make an expansion to look at the low temperature [INAUDIBLE] for systems with n greater than 1, we have to make an expansion involving Goldstone modes and, as we will see, interactions among Goldstone modes. So you can no longer regard at that appropriate level of sophistication as the Goldstone modes maintain independence from each other. And very roughly, the main difference between discrete and these continuous symmetry models is captured as follows. Suppose I have a huge system that is L by L and I impose boundary conditions on one side where all of this spins in one direction, and ask what happens if I impose a different condition on the other side. Well, what would happen is you would have a domain boundary. And the cost of that domain boundary will be proportional to the area of the boundary in d dimensions. So the cost is proportional to some energy per bond times L to the d minus 1. Whereas if I try to do the same thing for a continuous spin system, if I align one side like this, the other side like this, but in-between I can gradually change from one direction to another direction. And the cost would be the gradient, which is 1 over L squared times integrated over the entire system. So the energy cost of this excitation will be some parameter j, 1 over L-- which is the shift-- squared-- which is the [? strain ?] squared-- integrated over the entire volume, so it goes L to the d minus 2. So we can see that these systems are much softer than discrete systems. For discrete systems, thermal fluctuations are sufficient to destroy order at low temperature as soon as this cost is of the order of kt, which for large L can only happen in d [INAUDIBLE] 1 and lower. Whereas for these systems, it happens in d of 2 and lower. So the lower critical dimension for these models. With continuous symmetry, we already saw it's 2. For these models, it is 1. Now we are going to be interested in this class of models. I have told you that in 2 dimensions, they should not order. So presumably, the critical temperature-- if I continuously regard it as a function of dimension, will go to 0 as a function of dimension as d minus 2. So the insight of Polyakov was that maybe we can look at the interactions of these old Goldstone modes and do a systematic low temperature expansion that reaches the phase transition of a critical point systematically ind minus 2. And that's what we will attempt in the future.
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https://ocw.mit.edu/courses/3-020-thermodynamics-of-materials-spring-2021/3.020-spring-2021.zip
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PROFESSOR: All right, so it is Friday, and we are where? Problem set is due today-- problems that will go out later today. And we're going to continue today with binary phase diagrams. We're still very much firmly in DeHoff chapter 10. And today, we're going to talk about peritectic systems and intermediate phases and start spending more time with the software. OK, so let's talk about peritectic reactions. OK, so eutectic reaction we defined earlier, and now we're going to do peritectic. And this is a peritectic reaction. So I'll define it formally, and then I'll show you some examples, just as we did for the eutectic case. So a peritectic reaction is a liquid and a solid at high temp transforming into a solid. That is our peritectic reaction. So as before, we have the Gibbs phase rule gives us c minus Ph plus 2 equals 1. So just as with the eutectic case, this is a line in T P composition space. As for eutectic-- as for eutectic case. And by the way, speaking of eutectics, I hope some people were able to check out that video demo of the indium gallium system. And as with eutectoids, we have peritectoid, which is kind of like a peritectic, but not quite. And this is a peritectoid solid 1 plus solid 2 going to solid 3. So let's see some examples because I know this is kind of dry. So here's the copper-zinc system. We have, again, brass-- and so we've seen this system before. I like to come back to systems that you've seen, just from familiarity. So we have pure copper and solid solution of copper with zinc here. That's the alpha phase. And then we have several other solid phases. So there's a beta phase and the gamma phase, delta phase, epsilon phase, and then there's pure zinc, which is eta. And what I want to highlight now are these transformations. So I'm going to flip back and forth here. This region here is bounded by the liquids, the delta phase, and two isoforms here, one at 697 and one at 594. And you could see that this region here is what? Somebody told me, please-- what is the phase composition in this region that I'm working? AUDIENCE: They'll turn in liquid? PROFESSOR: Delta on liquid, right. Two-phase region-- delta on liquid. And you see, you can cool down and enter this epsilon solid solution. So right there is a peritectic. So just as your eye can start to pick out eutectics with this [INAUDIBLE] shape, your eye can start to pick out peritectics. It looks like a beam balancing on a triangle. There's a triangle, and there's a beam. And visually, I can start to pick these out. So there's a peritectic. Here's another one. So right here is beta on liquid two-phase region. This little sliver is beta on liquid two-phase region. And then down here is a gamma. And so there we have, again, a little beam balancing on a triangle. And there we have another peritectic. And there's another one there. And I picked out five peritectic points here. And just so that you've seen it, these are sometimes called invariant points, same as with eutectic. It refers to the fact that although they are, in fact, lines, on a binary phase diagram at fixed pressure, they appear as points. So they occur at only one given composition and temperature as long as you're at 1 atmosphere. So peritectoids-- here, I have chosen the copper tin system now. The copper tin system is definitely complicated. I'll start with these peritectic points here, for instance, is copper and liquid. Two-phase region here-- copper solid solution and liquid. And if you cool down, you enter this nice, purple solid solution, which is labeled as-- it's got this funny label-- high temperature phase of copper 0.85 tin 0.15. So that's a peritectic. Here's another peritectic point here. Here's a nice, typical peritectic. Here's the beam balancing on a little point. So up here, we have a coexistence between a solid and the liquid. And down here, we have a solid solution phase. Those are peritectics. But we also have peritectoid, and I don't want to dwell on this for too long because they're less common. But if you look, I zoomed way in here. And if you look in this little triangle region here, this little triangle region is a two-phase region where we have coexistence between this phase, which is copper 3 tin high temperature, and this phase here, which is copper 3 tin room temperature. In this little triangle region, at equilibrium, the system will phase separate into solid 1 plus solid 2. And if we cool down right into this tiny little sliver, then we enter a single-phase region, which is copper 10 tin 3 high temperature. So for every one of these transformations that we can find in a phase diagram, there is often a type-- there's a label, eutectic, peritectic, eutectoid, peritectoid. We have others that we haven't discussed. I think that's all I have to say on peritectics and peritectoids. Consider it a series of labels to characterize common visual motifs in these phase diagrams. There are pretty substantial implications of these types of transformations for materials processing, but that really goes beyond the scope of 020 and veers into kinetics and, of course, materials processing, which you take later on. Questions on peritectics, please, before we move on? I'll move back to the board. We're going to now talk about another feature of binary phase diagrams, and that is intermediate phases. Intermediate phases so an intermediate phase is a phase that is stable for some intermediate composition, but not for pure components. So it's a phase that does not exist in the pure component case, and it is structurally distinct from reference space. So let's show an example-- chromium iron. So here is chromium iron. And this is generated using Thermo-Calc. And chromium iron is a kind of a fun system. It has this weird-looking thing, which will come to. But what I want to focus on now is this intermediate phase. So you see, there is something that looks like a spinodal. This whole region here is BCC. This whole region here is a BCC solid solution. And it looks as if chromium iron are fully miscible in BCC, except that miscibility is interrupted. And there is this teardrop shape here, intermediate phase, which, unlike BCC, it is structurally distinct. And-- how come I can't get my laser pointer here-- so we have BCC, and then this region here is a two-phase BCC and sigma coexistence. This is a sigma solid solution, intermediate phase, and then if we keep on moving to the right, we have a sigma and BCC two-phase region again. And if we drop sufficiently low temperature, sigma is no longer stable, and we have a very typical spinodal pattern, which is chromium-rich BCC and iron-rich BCC coexisting in this wide two-phase region. And these are, indeed, very distinct structures. So BCC, body centered cubic, is a structure that you know. It has only one site. There's only one type of site in a BCC lattice, and has eight-fold coordination, whereas this intermediate sigma phase-- I had to look this up-- this is a structurally very complicated material. There are five inequivalent crystallographic sites. And they have different combinations of 12, 15, 14, 12, and 14. So yeah, this is kind of a complicated thing, and this structure does not form for pure chromium, and does not form for pure iron. But it does form for this chromium iron solution. As you might guess from the coordination numbers, do you think this is a more closely packed structure than BCC or less closely packed? Well, I'll ask what is the coordination number for a close-packed either FCC or HCP? Who remembers this? Nobody remembers this from Structure or even from 3.091, the coordination number for a closely packed lattice? AUDIENCE: Is it 12? PROFESSOR: 12, right. So this crystal structure here has coordinations which are 12 and higher. And, as you might guess, the way it does that is it has atoms of two different sizes. You can't get coordination number higher than 12 if you only have one type of atom. But you can find ways to pack in more atoms if you have atoms at three different sizes. And you might guess that this phase becomes more stable at higher pressure. It seems more dense, and you'd be right. So you might guess that the range of stability of this solution might be expanded as you increase the pressure. But right now, we're just sticking at 1 atmosphere. So let's go back to the board. So how did the free energy composition diagram look? So, for example, chromium iron is a case of a spinodal interrupted by an intermediate. So let's draw a free energy composition diagram for this. So we know what a spinodal free-energy composition looks like. Let's draw it in a region of phase separation we have here. OK, that's a typical solution model for a system that undergoes spinodal decomposition. But now let's say that I'm at a temperature for which the sigma phase is stable for intermediate compositions. Could somebody please tell me, qualitatively, how I should draw this sigma phase solution model? So this is the BCC phase model. How should the sigma phase model look? We know it's going to be stable for some intermediate competition. So somewhere in the middle of this diagram, we want that sigma model to be a stable solid solution. And we know that we need this BCC phase to be stable on the left-hand side of the diagram and we need this BCC phase to be stable on the right side of the diagram as well, and that there's going to be regions of two-phase coexistence. So we need some common tangents. You think about the taut rope construction. How should I draw a sigma? AUDIENCE: Would it be a parabola that's like a U-shape? PROFESSOR: Yeah, it's like a plunger. Imagine the taut rope. I'm going to I'm going to risk really messing up here. Imagine the taut rope like this. Don't worry, I ordered more sharpies. They were delivered to my office. So we will be back in business. Imagine that this is a-- man, I really messed up. Imagine this is the taut rope. I should have drawn this more straight. Imagine that there was no sigma phase. In that case, I would have the taut rope like this-- single-phase, spinodal decomposition, single-phase. Does that make sense? Did I draw that decently? And now I'm going to introduce a sigma phase, and the sigma phase is going to be like a plunger that comes down. And it pushed my taut rope into in a new configuration. So now my rope got pushed down, around. So this phase came down and pushed the taut rope, and now I have what I want to have, which is I have solid solution BCC, two-phase region, solid solution sigma, two-phase region, solid solution BCC. Thank you for that. So you can see how introducing the intermediate phase disrupts the spinodal. And so we're going to spend some time now in Thermo-Calc. So what I want to do is I want to share my Thermo-Calc screen. OK, can you see Thermo-Calc? Is it open? AUDIENCE: Yep, we can see. PROFESSOR: OK, wonderful. So what I'm going to do is I'm going to generate the iron chromium phase diagram. And by the way, this stands in as a bit of a tutorial on using Thermo-Calc. And there are also tutorial videos on Thermo-Calc available on the course website that I shot previously. So if you find yourself stuck on just how to navigate this software, there are resources available. So what I did is I clicked on Phase Diagram because that's a good template to start with. All right, here I am. So I was able to define the system-- system definer. You see this project window here? I'm going to keep iron demo database, and I'm going to click on chromium and iron. And so the generic project has the system definition, an equilibrium calculator, and a plot renderer. And the equilibrium calculator is configured to calculate the phase diagram at 1 atmosphere as a function of temperature and composition. So the first thing I'm going to do is just run this and generate a phase diagram. And I did save a version of this project with everything already computed, sort of like on a cooking show. I've got the cake baked in the other oven. But I want to at least step through some of this in front of you so that those of you who have not seen Thermo-Calc used in certain ways will feel a little more comfortable with it because we're going to be using it a fair amount on the upcoming problem set. And so what it's doing is it is solving the common tangent construction. It's solving the taut rope construction, which is also called the convex hull construction. If we were getting into the numerics of this, it would be the convex hull-- solving for that convex hull solution for the system. And although the iron chromium phase diagram is already known, Thermo-Calc does everything from scratch. So here we go. Here is the iron chromium system. Just as promised, it has this intermediate phase. Now you've probably discovered that sometimes mousing over is effective and sometimes it's ineffective. Right now, it's telling me I'm in BCC A2, if folks can see that-- BCC A2. So it's a BCC phase that Thermo-Calc has labeled as A2 for some reason. And it's this wide rating of solubility. And here is another solid solution. And for some reason, Thermo-Calc says insufficient information, but that's not true because all I need to do is mouse over the two-phase region. And it tells me this is BCC A2 plus sigma. And that's what I was expecting to see. This is a BCC and sigma two-phase region, and here is also the BCC and sigma two-phase region. And down here is a BCC and BCC region of spinodal decomposition. There's one more thing which I want to highlight, which is kind of interesting. For pure iron, pure iron transforms from FCC at low temperature-- from BCC at low temperature into FCC and intermediate temperatures and back to BCC at high temperatures. Iron is a funny creature in that way-- BCC, FCC, BCC. And we've reminded a number of times in this class that if you look along the y-axis of binary phase diagrams for pure components, you're looking at a slice of a unary phase diagram. So I want to show you-- quickly, I'm just going to switch over to PowerPoint-- how we are, in fact, looking at a slice of the unary phase diagram. So here's the unary phase diagram of iron. Unaries are able to plot temperature and pressure now on one two-dimensional sheet because we have no composition variable. And if we focus here at 1 atmosphere-- that is, 1 bar-- we see alpha iron BCC, gamma iron FCC, delta iron back to BCC. So BCC, FCC, BCC, liquid-- BCC, FCC, BCC, liquid. So iron is a funny creature that way and, of course, that has implications for steel making. And anyway, back to the binary, you can see that behavior by tracing pure iron along the right-hand side of the binary phase diagram. OK, questions on that? I want you to be able to look at binary phase diagrams and read off some information about the energy systems. For instance, chromium appears to melt at around nearly 2,200 Kelvin. You can just read up the y-axis and say BCC, BCC, only one structure of chromium, until it melts-- becomes a liquid. So you can get a lot of information from these. But I wanted to show you for energy composition diagrams. So what I'm going to do is I'm going to ask Thermo-Calc to plot for me, for energy composition diagrams at three different temperatures, in this region where it's just a simple spinodal, in this region where there's an intermediate phase, and this region, right above the end of that intermediate phase region. So I'm going to plot 700 Kelvin, 900 Kelvin, and 1,114 Kelvin, right there above the end of that region. And so I want to show you how to do that. First, I'm going to rename these modules, just to make things visually simpler. You don't have to do this, but you can do this. So this is Phase Diagram Calc. That's the calculation of the phase diagram. And I'm going to rename this Phase Diagram Plot. So now I want to create a new successor to the system definer. It's going to be an equilibrium calculator. I want to calculate the equilibrium of this system. I'm going to rename this. I'm going to call this 700 K Calc. I'm going to calculate some properties of this system at 700 Kelvin. Over here, to my setup window, I'm going to make this a one-axis calculation. So I'm going to be at 700 Kelvin, I want to plot mole fraction, and I want my access to be mole fraction from 0 to 1. So what I've just done is I've set this up to calculate the equilibrium properties of the system at 700 Kelvin as a function of mole fraction iron for iron going from 0 to 1. And, by the way, while I'm at it, I'll go over here to the phase diagram calculator and change this into mole fraction so that I can deal with mole fraction instead of mass percent. So this calculator here is going to calculate properties of the system at 700 degrees. Now I want to plot that when it's done. So I open up a plot renderer, and I'm going to call that plot renderer 700 K Plot. Again, you don't have to name these things. I can name them Apple and Banana. But it helps me keep it clear. So the 700 Kelvin plot, I'm going to have the x-axis be composition mole fraction iron. On the y-axis, I want free energy, Gibbs energy per mole for all phases. So I go back up here. I'm going to perform the tree. And it's going to do all the calculations in this tree. And again, if I apologize once, I'll apologize 100 times for turning these parts of lectures into software tutorials. I don't like being subject to the whims of software and waiting for software to run, but I think this is important, not only because we ask you to do these things on P Set 8, but because it really is quite an important job function for people that are designing and making materials for a living, which is where many of you are headed. And I will also mention that one of the originators of this approach to designing materials is Professor Greg Olson, who will be giving a guest lecture in this class next week, talking about some of his experiences in industry using CALPHAD software to design really cool real-world things. All right, with that justification, it should be finishing-- wonderful. So now I have two plots. I have the phase diagram plot, which for some reason is still in mass percent, probably because I still have it plotted as mass percent. And then we have the 700 K plot. This is a plot of free energy, has the function of mole fraction of iron at 700 degrees for all phases plotted. And so you're a little bit familiar with these sorts of plots now from P Set 6, I believe. This is the free energy of chromium-rich BCC, and then the data kind of flatlines. And the data flatlines not because this suddenly becomes composition independent, but simply because the software-- the database doesn't have data for this region. And so it extrapolates. It does a weird thing. The programmers decided to make it flat. This flat region is not real data. And you could say the same down here. This is iron-rich BCC. There's a solution model. It looks real. It's got some curvature to it. It looks as we expect for, let's say, a regular solution model. And then that flatlines. So what's going on? What do we actually expect here, between these two regions of the solution model? I'm moving my mouse-- AUDIENCE: A common tangent? PROFESSOR: A common tangent. So one funny feature of Thermo-Calc is if, instead of plotting all phases, you plot system, it will stop plotting the different phases with different colors, and it will just plot the free energy of the system as a whole, as one line. And so although you lose the common tangent construction-- it becomes a little bit not obvious where the tangent starts and stops-- you do now get an accurate plot showing the free energy of the system as a function of composition. So this is a straight line right there. That's your common tangent. And we can flip back and forth between these views, between all phases or system. Another funny feature, or you might call it a bug, in Thermo-Calc is that if, instead of plotting per mole, you plot for no normalization-- the actual numbers don't change, but the way that it extrapolates to regions of no data does change. So the actual data has not changed. This is still the solution model for chromium-rich BCC. This is still the solution model for iron-rich BCC. And it's the same data. But for some reason now, when you plot no normalization instead of per mole, it decides to add the zero point and makes a big straight line. And these crossing straight lines-- they're not real data. You should not pay any attention to them. I won't make excuses for Thermo-Calc. It's the way it is. So getting back to the actual thermodynamics, at 700 degrees Kelvin, we expect a spinodal-like system of phase separation between chromium-rich and iron-rich BCC. And that is just what we see. Now let's do it at 900 Kelvin, where we have an intermediate phase. So equilibrium calculator-- I'm going to rename this. I'm going to call it 900 K Calc. And I'm going to make this calculation at 900 Kelvin, one axis, and again, I want my axis to be iron composition. I'm going to change my composition unit to mole fraction. And I'm going to make sure I can plot the results. So I'll call this plot renderer the 900 K Plot. And I'm going to plot composition and mole fraction on the x-axis, and the y-axis, again, I'm going to plot Gibbs energy per mole for all phases. Let's see what happens. Looks like the point density is a little low. Instead of plotting 50 points, let's plot 100 points across the axis. I have to start this again. That looks a little better, a little smoother. So here I have a solution model. Point density is a little low-- lack of data. In intermediate phase, lack of data and another solution model. And if I plot this for the system instead of all phases, you now see the solution model is connected with tie lines. And looking back at the phase diagram for 900 Kelvin, solid solution two phase-region, intermediate phase solid solution, two-phase region, solid solution. I was going to finish this and plot it at 1,114 Kelvin, where you can just see that it becomes a single solid solution almost across the entire composition range. I'll do that in a minute if time allows. But I want to pause and ask for questions, either on the thermodynamics or on the software. OK, I want to show you one more neat thing before I come back to this at 1,400 Kelvin. I mentioned, I think in the last lecture, what happens if you were to imagine the nonexistence of a particular phase? I think it was a liquid. I said imagine that we could turn the liquid phase off, remove it from consideration. How would the phase diagram change? Now I'll ask you, imagine if the sigma phase didn't exist. How would this phase diagram change? What would you think would happen? AUDIENCE: Would you just get regular BCC-BCC separation, like for the lower section? PROFESSOR: Right. Let's see that. I want to show you how you can do that. Go up to My Project and go to the System Define. The system definer, we chose the atoms-- chromium and iron. But there's a bunch more information here. Species-- chromium and iron. There's also phases and phase concentration. And these are the phases over which the system calculates equilibrium. Each one of these phases is a solution model for which Thermo-Calc has at least some data. So here is kind of like the guts that we've been playing with, the solution models. And I'm just going to uncheck sigma. So now what the system is going to do is everything it did before, except it's not going to include the sigma phase. So you can kind of play God in going to perform the tree. So we're solving the taut rope construction here for a world in which the chromium iron system does not have a sigma phase. And so Shane gets that this would look like a BCC region of phase separation, or I'll just call it a BCC spinodal. So let's see. It might take a little bit of time. Oh, look, there it goes. Look at that. So here is the phase diagram of iron chromium in an alternate universe in which the sigma phase does not exist. And so you see we have a spinodal. We didn't get rid of the FCC phase. We didn't get rid of the liquid. But we did get rid of that sigma phase. And what do we think the free energy composition diagrams look like? They look a little more boring. Let's have a look. 700 K? There's a two-phase region phase separation. 900 K-- same thing. Well, the point density is a little poor, but common tangent phase separation. So on the problem set eight that goes out later today, we have a number of tasks for you. About half the tasks are here in Thermo-Calc and involve not the iron chromium system but a different system, where we ask you to generate these plots, generate data, then extract the data. Then we ask you to analyze data in a data analysis software of your choice-- Excel or MATLAB or Mathematica or what have you. And then the final part of the problem set is you're going to then simulate the free energy composition diagram and the resulting phase diagrams using a phase diagram simulation tool which we've built for this class. and that I'm going to introduce on Monday. So if you look at the problem set, the first part you could start on this afternoon. The second part, I advise that you could wait and start on after Monday when we introduce the phase diagram explorer. But what it's going to boil down is going into Thermo-Calc, looking at a phase diagram, then reverse engineering the phase diagram, understanding what models are underlying the phase diagram, what data do I need to populate those models, and then how do I simulate those models in another piece of software. And by the time we're done with us next week and this next problem set you will all be experts at not just reading phase diagrams, but understanding how they're constructed. So that is almost everything I wanted to get through today. I did not have time to talk about more general free energy composition diagrams for intermediate phases, mainly because the software is too slow to load. It's not going to be terribly important for this P Set. So I'm not too concerned about it. But if you want to know what you're missing, I was going to walk through-- where is it? There's a nice schematic in DeHoff of intermediate phases. And the free energy composition diagrams, the intermediate phases in general produced, and I reproduce the picture right here in DeHoff-- so DeHoff Figure 10.19. A general binary phase diagram with an intermediate phase. And here's, flipbook style, the way that the free energy composition diagram evolves to give you these intermediate phase diagrams. So just further views of the type of phenomenon which we saw about half an hour ago.
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https://ocw.mit.edu/courses/5-95j-teaching-college-level-science-and-engineering-fall-2015/5.95j-fall-2015.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation, or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: And before we get into the topic, which is educational teaching with educational technology, I just want to go back to some of the previous assignments, a little bit of logistics, et cetera. So you should have feedback on all the assignments that you can be given feedback on. The assignment where you gave a presentation to a friend, I can't really give you feedback on. I looked over it. You all made nice observations, good pointed observations, and I hope that you can see how-- and I think I made this comment on the wiki-- that you can see how just your presence, and the choice of words that you make, can have a big effect. And that was the whole point of that assignment. So that's my feedback on that assignment. The previous one, the last one that I feel like I can give you substantive feedback on is the active learning, how to incorporate active learning. As I mentioned last time, you guys did such a great job of incorporating active learning in the first time you'd submitted that, where it was just creating a course outline, that there wasn't a lot extra to add. But I added comments whenever possible. One thing that I commented on at least two people's homework, the one about active learning, was the idea-- you mentioned you doing demos. So there's a very interesting piece of work out of Eric Mazur's group at Harvard about the use of demos. And that paper is called "The Crouch and Mazur Paper." And it's a optional reading in, I believe, it's the active learning session. So this is the active learning class, which is class 5, right? Am I right about that? Or it was 6? AUDIENCE: It was 6. PROFESSOR: 6? AUDIENCE: Yes, 6. PROFESSOR: 6. AUDIENCE: [INAUDIBLE] PROFESSOR: Oh, OK. So it's 5. Yeah, so that's a really interesting paper. Because I think sometimes we say, oh, I'm going to do a demo. Apriori, a demo isn't active. I mean, it's active for you. You're doing the demo. But it's not necessarily active for students. And if we enter into the enterprise of doing a demo thinking we're being-- well, thinking they're being active-- they might not be. And Croutch and Mazur did an interesting study where they looked at-- they were doing a freshmen mechanics class. And they had amasses on an air track, when you have like a massive 2M, and you slam into it with a mass of M. What happens to the mass of 2M? What happens to this mass of M that was coming in, that kind of stuff. And so that it would show the students these demos. Oh, look what happens. Here comes a mass of M at a velocity, da, da, da. And then when they asked students in the same class at the end of the same class period to say what happens when a mass of M collides with a mass of 2M, a mass of M going x miles an hour collides with the mass 2M, what happens, students got it wrong. They had just seen the demo. They had just seen the demo. And they still got it wrong. And I think that's a really profound finding. Because as experts, we think, of course, they're going to see this. It's going to be emblazoned into their brain. They'll never forget it. But they never learn it, basically. So it's more evidence for active learning for sure. But what they found out was they forced students to make a prediction, to discuss that prediction with somebody else before they saw the demo. And when they did that, the scores on these sort of post-demo quizzes, if you will, went way up. So it's a small tweak, but it's a profound one. Because you might really just be wasting your time if you show a lot of demos and you don't ask students to make these predictions beforehand. This goes back to the mathematician George Polya, who wrote in the '50s about asking people to make predictions before you gave them the answer. I'm not sure he had any data for it. But this was his thing. This is pretty good data suggesting that you really should ask students to make predictions before showing the demo. So a couple people brought that up. And I wanted to make sure everybody heard that. OK? If you can't find this reference, let me know. It might be hiding somewhere else. I'm pretty sure it's in this-- it is? OK. OK, so that was that. But overall, a great job on that assignment. And then I briefly looked over the assignment it was due today or whenever creating prompts at various levels of Bloom's Taxonomy or a taxonomy of your choosing. Again, I thought you did a great job. I'd like to hear a little bit of reactions to that assignment. Was it difficult? Was it easy? Do you see how it could be useful? Does it seem like a waste of time? AUDIENCE: I think it was very, very useful. Yeah, it lets you see what the students will go to when you [INAUDIBLE]. Otherwise, because it's like you are solving too. So you kind of feel and see what level to take them to. It's very good. PROFESSOR: Great. AUDIENCE: Otherwise, you don't know. You just [INAUDIBLE]. And then you don't know how [INAUDIBLE] PROFESSOR: Exactly. And I mean, it's happened to me when I first started teaching intro materials. [INAUDIBLE] was like, OK, they'll calculate the stress on this plane, and blah, blah, blah. And you're like, well, why are they doing that? What is that getting them? And it's just this reality check for you, so that you don't go off into the weeds sort of. So, yeah, [INAUDIBLE]? AUDIENCE: I think [INAUDIBLE] most of very good question that [INAUDIBLE] They're [? setting ?] very good questions and [INAUDIBLE] more than just one, [INAUDIBLE] three at the same time. So I think it's really very good. It's a good experience. PROFESSOR: Right. And also to make sure you're aware of that, and then also decide, well, if they don't remember the value of the universal gas constant in these units, but they do the apply part fine, they just use the-- so it's important, we always have to think about that when we grade questions. But I think by parsing it out like that, it makes you more cognizant of those issues that might come up, in terms of how the students respond to the question. AUDIENCE: I thought it was actually difficult to make a question that only satisfied one of-- like applied [INAUDIBLE] Because everything I was asking [INAUDIBLE] to try to apply, was also analyzed. Because I tried to-- I guess i could have just made one that did both. But I was sitting and I was like, no, OK, not this one. But really, I almost left it blank and said, I'm stuck. I finally come with something that I think just answered the [INAUDIBLE]. But it was interesting to think about. PROFESSOR: Right, right. And you know, in real life, you don't have to just have one. You can embed them together, but you know, recall and apply, or whatever. And I think for some topics, it might be easier to add to silo-ize it a little bit. Given the data, construct a whatever, so that's more of an apply. But there may be classes of subjects where you can't really do that quite the same way. Other reactions? AUDIENCE: I was also thinking of it in terms of grading. Like if you have a rubric or you have assigning the number of points based on which level you were asking. Like as opposed to say, OK, five points for each question, to say, OK, you get one point for the remembering and then you get 10 points for the creative. PROFESSOR: Right. Right. I would be careful-- I mean, I think that's a fine idea. I would be careful to make sure that students know that, so they don't kill themselves memorizing pi out to 25 digits or something, right? So if you do have these sort of, this is what I think is important, you'll need to specify that. I mean, and a rubric could incorporate that for sure. But you do want to make sure students know what your logic is. AUDIENCE: I think also there are some questions you said that you give the formulas to the equation. Just to make sure that the student [INAUDIBLE] PROFESSOR: Yes, yes. Right. And in most classes, that's totally reasonable. You know the joke I make about it's rare to real life that somebody comes in and says, I need this answer right now! Don't look up anything. I mean, it's kind of a weird pretense. So it's usually reasonable to just put the equations at the bottom, or write it in the problem statement. Anything else on that? OK, so today we're going to talk about teaching with educational technology. I had written on the website to bring a web-enabled device. If you didn't, I have an extra laptop. And we can share. It's shouldn't be a problem. As part of this, I hope that you'll be able to identify appropriate educational technologies to advance an intended learning outcome. And I want to get back to that in just a second. And maybe describe best practices and potential pitfalls of particular educational technology, so somewhat modest goals. I think it's a really interesting thing to think about technology-- and I didn't make this up. I heard this at a conference. So they said, when you walk into Home Depot or one of these big home improvement stores, if you happen to be blessed by having an employee come up to you-- because sometimes it's hard to find an employee in those stores-- what do they ask you? AUDIENCE: Is everything OK? Is there anything I can help you with today? PROFESSOR: Right. OK, is there anything I can help you with? If and you say, yeah. Yeah, you can. And then they say? Usually, it's, what are you working on? What are you trying to do? They don't say, what tool do you want? I mean, they might. But if they're good, they don't. And I think that's a word the wise when we think about educational technology. It's not, what looks really cool and fun, and what tool do I want? It's, what am I trying to build? What do I want my students to build? What am I trying to build? What am I trying to accomplish? And then I can step back and say, which of these technologies is going to help me do it? And it's easy to get wrapped up-- I mean, we're all kind of techie, geeky. I mean that is a compliment. But it's kind of fun. Like, oh, cool, look what this can you. And then we go walking around looking for something to do with it. And that might be OK, but you have to check your assumption. You have to check your assumptions about why you're doing it. Are you doing it just to be cool? Or are you doing it because it's really going to help student learning, or it's going to help advance the learning outcome? We can use educational technology as some kind of a scapegoat, but that's true for anything. It's true for that amazing exam question that you thought up in the shower. Like, I know what I'll do. They'll do this, and do that, and blah, blah. Maybe it's cool, but maybe it's not going to get you anywhere. Maybe it's not consistent with your learning outcomes. So it doesn't have to be technology. But often, technology is kind of pretty shiny. And we tend to gravitate toward it. But it's remembering that we're using it to advance some particular learning outcome, OK? So you guys did the readings. What are educational technologies? What does it mean? Can we come up with a definition? AUDIENCE: Yeah. I think from what I can understand, educational technology is a tool which, perhaps, the student and also have the teacher. So that means something that can enable inter-communication between the teacher and the student. And then the student and the student. PROFESSOR: OK, so we have a lot of ideas in that statement. So it's a tool. You said, it enables communication. AUDIENCE: Yeah, students connect. And then the teacher [INAUDIBLE]. PROFESSOR: And I think you had something else in there. Did you? Or no? Somebody else? [INAUDIBLE]? AUDIENCE: Education technology is like a tool to enhance students' learning-- enhancement of student's learning. And enhancement of instructors' teaching as well. AUDIENCE: The technology part is like a electronic device. PROFESSOR: So it has to have electricity? It has to involve electrons? No, that's everything. Wait, let's let Michelle-- what should we say about that? AUDIENCE: I think that's kind of correct to say it has to be an electronic device, right? Like if it's a clicker or if it's a-- [INTERPOSING VOICES] PROFESSOR: It must be electronic. That sounds so like 1950s. I think we had-- yeah, Gordon? AUDIENCE: I would just adjust it a little bit and add in front of the two [INAUDIBLE]. AUDIENCE: A kind of tool that's created [INAUDIBLE] by time or place. PROFESSOR: So it lends itself to asynchronicity? AUDIENCE: Yes, something like that. PROFESSOR: Most-- Anything else? AUDIENCE: I think it's also enhancer, easier use of materials. PROFESSOR: Facilitates-- AUDIENCE: Yeah, materials like. It has to [INAUDIBLE]. PROFESSOR: Right, a distribution, aids distribution? AUDIENCE: [INAUDIBLE] PROFESSOR: Did I hear something else? AUDIENCE: [INAUDIBLE] PROFESSOR: Efficient, that's a loaded word. Anybody have anything else? Or does somebody want to make a comment on this collection of attributes? AUDIENCE: [INAUDIBLE]. PROFESSOR: I don't think I have any cynics in this-- well, the cynics haven't said anything. AUDIENCE: I think all the comments are positive things. Like if you're doing it well what should happen. But they aren't necessarily what actually happens. Because most of the time, it's probably not done well. Using technology is not optimal. PROFESSOR: OK, that's one observation. It's sort of, you know, that rhyme, there was a little girl who had a little curl right in the middle her forehead. And when she was good, she was very, very good. And when she was bad, she was horrid. So it's a little bit like this, right? When it's good, it's very, very good. And when it's bad, maybe it's even worse than if you hadn't bothered. But-- Alex, you had something to say? No? AUDIENCE: Well, on the side of the cynics, I thought that students are often disinterested when they don't see the point, or don't see them using it later. It's not necessarily a definition of it. PROFESSOR: Right, right. No, but that's a very good point. And somebody had quoted this idea of an app, that if an app doesn't load in something like seven seconds, people stop using it. Like if you try to get an app to open on your phone, and you're like, what's going on here? Come on, 1, 2, 3, 4-- OK, forget it. And that's a similar idea, that if it doesn't look like it's going to help you-- and whatever that means, help, it's self-defined, help. But if it doesn't look like it's going to help you, you're not going to buy into it. Some other thoughts about this list? Personally, I don't think that to be an educational technology it has meet all these criteria at once. That's an important thing to note, that it might improve productivity. The old school learning management systems that you put up problem sets, and you put up-- the grade book was in one place, and that kind of thing. That might help your productivity. You're not looking around for stuff. But it's not necessarily promoting learning at any level. It's a delivery system, a materials delivery system. So it kind of does this. And it does this. But it might not really be part of the learning enterprise. It's a store. It's a repository. And that's fine. There's nothing wrong with that. But that's one type. If you're saying that it enables communication, well, that might be part of this. But it might be something else. If I say, all right, it's a wiki. And I want you guys to read what everybody else wrote, and to make a comment on at least one other persons' posting, well, I have facilitated communication between you in an asynchronous way. I didn't tell everybody to do it at noon on the 29th of October. But it does facilitate communication. I think this one is a loaded one, for sure. And it gets to Michelle's comment about if you did it right, it would do that. But if you did it wrong, it's likely to not do that at all. So these are attributes that you could use to describe some educational technologies. I do have a definition from the reading book. But before I put that up, any other thing you want to add? It's interesting to think about it, like does it have to be electronic? So I have had classes in the past where people argue, OK, well, a blackboard is an educational technology. I mean, you could do that without some sort of human-made product. And without it, I would have to be walking around speaking very slowly and hoping you took notes, or walking around with-- I don't know-- a piece of paper that I wrote on, and sharing it with you or something. So a blackboard does fit a lot of these categories, right? It's an efficient delivery system. At some level, it enables communication. It's a distribution of learning materials. So we want to think about whether it means it has to be electronic. And you may have guessed by now, there is not a single definition, so that's the other issue. Comments? Yes, go ahead. AUDIENCE: Just a comment. What about if it's not well-moderated [INAUDIBLE]? PROFESSOR: Yeah, that's the efficient one, that we've all gotten sucked into websites, or sucked into, perhaps, an online discussion, or commenting on something. And it might be interesting. You might learn. But it may not be efficient at all. Or you might just be on the screed end of things, where people are just either ranting or they're reinforcing misconceptions or whatever. So yeah, you do have to kind of make sure people are using it carefully. But that's true for virtually anything you do. Any other comments? AUDIENCE: And I think, as well, that the fact that these tools are valuable. A student can decide to stay home in the bad weather since he knows this is available. Then he can go [INAUDIBLE]. That could be good and bad. Because the student would continue to learn on his own and become a good student. And then he could miss out on some critical points as well. PROFESSOR: Right, right. But that implies some sort of like web-housed delivered repository of content, or of video lectures, or whatever. You could have a course that used clickers, let's say, that didn't even have a website, but they used clickers. So that course might be using educational technology, maybe in a very, very good way in the classroom, but they don't even bother with a website. So in that case, the student has to come to class if they want to experience that particular use of technology. I think you're right, that if you do have a website, or you do have lectures online, or you do have rich content online, then it does sort of, perhaps, make being in the classroom a little bit less necessary. I will say, if you employ active learning methods in class, group discussions, et cetera, that being in the classroom is, certainly, I would argue, is value added for the student. So it's not a replacement experience, but it might be better than nothing, let's say. It might not. Yes, Gordon. AUDIENCE: I found something written very interesting when it said that if you use clickers-- I would love to do that. To take attendance of the students, that it might [INAUDIBLE] the students from using technology. But it's only [INAUDIBLE]. It might actually be good to have technology they can use non-evasively to find out if the student is actually coming to your class. AUDIENCE: What if they gave their clicker to their friend? PROFESSOR: Yeah, it's true. AUDIENCE: One person will come with five friend's clickers, the clicker won't-- AUDIENCE: If you have like RFIB technology or something like that that can use to [INAUDIBLE] come to the class to know that, OK, this person is the one using [INAUDIBLE] [INTERPOSING VOICES] [INAUDIBLE] AUDIENCE: So in certain courses you have to do that. AUDIENCE: Yeah, you must do that. AUDIENCE: You have to show up? AUDIENCE: Yeah, you should show up. But if you decide it's difficult to take attendance, and if you have this wonderful way of getting the students to [INAUDIBLE]. AUDIENCE: [INAUDIBLE] PROFESSOR: Well, I don't know if there's an answer to that. For some students, it's never a problem. For some students, it's awesome. For some students, it's just more information. And it's just another path. And they're learning. And they're learning. And they're learning. And it's supplementing what they're learning in the class. It kind of depends on-- AUDIENCE: Taking attendance that Gordon is talking about, so is it [INAUDIBLE]? PROFESSOR: Well, yeah. And actually that's an important point is that I think I showed these slides about clicker use in 511 1. I showed when we did active learning. And the first year they used them, they used them more or less to take attendance. And students figured that out pretty quickly. They did all sorts of things, sharing clickers, clicking and then leaving, or coming in at the end and clicking, but not being there the whole time. And if you ask students, well, what did you think about the clickers, they're like, fu. This is crappy. This didn't help me. This is terrible. And probably it didn't. There's an attitudinal thing that says at some point, if you think it's not helping, it's not helping. So the next year, they made it a point to really incorporate it into the class, to ask good questions, rich questions, questions that had interesting and informative distractors. So the wrong answers told the instructors what the problems and misconceptions were. And this was made clear to the students. And in that way, if you ask students what they thought about the clickers, they were not at all resentful of the use of clickers. And they did not see them as a way of just taking attendance. They didn't think they were sort of baby or whatever. And they were therefore more effective. [? Rachel, ?] you had a comment. AUDIENCE: Yeah, I was just going to say going to 802. PROFESSOR: That's the E and M, Electricity and Magnetism freshman year. AUDIENCE: The class I took it in [INAUDIBLE] So we did all the online [? courseware ?] and then the idea was that you come to class and you just learn for two hours from the teacher using group discussions, and all these other things. And it was one of the worst classes I've ever [INAUDIBLE]. Because the professor wasn't prepared to do these two hour discussions, like group discussions, and group work, and all these different things that you could tell he wasn't used to doing. And it was just a giant mess. So I think they're really useful tools. But they also require the professor to really put something into it for the students to get something out of it. PROFESSOR: Right, right. And that's a consideration. For some things, you can adopt them. There's kind of a low bar for adoption and a low bar for solid use, maybe not innovative, maybe not groundbreaking, but a solid pedagogical use. For others, if you're going to use them effectively in the classroom, you're going to have to spend some time engaging with the technology, making sure you understand it, making sure you understand what you can do with it. Otherwise, it's not going to work. A couple stories, one is the idea of-- well, maybe I'll hold that one. I'll hold that one. I'll bring it up later. Remind me. Any other comments about this? So this is a definition that I got, I think from one of the [? reasons, ?] so technological processes and resources that are used, created, or managed for learning and/or improving performance. So that's crazy broad. And then in that definition, there's this technological, which gets around the electronic, but brings back the idea of well, a blackboard is probably a technology at some level, or at least it was at some point. So it's admittedly an incredibly difficult concept to pin down. And maybe at some point, you don't necessarily have to. You have to decide which slice of it you want to deal with. And so the other important thing is that it's-- and it doesn't have to be, but it can be-- an integral part. Or it should be well-integrated within the course, and or it could be a supplemental learning aid. So it could be a way of delivering content outside of class, or the idea of an enrichment of the material. And it could be something like hardware, really just like hardware. I'm going to come back to how we chop this up in just a second. Remember that we don't want to do any thing-- this is the idea about walking into Home Depot. We don't want to do anything without thinking first about our intended learning outcomes. What intended learning outcome is the technology going to advance, not, what am I going to do with this technology? And so in terms of categorization, this is something that I came up with just after grappling with things. So this is just my thing. And we're going to talk more about different ways to think about it. So this is just one slice of it. You could think about technologies that you use in the classroom, so that would be clickers, that would be using some sort of projection system, using a smart board, using some sort of simulations that you show students, et cetera, like virtual demos. It could be a way for students to engage with the content and with other students outside of the classroom. So it could be some sort of a wiki, some other collaboration tool, and/or it could be something that delivers the content to the student outside of class. And there may be an overlap. You may show a simulation in class, and then decide students should have access to it outside of class. So they're not mutually exclusive. And so the engagement with the content can include these collaboration tools, extra materials, content enrichment materials. And then the third one might be student assignments. So the idea, maybe you're going to say to students, OK, I want you to create a website for a company that you're going to create, and this parameters. And you want to make sure you're explicit about what the company sells, and how you deal with customers, or whatever. The assignment involves the technology. Go out and make an audio recording of people using a particular technology. Or if you're going to design something for use in some community, interview the people in the community, and submit this audio recording. So it could be that technology could be used in the student assignments. And you could back that all the way up to online problems as they use in MITx. And then for assessment and feedback, the idea that you can either use clicker questions, which would be for the formative feedback. Or you can use it as some sort of an online quiz, or some other way of giving feedback to the students. So that gets to the summative and the formative idea. And then the idea of just using a learning management system, just to administrate the course. That's one slice. There's another couple other very interesting ways to slice this up. And I have a handout here, because these are woefully tiny, these schematics. So if you just take one and pass it along. One is called the SECTIONS framework. And the other is called the backward Ed Tech Tool Flowchart. So these are different. These are not mine. And these are different ways of thinking about technology. So the first one is this Ed Tech chart. And I know you can't read that here. So that's why you have the handout. And I like it, because it kind of starts with the activity that you want students to use. So it starts with this, what do you want students to do? And that's what the purple box are, the student tasks. And then it kind of creates these sort of guiding questions that sort of reroute you. And then it gives you a suggestion for a tech tool. So I'm going to give you a few minutes just to parse that, just to engage with it a little bit. AUDIENCE: It's very beautiful. PROFESSOR: I like this one. I like this one. I think it's very useful. Any other comments or thoughts about it? Yeah, Gordon? AUDIENCE: I think it's [INAUDIBLE]. PROFESSOR: OK, Gordon doesn't get it. Somebody want to-- maybe nobody gets it. But do you want to explain it a little bit? Out loud, David, please. AUDIENCE: Yeah, [INAUDIBLE]. PROFESSOR: But can you explain to everybody? AUDIENCE: [INAUDIBLE] PROFESSOR: Oh, sure. AUDIENCE: OK, [INAUDIBLE] PROFESSOR: It's certainly not exclusive. I mean, it doesn't have everything on here. But it kind of gives you, I think, a way of thinking about, OK, I start with what I want students to do, which most of us have a pretty good handle on. And then it helps you just think backwards to the technology. [INAUDIBLE]? AUDIENCE: Yes, I just want to add that it's only about what you want to do, and how can you did it. PROFESSOR: Yes. AUDIENCE: That's what it's all about. And secondly, [INAUDIBLE] but what you want them to do, and how do you have them doing it, that's what [INAUDIBLE]. PROFESSOR: Yeah, Yeah. No, it is great. Other thoughts? And yeah, we'll get to get back to the idea of the taxonomy shortly. Other thoughts? Let's see, there's a second one, which is on the other side of your handout. And so these are just-- I know you can't see this. That's why you have the handout. These are just ideas. Again, there's no rules really. But depending on what you're trying to do, it just may help you start to think about how to get the right tool for the job. So it's called sections, because each of those boxes with the red outline, S-E-C-T-I-O-N-S. The S is student characteristics. The E is ease of use. C is cost, teaching and learning, interaction to promote active learning, organization, novelty, and speed. And this chart, I think-- I'm going to let you engage with it for just a minute. But it brings up the idea of the learning curve for the instructor, which you don't want to lose sight of, that you want to factor that into your choices about what technologies to use. So again, I'm going to shut up for a minute or so, so you can look at this. AUDIENCE: [INAUDIBLE]. PROFESSOR: Right, right. But what's the difference between this one-- I mean, there's many differences, obviously. But what's kind of a utility difference? AUDIENCE: One's a flow chart that points you to the answer. So it asks the yes or no questions. The other just gives you the-- then you have to select from a-- it doesn't guide you directly. PROFESSOR: Right. Exactly. It kind of helps you think about it. Oh, my gosh, I have no money. So that's certainly going to rule out a whole set of options for you. But it doesn't say, oh, you have no money, think about these. It's not a flowchart. It doesn't point you to an end result. But if you went through this exercise and answered these questions, I think you'd be in a better place, with respect to choosing the particular technology. Gordon? AUDIENCE: Yeah, also I think that second one is not [INAUDIBLE] if you want to use technology and you have the money, [INAUDIBLE] the things you can think about. Because if you don't think of everything [INAUDIBLE] PROFESSOR: Right. It's some things to just keep in your head when you're making these choices. It's not necessarily a formalism for making a choice. So one other slice of all this technology is kind of a classic, I would say classic, model, which is called the SAMR model, S-A-M-R. I know I'm throwing a lot of models at you. But you have to kind of find the slice that works for you. The SAMR model, I think, is very useful to keep in mind when you're deciding why you're gravitating to a particular technology. And it's this idea of, is the technology just a substitution? So back in the day, when everybody said, oh, courses need web pages. You have to have a course web page. Well, what was the course web page? It was an electronic version of your syllabus, basically. It had the name of the course. It had some picture. And then there was a link. And you could click on it, and the students got the syllabus. They got a text version of the syllabus, a soft copy of the syllabus. But then they printed it out and brought to class. So it was a one-to-one substitution. Instead of seeing the syllabus in a piece of paper, they see it online. It was complete substitution. It didn't do anything extra, except maybe made it a little easier for students not to lose the syllabus. But it didn't really have any functional difference. Then the next level-- that's the S. The A is that there's an augmentation. So there's a substitute, but with a little bit of an improvement. So maybe students could hand the homework in online. They're still handing homework in. They're still probably writing their homework out, scanning it, and handing it in. But it's a little easier. It's a little bit easier. Maybe you don't lose their homework, that kind of thing. So that's the augmentation. The next level will be a modification, so some sort of serious redesign of a task that you couldn't do before that technology. And then the final would be it just redefines the tasks that you do, the tasks that students do, in a way that's completely different. So I'm going to ask you to think about that for your own courses in about two minutes. There's a little bit of a schematic here that might help. These are sort a little bit baby. But the idea of substitution, so the idea of using a word processor instead of a typewriter, that was a substitutional technology. So then augmentation, you get spell check. Once you're in a computer, you get to have spell check, and formatting, and cutting and pasting. And that's different than what you can do on a typewriter. And then so modification, well, there was like an email, or you could put graphs and images right into the document. It changed sort of the way we viewed a whole document. And then the idea of having hypertext, or some sort of a web page that students navigated in a nonlinear, or a non-prescribed way, that would the redefinition, let's say. So that's some sort of baby examples of that. And then this next one, which is pretty fun, is there's this woman, Kathy Schrock. And if you Google Kathy Schrock, she comes up. She's got a ton of stuff. And this to get back to Bloom's Taxonomy there, is that she's taking Bloom's Taxonomy, remember, understand, apply, analyze, evaluate, create, and then she said, OK, well, here's some apps that can facilitate those things, facilitate the act of remembering, or facilitate the act of evaluating, or creating. So it's not mutually exclusive. It's not exhaustive. But it is important to remember that some of these tools are supporting remembering, or they're supporting some other level, understanding. But they're not necessarily at the highest level. Or they're not at the lowest level. It's good to be aware of that. Before I charge you with your assignment, I did want to demo a few other technologies, just so you can take a look at them. And there was that list in the reading, which I hope you checked out. I invited you to this-- well, I thought I invited you to-- oh, wait a minute. Huh. Hang on. Well, let's just start up a new one. So I can create this picture here. And I can share this, then, with you. And I can invite the board. And so this is the link. And if you have a device and want to log in there, you can check it out. And then you can write on it. Oh, somebody wrote on it. Somebody erased my picture. I don't know who that was. Oh. AUDIENCE: Is that a [INAUDIBLE]? PROFESSOR: U-Q-P-L. AUDIENCE: So that's different from the link that was in the wiki. PROFESSOR: I know. I just created a new one, because I didn't want to go look for the wiki link. AUDIENCE: But this is free? PROFESSOR: This is free, yeah. So we're working a little backwards here. I'm showing you the tool. And now I'm asking you what you might want to do with it, which is not necessarily best practice. But I have a use that I-- you can change the color. This pen here changes the color. No, something changes the color. Maybe not. Oh, somebody got blue. When I taught thermodynamics, we would often do sketches. I'd say, OK, sketch the Gibbs free energy versus temperature, or Gibbs free energy versus pressure. Tell me what the first derivative of g, with respect to t, is, things like that. Sketch the slope, sketch the curvature, et cetera. And so this kind of application is you could have students log in, do that kind of a sketch. And then people can evaluate it. It's anonymous, you can see. I can't tell who said what it, except David, who wrote his name. But I can't see who did what, so that it could work pretty well in a small class. If you had a class of 400 people, I'd be afraid. I'd be very afraid. But in a small class, you can come up with some interesting uses of it. The good thing about it is that you don't have students coming up to the board. It minimizes that motion, the sort of getting everybody unsettled and settled again. It's not a cure. I wouldn't say this redefines anything. In terms of the SAMR model, I'd say it's probably a modification, maybe an augmentation, maybe. I don't know. But it's certainly not and R. So I wanted to just demo that, because it's kind of fun. The other thing I wanted to do is, I think you guys saw this list? Did you see this list from the readings? This is just some things that I've compiled. It's kind of a mishmash. But there's some interesting things on here that you might want to explore, depending on your learning outcomes. And we've used Backchannel. Today's meet is pretty similar to Backchannel. Backchannel chat is pretty similar Backchannel. So those are some things that are pretty similar. We just looked at the web app. We've used Plickers. We've talked about Socrative, which is just a cell phone-based clicker. But there's a lot of interesting uses. And you might want to just play around with them. And I'll augment this list. So if you click it again from the wiki, you'll get a augmented list of things, a few things I want to add to it. So now, given your experience in life, and the readings, et cetera, is that I'd like for you to think about a technology that you haven't used for-- actually, what you want to first is think about a learning outcome. But you want to think about a learning outcome. You want to think about a technology that might advance that learning outcome. And then just jot down why you chose this technology, where does it fit in the SAMR model, how you would use it in your class, any kind of difficulties. And then there's a Google doc. I'm pretty sure everybody can access it. We can only hope. And if you can type your responses into that. So we can give you about five, seven minutes for that activity. And then we're going to do a lightning round based on it. So use whatever. Use the handouts. Use your knowledge, use, et cetera. And then somebody can tell me really quickly if they can get into the Google doc. Yeah, you guys are good. I was going to say, write your name down. But it doesn't really matter whether your name is up or not. So I want to make sure that you take a look at everybody else's some of the learning outcomes. We're going to do this lightning round where you're going to be talking to other people about the technology they chose, their learning outcome, they chose it, et cetera. And you're going to be trying to help your partner troubleshoot, or perhaps, implement the technology in a better way, solve some of their problems with the implementation process, or perhaps, think about a different technology altogether given their desired learning outcomes. So let's just take a minute, and either read it on your own device or read it up here, so that you kind of get a sense for what you're going to be talking about. And if you haven't read over it, please do that now. So there's some really interesting ideas up here, I think, and a broad spectrum of ambition. We had some pretty small-- like one assignment's worth of technology, perhaps. And then we also have kind of a whole course design. So that's another thing we didn't really talk about. I didn't narrow the scope in any way, or define the scope in any way. So it could be for a whole class. Or it could be for one particular assignment, or one particular class activity. And that's fine. That's OK. So did everybody get a chance to look over the other posts? AUDIENCE: I'll say, I've have had [INAUDIBLE] AUDIENCE: So it gives a follow up of all of these things is that when the AWW where everyone was drawing, I could draw here. But it was never showing up there. PROFESSOR: Interesting. AUDIENCE: Oh, yeah, my thing I drew never showed up. AUDIENCE: Yeah And then the other thing with this now, with the Google doc, I tried to edit it. And it doesn't let me click on it. It says, do you want to use the app? So then I had to download the app. And now I say, yes, use the app. And then it says something about the server not being recognized. But then if I go back and say, no, thanks, and I try to use it without the app, then I can't put anything in. PROFESSOR: Interesting. OK, so who had trouble? AUDIENCE: Yeah, I downloaded the app and used it instead. PROFESSOR: All right, did anybody that was using a smartphone not have trouble? AUDIENCE: I didn't have trouble. PROFESSOR: You didn't have trouble? AUDIENCE: No, I downloaded and use it. PROFESSOR: Oh, OK. And were you an Android, or Android? AUDIENCE: [INAUDIBLE] PROFESSOR: Android, iPhone, iPhone. Who else had trouble? It was more than just the two of you with this one? AUDIENCE: With this one. PROFESSOR: With this particular one? AUDIENCE: Yeah, [? Rachel ?] had problems with it. PROFESSOR: So for now, if you wouldn't mind, just-- let's see. We don't have to complete this right now. But maybe if there's a lull, just come on over here. And you can type your app in, just so we have a complete record-- your technology in. But that's interesting. It's a good thing to keep in mind that access might be an issue. I did try this out on two computers. But I tried it out on two computers. And I didn't try phone. AUDIENCE: [INAUDIBLE] Couldn't do. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, after is fine. So if we could get up. And anybody else that wasn't able to input their ideas, if you will do it later, that's fine. So we're going to do the lightning round. And if I could get you gentlemen to come around and fill in the back. Just fill in up to here. Great, so what we're going to do is I'm going to ask the row in the back, this row, to share their technology, their intended learning outcome, their concern. And then this row will give their advice. And it's two minutes for the whole process. And then we'll switch. OK? AUDIENCE: So what do I do? PROFESSOR: So you just have to tell me more or less what you wrote on the screen. OK, but hold on. I got to set the timer. OK, go. [INTERPOSING VOICES] PROFESSOR: Is going to share their ideas with the other side. [INTERPOSING VOICES] PROFESSOR: It's moment you've been waiting for. So we're going to do this two more times, this time and one more time, OK? All right. [INTERPOSING VOICES] PROFESSOR: Yeah? AUDIENCE: For the course which I want to teach using educational technology, so it was a big challenge, because the head of department wanted to say no. But then it continued because many of the professor already preferred it. PROFESSOR: Wow, so does [INAUDIBLE]. [INTERPOSING VOICES] AUDIENCE: Anybody can do anything. And you can reconfigure it. PROFESSOR: Alex, if you come around. And then, [INAUDIBLE], if you come around, please. Everybody slide down. AUDIENCE: We already went that way. PROFESSOR: Oh, you went that way. AUDIENCE: We already saw those people. PROFESSOR: So, [INAUDIBLE], if you come on over. [INTERPOSING VOICES] PROFESSOR: So I know in my conversations, I heard some really interesting ideas, some fascinating uses, and some incredible obstacles. I'll just point out David's obstacle is on the spreadsheet. But he wants to use his piece of software. And it costs $12,000 a year. So it sounds fabulous. It sounds really fabulous. And your department actually bought it for you? AUDIENCE: Yeah. Have it for all faculty [INAUDIBLE] PROFESSOR: Yeah, so that's great. But that's a huge hurdle. So that's an example of trade-offs that you might not always be able to make. So that was my interesting-- but I did hear some other wonderful ones too. So I hope that you take a look at it. You'll note that I always give you time after this lightning round. Because it's a very noisy kind of little crazy exercise. And I always like to give you a few minutes of silence afterwards to sort of process it a little bit. Because it can be a bit of an overload. So that's sort of the part of the design or the planning for this exercise. Anything anybody want to share about something they learned, or some issue or problem that was solved, or still unsolved? All right, I see some people are still typing. I'm going to go on. Feel free to keep adding to this. And again, please take a look at it. Because we only got so many pairings. And you can see virtually everything from the Google doc, assuming everybody gets in. So what I wanted to do is flash back a little bit to what we talked about the second class that we met. We talked about all these different learning theories, about behaviorism, cognitivism, constructivism. I think we've done some other exercise with these three learning theories before. But it's a good chance for you to reflect back, to do a little backward transfer, and to reinforce some of the concepts from the beginning of the semester, and to also think about how technologies might be-- so we talked about how technologies might fit at different levels of Bloom's, or supports different levels of Bloom's, but also, different technologies-- different anything you do in the classroom. But use of different technologies can actually be more behaviorist, cognitivist, or constructivist in nature. So I think it's useful for you to think about that just as a way of just getting a bit of a marker. About again, it's this idea about, what are we trying to do with this tool? And so I'm going to divide you up. And we have nine people left, so three groups of three, which works absolutely perfectly. Yes, three groups of three. And we're going to have the behaviorist group, the cognitivist group, and the constructivist group. What I want you to do is think about a particular technology that sort of would be consistent with this-- ooh, OK, one group will have four-- a particular technology that's consistent with that particular learning theory. And so this will take about maybe four minutes. And think about, perhaps, even think about a learning outcome that that tool advances, but at that level, so the behaviorist level, let's say. And then what I'm going to do is the activity is called a jigsaw. And so first, you're going to be in these homogeneous groups. So one group talks about constructivism. One group talks about behaviorism. And one group talks about cognitivism. And then I'm going to mix you up. So each group then has one cognitivist, one constructivist, and one behaviorist in it. So this is an active learning technique. And I wanted to model it for you. So let's just create some groups. [INTERPOSING VOICES] PROFESSOR: All right, folks. I don't know how far you got here. But now each one of you is going to be expert in your particular field. So I'm going to take-- and this is the constructivists. So here's three constructivists. I'd like one cognitivist and one behaviorist to find a constructivist, and to make a new group. All right, so now each one of you is the content expert for your particular theory. And just sort of explain what you came up with to your peers. So you'll just go around the table and talk about it. [INTERPOSING VOICES] AUDIENCE: They're still learning. But they're having an active roll in participating in when they're going to learn about, rather than just being passively taught that. AUDIENCE: And that's? AUDIENCE: Constructivist AUDIENCE: OK, so what technology [INAUDIBLE] AUDIENCE: Well, I think that a cool way to do it is to have computers set up throughout the room to have internet access. PROFESSOR: So the class is officially over. I mean, it's five of. I don't know if people have to go other places or not. So we can take this up beginning of class next time if there's things you want to finish up. I'd like to hear some of what you came up with for sure. So can we just plan to do that, and just tie up the exercise? Not the best of the best pedagogical strategy, to make you wait a week, but it is what it is, so. OK, thanks very much. And I look forward to seeing your posts on the assignment.
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https://ocw.mit.edu/courses/7-91j-foundations-of-computational-and-systems-biology-spring-2014/7.91j-spring-2014.zip
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ANNOUNCER: The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK. Welcome back to computational systems biology We have the honor today of having Professor Ron Weiss visit us. As I told you on Tuesday, he's going to talk about synthetic biology. And Ron is probably from both the department of biological engineering and the Computer Science department and also a founding member of the Synthetic Biology Center at MIT. And now, thank you, Ron. PROF. RON WEISS: Thank you, Dave. Thanks for inviting me here. Did you mention our background? Dave was actually-- advised me when I first came to our graduate school at MIT. And at the time, I was working on digital video and information retrieval. And Dave started getting into the business of biology. This is back in the early '90s. And I was like, this is cool stuff but it's really messy. How can you engineer with these molecules? PROFESSOR: And we've got the answer to that. PROF. RON WEISS: Yeah. So we'll see. So let's see if the answer is-- it does actually work. So yeah, so I-- after being a non-believer-- I don't know if non-believer, but just-- I didn't feel I quite had the engineering capabilities. Towards around '96 or so is when I decided to actually make the switch. At the time, I was working-- around '96, I was working at this notion of how can we use what we know in biology to understand how we program computers and especially situations where you have lots and lots of computing elements like-- things like smartDOS. I don't know if people have heard, but amorphous computing. So back in the mid '90s or so, this notion that we would be able to embed computation everywhere was kind of an exciting notion. And I thought to myself, where-- how could I get inspired? And I thought, well, biology obviously could serve as great inspiration. Because that's a situation where you have millions or billions of little computing elements that don't have too much power, kind of interact locally. But they still perform very robust operations. And so I performed a variety of simulations, for example, of embryogenesis and other processes to try to understand what happens in biology and can I again use that to program computers or little tiny computers. And I remember one day, I just decided to flip the arrow and basically rather than trying to use biology to understand or program computers, I decided, let me use what I know in computing to actually try to program biology, OK? And so now that field is basically called synthetic biology. So I've been in that field-- it's hard to count now, but I guess maybe 18 years or so. And it's been fun and has not been easy. But I think at least we're starting to make some-- we're making some progress. So I'll try to tell you about some of our efforts there. And I certainly encourage you to ask me questions. So please interrupt me at any point in time. Any kind of question is fair game. Dave promised me that you guys are a tough crowd. So let's see. And you can always stump. Let's try to have that happen. So when I look at this, I get excited as an engineer, OK? And I think to myself, wow. This could be really cool to be able to program something like this, again, in the same way that we may program computers. And so this notion of genetic engineering in a direct way, in a way where we can create new DNA, certainly has been around since the '70s or so. And so this notion has allowed us as a community to create various mechanisms that control what the cells do. So for example, transcriptional regulation, translational-- so being able to regulate things in a cell, be able to create genetically encoded sensors, cell-cell communication mechanisms, synthesis of various interesting molecules-- biofuels, pharmaceuticals, and control physical aspects. And so those capabilities have been around before synthetic biology. But if you were to ask me what is different about synthetic biology, I would say it's really the emphasis on systems level engineering. OK, so this notion that we are not just trying to engineer over expression of this gene or that gene or a couple of genes, but really trying to understand how to create systems of interactions. So in the same way that systems biology has come to the forefront with this notion that you can't understand a cell by understanding what is the exact purpose of this particular gene-- we always have to think about it within the context of a pathway within the context of the entire organism-- in the same way, when we want to be able to get cells to do interesting things, we have to think about the system as a whole. And to get the sophistication that we need, we need to understand how to connect these various elements, regulatory elements, all these kinds of elements, in reliable, predictable ways, efficient ways and so on to be able to get the cells to be as programmable as computers. So that if you were to ask me what synthetic biology is, that would be my answer. And so now you know it and you can tell all your friends that it's a completely defined notion and so on. Maybe if yo ask other people, they'll give you slightly different answers. But there you go. So how do we develop an engineering discipline out of that? OK, so that's really how can we get undergrads to come in and take Synbio 101 where it's really a well defined mechanism and set of methodologies and practices that allow us to do this reliably. OK, and so we often try to get inspired by how other disciplines approach the engineering of complex systems. And so kind of an obvious one would be of computing or robotics where there's this notion of, for example, bottom up assembly. And so you start with basic devices and think about how to create modules that have specific behaviors in them and then put those modules and integrate those modules to create these autonomous entities such as robots. And we often think about how to create communities of interactions, communities of robots in this case, and so on. And so that's worked quite well in a variety of different other engineering disciplines. And so we often ask the question, can we import these mechanisms into the world of biology? So can we take basic mechanisms of regulation-- it could be transcriptional, it could be other modes of regulation-- and then wind these things up to create customizable pathways that we then embed into cells? And then we can create programmable communities of bacteria. We can create programmable tissues of mammalian cells and so on. And so the question is, is this a useful and efficient way to approach things? So for example, how are these different approaches similar? What can we borrow from here that make sense to push on over there? And I will say when I started working in synthetic biology, most of my efforts were really focused on adapting and implementing these things-- so adapting them from other disciplines and trying to understand how to implement them into the world the biology. But as time has gone by and as we've started to understand and appreciate the cell more and more, we are also quite interested in how these things are different as well. So what makes engineering biological systems a truly unique, new engineering discipline? What would you do in the world of biology that might be different than what you would do with computers or robots or building bridges and cars and planes and so on? And so that that's become more and more of an important focus in my lab and I think in the community as a whole although not yet everywhere. And you often see situations where people come in from other disciplines and just think, oh, we'll just program it, engineer it, just like we do in computing and so on. And it doesn't just work like that. So when we approach these tasks of programming the cells, we usually divide things up into modules of sensors, processing, and actuation. So for example, we would want to develop sensors that can detect in live cells levels of microRNA messenger and then proteins and then connect them to synthetic regulatory circuits that we embed in the cells, OK? So it's important that these sensors not just, for example, give us fluorescent readouts, but it's important that these sensors then connect to the regulatory networks that we have in mind and so that these regulatory networks can then integrate multiple pieces of information and make decisions about actuation. So how do we turn on specific proteins that will then influence that particular cell or even the environment in a programmable fashion as dictated by the levels of particular sensors as well as by other mechanisms or, for example, from looking at historical information that the cell itself has processed as well? And so this is, I would say, represents the paradigm for most of the things that do take place in synthetic biology. And so why do we want to do this? It's not the program the next version of the iOS or iPhone or something like that. even though that initially that was one of the things that was discussed, it's not just for the sake of computation, but really for the sake of specific applications. So for example, if we have really slow logic gates that work on the order of hours or even days, that might be fine if the application, for example, is a tissue engineering application, OK? And so in synthetic biology, initial emphasis is really been on what can we do with microbial, let's say, communities or individuals, for example, for synthesis of high value compounds? I mentioned bioenergy, environmental applications as well. So that, I would say, was most of the emphasis there. But over the last few years, there's been a growing interest in health-related applications. And so my lab in particular looks at mostly health related applications. And so I'll give you examples of those today, OK? And those include things involved with cancer, diabetes, in tissues by design. And in order to do this, in order to have this programmability, you want to think about scales. So you want to think about how much DNA does it take to do X? And to a large extent, that controls the sophistication. It really is an important defining element what we do is the scale of the DNA that we can actually engineer reliably quickly, efficiently, predictably, in high throughput fashion, and in inexpensive ways. So we would start with things on the order of genes where I would say that that's really the basic elements. I would say that a single gene that you overexposes or a few genes than you inducibly express, I wouldn't count that as synthetic biology. But when it gets kind of interesting for synthetic biology is when we have this circuitry, where we now embed interactions that didn't previously exist in that particular cell context. And so most of synthetic biology has been really at this level right here-- actually, mostly from here to here in terms of the scale of the DNA and now trying to go beyond that-- more along the lines can we create something that's 20,000 bases, 50,000 bases of DNA? OK, is this something that a graduate student can come into the lab and say, I want to design something that will take 20,000 to 50,000 bases? Is this a reasonable thing to consider? OK, and then the question would be, what kind of power does that provide to you? What things can you do with that? Beyond that, people have explored this notion of minimal life and even full genome rewrites. I would say at this point, this is-- what some people clump that in with synthetic biology, which is fine. We don't have, at the moment, really good ways of being able to engineer minimal life from scratch or even in a really fundamentally different way. So most of the efforts on minimal life would be take an organism and try to figure out what to knock out, right, as opposed to saying, I'm going to engineer this new minimal organism. And I'm going to define what reactions to put in there from scratch. And I'm going to create a whole bunch of new ones that didn't exist before. OK, in the future, will we be able to do this? Hopefully, OK? Not quite yet-- this is really where the action of right now. And again, driving force for this is how inexpensive is DNA synthesis. And so we're following some kind of Moore's law with respect to dropping costs in terms of DNA synthesis. And this is one of the enabling-- I don't know if it's-- it's not the only enabling technology. But is one of the most important enabling technologies is the fact that it is less and less expensive to be able to order longer and longer sequences of DNA. And so this notion that, for example, you'll be able to design something that's, again, that's 20,000 to 50,000 bases of DNA and just go online and order to that and have your advisor willingly pay for that-- not at the level of 20,000 to 50,000 bases yet. But that's going to change. And that's going to get to the point where those really become available to everyone. And I think that's going to fundamentally change how we do business in biological engineering and how we do, I would say, almost everything in biology as a whole. So if you have-- even if you don't care about engineering new biological functions but you want to understand biological systems and your adviser told you, well, just design a whole bunch of circuits that will allow you to regulate things in arbitrary ways to learn something about the underlying networks that control a natural systems, again, I think that that fundamentally changes what kinds of questions you will ask. OK, so I'll talk about basic design. I'll talk about scalability. So how do we go from these basic elements to bigger and bigger things? And then I'll talk about some recent things that we're doing where we're building this foundation. But we think that this foundation then can matter. I think this foundation can change how you approach things that really don't have the greatest of solutions. Now, they really change the paradigm, for example, for cancer, for this notion of building tissues by design on chips and for diabetes and so on. OK, so we start with parts. So just about everything that we do, we define what are the basic parts that are available in our toolbox. And so these will be transcriptional regulatory parts. We do things at the translational level. We do things also at the protein-protein level. One of the things we often do, not always, is engineer cell-cell interactions. Could be by means of cell-cell communication. We often want to find out what's going on in the cell. So just like when we program-- where we create a new software, new computer program, we have debugging outputs that tell us what the program is doing. Usually, the way we do this is with fluorescent protein. It could be with dyes too. So they tell us, you know, here's how your circuit is behaving. Here's how the cell might be behaving. And another set of parts would be ones where we want to be able to create sensors and actuators inside the cell. What are specific biomarker levels? How can we affect what the cell is doing? For example, one that gets used a lot is kill the cell is one of the favorite actuators that people are using. Another one would be, let's say, tell this stem cell to differentiate into a different cell type. That would be another kind of actuator. A different one might make the cell-- make this high value compound that would be relevant for some application. And so right now, if you're looking for parts, they actually used to be stored in Stata up until-- or big libraries of synthetic biology parts were stored in Stata that up until about two years or so. So I don't know how many people know about iGEM. Any folks know about iGEM? So iGEM was started at MIT, was headquartered, as I mentioned again here, in Stata. There are these couple of big freezer that were on the, I think, the fourth floor here. And they stored 5,000 to 10,000 parts that word commonly used by synthetic biology folks. OK, so now they moved over closer to Cambridge brewing company. And they're not affiliated directly with MIT anymore and they have 15,000 parts or so available. So if you want to get started in synthetic biology, this is one quick way to do that. You can contact iGEM headquarters and say, please send me 1,000 parts, OK? And as long as you're credible and not from one of those blacklisted countries, then they typically will send it to you. So that's a good way to get started, OK? So what are these parts? So this is actually going back to my Ph.D. here. This is one of the parts that I characterized. So this notion of an inverter-- so digital logic convert. So I assume people here-- everybody is familiar with logic gates. Is that true? OK, raise your hand if you are familiar with it. I just want to see- oh. Just trying to calibrate-- and again, ask me questions. So this notion that you have a single input, single output device that works on binary values that has-- basically inverts the signal. So you have zero on the input. You have one on the output. One in the input, you have zero in the output. And so one of the ways in which you can implement this in a biological system is just use transcriptional repression. OK, so if you have no repressor present, then you have a high level of output protein. If you have a repressor present, it represses the production of the alpha protein. And so in theory, you should be able to use this as a digital logic gate. OK, and so that-- sounds-- looks pretty simple here. But for my Ph.D., it took me about three years to do something like this just to give you an indication. Now it's a lot faster. Now you can do this in-- you can do many of those in a day. So there has been progress. Here's another one of those gates that I used for my Ph.D. And so this is now not just a repressor, but a repressor that can be inactivated by a small molecule, OK? And so the way it works is you have this repressor that works as before. And then when a small molecule comes in, it prevents a repressor from binding the promoter. And as a result of that, even if the repressor is present, you can have activation of the output protein, OK? So this is what's called a not x or y or it implements the implies logic function. How many people use the implies logic function to do anything? OK. So it's not a commonly used logic function. And you won't find it-- there's no logic gate that does the implies logic function in a typical computer. But this is a useful logic function that we can implement in cells. And it allows external control of gene expression, OK? And so this is a simple way-- and so once you can do that as a user, essentially you can interact with the cells and modify what's going on inside the cell, OK, using a pretty simple looking mechanism that predates synthetic biology, if you will. But I don't know if it was called the implies logic function before. So anyway, so then logic gates-- can we build logic circuits? This is where I would say synthetic biology starts kicking in. And so this is one of the first logic circuits that we built. And the question was, OK-- looks nice to have this logic gate representation. In biology, does this make any sense at all? Can you really do digital logic inside cells? Can you take noisy biological components and actually implement reliable digital computation in cells? And it wasn't an obvious thing, I would say. Is it 100% obvious now? In some situations, I think we can claim that we can build digital logic that's reasonably reliable. So in this particular case, I'm showing you this implies logic function that allows us to have small molecule induction of a cascade of not logic gates or transcriptional repressors. And so the nice thing about this in particular is the fact that this is the input output steady state is that as the circuit so goes from blue to black to this yellow color here is as the circuit gets longer, as a cascade gets longer, it actually becomes more digital. It actually becomes more step-like, OK? More on off. So we're going from this blue input output function to this yellow. So now we have over 1,000 fold change in the output in response to two to four fold change in the input. OK, and then we have good noise margins, good signal restorations, all these good things that we need to have for the creation of larger and larger reliable digital circuits. So the basis of digital computation is that you have-- and the reason why you can actually create computers is that you can have logic gates that do signal restoration-- that the output is a better representation of the digital meaning then the input. So as the signal , propagates this analog signal could be voltage. But it could be protein concentrations. As it traverses through the logic gates, it needs to actually become cleaner in order for us to be able to have reliable digital computation. So people have figured out how to do this with electronics a long time ago. We figured out how to do it with synthetic biology, let's say, 10 to 15 years ago. And nature has figured out how to do this billions of years ago, OK? So things like cooperativity-- so I assume you've looked a little bit on cooperatively in, let's say, gene regulation. OK, so that is a situation where you get a nonlinear response in a system that biology has figured out is a useful mechanism so that signals that come in actually result in some kind of actual digital behavior. So you get non-linear signal processing in these regulatory elements. And at the end of, let's say, a signal transduction cascade, the output is either high or low. There's no-- for the most part, there's no in between. The transition between high and low is super fast. OK, so in a sense, that's creating digital or discrete outputs. And that's really critical for many situations-- certainly in synthetic biology, but many situations in biology as well. So one example would be, let's say, stem cell differentiation. You want the cells to be able to make a discrete decision. Should I make-- should I become a kidney cell or a liver cell or a muscle cell and so on. So those are discrete decisions that have to be made by the cells. And so the cells have come up-- or nature's come up with mechanisms to guarantee that. And so we've now figured out how to do that ourselves in a synthetic fashion as well. It's important to note that when we engineer these systems, we don't just think about digital behavior. So we spent an equal amount of time perhaps thinking about how to implement things that have transient properties or things that have more kind of analog behavior to them. And that's absolutely critical to be able to program cells to do whatever we want. So this is an example where we have engineered cell-cell communication where sender cells make a small diffusible molecule which then goes to receiver cells. OK, so now the receiver cells don't just have an on response, but rather they have a pulse, OK? So a signal travels from the sender to the receiver cells. And the cells, what we engineer them to do is have a pulse response. And the idea is to have GFP go up-- a Green Fluorescent Protein go up-- and then go down. And so to be able to do that, we engineered a feed forward motif where we have binding of the small molecule to this activator which activates two things simultaneously-- a green fluorescent protein and a repressor which then represses the green fluorescent protein. And then-- so the idea is that the green fluorescent protein goes up. And then eventually, the repressor builds up to sufficient levels to repress the green fluorescent protein. So again, one of those simple looking motifs. This is about three years to actually make that happen around the 2004 frame. Looks simple. If you study a naturally occurring system that has this motif, you say, oh yeah-- no problem. Yeah, we have this feed forward motif. And obviously, you can do this kind of information processing function. Let's move on to another motif. You actually try to build this in a lab in a new organism, I'm not sure if it can drive you insane. But it is not trivial to actually make it work. It's much easier now than it was 10 years ago. But you still have to pay attention to a lot of things-- rate constants, threshold matching, and so on to actually make it happen. But eventually, after looking-- creating-- so this is our first attempt at this was this blue line right here. So a completely flat line, OK? Input comes in, nothing happens. So I would say that pretty much typifies synthetic biology maybe up until today. You build something. You think it's going to work. It doesn't work. And then you stop crying after a little while. But then you have to think about how do I fix this. And so this iterative design debug cycle is absolutely critical. So what you normally do is you create computational models that tell you how different rate constants in the system affect the behavior of your circuit, OK? For example, you could do sensitivity analysis. Which rate constants have the most influence on the performance of your system? And so we did some sensitivity analysis here and learned that, for example, the degradation of this repressor makes-- one of the things rate constants makes the biggest difference is on the performance of the system or its affinity to the binding site on its respective promoter. Yes. AUDIENCE: Just knowing the sheer [INAUDIBLE] entire circuit, it started off with the [INAUDIBLE] constant? PROF. RON WEISS: No. I wish it was. Because that would make life a lot easier. And we are trying to get better at that. So we're trying to-- so here's maybe two ways of thinking about that. One challenge would be somebody comes in, gives you DNA sequence, and you have to predict the rate constants. OK, so I would term that person an adversary, not your friend. It's just too hard to do that. Now, an easier task would be give your adversary or friend limited choices and say in the freezer, I have these DNA sequences that consist, let's say, of specific promoters, specific ribosome binding sites, specific proteins with specific degradation tags on the proteins, OK? And that's-- you're allowing that adversary or friend to only use those elements in the design of a circuit. And then they come back to you and they say, now predict what the circuit will do. You still-- it still doesn't work yet. But I think-- but I would say that's how we would phrase the challenge, OK? Stick to things that we know and allow us to even characterize those things ahead of time. What we have that-- unfortunately, I don't have that here. But what we have done-- so people can get a kind of a general characterization. So they can say it'll be roughly this input output behavior. And when I say roughly, the errors could be on the order of five to 10-fold. That's approximately what's been published so far. Now, five to tenfold depending on your perspective could be great because it's biology or could suck if you're an engineer. It just depends on if you're trying to do something like get green fluorescent protein to turn on and off, tenfold is probably great if you're doing this in a Petri dish. If you're trying to create a cancer classifier circuit you put into humans to kill cancer cells but not harm healthy cells, tenfold is probably not great I wouldn't take a circuit like that into me, especially if that circuit controls, for example, the production of a killer protein, which I'll try to show you a circuit that does that. We recently have been able-- we're in the process of submitting a paper about this-- been able to show that if you have really good characterization of regulatory elements such as these repressor devices-- and you have to do a lot more characterization and you do the-- we can get within 20% on average on predicting the behavior in mammalian cells, actually. And so as an engineer, I would be happy with 10%, 20% percent for many, many applications. So I think we've gotten better at it. But it's not quite perfect yet. One of the things about that approach is that we don't necessarily know the rate constant for everything. What we do know, however, is a very detailed behavior, both steady state and dynamic behavior, of a repressor promoter pair. So we don't know, for example, what's the binding affinity or what's the rate constant for the repressor binding the promoter, what's the rate constant for RNA polymerase binding that promoter, what's the exact translation rate or transcription right too. But we do know what's the input output behavior. And that's actually been enough to get really good predictions. But I would say these kinds of predictions are one of the most important aspects and challenges and bottlenecks of synthetic biology. So those include-- again, the challenges include how fast can you build DNA. But wouldn't it be nice if you can actually predict what the DNA does? So it's just as important, if not more. And also having-- so those are two of the important challenges. I'd say another one would be-- OK, so if you can predict things how many parts do you have in your freezer that you can actually put together and they're well-characterized actually now build the circuits? Probably three of the most important challenges. And so if you go back to what we'll call the post generator, this is a loop tape of bacteria that now respond to the pulse. So sender cells that then secreted the small molecule then went into receiver cells. And they light up. So one of the things to note here is that it works. The other thing to note here is that it's not perfect, right? And so the amount of heterogeneity here I think is quite astounding. So if you take the average behavior, it's actually quite predictable. But if you now start looking at the distribution in the response, it's staggering. And we quantified that. And so we quantified what's the distribution in terms of the fluorescence levels that-- the peak and also the buildup and so on. And we then correlated that also-- we created the stochastic simulations that then correlate reasonably well with the system. So we can get simulations to generally correspond with what we're seeing at the population level. But I do want to bring up this point that when you think about engineering biological systems, don't try to figure out how to engineer a single cell to do something reliably, OK? So you always want to think about kind of statistical engineering. You want to think about, I'm going to create a circuit. And when I put this circuit into a population of cells, this is the distribution of behaviors that I'm going to get. Because if you're trying to depend on any individual cell giving you exactly the behavior that you're looking for, it is just-- it's going to fail. So you have to really think about distributions. And that I think changes things a little bit. So that's not normally the way you think about-- maybe that's the way Microsoft thinks about programming. So if 90% of the time, the computer doesn't crash, that's pretty good. Probably Bill Gates agrees with that, right? But that's not what we want. That's not what we typically do with software. So to kind of further think about this in terms of populations, we program something else which was a pattern formation. So now we have the desire to create senders and receivers where the senders send the same message to the receivers. Now we have a longer feed forward motif. OK, so this feed forward motif has two branches. And so these two branches actually have a different impact on the final output. One has-- there's two repressors meaning that input comes in. It activates a fluorescent protein and another one represses the fluorescent protein. OK, so it's an incoherent feed forward motif. And so what we use that here is not for post generation, but rather to define a range of concentrations that would turn on the final output, OK? So it would be activated-- the range of concentration would be activated starting with this branch right here and then ultimately repressed by this. So this defines the low threshold and this defines the high threshold. So under the low threshold, nothing gets activated. Whenever you have just the right amount, it activates this which represses this which allows this fluorescent protein to get turned on. OK, so this branch right here is more sensitive. So it defines when this thing goes up, when the response goes up. And then this is less sensitive. So this defines under high concentrations when the output goes down. So we basically have a non monotonic response to the input, which is low then high then low. So that's the design that we had in mind. And the idea is that whenever you put, let's say, receiver cells everywhere in a Petri dish and you put senders in the middle, then the communication signal basically builds up. There's a chemical gradient. Each cell interprets the chemical gradient and then decides whether to make a fluorescent protein. And then only-- because there's this steady chemical gradient due to diffusion and decay of the signal, then you would get some kind of a bullseye pattern. So that was the hope at least. And so to give you again a timescale, so it took me about I think three hours on a plane to make the slide. It took us about three weeks to create the computational model. And again, for whatever reason, three years was the magic number to create the actual functional circuit. So that was an older version of PowerPoint. But I haven't tried it on the new. But anyways, so we created this. This is a computational model. We actually used a computational model to predict how changes in rate constants would affect this band detect, the region where we're actually responding to the signal. And we used that to engineer different responses. And so we created eventually three different responses input versus output. And we put different fluorescent proteins on them-- a red fluorescent protein and a green fluorescent protein. This is the experimental set up over here. And after 16 hours of waiting, this is basically what we got. So we got a lot of bacteria to make all kinds of patterns. And we were very happy about this. We danced around in the lab a little bit-- you know, yay! So this was fun on those rare occasions where things actually work. So we said, let's have some more fun. And so we put senders in other configurations. And so we have programmable patterns of bacterial communities. So I'm not sure of is this useful. I'm not sure by itself besides having some fun with it. But one of the things we're using this for right now-- and depending on time, I may get to that later-- is this notion of embedding these circuits in mammalian stem cells or actually also in human IPS cells so that we engineer these human IPS cells to communicate with one another to make decisions. And then those decisions actually lead to differentiation patterns, right? So you can imagine in principle, if you can create three dimensional versions of these and use those to cause the cells to make differentiation decisions so that red would mean make neurons, green would mean make muscle, different colors-- yellow would mean make bone and so on. So in principle, you might be able to create tissues by design. OK, so that's something that we are working on actively in the lab right now. So we don't quite have a working heart in a Petri dish yet. We won't for a little while. But we taking some baby steps along the way. And so we have been able to get cell-cell communication to work. We've been able to get programmed stem cell differentiation to work. And hopefully, I'll be able to show you some images that we have of some recent examples where we take human IPS and actually created these embryonic liver buds that have lots and lots of interesting-- and actually all the cell types that are known to exist in the embryonic liver. So there are some progress along the way. Now, we don't anticipate to replace your liver, you know, any time soon. So don't destroy it. So actually one near term application that we're specifically looking at is if we can take-- imagine taking your own fiberglass, de-differentiating them into human IPS cells-- those are your human IPS cells-- and then differentiate them into, like, this liver-like environment and put that in a Petri dish and then test out the effect of drugs on your mini liver. OK, so maybe it's a good idea to test drugs on things that resemble human tissue as opposed to some random mouse that may or may not be as correlated with what the drug would actually do to actual human cells. And if we actually even do it in the patient specific manner, I think that really changes the way drug development would actually work. And that's something that I think within the next few years could become a reality. They're talking about the next-- beginning to do that within the next one to three years in a laboratory setting. So I think that is near term and realistic. So one of the things that we did notice is that when we engineer the small systems, its intuition works quite well. So I can look at this circuit design and say, if I modify this, this is what's going to happen. If I modify this, that's what's going to happen. So you can use intuition and it works reasonably well. And what happened in I would say the first 8 to 10 years of synthetic biology, every paper would have a computational design. But most of those would build something. And in order to publish, we also tacked on a computational model that correlates really well with the experimental results. And we are just as guilty of doing that as anyone else, OK? So it wasn't critical to have a computational model to create something successfully in the lab. And I think that that is changing. So we have examples right now of designs where-- I'm not sure if I'll get to that today, but we have published on that-- designs where it involves about 20 to 25 components. And it's related to a diabetes system the retired engineer where we can have intuition about it. But our intuition doesn't work great anymore, guys. So we might have some intuition about the system. But the computational analysis would all of a sudden shed light and provide insight that is very difficult to get this by drawing this thing on a blackboard. OK, and so I think that computational design tools are becoming absolutely essential. They can provide insight into system behavior that you can't get just using intuition alone. But in another aspect of computational design is one where imagine being able to specify want this behavior. And then the computational design tool says, here's 1,000 different versions of circuits that you should build and test. OK, and so it's still difficult for a human to generate easily 1,000 different versions of a circuit to build and test. It is becoming easy to actually build-- I wouldn't say-- maybe easy is a strong word-- feasible to generate 1,000 versions of a particular circuit. I'll give you an example. Very recently in my lab, one of the graduate students has come up with a framework that-- he is generating 200 versions of a circuit in three hours, OK? And they're pretty much gotten to the point where they're all correct. In three hours-- so that really, I think, changes what you do in synthetic biology. And so again, having that be connected to a computational design tool that tells you which ones to build would be rather useful. And so specifically recognizing that-- this is a collaboration with some folks at BBN. And actually Jake Beal was one of my former graduate student colleagues. So he was in the same lab as me. And then Doug Densmore is from Boston University. And so the notion here is that this is what we want synthetic biology to look like. OK, so if you're trying-- or maybe all of biological engineering. But we'll start with synthetic biology. So if you want to program biology, should you really care what the ribosome binding site is for the lambda repressor? You know, hopefully not, right? What you should do, just like when you program your simulations in MATLAB, you don't think about, well, here's the shift register in this Intel Pentium chip. And this is how it's working to simulate this ODE over here. That's just not the level at which your program. So you think really at a high level. And then you have compilers and lots of infrastructure that takes care of everything in between. And so maybe someday in the future, the graduate students maybe five to 10 years will look back at synthetic biology graduate students now and would just have a lot of pity for them and, oh my god. You actually had to know what elements you were using in circuit and actually build them by hand? You know, wow. And so here's a notion that we start with a high level description. And by the way, this is now-- there's a website that you can go through right now and get a free account and then type in a high level program. And it will actually create a low level genetic circuit representation. It will also give you MATLAB simulation files of this. And so, in this course I'm teaching to undergrads right now that many of them didn't even hold pipettes before they started the course, one of the first things that we taught them was this tool. So before telling them, for example, this is the way the lac repressor works by DNA looping, we said, there are these things called repressors. And when there's more of them, there's less output. Now let's design with that. This is, I think, heresy to biology as a whole probably. I don't know that-- this is the first time that we've actually tried to do that. And I think it's actually worked out OK. But teach them enough so that they can move forward. And then, yes, let's simultaneously be teaching them about the underlying biology and mechanisms as much as they need to know. But what can they do if they just know that there's this thing called a transcriptional repressor? And then the bio compiler will figure out everything that needs to happen. And so we're actually-- so they did a whole bunch of designs. And starting next week, we're going to be testing them out. So we will find out whether that was a useful way to teach biological engineering. But I think so. I think it is. Because they seem to understand what design means, OK? Because that's a thing that we focus on-- design as kind of a first class object. OK, so what you do is you write code that looks like this. Has anybody programmed in code that looks like this? It should be-- so Lisp. OK, one person only? Any computational? OK, we usually get one or two people. But I was hoping for more here. So anyways, for the people that should be ashamed of themselves and haven't programmed in lisp, this is a simple program. This is, if the input is high, it produces cyan fluorescent protein. Or else, it will produce a yellow fluorescent protein. And so the bio compiler then automatically takes that and first translates that into a data flow. So you have a data flow-- and I'll actually go through an example of that-- and then creates an abstract gene circuit and then looks at essentially what you have in the freezer and says, well, this is the actual DNA sequence that would implement this. And it creates robot instructions to assemble this so that God forbid you would have to actually touch a pipette to build this. And then we have a robot, liquid handling robot, that does most of the assembly. Now, this pipeline is not fully end to end yet. So it's not-- if you came to my lab right now, the still-- I could tell you that this works but then I would be slightly lying. Is-- mostly works. But it's actually-- there are companies right now that will go from this level-- about this level-- not this level, but this level. And actually, this is mostly automated. So-- to the point where the only thing that's not automated is somebody, not necessary your op, but maybe a technician goes from a robot and takes a plate and puts it in another robot. And, like, everything else is essentially automated in DNA assembly So those companies have, I guess, more money than us. But eventually, this is the way DNA assembly will happen. So we're collaborating with some people. I'm doing this with microfluidics. The problem with this robot-- it's $150,000. So we can't put it on everyone's bench yet. But we are collaborating with Lincoln Labs to have microfluidic devices that would cost around $3,000 that would do this. And we've already demonstrated with the microfluidic devices we can do DNA assembly of large circuits. So this is not that far away that everyone will have microfluidics. They program here. And they get the DNA assembled. And then they realize it doesn't work. But at least there'd be a lot faster to realize that things don't work, which is good. So let's look at how this compiler, bio compiler, works to see that it's not magic. So this is saying green if not IPTG. So IPTG is a small molecule that we have a sensor for. And then the information gets routed to an inverter which then gets routed to activate a green fluorescent protein. So you can take a program specified like this and automatically convert it into a data flow graph. And then the data flow can be translated-- again, this is automated and this is also automated in a rather simple way-- to an actual portions of the gene circuit. So IPTG sensor is simply this motif. So you have a repressor that responds to IPTG. And then it basically inactivates the repressor. So more IPTG, more output. And so this is the IPTG sensor box. And this is how you implement this. Not gate-- so I showed you how a not gate already works. So a repressor represses the output. Green fluorescent protein, you just have to have an activator that activates a green fluorescent protein. So every data flow box can automatically be converted into a small motif. And then what we're missing is the glue. And so the glue is just these transcriptional activators and repressors. And so once you put them in there, then this is a circuit. And so this is an automatic way to go from here to here. And it can do rather complex circuits and logic. It can't do everything yet. But it can do an interesting set of things. OK, so this is-- again, it's not completely finished in the sense that I haven't told you what A is. I haven't told you what B is. So there's a whole bunch of things that still are-- we have-- either published on or still need to be designed. But there is, for example, tool called matchmaker which will decide what's a good protein A, what's a good protein B. So things like that-- so those things already exist. Anybody look at this and figure out why this is not perfect? So we have an input that represses a repressor, which means that more input, more activator-- sorry. More input, more protein here, which represses B, which activates GFP. So the compiler spits that out automatically. But you may not want to build this right away. Any ideas why? I'm sure John would. AUDIENCE: So basically, you have this A can directly repress object, so you activate B? PROF. RON WEISS: So this is what it-- the first version of B. But A represses B. B activates GFP. So why not just hook A up to regulate GFP directly? So this seems like a non-optimal solution. And so you can get the compiler to figure that out too. And so what you do is you say copy propagation. So these are actually just tools that are available. These are mechanisms that are part of normal compilers-- normal software compilers. So they do something called copy propogation-- means that A, if it sees that A-- the compiler sees A regulates B, A represses B, and B activates A. So you can just say, well, let A just directly regulate this. And then the compiler realizes, well, B doesn't do anything. Let's get rid of it. And then the promoter doesn't do anything. Let's get rid of it. And now the compiler figured this out. And that's using just basic compiler technology. OK, so it's able to do that. So the difference in your life between four and three may not be huge. But the difference in your life between 15 and five could be the difference between getting your Ph.D or going insane, dropping out, and then starting a company and becoming a billionaire. OK, so-- but this is what the compiler can do. And that does make a difference. And it may be able to come up with optimizations that you wouldn't be able to really easily come up with or even at all under a reasonable amount of time. So the compiler can do combinatorial logic. It can do state. There's also some aspects that can do spatial things as well. And again, this is available online right now. I'm going to-- maybe I should skip a few things. So very quickly, I'll tell you. So in the lab-- so it's nice [INAUDIBLE] But if you can't do anything in the lab, then nobody believes you in the world of biology or synthetic biology. So we can build big things. So here, you can-- there's a library of promoters and genes that we have available. And you can decide, I want to build a new circuit has these promoter gene pairs. And within five days, you can create in the lab that circuit. And we've been able to demonstrate things that are 61, 64 Kb that you build in five days. And you build them efficiently. And then the undergrads that we're teaching also have been able to-- again, these are people to barely knew what pipette are in the beginning a semester can now efficiently build large circuits. So that's become an easy to use technology. And then I mention this notion that you can build this one at a time. But this very recent development where you could build 200 versions of these at a time. OK, so you can build them. We can then-- again, I'll skip this part. You could build them. You can put them into mammalian cells. These things work. And we have lots and lots of parts-- regulatory parts. And so let me skip this particular example. So this is-- the two sentence explanation is, OK, you build lots of parts. You can build modules. And then you put modules together. And guess what? This is biology. So these modules actually affect each other, potentially in undesirable ways. So they can place things like load. So whenever you have a module that works really well-- maybe I will show you one slide. But I won't show you how we solved it. But one slide-- so this is a circuit. Any idea what this circuit might do? So this is a regulatory circuit. They have an activator activating itself and a repressor that represses the activator. Have you seen this motif? So it's a-- some people are whispering to themselves, doing this. So this is a relaxation oscillator. When you do it by yourself, it works great. So these are simulations. But then if you connect it-- so it's nice to have an oscillator in a cell, again, if you want to have blinking bacteria or mammalian cells which is, again, fun. But then you typically want to connect it to something. So you spent three years building an oscillator, got it to work. And now your adviser says, OK, let's just get a paper on this. But before we get the paper, I want you to connect it to something meaningful. Because we want to go for a high impact journal. And then you connect it to something. And you realize that it doesn't work anymore. I think you're really pissed off at-- you will be pissed off at your adviser. I don't know if your adviser would want to be mean to you. But anyways, so that's a real problem in biology or synthetic biology is that these things have impacts on each other that, first of all, are going to be undesirable but because of unique aspects, for example, of the substrate. Now, they're not as unique as you might think because these load issues also come to play in electronic circuits. And so in electronics circuits, these things have been solved decades ago with these notions of load drivers. So if you have a module that has, for example, high fanout and it controls many things, then guess what? Those things that it's trying to control, even though the arrows are pointing one direction, they actually have an upstream effect too. So those modules that you think you're-- the downstream modules that you're just controlling, they actually have an impact on the upstream. So what you do-- in electronics, you just build a load driver. And it basically takes care of things. So now there's no kind of parasitic effect. And so we've demonstrated that we can also build-- so this is a notion of retroactivity. And this is work with Domitilla Del Vecchio here. And we've demonstrated that we can-- so I wont' go into the details here-- so we can actually solve this. So this is a real problem even in simple circuits experimentally. And then it uses this cool notion of timescale separation to solve it. But we can take these things that are highly affected to go from black to red and put a load driver in and then it fixes it. So this should hopefully lead to the generation of much more predictable circuit construction targeting one of the real challenges in scaling going from simple toy modules to large scale systems. So I'll skip that and then move on to another example-- oh, sorry, an example of an application. So in this particular case-- again, besides turning GFP on and off, what can we do with synthetic biology that we can't necessarily do without synthetic biology? And so one of the most important challenges in cancer therapeutics is specificity-- perhaps the most important. There are other things that are important such as delivery of a therapeutic agent. But as you improve specificity that it can actually change how you do delivery of a therapeutic agent. So if you have a therapeutic agent that's much more specific and has no side effects, you can deliver lots and lots more. So these things are highly related. So imagine a therapeutic agent that recognizes something on a cell surface and then says, this is a tumor cell. Kill it. So there are actually a lot of efforts ongoing that have this particular approach. So whether it's small molecules, whether these vesicles that contain various cell surface-- various molecules that bind to cell surface receptors, a really hot area right now is these engineered killer T cells. So this notion that you can actually engineer your immune system by placing various receptors on these killer T cells which then go and then bind tumor cells by recognizing something on the cell surface and then kill those. Sounds great, maybe, depending on your perspective, or not so great, depending on your perspective. So what's happened with those is they're really sometimes great at eliminating your tumors. But then the side effects can be horrendous. And so what happens is that those cell surface markers that exist on the tumor cells-- guess what? They're also present on healthy cells as well. And so the side effects for those killer T cells approaches have been, I mean, just terrible in various patients. So one-- this should not be a shock. One marker is typically not enough to distinguish a cancer cell versus a healthy cell. Seems pretty obvious. So actually, what they've been doing with these engineered killer T cells is now just saying, OK, has to be this cell surface receptors, but also cannot be this. So it's like, and biomarker one and not-- you know, biomarker one, and not biomarker two. OK, so they're starting to engineer more logic into these, more multi-input logic into these things. And so they haven't done any clinical trials. But I saw a couple weeks ago where they've actually made progress in Petri dishes. So we recognized this as an issue several years ago. And this is work with Coby Benson. And so really, all the information that you want is actually inside the cell to make a highly precise decision about whether this particular cell is cancerous or not. OK, so the idea is when the therapeutic agent, does the computation by integrating multiple pieces of the sensory information and then decides whether to express a killer protein or not. Now, even if this is not the cure for cancer-- can't really guarantee that, right? But even if this is not, just this notion of being able to create multi input circuits to go into cells and analyze in live cells real time what's going on in the cell with various molecules that might be interesting-- that has applications I would again say just about anywhere you could imagine in biology. So one of the things, for example, we're looking at-- I mentioned this notion of organs on a chip or programmed organs on a chip. So one of the things we're looking at now is placing these types of sensory circuits into cells within this organ on a chip. And so the idea would be expose cells to these drug candidates. And then the cells light up in different colors to tell you how they're responding to it. So for example, certain color green would mean just fine. Green with red and yellow would indicate that some apoptotic pathway is being expressed or this drug is affecting this proliferation path or this pathway or that pathway. And so these mechanisms can tell you in real time in single cells-- also specially-- what kind of impact there is to, let's say, the drug candidate you're looking at. So again, this is something that I think would have an impact today if it was available, but realistically making prototypes within the next year to three years. OK, so we started with looking at HeLa cancer cells as an example. And so we did some bioinformatics. And we saw that based on microRNA profiles that were available to us, this would be a bio program that would distinguish HeLa cells from all other cell types. And so some of you are familiar with this way of annotating logic. Pretty ugly if you're not used it, but it's simple. This just says, these microRNAs have to be low and these microRNAs have to be high. And that's a HeLa cancer cell. Because that's the basic logic in this. So it's a rather simple logic statement. What we would argue is that every cell type would have a logic statement that would be true of that cell type and not true of other cells. It's almost by definition. There's something different about that cell than other cell types. Now, it gets interesting when you start talking about heterogeneous population. So for example, and that's-- I won't get into details about that. But there is heterogeneity. There's heterogeneity in tumors. So the cool thing you can do-- this is a six input and gate. So if you have heterogeneity, where you can do is you can have a six input and gate with an or operation. So you can identify this sub population of the tumor has this microRNA profile. And this cell population has a different microRNA profile. And all you have to do is create a new logic circuit and just combine them as like a cocktail drug as an or gate. So this is a really general approach that I think should be relevant, again, for any kind of population. And you can also set the thresholds and have all these fun things that you can do with it. So I think the question really is, is that microRNA expression profile sufficiently small so that it can be encoded on a circuit that you can deliver into cells? And can you do it reliably? That really is the challenge. OK, so the idea is that you have a therapeutic agent, goes into a cell does the computation, and then decides whether to make a killer protein or not. OK, so how does this actually work? So how do we implement an and gate with inverted inputs? OK, so that's one portion of the circuit. So it's actually-- for microRNA, it's actually pretty easy. So what you do is you put microRNA target sites on the gene of interest or the output gene. And so the idea is that the only way that this gene is being expressed, this output protein is expressed, is when this is low and this is low and this is low. So in principle, this is pretty easy to do. OK, so that's how we have a three input and gate with inverted inputs. And now we can add logic to that. So now if we want to make sure that the output is high also when this input as high as well. And so what we do-- so remember the first circuit that I showed you was this cascade. And the cascade was a bunch of repressors that would then cascade that did the not not operation. So when you think about the not not operation, it doesn't seem like a very useful operation. But this here, the not not operation, is actually useful. So it can convert a microRNA sensor into something that can be integrated with other sensors to create the four input logic function. So this microRNA has to be high in order to repress this repressor which then represses the final output. So the only time that the final output can be high is when this is high and these three are low, OK? And then you can continue. And you can add another one. This is slightly more complex in the sense that now, essentially the sum of these microRNAs has to be high in order for this branch to allow expression here. So it's slightly more complex in the sense that it's not just-- it's like a plus-- again, like a plus operation, which we found to be actually the relevant operation here. And so the only time the output high is when this microRNA is high, combination of these two is high, and these are low. Anybody see a potential problem here? A potential crosstalk? So everybody here understand the circuit? Raise your hand if you actually understand the circuit. OK, that's not a lot of people. OK, maybe I should go back for a second. Everybody-- raise your hand if you understand this. OK, so if microRNA is high, they repress-- they result in degradation of the RNA. So all of these have to be low in order for this to be high. OK, now if we had this, this RNA-- let's focus just on this maybe to simplify. So this microRNA represses-- this is a repressor which represses this, OK? So if this microRNA is low, then this repressor as high. And as a result of that, this repressor represses this promoter regardless. So it doesn't matter what these are. If this is low, this repressor is high. And then there's no way that this output can be high. Now, if this is microRNA is high, then this becomes low, allowing this to potentially be high. Doesn't guarantee it, but it has a potential of doing. OK, now raise your hand if you understand that. OK, more. OK, cool. That's progress. By the way, this was not a trivial circuit. There's a lot of connections here. And then we connect the same-- now we have the same repressor. So this microRNA-- these set of microRNAs repress the same repressor which is now repressing this. So why is it the case that you need-- let's just assume this is one microRNA, this is another micro. Why do you need both microRNAs to allow high output? And then I will not let anyone leave until you answer the question. And I can sit down. So why-- and then we'll stop there. And everybody will believe me that the cure for cancer, you got it figure it out. Otherwise-- so there's-- oh, thank you. Letting-- you will be the hero. AUDIENCE: So if the microRNAs are not present, then you're going to get an expression of Lacks E. So-- PROF. RON WEISS: Lack I. AUDIENCE: Yeah, like eyes. So only if both the microRNAs are high and present do you repress both of the lack I's which then allow for the expression of the-- PROF. RON WEISS: Exactly. And what happens if one of them is present and the other one is not? So if this microRNA is present and this one is not? AUDIENCE: Then the lack I still gets expressed because the other one is still not repressed. PROF. RON WEISS: Exactly. So the only way that you can actually have no expression of lack I is when this is high and this is high. So this is essentially like a two input and gate. These two have to be high to allow this promoter to potentially be high. And in the other three logic cases where one of those is low, lack I is expressed and hence does not allow the output protein to be expressed, OK? Yeah. AUDIENCE: So why do you have the partition? All three are acting the same way. PROF. RON WEISS: Ah, great question. So what happens if microRNA 21, 17, and 38 were all here? So that's a great question. So that's a question, right? So if you put microRNA 21 here, 17, and 38? What is the logic function here? Yeah. AUDIENCE: Then you only need one of them in order to degrade the RNA. And then you won't get the same logic. PROF. RON WEISS: Great. So then it becomes essentially an or. This will be an or operation of the microRNAs where what we want is an and operation on the microRNA. So that by having two separate paths that have the same repressor where they converge on the same repressor, we achieve the and operation. Where we have them on the same repressor, they do the or operation. That's a great question. OK, so this works. And maybe I'll say thank you. I'll stop there.
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https://ocw.mit.edu/courses/5-07sc-biological-chemistry-i-fall-2013/5.07sc-fall-2013.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. BOGDEN FEDELES: Hello and welcome to 5.07 Biochemistry online. I'm Dr. Bogden Fedeles. Despite the staggering biodiversity we see in nature, the types of chemical reactions employed are only but a couple of handfuls. And these are used over and over again very efficiently and with conserved mechanisms. As you might recall from Organic Chemistry, one of the most versatile chemical groups is the carbonyl, C double bond O. Not surprisingly, carbonyl chemistry is well-represented in biochemistry. In fact, the carbonyl chemistry allows formation of carbon-carbon bonds. It's one of the very few ways in which enzymes can start with small molecules and put them together into a macromolecule, or start with a macromolecule and break it down into smaller pieces during metabolism. This video summarizes some of the most important carbonyl reactions you will encounter in 5.07. In this video, we're going to be talking about carbonyl chemistry. And as we will see, carbonyl chemistry is fundamental for some of the carbon-carbon bond formation and cleavage reactions. As you recall from organic chemistry, carbonyl contains a C double bond O. And all the properties of the carbonyl derive from its ability to polarize this bond, so that we can draw a resonance structure where the carbon has a positive charge and the oxygen a negative charge. As you recall, there are simple carbonyls, such as aldehydes and ketones. Also we have acyl derivatives, compounds in which the carbonyl is attached to a heteroatom. x can be oxygen, nitrogen, sulfur. So here, respective, we have esters, amides, thioesters, and of course, we have an OH group here, carboxylic acid. Here is a summary of the reactions that we're going to be talking about. First, we're going to be discussing nucleophilic addition. Here, the good nucleophile reacts with the carbonyl, adding to the carbon that the carbonyl can generate. This tetrahedral compound. Next we're going to be talking about enolization. This is the property of carbonyls that contain an alpha hydrogen, which can rearrange to form enol. Next we're going to introduce the aldol reaction. This is the reaction in which a carbon-carbon bond is formed and occurs between a carbonyl that acts as electrophile and a enolizable carbonyl, which acts as a nucleophile. In the aldol reaction, a bond is formed between these two carbons, generating an aldol. We're also going to see that the aldols can dehydrate. The aldols we saw above can lose a water molecule to form an alpha, beta-unsaturated carbonyl. Now about the acyl derivatives, we're going to be talking about acyl transfer reactions, where an acyl derivative can convert into a different acyl derivative with the appropriate nucleophile. A variation of this reaction is Claisen reaction, where similarly to the aldol reaction, we have an enolizable carbonyl reacting with an acyl derivitive and generating a beta-keto carbonyl. This reaction also forms a carbon-carbon bond, which is right here. Let's talk in more detail about the nucleophilic addition. The general reaction scheme is as we saw before. Here is a carbonyl compound reacting with a nucleophile and forming a tetrahedral intermediate that contains an alkoxide or an alcohol. Now let's take a look at two different reactions. One is the reaction of alcohols with carbonyl compounds, where we form a compound that looks like this. This is called a hemiacetal. Now this reaction is reversible. And, in fact, it reaches equilibrium because delta G naught is approximately zero. This reaction can be acid or base catalyzed. Let's take a quick look at that mechanism. If it's based catalyzed, the base will first deprotonate the alcohol, which will form the alkoxide, which is then a very good nucleophile to attack the carbonyl, which forms this alkoxide version of the hemiacetal, which can be then protonated. In acid-catalyzed mechanism, we have to activate the carbonyl first, so the protonation of the carbonyl is the first step. All right, so this activated carbonyl can then be attacked by our alcohol. Which, this product is just one proton transfer away from our hemiacetal. All right, the second reaction I want to include here is the formation of Schiff bases which is the reaction of a carbonyl with an amine. Similarly to the hemiacetal formation, this reaction generates first a tetrahedral intermediate, which is, however, unstable, and loses water to generate the imine, with a Schiff base. Let's take a look at the mechanism. As you notice, the reaction-- because the amine group is a good nucleophile, the reaction can occur even in neutral conditions. We don't need, necessarily, acid or base catalysis. The first step, the imine attacks the carbonyl, forming this compound with split charges. Now proton transfer happens to generate our intermediate. Then water is eliminated. And this is the imine. You'll notice the imine nitrogen can also be protonated, to generate this iminium ion, which, as we will see in other situations, it's an activated version of the carbonyl group. From these two examples, we can get some idea of how the nucleophilic addition occurs. So let's take a look at what kind of nucleophiles we can add to the carbonyl group. We have some good nucleophiles. And here we have things with negative charges, such as alkoxide, or hydroxide. I have the thiolates. And other things such as amines. And we also have some OK nucleophiles. And here we have alcohols, even water, and thiols. As you saw in these couple of mechanisms, the OK nucleophiles don't react very well, unless they are deprotonated to form good nucleophiles, such as the alcohols. Or the carbonyl gets activated, either by protonation in a strong acid, as we saw here, or it becomes an activated carbonyl, for example, in an iminium ion. Another important nuclear force that we're going to see is the, what we're going to call, a C minus. Basically a carbanion In our case it's going to be enolates, which can also add to the carbonyls. And these will form the basis for the aldol reaction. The second reaction we're going to be talking about is enolization. Here, a carbonyl that contains an alpha hydrogen can rearrange to form an enol. We're going to call this the keto form and this the enol form. An equilibrium between a keto and an enol form is called tautomerization. And this is a very important reaction in many biochemical systems. Turns out, the delta G, for the reaction as written, it's very high, 30 to 50 kilojoules per mole. That means that equilibrium strongly favors the keto form. However, in certain cases, the enol can form and get stabilized. The mechanism of enolization, it's very straightforward. All we need is a decent base that can remove the alpha proton. And it will form this enolate. Now, enolate is able to form because it has resonance stabilization. We can draw another resonance structure, as such, where we see the negative charge is on the carbon. So it is in fact a carbanion. We're going to call it a disguised carbanion. As the carbon is not very electronegative, having such a high electron density on the carbon would make it a very good nucleophile. And in fact, this enolate is the nucleophile that executes reactions such as the Aldol reaction and the Claisen reaction. Something to keep in mind, well, how acidic is this alpha hydrogen? We can compare it with a hydrogen in an alkyne. The pKa of such a hydrogen is close to 50. It's extremely hard to remove a proton. Now if we look at an alpha hydrogen next to a carbonyl, the pKa is 18 to 20. So it's 30 orders of magnitude more acidic, and this is because, as we saw, when we removed this hydrogen, we formed the enolate anion, which is resonance stabilized. The more extreme case of this, if we have two carbonyls, alpha to the same proton, the pKa drops even further, around 9 to 11. This is because we can draw even more resonance structures to the enolate that's formed. This is one. This is another. And another. As we saw before, the charge here is delocalized between the oxygens and the alpha carbon. So it is this beta keto carbonyl in its enolate form will behave as a carbanion and it can act as a good nucleophile. The Aldol reaction. This is a very important reaction in biochemistry because it allows formation of carbon-carbon bonds. Or, if the reaction runs in reverse, cleavage of the carbon-carbon bonds. The Aldol reaction is the reaction between an enolizable carbonyl, as we show here, a carbonyl that has an alpha hydrogen, and another carbonyl. And what happens is, a new carbon-carbon bond forms between alpha carbon and the carbonyl carbon. The product of the Aldol reaction, it's called Aldol as a contraction between aldehyde and alcohol, as in some cases this carbonyl will be an aldehyde and this would be Aldol. It's essentially a beta hydroxy of carbonyl. Now, this reaction has a delta G naught close to 0. That is, it reaches equilibrium. And it can be catalyzed by acid or by base. Let's take a quick look at the mechanism. Given the previous mechanistic insights-- we looked at the nucleophilic addition, and enol formation, then the mechanism of the Aldol reaction should be fairly straightforward. If it's base-catalyzed, the base is going to help us form the enolate, as such. And as we discussed previously, the enolate is a good nucleophile, and can react via nucleophilic addition with the other carbonyl. And one proton transfer to generate the Aldol product. The reaction can also be acid-catalyzed. Again, formation of the enol in acid catalysis involved first protonation of the carbonyl. Now this activated carbonyl, it's a much better electron sink, and stabilizes the enol formation. Now, in the second step the enol can react with the other carbonyl, to generate a protonated version of the Aldol, which is one proton transfer away from the Aldol product. In biochemical systems, the enzyme that catalyzed the Aldol reaction is called aldolase. And there are actually two kinds of aldolases. Class one, and class two. The distinctive feature of these enzymes is the way they catalyze the reaction. Class one uses an active site lysine to form a Schiff base with the carbonyl, which activates the carbonyl, and allows for the enol formation. Class two uses a metal ion, such as zinc, to accomplish the same thing. So here is how the mechanism for the class 1 aldolase would look. So here is our enolizable carbonyl, and here is our active site lysine. As we saw before, an amine reacting was a carbonyl will give us a Schiff base. The reaction goes via a tetrahedral intermediate, which we're not going to draw here, but what we form is this iminium ion. Now the carbonyl is activated enough that an active site base can remove an alpha hydrogen to form the enol. Which is now well-positioned to attack the other carbonyl. This generates the Aldol product, in its imine form, still attached to the enzyme. And now the hydrolysis of imine is going to release the Aldol. Now, class two enzymes use a zinc ion. As the ion approaches the carbonyl, it's going to draw some of the electrons from the carbonyl, and make the proton in the alpha position a lot more acidic. So you can imagine, some of these electrons get de-localized. So that a base can remove the proton and form the enolate. Which, in the second step, it reacts with the carbonyl, which will generate the Aldol product in the active site of the enzyme, still bound to the zinc, and now which can dissociate, and generate the final-- and release the product. Now, a very important consideration for the Aldol reaction is that it can occur in the reverse fashion. For example, to cleave a carbon-carbon bond. So the bond that will be cleaved, as we see here, is the bond that got formed, which is the bond between the alpha and beta carbons. The aldolase is one of the key enzyme in glycolysis, that allows us to break a six carbon sugar into two three-carbon sugars by cleaving a carbon-carbon bond via the Aldol reaction. As the mechanism catalyzed by the aldolase, we can see that the reverse pathway is pretty straightforward, where the Aldol binds to the enzyme, say in class one, forms an active site, covalent attraction, a Schiff base with the lysine, from which the chemistry occurs to break the carbon-carbon bond, and leads to the release of one carbonyl molecule, and then the other one will be still bound to the enzyme as a Schiff base and hydrolyzed. For the class two, the Aldol will interact with the enzyme by forming an interaction with the zinc, and this activated carbonyl allows the chemistry to occur exactly in the reverse manner, as shown here. One other reaction involving Aldols is Aldol dehydration. Here's an Aldol, beta hydroxy carbonyl. Now, if an Aldol has an additional alpha hydrogen, it can lose a water molecule to form an alpha beta unsaturated carbonyl. Now this reaction is favorable thermodynamically. The delta G naught is approximately 0. And this is a reaction we're going to see in a lot of biochemical pathways, for example, in the biosynthesis of fatty acids, going left to right, or in the catabolism of fatty acids, going right to left. Here's a quick insight on the mechanism. Once again, it can be base- or acid-catalyzed. This reaction works because the alpha hydrogen here is next to a carbonyl, and therefore can form an enol. So if a base can remove this hydrogen to form the enolate, then we can envision how this electronic movement will allow for a water molecule to be eliminated, forming our alpha beta unsaturated carbonyl. The acid-catalyzed mechanism goes along the same lines. As you remember, in order to form the enol in an acid-catalyzed context, first we have to protonate the carbonyl. All right. Now a base can remove our alpha hydrogen, forming the enol, which can kick off a water molecule, generating these pieces, which is just one proton transfer away from our final product. So let's talk now about acyl derivatives, and acyl transfer. As we mentioned, acyl derivatives have a carbonyl attached to a header atom. And this header atom can be oxygen, nitrogen, sulfur. As all these header atoms contain a lone pair of electrons, one of the key properties of the acyl derivatives would be resonance between the header atom and the oxygen. Now the properties of the acyl derivatives will be dictated by how easy or how difficult it is to adopt this minor resonance structure. In other words, how likely is it for the header atom to participate in these electron conjugations. Let's take a look at a couple of acyl derivatives. This is a carboxylate. If the header atom a nitrogen, we have amide. If the header atom is oxygen, we also have esters, or carboxylic acids. And when the header atom is sulfur, we have thioesters. The order in which I wrote them here is not actually random. It turns out for the carboxylate, because it has already a negative charge, the ability to adopt this resonance is greatly increased. So it's very well resonance-stabilized. The ability to form these resonance structures, it's also great for amides, and this dictates the chemistry and the biochemistry of the amide bond, which is explored in greater detail when we talk about protein. Esters can also adopt these resonance structures. However, thioesters, because the sulfur is a third-row element, so the p-orbitals of sulfur are much bigger, they don't overlap very well with the p-orbitals of the carbon, the ability to adopt these resonance structures is greatly diminished. Therefore, thioesters behave a lot more like ketones, where the electrons of the carbonyl bond are localized between the carbon and oxygen, and not so much between the carbon and sulfur. So therefore, thioesters are the least resonant. And this is the trend. And this trend inversely correlates with the reactivity. Carboxylates are least reactive, whereas thioesters are the most reactive. Now, when we talk about acyl transfer, we talk about the reaction between an acyl derivative with another nucleophile, which will replace the x header atom with the y header atom. So this reaction always occurs via a tetrahedral intermediate. When both substituents are attached to the carbon. Now from here, this tetrahedral intermediate can fall apart by kicking off the YR to regenerate the starting material, or it can kick off the XR group, to generate a new acyl derivative. Let's now talk about the Claisen reaction. This is a very important reaction in biochemistry, related to the Aldol reaction, in which we form or cleave carbon-carbon bonds. The Claisen reaction happens between an enolizable carbonyl and an acyl derivative. Let's pick in this case an ester. And during this reaction, a carbon-carbon bond is formed between the alpha carbon of the enolizable carbonyl, and the keto carbon of the acyl derivative. The product of the Claisen reaction is a beta keto carbonyl. Let's look at the mechanism. As with all carbonyl reactions, when we form a carbon-carbon bond, we need to form an enolate. So this is the first step. A base will form, remove the alpha proton, and form the enolate, which is now poised to add to the acyl derivative in an acyl transfer reaction, forming first a tetrahedral intermediate, which can spontaneously fall apart by eliminating the header atom group, to form our beta keto carbonyl product. Now, in biochemistry a preferred substrate for Claisen reactions is a thioester. One of the most common thioesters we're going to encounter in this course is acetyl-CoA. CoA, or coenzyme-A, it's a thiol that can form thioesters with a lot of acids, for example, acetic acid here. Acetyl-CoA can undergo a Claisen reaction with itself, and therefore acts both as an enolizable carbonyl and as an acyl derivative. From when we were talking about thioesters, because of their limited conjugation with the carbonyl, they are very reactive, and they allow the formation of the enolate. Here is the acetyl-CoA enolate, which can react with another acetyl-CoA molecule. It will generate a tetrahedral intermediate. Let's draw this molecule first. Which can lose one of the CoA molecules, to generate this beta keto thioester, acetoacetyl-CoA. As we will see later in the course, this is a precursor to formation of ketone bodies, one of the ways in which acetyl-CoA can be used to store energy. Now, what is coenzyme-A, often abbreviated CoA? We mentioned it's a thiol. That means it has an SH group, which it turns out, is on a very long linker. There you go, this is coenzyme-A. You might recognize this part of the molecule as being adenine bound to a ribose bound to two phosphates. It's essentially ADP. But notice there's another phosphate in the three prime position, so it's an ADP with a three prime phosphate. This portion of the molecule, If we squint, resembles the amino acid cysteine, but without the carboxyl group. And this middle portion of the molecule, it's something that looks very difficult to synthesize. Notice this carbon that has two methyl groups attached, and two other carbons attached to it. So it's like a tetravalent-- a carbon attached to it, four other carbons, that's it. Fairly rare sight in biochemistry. This portion of the molecule is called pantothenic acid. Pantothenic acid is an essential nutrient, also known as vitamin B5. In this video we talked about carbonyl chemistry. Carbonyl is the C double bond O, and a lot of its properties are due to the polarizability of this bond, where the carbon has a partial positive charge, and oxygen a partial negative charge. We talked about reactions to simple carbonyls, such as nucleophilic addition, enolization, Aldol reaction, and the Aldol dehydration. And acyl derivatives, where the carbonyl is next to a header atom, such as oxygen, nitrogen, or sulfur. And we mentioned the acyl transfer reaction, and the Claisen reaction. We saw in this video the nucleophilic addition, where a nucleophile attacks the carbon of carbonyl to add and form a tetrahedral product. For example, alcohols can add to carbonyls to form a hemiacetals, and amines can add to carbonyls to form imines, or Schiff bases. And we reviewed that good nucleophiles are the ones like alkoxides, thiolates, amines, or C minus enolates. Whereas OK nucleophiles like alcohols and thiols, they need to be activated first to undergo nucleophilic addition. We also talked about enolization, the ability of a carbonyl with an alpha hydrogen to rearrange into a hydroxyl bound to a double bond, which we call an enol. Now this equilibrium, called tautomerization, favors strongly the keto form. However, it does form to a sufficient extent to allow chemistry to happen. For example, when we remove the alpha hydrogen, we form an anion called enolate, which is a disguised carbanion which is a very good nucleophile. Next, we discussed the Aldol reaction, a very important carbon-carbon bond formation or cleavage reaction in biochemistry. This reaction happens between an enolizable carbonyl and the regular carbonyl, and a new carbon-carbon bond is formed between the alpha carbon and the keto carbon, as shown here. The mechanism can be both base-catalyzed and acid-catalyzed. And the enzymes that catalyze this, called aldolases, use either a lysine in the active site to form first a Schiff base, or they use a zinc in the active site to polarize the carbonyl and allow for the enol formation. We also saw that Aldol products can dehydrate to form alpha beta unsaturated carbonyls. The mechanism could be both acid- and base-catalyzed, and involves in both cases formation of an enol. Next, we also talked about acyl derivatives, and acyl transfer. As we show here, the resonance in the acyl derivative dictates there how well they react. Carboxylate and amine are the most resonant stabilized, and therefore are the least reactive, whereas esters, especially thioesters, are the least resonance stabilized, and therefore most reactive. Finally, we discussed the Claisen reaction, a reaction similar to the Aldol, between an enolizable carbonyl and an acyl derivative, which generates a beta keto carbonyl. We introduced the acetyl-CoA, a very important thioester, that can undergo Claisen reaction with itself to form acetoacetyl-CoA. And we also introduced the structure of CoA, which is built around vitamin B5, an essential nutrient.
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https://ocw.mit.edu/courses/5-111sc-principles-of-chemical-science-fall-2014/5.111sc-fall-2014.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality, educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. CATHERINE DRENNAN: If you haven't clicked in yet, please pay attention to the clicker question, and I'll give you a little bit more time to click in. And I'll give you a 10-second warning. All right. Let's just go ahead and take 10 more seconds. All right. Does someone want to say how they got this one right? Anyone willing to say how they got it right? We have a American Chemical Society, whatever these are called that hang around your neck and clip things to them. Yes. There's the word. No? Oh, there we go. AUDIENCE: Thank you. So this molecule is the phosphorus has five valence electrons and then each hydrogen has three so it's per total of eight. So it has three bonding and then one lone pair, which makes it a tetrahedral, but then the lone pair pushed the other bond, so it's less than 109.5 degrees. CATHERINE DRENNAN: Excellent. Yeah. So I see that some people just decided that there were three things bound to it. And so then they decided either 120 or less than 120, but you really need to do the Lewis structure and see how many loan pairs there are first. So once you do the Lewis structure, then you figure out the parent geometry, Sn forces sum of tetrahedral system. All right. So more practice on these coming up. All right. So we're continuing on now with molecular orbital theory, and we started the course talking about atomic structure and then we talked about atomic orbitals or wave functions. And then we moved on to bonding, that atoms can come together and bond and talked about the structure of molecules, and now we're going to molecular orbitals. And then we'll also talk more about the structure of molecules based on those molecular orbitals on Wednesday. So we're really coming all the way around, we're using a lot of the ideas that you've seen before, but now we're applying them to molecules. And then, to me, this is really an exciting part. I love getting up to the molecules and talking about structures and molecules and how orbitals really play a role in those properties that molecules have. And then, to me, the really exciting-- I like reactions. So after we finish with the structures, we're going to talk more about how molecules react. And so on Friday, we're going to be starting that, reactions of molecules, and getting into thermodynamics. So we're sort of winding our way away from orbitals for a while. We will come back to d orbitals around Thanksgiving time, but we'll have a long time before that where we're talking about reactions. So we're about to do a transition in the type of material. But before we do that, more orbitals. But these are super cool because these are molecular orbitals. So we're going to talk today about MO theory, MO for molecular orbitals. So molecular orbital theory presents the idea that these valence electrons are really going to be delocalized around these molecules and not just sitting on individual atoms. So to think about this electron do localization, we need to think about molecular orbitals. Molecular orbitals, as we've learned, another word for orbital is also wave functions. Wave functions are orbitals. Orbitals are wave functions. You need to consider the wavelike properties of electrons to think about where the electrons are going to be, what is their probability density, how are they going to be arranged with respect to the nucleus. And so when we take atomic orbitals and we bring them together as atoms come together to form bonds, atomic orbitals come together to form molecular orbitals. So we're going to be adding superimposing, atomic orbitals to form these molecular orbitals. And this is called a linear combination of atomic orbitals, or LCAO. And so we're going to bring those atomic orbitals together and create molecular orbitals. And we're going to create two types of molecular orbitals. We're going to create bonding and antibonding. And some basic math principles apply here, and that is that you can create N molecular orbitals from N atomic orbitals. All right. So that's really the basis of molecular orbital theory, and now let's apply it to our friends, the s orbitals. All right. So we're going to think about really simple molecules, bringing together two atoms that are identical with each other and what happens to their s orbitals when this happens. So first we'll talk about bonding orbitals. So bonding orbitals, again, arise from this linear combination of atomic orbitals, the LCAO. And if it's a bonding orbital, it's going to arise from constructive interference. So we talked before about the properties of waves, and one of the great properties of waves is that they can constructively interfere, destructively interfere. And orbitals are wave functions so they can constructively interfere and destructively interfere. Bonding orbitals are generated by the constructive interference. So let's look at two atomic orbitals. And so here we have an orbital. The nucleus is in the middle. It's a little dot. It's a nucleus. And these two atomic orbitals are going to come together. There's going to be a bonding event. And so we have a 1s orbital from atom a and a 1s orbital from atom b, and they're going to come together, and they're going to form a molecular orbital, an ab molecular orbital, and this is called sigma 1s. So a sigma orbital is symmetric around the bond axis, so the bond axis here would be just a vertical, a line between these two nuclei here. And so this molecular orbital is symmetric around that bond axis. There are no nodal planes for something that is symmetric. There are no nodal planes for our s orbitals, and so there's none for the molecular orbital either. And we can also write this as 1sa plus 1sb. So the atomic orbital from 1sa, the atomic orbital 1sb coming together to form sigma orbital 1s. That is a bonding orbital because it's constructive interference. It's a bonding MO, or molecular orbital. So now let's consider the wavelike properties and think about these atomic orbitals coming together as waves in what is happening. So here we have the same equation up here. We're bringing together 1sa and 1sb, but now let's think about this as a wave function. So there is an amplitude associated with the wave function for 1sa, and there is an amplitude associated with the wave function for 1sb. These wave functions come together. Here is one nuclei, here's the other nuclei. And for a bonding orbital, it will be constructive interference, and so the amplitude where these atomic wave functions overlap will be increased when you have constructive interference. So our sigma 1s now has increased amplitude between these two nuclei due to that constructive interference. So an increased amplitude between these two nuclei, again, this is the bonding axis-- here is one nuclei, here is another-- so an increased amplitude here corresponds to an increased or enhanced probability density. Remember our wave function squared is probability density. It suggests the likelihood of finding an electron in a certain region of volume and if it's enhanced by this constructive interference. So if that probability density is enhanced, you have a greater chance of finding an electron between these two nuclei, which will be attracted by both nuclei. So now why don't you tell me what you think is going to happen to an electron that is in this region of constructive interference, this increased area of probability density. OK, 10 more seconds. Interesting. OK. So we should have probably not put up the answer there and re-pole. Yes. So here, if we have an electron that's attracted to both nuclei, then we want to think about whether that's going to be lower or higher in energy than something in an atomic orbital. So if it's attracted to both, it's going to be more stably bound to those, It will be harder to remove that electron, which means that it's lower down in energy. And so we should look at that, and we're going to. So the answer is it should be lower energy, more stable, harder to remove that electron. It's in a sweet spot. It has two positively-charged nuclei, and it's hanging out right in the middle. It's very happy. It's going to be more stable, and that means lower in energy. All right. So let's take a look at that. So the electron is lower in energy, and the bonding orbital energy is also going to be decreased compared to the atomic orbitals, and that has to be true if that's where the electron is. So let's look at the atomic orbital from a and the atomic orbital from b. And now, I'm going to put the bonding orbital at a lower energy level. First, I was going to put the electrons on, and now, I'm going to put the orbital at a lower energy. So remember, this is increasing energy here, so the atomic orbitals would be up here, whereas the molecular orbital is down here. The molecular bonding orbital will be lower in energy. And now we'll put our electrons there. So we have one electron up here and one here, and so when they come together, we're going to have two electrons in this molecular orbital. So when you have these two electrons and they both occupy the bonding orbital, and this is the case for H2, each H atom is bringing one electron, H2 has these two electrons, and that's going to make H2 more stable. And we saw that before that to associate the H2 bonds, you have to add energy into the system. H2 is more stable than free H plus H. So when you bring atomic orbitals together and you have constructive interference, an increased probability of electrons between those two nuclei, that's a sweet spot. Those electrons are going to be very happy there, and that will result in a more stable, lower energy structure. That's bonding. But whenever there is a positive event like this, there's always a negative event because that's just how life works. So we have bonding orbitals, but we also have antibonding orbitals. So antibonding arise from the linear combination of atomic orbitals, LCAO, through destructive interference. So here these are going to be destructively interfering, and that will generate a molecular orbital that is an antibonding orbital. So here are our little nuclei again, and this antibonding orbital is called sigma 1s star. So we can write an equation for this as 1sa minus 1sb equals sigma 1s star. That is an antibonding molecular orbital. And let's just think about how this kind of shape arises considering the wavelike properties of these atomic orbitals. So I'm going to now move this equation up to the top. And now, I have my 1sa here, my 1sb here, and now it's destructive interference. So we have opposite phase of the wave function. And we'll put up a wave function there. Now, the next one has the opposite phase, so they're going to destructively interfere. There is overlap over here, but when you have destructive interference, then the amplitude is going to decrease. So now, when we consider this destructive interference between these two orbitals, you see that you have what arises between them. Instead of enhanced probability of finding an electron, you actually get a node. So you have decreased amplitude translates to decreased probability density between these two nuclei-- one here, one here-- and that results in a node between the two nuclei. So in the antibonding orbital, there is a much lower probability that it will be in this sweet spot between the two nuclei that have this nice, positive charge for it's little negative charge, so there's pretty much no shot at being in that nice, sweet spot. And so what that ends up meaning is that an electron and an antibonding orbital is pretty much excluded from that internuclear region, that region between those two nuclei, and that results in a molecular orbital that has higher energy than the atomic orbital. There's just no chance of being in that wonderful spot. It's really very sad for the poor electron that has to occupy an antibonding orbital. So now, let's put this on our energy scale. So we'll go back to our energy scale. And we saw before that when we had 1sa and 1sb and you had a sigma 1s, a bonding orbital, that was lower in energy. Electrons that occupy it are more stable compared to their positions in the atomic orbital, but we also now have an antibonding orbital from destructive interference between the wave functions of the atomic orbitals, and that's higher in energy. And so that's up here. So this is what our diagram is going to look like that brings two atomic orbitals together to create two molecular orbitals. So the antibonding orbital up here is raised in energy by the same amount that the bonding energy is lowered. And so that gives rise to this diagram. And importantly, as I mentioned, we have N atomic orbitals forming N molecular orbitals. So if we have two atomic orbitals, we generate two molecular orbitals, one is bonding, lower in energy, and one is antibonding, higher in energy. All right. So let's take a look at some examples. So always start with hydrogen. So hydrogen has how many electrons? One hydrogen, hydrogen atom. AUDIENCE: One. CATHERINE DRENNAN: One. So we have two hydrogen atoms, and so we have two 1s orbitals, 1sa, 1sb. And now where do we want to put our electrons? In the highest energy possible, lower energy? Where do we want to put them? AUDIENCE: Lower. CATHERINE DRENNAN: Always start with lower energy. So we're going to put them down here. And so this is now the MO diagram for H2. And we can write the electron configuration for the MO diagram, which you'll note is a different electron configuration than you were writing when you were looking at the periodic table because now we're not writing it in terms of 1s 2, 2s 2, we're writing it in terms of molecular orbitals. And there are some of these on the problem set, and I tried to indicate example to make sure you know what kind of electron configurations we're talking about. So for MO diagrams, when it says electron configuration we're talking about where the electrons are with respect to the molecular orbitals. So the answer to this would be sigma 1s 2. There are two electrons in the molecular orbital, sigma 1s. Good. So that means that you can do the same thing for dihelium, and that's a clicker question. So let's just take 10 more seconds. OK, you can vote. All right. So does someone want to tell me-- that is correct-- what's number two for? For hydrogen. What about number three, what was wrong there? AUDIENCE: [INAUDIBLE]. CATHERINE DRENNAN: OK, everyone's doing really well. Yeah. So all of the electrons were parallel in there, and what does that violate? What would be true in that case? AUDIENCE: [INAUDIBLE]. CATHERINE DRENNAN: Yeah. And so they'd have the same four principle quantum numbers. This is not allowed. So that one's not good. And what's wrong with four? AUDIENCE: [INAUDIBLE]. CATHERINE DRENNAN: Yeah, star was on the bottom. So they were flipped around. Antibonding was lower in energy. Great. So we'll just put those in over here, and so we had two electrons. Helium brings two. So two went into the bonding, and two went into the antibonding with opposite spins. And we have the bonding, lower energy; antibonding, higher energy. All right. So then we can put in the electron configuration here, and we have sigma 1s 2, sigma 1s star 2, so that's the electron configuration. Now interestingly, you see, you have two at lower energy and two at higher energy, so there's no net loss or gain in energy of H2 compared to just two helium atoms. And so that raises the question, does helium 2 exist? So what would molecular orbital theory tell you about whether it exists, and it would actually predict that it does not exist. And the way that molecular orbital theory gives you these predictions is through the calculation of something called bond order. So bond order is half of the number of bonding electrons minus the number of antibonding electrons. So let's just write out what the bond order for helium would be. So helium 2, our dihelium molecule. So we have a bond order equals 1/2. There's always 1/2. And now, the number of bonding electrons, so how many bonding electrons do we have for helium? AUDIENCE: Two. CATHERINE DRENNAN: Two. How many antibonding electrons do we have for dihelium? AUDIENCE: Two CATHERINE DRENNAN: Two. And can someone do this math for me? AUDIENCE: Zero. CATHERINE DRENNAN: Zero. Right. So that would suggest a bond order of 0, which means no bond. And let's just compare that to hydrogen, H2, which should have a bond order equal to 1/2. It's always a 1/2. In the hydrogen, how many bonding electrons did we have? AUDIENCE: Two. CATHERINE DRENNAN: We had two. How many antibonding electrons did we have? AUDIENCE: Zero. CATHERINE DRENNAN: Zero. And the math? Bond order is? AUDIENCE: 1. CATHERINE DRENNAN: 1. So that means 1 bond or a single bond. So MO theory would predict that dihelium, has no bond, i.e. it's not really a molecule without a bond, but H2 should exist. So let's take a look at what experiment tells us, and it does exist, but only really not that much. So it was not discovered until 1993, which for some of you might seem like quite a long time ago, but since we've been mostly talking about discoveries that were made in the 1800s, it took a long time before someone could prove that dihelium existed. And if we look at the bond association energy for H2, 0.01, and compare that to H2, 432, dihelium really doesn't exist very much. 0 is a much better approximation of its bond than 1 would be. It really is not a very good molecule. So I would call this a win for molecular orbital theory. It correctly predicts that our helium 2 is not going to be a very good molecule, but H2 will be a good molecule, and that works. All right. Now, let's consider 2s. So 2s orbitals are analogous to 1s except that you have to remember that they're bigger. So we have our 1s and our 2s, but for the purposes of this, it doesn't really matter. So let's look at a diagram now that has 1s and 2s. So we have lithium. So how many electrons does lithium have? AUDIENCE: Three. CATHERINE DRENNAN: Yep. So we'll put on lithium, dilithium. We'll put on lithiums has two in 1s and one in 2s, and so we have one lithium here and one lithium over here. And our 1s orbitals are going to be lower energy, so they're down here. Our 2s orbitals are higher in energy. So that goes for both the atomic and the molecular orbitals that are generated. So bringing together 1s with 1s, we get sigma 1s and also sigma star 1s. And so we can start putting our electrons in. We're going to start with the lowest energy and move on up. So we have four electrons that we need to put in, so we fill up everything here. And now we go up to our 2s. The 2s orbitals will generate signal 2s and sigma 2s star. We have two electrons, so they both can go down here. So here is what our MO diagram looks like. We can write out the electron configuration again, based on this diagram. So we have two electrons in sigma 1s, so we have a 2 there. We have two electrons in sigma 1s star, our antibonding orbital, and we have two electrons in sigma 2s. And now we can calculate the bond order, which is another clicker question. OK, just 10 more seconds. Excellent. 84%. So if we do out the math here, we always have our 1/2. We have four bonding electrons. We have two down here and two up here. We have two antibonding electrons in our sigma 1s star, and so that gives us a bond order of 1. And in fact, the dissociation energy does suggest there is a bond. It's not a great bond. It's 105 kilojoules per mole, not necessarily enough to power a starship, but still this molecule does exist. All right. So let's look at beryllium now, do another example. So how many electrons is beryllium going to have? It will have four. So we'll put those up. Now, we're going to start with our lowest energy orbital, and we're going to put some in the antibonding as well. And then we'll do this, and then we'll do that. And we filled up everything. And so we can write out our configuration. We have two in our sigma 1s orbital. We have two in our sigma 1s star. We have two in our sigma 2s, and two in our sigma 2s star antibonding orbital there. And now there are two different ways that I can calculate the bond order here, and we'll do both of them and show you that they come out the same way. One way that we can calculate this is to consider all electrons. So if we consider all electrons, our bond order, and that's often just said B.O., is 1/2 of our bonding. And now we can count up both the 1s and the 2s. So how many bonding electrons do we have? We should have four, so we have two here, two here, so four. How many antibonding? Also four, so that suggest a bond order of 0. Or we could just consider our valence electrons, so that would be the electrons in 2s, and if we do that, bond order equals 1/2. It's always a 1/2, so how many valence electrons do we have in bonding orbitals? Two. How many in antibonding orbitals? Two. And that gives the answer of 0. So it should always work unless you do something very strange. You should be able to do it both ways and get the same answer. So if this is a complicated problem, you might want to only consider the valence electrons. In fact, on a test, you may only be asked to draw the molecular orbital diagrams for the valence electrons, and you don't have to do the other ones. So that should work both ways. And we get a bond order of 0, and in fact, the dissociation energy is only 9 kilojoules per mole. It's a little stronger than dihelium, but this is an exceedingly weak molecule. So when you have the same number of electrons and bonding as antibonding, it doesn't lead to a very strong molecule. All right. That's the orbitals. Now, it's time for molecules that have p orbitals as well. So in this case, now I have my p orbital and another p orbital. And we have our nuclei in the middle, and we're going to bring these together, and we will have constructive interference. And so here we're bringing together 2p x of a and 2p xb or 2p ya and 2p yb. So just x and y we're considering right now. And if we bring those together with constructive interference, then we're going to form a bonding orbital that has enhanced probability density in both cases and nodal plane along the bond axis because we had nodal planes along there to begin with. So if we think about this and we have both of these, we're bringing them together and they're going to interfere constructively, enhance probability density here, enhance probability density down here, but we still have our nodal plane because we started out with one. And so if we have a nodal plane, this cannot be a sigma orbital. It has to be a pi orbital, because sigma orbitals are not going to have a nodal plane along the bonding access. All right. So we could generate pi 2px or pi 2py this way. So a pi orbital is a molecular orbital that has a nodal plane through the bond axis or maybe I should say along the bond axis, so here is our nodal plane right through the bond axis. We can also have antibonding, which means destructive interference. So now, I'm going to be subtracting one of these from one of these, and I'm going to get something that looks like this, and it will be pi 2px star or pi 2py star, and it will have two nodal planes. So let's look at this. So this is destructive interference. I'm subtracting one of these from one of these, so the phase has to change. So I'm going to change the phase on one of them and then bring them together. Now, we're not going to have that awesome, constructive interference increased probability density. These are negatively interacting with each other. And this generates a nodal plane between the molecules. They really look much more like this now, so we have still our nodal plane through the bond. We had that before. We're always going to have that. But now we have an additional nodal plane between the nuclei. So in one case, we have enhanced density, probability density again. And the other case, in antibonding, we have another nodal plane. All right. So now, let's look at what happens to the energies of these pi orbitals. And the diagrams that I'm about to show you, we're only talking about px and py now. We have for the moment forgotten 2pz. So these diagrams are rated I for incomplete. Warning to the viewer, people come to me and go where are the 2pz? Yes, they're not in these diagrams, but when you're asked for a complete diagram, you will always have to put those orbitals in. And in fact, completing these diagrams could be a question you get later. But for now, we're going to have 2pz here, but it's not forming a molecular orbital in this diagram. This diagram is, thus, incomplete, but we're going to start simple and build more complicated. So only 2p orbitals first, and then we'll add the third one. Because this compound doesn't actually need that orbital, so we're good for now. All right. So we have moved on to the first element that has a p electron in a p orbital. We have boron, and we have two of them. So now I dropped off 1s to simplify this. Now, we just have our valence electrons, so it's a good thing we can calculate bond order just using our valence electrons. All right. So we have two in 2s and one in 2px or 2py. I could have put it in either place. So let's put in where they would go. So we have 2s orbitals. They'll go into bonding sigma 2s first, then into antibonding, sigma 2s star, and now we have p electrons, and we're going to put them into our pi orbitals, our molecular orbitals. I'll put one in, and where am I going to put the other one? Am I going to put it next to it or over here? What do you think? AUDIENCE: [INAUDIBLE]. CATHERINE DRENNAN: Yeah. So we're going to do that because, again, when you're going to sit on a bus, you want to have if they're degenerate in energy levels as they are here, you're always going to put one electron in each orbital of the same energy first before you pair them up, and they'll have parallel spins. So we're reviewing things we learned before. I love doing that. So now, let's see what our electron configuration is, and this is just the valence electron configuration. We're simplifying. We're not going to consider the 1s orbital, and we can write this down. So we have two electrons in sigma 2s, and we have two electrons in sigma 2s star, our antibonding orbital, and we have one in pi 2px and one in pi 2py. And I can put a 1 or not put a 1. If I don't put anything, for a 1, 1 is assumed. And we can calculate our bond order as well. So we have 1/2, and again, we're just using our valence electrons, but that's OK. We have four now who are bonding, two down here, two up here, these are bonding orbitals, and we have two that are antibonding. And notice for our pi orbitals, this is what we saw before bonding are lower in energy, antibonding are higher in energy. With the bonding orbitals, we had constructive interference, enhanced probability of the electrons near the nuclei, and so that's lower in energy. But in our antibonding ones, we have a nodal plane in between our nuclei. So we don't have any probability that electrons are right in between there because there's a nodal plane, so those are higher in energy. So here is our B2 diagram. So now let's try the same thing for carbon, C2, and that is a clicker question. All right. Let's just take 10 more seconds. So let's take a look at that over here. The easiest thing to do to answer this question was to fill in the diagram in your handout. And so if you did that, you would have put two down here and you would have put two up here. Then you would have put one here, one there, another one there, and another one there. So now we have used these up, and so our configuration is sigma 2s 2, sigma star 2, pi 2px 2, pi 2py 2. And the bond order is 1/2. There are six bonding electrons-- 1, 2, 3, 4, 5, 6-- and two antibonding electrons, and so that adds up to a bond order of 2. And so sometimes on a test, they'll be a simple question what is the bond order, but to get there you have to draw your whole molecular orbital diagram and figure out how many bonding and how many antibonding, so these are not really that fast questions. And it's nice. Sometimes we give you like a little space, and you see this whole little molecular orbital diagram fit in there to answer the question. All right. So let's just compare these two diagrams for a minute. So in both cases, we had 2s orbitals, two atomic orbitals for 2s, and they both generated two molecular orbitals, a bonding and an antibonding. The bonding is lower in energy, and the antibonding is higher in energy. We also had two 2px atomic orbitals. They generated 2 pi 2 px orbitals, one bonding, one antibonding. And the same for our two atomic orbitals for 2 py. We had two of those, and they generated one lower energy bonding, pi, bonding and one pi star antibonding. So you always have N atomic orbitals generating N molecular orbitals. So the stability of the resulting molecules in these cases depend on how many of the electrons are bonding, how many are in energy lower as a result of formation of the molecule, and how many are at higher energy as a result of formation of the molecule. And if the net result are more electrons in lower energy, more electrons in bonding orbitals, then that molecule is more stable. If there's a very slight or no difference, then that's not a very stable molecule. So now let's just compare these two and think about which of these is going to be more stable. So we have our configurations again. So in the case of B2, how many electrons are in lower energy or bonding orbitals? AUDIENCE: [INAUDIBLE]. CATHERINE DRENNAN: Yeah, we have four-- 1, 2, 3, 4. How many in higher? AUDIENCE: Two. CATHERINE DRENNAN: Two. Up here. For carbon, we had six-- 1, 2, 3, 4, 5, 6-- two in higher. And so the bond order here was 1, the bond order here was 2. Which is more stable? Higher dissociation energy, which one do you think? AUDIENCE: [INAUDIBLE]. CATHERINE DRENNAN: Carbon. Right. Has a bond order of two. It has more electrons in lower energy orbitals. So it cannot really well out of this bonding deal, and the dissociation energy for B2, 289, whereas, for C2, 599. So when molecules come together such that more of the electrons are in lower energy or bonding orbitals, you form a nice, stable molecule. When molecules come together such that more of their electrons are an antibonding or higher energy, that's not a happy molecule. So I'll just and with one way to think about this. In this cartoon molecular, break up lines, sometimes two atoms just have an incompatible number of valence electrons. And there are just too many-- just here this molecule saying, I'm sorry. Too many of your electrons are in my antibonding regions. I don't know how many times we've all heard that, but it's time to dissociate, but our atomic orbitals, well, they can still be friends. OK. See you on Wednesday. Take a look at the clicker question. All right. Let's just take 10 more seconds. All right. Let's just go through this one. So this is a review of where we were last time. So the correct answer is 1. So sigma orbitals are cylindrically symmetrical. Let's quiet down a minute. You can hear the answer. So this one is true. The second one is not true. A bond order of zero doesn't mean that you just have antibonding orbitals. Whenever you bring together two atomic orbitals, you have to make two molecular orbitals. So it isn't that sometimes you make bonding orbitals and sometimes you make antibonding orbitals. Every time you bring together two atomic orbitals, you make two molecular orbitals, one that's lower in energy, and that's the bonding orbital, and one that's higher in energy, and that's the antibonding orbital. So that is not what that means. A bond order of zero means that you have equal numbers of electrons in your bonding and antibonding orbitals. So there's no net stabilization due to the formation of these bonds. So here bonding occurs when you bring together two atomic orbitals to make two molecular orbitals that are both of lower energy. No. Every time you make the two orbitals, one is lower energy, one is higher that energy. You can't make two that are both lower in energy. And bond order of 1 means constructive interference is generated at one bonding orbital. That's not what a bond order of 1 is. And again, every time you generate a bonding orbital, you generate an antibonding orbital. And 1 means that you have twice as many electrons in your bonding orbitals as antibonding orbitals because the formula is 1/2 the number of bonding minus antibonding. But that's good, and it's important to remember that sigma orbitals are symmetric around the bond axis. All right. So we had these diagrams for boron and carbon, just talking about the interactions of the px and py, and so we had forgotten about our pz. And you can't do that on a test. You get into trouble, so I'm always telling people for these two handouts, you must include the molecular orbitals that are derived from p to z. So on a test, you need to put them even if they're empty. Even if they don't have anything in them, they need to be part of your molecular orbital diagram. We didn't have them in the diagram because we hadn't talked about them yet, so now we're going to talk about them. So two pz orbitals, again, this is on Monday's handout. You have this linear combination of atomic orbitals. And all are p orbitals. They all look the same as each other, they're just different in orientation. You have one along x, one along y, one along z, but they're the same. So now we're going to bring our two pz orbitals together, and we're going to do it along the bonding axis. So we've defined this as the bonding axis in the class, so we'll bring them together, and they'll be constructive interference with our bonding orbitals. There's always constructive interference that generates bonding orbitals. And so we're going to create a enhanced amplitude as the wave functions come together, and it's going to be cylindrically symmetric. So what type of orbital do you think this is going to be, sigma or pi? AUDIENCE: Sigma. CATHERINE DRENNAN: It'll be sigma because it's cylindrically symmetric. So we do not have any bonding plane along the bond axis, and it's symmetric around. And we have enhanced probability density, and we have the wave function squared enhanced probability of having an electron between the two nuclei. And so this is a sigma 2pz. So p orbitals can form sigma molecular orbitals. So we do have nodes passing through our nuclei. Here are our nuclei again. We do have them. They were here before in our p orbitals. There's the nodal plane in our p orbitals, but we do not have, in this case, a node along the bond axis. So that is our bonding. So whenever you generate a bonding orbital, which is going to be lower energy, we're going to have our increased amplitude between the nuclei, again, our increased probability density and therefore, lower energy. So whenever you have constructive interference generating a molecular orbital of lower energy, you got to create something of higher energy. That's just how life works. So we also are going to have antibonding orbitals, which are generated by destructive interference. And again, these orbitals can be thought about as wave functions, and a property of waves is that they constructively interfere and destructively interfere. So now we can subtract our two orbitals, which we're going to switch the sign and they're going to be out of phase. So they'll destructively interfere, and that's going to look like this. So now, you're going to generate a nodal plane between the two nuclei, but it's still symmetric around the bond axis. So this is a sigma 2pz star, so it's an antibonding orbital. So again, it's sigma, so it's still cylindrically symmetrical with no nodal plane along that bond axis, but you do have a new nodal plane that's generated. So nodes pass, again, through the nuclei, but also now between these two orbitals. So we have a new nodal plane that's generated that's between these nuclei, and that's a result of destructive interference. Generates that nodal plane. You have decreased probability density for an electron being found there. And so that poor electron is shut out of that sweet spot. The electrons like to be between those two nuclei where they have the two positive charges of the nuclei and then their little negative charge, and they can sit right there and be very happy in a low-energy state. But here there's really lower probability density, lower likelihood the electron will be found here, and that generates a molecular orbital that's antibonding or higher in energy. All right. So now, we have to go back to our MO diagrams and figure out where to put these new molecular orbitals onto our nice diagrams. And it's not as simple as it was before because where we put these new sigma 2pz molecular orbitals depends on what z is. So it depends on the value of z. So if z is less than the magic number of 8, then we have our pi 2px and 2py orbitals are lower in energy than our sigma 2pz molecular orbital. But if we are equal to or greater than 8, then the sigma 2pz orbital is lower energy than the pi 2px and 2py. So less than 8, pi is first. It's lower in energy as you go up your energy scale. And if z is equal to or greater than 8, pi is second and sigma is first. So how are you going to remember this? There could be many different ways one can remember it, but I'll tell you how I remember it. And my life revolves around my daughter and my dog. And so at Thanksgiving, we always have the question, can I eat pie first? So if you are under the age of 8, you always want your pie first. So if z is less than 8, pi comes first. Pi is lower in energy and sigma is higher. However, when you mature to the grand age of 9, say, or 10, if you were a kid, 10 is like the oldest you can possibly imagine being, very mature. And you can eat your Thanksgiving dinner and wait for pie. So that is how I would remember it, under 8 pi is first, equal to 8 or greater, you can wait till after dinner to have your pie, pi comes second. Note that the ordering of the antibonding orbitals is the same, so all you have to remember is down here depends on z, is pi first or is pi second? So let's take a look at an example. Let's look at our friend molecular oxygen that has a z equal to 8. So oxygen is at the old, mature age of 8, and so it's going to have its sigma 2pz first. It can wait for its pi orbitals until later. So let's start putting in our electrons. So we have each oxygen making up molecular oxygen, or O2. Brings two 2s orbitals to the Thanksgiving dinner table, and two of them go down in energy into the bonding sigma 2s orbital and two go into our antibonding, sigma star 2s orbital. Now, we have four electrons in our atomic pz orbitals, four from each molecule, so we need to put all of those in. So we always start with the lowest energy orbital. So we'll put two in there, then we'll go up. We have two more here, two more here. We'll put them in singly with their spins parallel, and then we'll pair them up in the lowest energy orbitals. And then we have two more left, so we're going to have to put those up in our antibonding orbitals. So they go into pi 2px and pi 2py star orbitals up here. All right. So now we can calculate the bond order for oxygen, and that's a clicker question. All right. Let's just take 10 more seconds. All right. So let's take a look over here. So bond order, again, is 1/2 the number of bonding electrons minus the number of antibonding electrons. And we have eight bonding-- 1, 2, 3, 4, 5, 6, 7, 8. And we have four antibonding-- 1, 2, 3, 4. So that gives us a bond order of 2. And the bond order equation is one that you do have to memorize. That will not be given to you on an exam. And with a bond order of 2, we have a pretty big number for dissociation energy. Again, that's energy you have to put into a bond to break it, to dissociate it, and that means it's a pretty strong bond if it's a big number. If you need a lot of energy, that's a strong bond. Another thing that you can see from this diagram is that O2 is a biradical. It has two lone pair electrons, so two unpaired electrons, which also makes it paramagnetic, or attracted to a magnetic field. So whenever you have unpaired electrons, you will have a paramagnetic species, and diamagnetic means they're all paired. And so there are some questions on problem sets and on exams, so you need to know the definitions of those. All right. So we talked about this when we were doing Lewis structures, if you recall. And we drew a beautiful Lewis structure of molecular oxygen that had two lone pairs on each oxygen and a double bond. But I told you that that was not really a complete description of molecular oxygen, that molecular oxygen was actually a biradical, but you would not get a hint of that from the Lewis structure. So if you draw the Lewis structure as a biradical, then you have a single electron here and a single electron there and a single bond. But we also see the bond orders 2, so this one describes the biradical nature, but doesn't really describe that double bond character that it has. So neither of these Lewis structures really completely describe molecular oxygen. And we need our molecular orbital diagram to really help us understand the properties of molecular oxygen, that it's pretty strong bond between it, it has double bond character, but it also is very reactive. It's a biradical. So this is just a bizarre molecule. And really that is why this diagram right here really tells us why our life, our planet is what it is, right here, that describes it. It's because of oxygen that we have the life forms that we did. Life was very different on this planet before molecular oxygen came about. You did have microbes that lived anaerobically without oxygen. But when oxygen came, everything changed. And so if I say, can you explain life to me? You can draw this, and there it is. This explains life as we know it because life as we know it exists because of this crazy molecule that is O2, nothing else really like it. It is an amazing molecule that allows us to live. So there you go. I was going to say you don't learn anything in chemistry. This diagram I just explains life as we know it to you, right there. All right. It's not just oxygen, there's a few other elements that, yeah, are pretty important, and one of them is nitrogen. We wouldn't really be much anywhere without nitrogen either. Oxygen is, no, nitrogen is pretty special too. This also helps. Maybe these two molecular orbital diagrams really sum up life as we know it. All right. So let's look at molecular nitrogen, N2. So we have two electrons in our 2s orbitals, so we're going to bring them together. We're going to put two in the lower energy orbital, our sigma 2s, and two in our antibonding orbital. And then we have three in our p orbitals from this nitrogen, three electrons from this nitrogen, and so we'll put them down here. We'll put in 2, 3, 4, 5, 6. So we didn't need to use any of these antibonding orbitals, and this is a z less than 8 case. So here we have our pi orbitals first because it is z less than 8. So let's look at the bond order here. And the bond order is 1/2 of 8-- 1, 2, 3, 4, 5, 6, 7, 8-- eight bonding electrons and just these two antibonding electrons, so 1/2 of 8 minus 2 is 3. And you have a really big number for your dissociation energy, 941. This is a very stable molecule, and you can draw the Lewis structure of this without much difficulty. This one works quite well, and you get a triple bond and two lone pairs. So this would be a diamagnetic molecule, no unpaired electrons, no radicals here. But this is a crazy, strong bond. It's really hard to split nitrogen because of this triple bond. And molecular orbital theory tells you it should be its triple bond. It should be really, really strong interactions between these nitrogens. And so this is a very hard thing. We want to do is industrially, and we'll talk more about this when we get a chemical equilibrium. How do you split the nitrogen bond. This is actually, currently, a big area of research, how you break that bond because we need nitrogen for life. And so how do you split it? There's lots of nitrogen N2 in the atmosphere, but we need it here, and we need it usable, so we need to break that bond to use it. So we'll come back to this in chemical equilibrium and how people are able to split the nitrogen bond. So forget man of steel. We should have man of nitrogen. That's strong. Nitrogen is strong. Forget steel. Man of nitrogen. One more thing to do before we move on to today's handout and this is really fast, you can also be asked to draw molecular orbital diagrams where both atoms are not the same. So if you were asked to do that, you can use the following rules. If z is less than 8 for both atoms, pi is first. If z is not less than 8 for both atoms, it's hard to know what to do, so you don't need to know what to do. And that's it. So we might tell you something about it and then you can do it, but otherwise, just worry about things when z is less than 8. So that's molecular orbital theory.
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https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/8.06-spring-2018.zip
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PROFESSOR: So here it is, connections formula, connection. So we'll take a situation as follows. Here is the point of x equals a. I'll put just a here. Here is the x-axis. And I will imagine that I have a linear potential. So we have a linear potential here. Why do we imagine a linear potential? It's because the thing we would really kind of want to think about is what is called the turning point, energy, and a potential. There's a turning point. But it's clear that near enough to the turning point, this is linear. And the problem for the WKB approximation fails, as we discussed, is precisely in this region. Now, I will make one claim that this is really what's going to connect here. These two asymptotic expansions that we've found here are in fact WKB approximations of our calculations. So our whole procedure here is going to be assuming that we have a linear potential here, and I will take the linear potential seriously. So I will imagine it even goes forever. And what we're going to do is find a way from the turning point. So the energy is going to be here. Here is v of x. x v of x, the energy. And then we have this region, which is a WKB region. And we have this region over here that is also WKB region. And the region in the middle is not a WKB region. It's a region where all solutions fail, WKB solutions fail. They're not valid in this region. So let's assume, we have here, therefore, a potential that is linear, and, therefore v of x minus the energy is going to be a number times x minus a, where g is a positive constant. It has some units. v minus e is a linear function that vanishes at x equals a. So this is a nice description of the situation. And then you could write a WKB solution. So what is your WKB solution? Let's write the WKB solution, psi WKB on the right, and then we'll write a WKB solution on the left, the right and left. So here is what we. Typical WKB solution on the right. That is a forbidden region. Square root of kappa of x, you would have a decrease in exponential, e to the minus a to x kappa of x prime, d x prime. And then you have b over square root of kappa of x, e to the a to x kappa of x prime, dx prime. So this is what you've been told is supposed to be your WKB approximation. Decay in exponential, a pre-factor, growing exponential, a pre-factor. And kappa of x is something that you know from this potential. So this you would write for any potential on the right or on the left as well. And here, your kappa of x squared is 2m over h squared, v of x minus e. That's the definition of kappa squared. You're in the region where v is greater than e, and therefore, this is equal to 2m g over h squared x minus a. So our next step is a little calculation. We should simplify all these quantities here. We should evaluate them, because we have a kappa. You have a linear function. We can do the integrals. We can put what kappa is. So everything can be done here. So if we do it, what do we get? So evaluate. We get the following. Now, I will introduce some notation to write this. So first, I'll write it, psi write of x WKB equals 1 over square root of eta. That's a symbol I haven't defined yet, u to the 1/4, e to the minus 2/3, u to the 3/2 plus b square root of eta 1 over u to the 1/4, e to the 2/3, u to the 3/2. OK, that's what you get. You'll recognize the a and the b's. And here is what we need to introduce a couple of variables to make sure we know what every symbol is. And here, they are. u is eta x minus a. And eta is 2m g over h squared to the 1/3. Just roughly k squared is proportional to u. So k is proportional to u to the 1/2. And therefore, square root of k gives you the u to the 1/4. The integrals also work that way. And you see already here, things that look like the array function. And that will become even clearer soon. So this is our solution for the WKB expression on the right, but let's do also the WKB expression on the left. The WKB expression on the left would be another wave of this kind, but with cosines or with exponentials that we've done before. So what did it used to look like? It used to look like k of x and integrals of plus minus i integrals of k of x prime dx prime. That was the way the WKB solution looked like. I'm now trying to write this expression for the WKB solution on the right. Now, I want to make a remark that even though I'm using the linear potential, in general, if your potential, even eventually becomes something different in this region, I'm good. So I'm moving away from the bad point sufficiently so that the WKB approximation is good. But I don't want to move away so much, so that the potential fails dramatically to be linear. I have to straddle limited region in fact. I don't want to go too far. So these were your solutions before in a region where you're in the classically allowed region. Now, this form of the solutions will not be practical for us. So I will write them, instead with sines and cosines. It will be better for us to write it with sines and cosines. So let me write it with sines and cosines. Psi left now WKB of u or x. Sorry. It's going to be c over square root of k of x, and I will write the cosine of integral from x to a. You see, I'm on the left. So it's natural to write integral from x to a. On the other hand, on the right was natural to write integrals from a to x. And that's what we did, cosine from x to a. We'll have here k of x prime vx prime plus d over square root of k of x sine of x to a k of x prime dx prime. And you could say look, that's good. You just traded the exponentials for sines and cosines, and that's fine. Still, for what we're going to do later, it's convenient to write our WKB approximations still in a little different way. I can add a phase here and a phase there. And that still is the most general solution, because they're both sines and cosines. And unless I would put a phase that differs pi over 2, and one becomes equal to the other, no. I'm going to shift both by pi over 4, and therefore, these are still different functions, and it's convenient to do that. So this is our onset for the WKB solution. We have to do the same thing with it here in which we expand and evaluate in terms of and eta in this quantity. So what do we get? We get c over square root of eta u to the 1/4 cosine 2/3 u to the 3/2 minus pi over 4 plus d over square root of eta u to the 1/4 sine of 2/3 u to the 3/2 minus pi over 4. So this is our psi left WKB of x. OK. We're in good shape. I'm going to raise this up. And now you could say, all right, you've done the WKB. But on the other hand, how about the exact solution? What is the exact solution to this problem? You've written the WKB approximation for the linear potential, the WKB approximation for the linear potential. Now, what is the exact solution for the problem? Well, your Schrodinger equation is minus h squared over 2n psi double dot of x plus of x minus the energy on psi is equal to 0. Well, we've written this already a couple of times. So this is minus h squared over 2m psi double dot plus g x minus a psi equals 0. That is our v minus e. And in terms of the u variable, the differential equation, of course, becomes the second psi, the u squared equal u psi. That's why the area function is relevant here. We get back to that equation. That is the equation for the exact linear potential, and those would be the exact solutions. So our exact solutions are psi being some linear combination of the array function and the other array function. And now let's do one case first before we get into something that is more complicated. Let's take psi to be the solution to be the array function ai of u. That's your solution now. If that is your solution, you would have this behavior, the behavior we've noted here for this solution. And now you compare and you look at what you got. You say, all right, I have a solution, and I know how it looks rigorously, how it looks to the right and to the left. That's how it looks, and this is the exact solution. This is the WKB solution, and I know how it looks to the right and how it looks to the left. And I see the correspondence. We see this term in yellow is a solution that this a decaying exponential, and we see it here. And we have the red term, the oscillations. And we see it also here. So here is that connection condition. In fact, when you derive this asymptotic expansions, you did the true connection condition, because you connected a single function from the left and from the right. Here it was to the right. Here it was to the left. And it's a connection formula because it's the same function. So here, we see that if we wrote the WKB solution, and we were aiming at maybe this solution, we would have that this term goes, connects with this term. And the coefficients differ by 1/2. That 1/2 has no other explanation than the asymptotic expansions. There's the square root of pi, these things, and there's the two functions. So if you put the 1/2 is there in the exponential, so you could put the 1/2 here, and 2a, if you wish. So you realize that in order to have a connection 2a would be equal to c. The two coefficients here are related in that way. There's a number here and a number here. It happens to be 2a and c in this case. So 2a is equal to c. So if we choose a equals 1, we would have c equal 2. So here is a matching condition therefore. We have that a WKB solution that has the c factor with c equals 2, so it would be 2 overs square root of k of x in here times cosine of this x over a k of x prime, dx prime minus pi over 4 is connected-- we'll see more on that-- with 1 over square root of kappa of x, this solution with a equals to 1, what we have here, e to the minus a over x kappa of x prime dx prime. So that is the first connection condition. We will discuss it further in a couple of minutes, but this is how you connect things. A single solution of the differential equation looks like this term and like that term on the left and on the right. Here is your WKB solution. This general solution looks like this term, and it then should match to something that looks like that if you have the array function. So in general, for not linear potentials, this is the term that gives rise to that, so we take that this term in general is matched to the top term in the blackboard here.
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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We've so far described abstractly what we mean by a dot product, by a definition of AB equals the magnitude of A times cosine theta, times the magnitude of B. But many times in physics problems, we actually have vectors in space and we want to see how to do this in terms of a Cartesian or any coordinate system in particular. So let's set up a coordinate system, we'll call it Cartesian, a and j. And now this is very important. Let's define a vector A here, and let's have another vector in a completely different point, B. We still can take the dot product of these two vectors. But in order-- this is our plus x, plus y. And I'm only doing things in two dimensions. So how do we calculate the dot product? Well, the first thing that we want to look at is our unit vectors. What are the dot products of unit vectors? Well, if we take i hat dot i hat, that's the magnitude of i hat times cosine of the angle 0, times the magnitude of i hat. And cosine of 0 is 1 and the magnitude of unit vectors are 1. So when you dot product the unit vector with itself, you get 1. And therefore, it's also true for the j hat. j hat dot j hat is 1. What happens when you dot product two vectors that are perpendicular? Well, in this case, this is 0 because the angle theta is 90 degrees. And remember that cosine of 90 degrees is 0. And anyway, when two vectors are perpendicular, there's no component of one vector that's parallel to the other. So these are the essential facts that we're going to need to calculate the dot product of two vectors that are separated in space. So the way we do that is we'll begin by drawing, writing down the vectors in Cartesian coordinates, where Ax A is a scalar and the vector part is and the unit vector. And we have Ay j hat. And likewise, we can write B as Bx i hat plus By j hat. And now when we take the dot product of these two vectors, we're going to write out all the terms here so that you see them. So here is our scalar or dot product. We use those words interchangeably. Now notice that we've already shown that the dot product distributes over vector addition. And also, if you multiply a scalar by a vector, you can pull the scalar out. So there's four terms here-- Ax i hat dot Bx i hat plus ax i hat dot By j hat plus Ay j hat dot Bx i hat. This is a little tedious to write out. By j hat. And now because these are scalars, we can pull them out and the only part of a dot product that matters is how the unit vectors dot. And that's why we have these two results. i hat dot i hat is 1, j hat dot j hat is 1, and i hat dot j hat is 0. So the first turn is Ax Bx. i hat dot j hat is 0, so we don't need that. j hat dot i hat is 0. And finally j hat dot j hat is 1. And so we get plus Ay By. And that's how we define the dot product in Cartesian coordinates of two vectors. Now, notice that if we dotted a vector, A dot with itself, that would just be Ax Ax plus Ay Ay, which is the components squared. And that's equal to the magnitude of the vector squared. And so we can say that the magnitude of the vector, of any vector, is you take its dot product with itself. You take the square root, but remember we always take the positive square root, because magnitudes are positive. And that's how we calculate the scalar product for vectors. And many times in the application of physics, when we have physical quantities that are vectors in different places, we use vector decomposition and we use this procedure. This is the Cartesian picture, we'll learn how to do that in polar coordinates when we need it later on.
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https://ocw.mit.edu/courses/7-01sc-fundamentals-of-biology-fall-2011/7.01sc-fall-2011.zip
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PROFESSOR ROBERT DORKIN:Hi, and welcome to a help session on recombinant DNA. Today we will be talking about the polymerase chain reaction as well as DNA sequencing. The polymerase chain reaction, also known as PCR, has many uses. One of the most common uses is to amplify a desired section of DNA. What you need for the reaction is your DNA sequence of interest, DNA polymerase, DNA primers, and then four different nucleotides. You combine all these together, and the first thing that you do is you heat the reaction up. What this does is that by adding heat to the system, you break the hydrogen bonds between the two different DNA strands. This results in two separate DNA sequences. Now what happens is that you allow the system to cool. As it cools, the DNA primers are able to hybridize to this separate strands. Now as you remember from lecture, when you're synthesizing DNA, you synthesize from the five prime to three prime direction. This means that the primers you design have to match the three prime end of your sequences of interest. So for example, if we were designing primers for these two sequences, one of them would be GGTA and the other one would be AGCT. Now, I've written four here. In actuality, these primers are generally longer, around 16 or so, 16 to 27. However, they can be a whole different variety of lengths dependent on numerous different factors. The next thing that happens is that the DNA polymerase is going to bind to the DNA sequence with the primer. Once the DNA polymerase is bound, it's going to take some of the free nucleotides in the surrounding area and slowly add them to finish up the strand. And so on and so forth. Once it's completed, we are going to have now doubled our original DNA sequence. We're going to have two strands that are identical to the first one. As you can see, by repeating the steps, heating it, allowing it to cool, allowing more primers to bond, and then allowing the DNA polymerase to elongate, we can double the number of sequences every round. And you can rapidly get a large amount of the desired sequence. PCR has other uses though besides simply increasing the total amount of DNA that you have. One of the uses of PCR is to sequence DNA. Now, if we look over here, normally DNA is form of deoxyribonucleic acids. You have the phosphate group on the five prime end. You have a hydroxyl group on the three prime end. This hydroxyl group is very important. That's because when a new nucleotide is added, this hydroxyl group undergoes a covalent bond with the phosphate on the new nucleotide and then adds a new nucleotide that way. So you can see you're adding the five prime the three prime direction. However, it is possible to create a dideoxyribonucleic acid. The dideoxyribonucleic acid, instead of having a three prime hydroxyl group, has a three prime hydrogen. This three prime hydrogen is no longer capable of forming a covalent bond with a phosphate. That means as soon as the dideoxyribonucleic acid is added to DNA, no further nucleotides can be added in the series. Let's go back to our example with the primers. What does that mean for here? Well, let's say you have a normal PCR reaction, but in addition to the four deoxyribonucleic acids you have, you also take a little bit of dideoxyribonucleic acids of one of the types. So let's say we add in some ddTTP. Now what happens is that your DNA polymerase will go along adding nucleotides as normal, but if it ever adds a dideoxyribonucleic acid, the polymerase will stop. So if it adds, say a normal T here-- continues down, continues down. If it adds a dideoxyribonucleic acid here, it's going to stop, and we're going to get a truncated sequence. And so you can see that at any position that we have an A, it's going to be possible to have a truncated sequence of that length. What this means that we're now going to, once the PCR is complete, have different DNA sequences of numerous different lengths. But the one thing they're all going to have in common is that they're all going to end with a T. So you can imagine doing this now for each of the four different letters. Then we can take them and run them out of a gel. Let's go look at such a gel over here. Here we have a gel. Each of these letters represents which dideoxyribonucleic acid was used for that experiment. And then the PCR was run out on the gel. As you remember, the strands close to the bottom are the shorter strands, and the strands close to the top are the longer strands. So if we look at this gel, we know that the shortest strand ends with a G. The next shortest strand ends in a T. Oh, sorry, ends in an A, excuse me. Then the next short strand ends in a T, then two A's, then a G, then a C, then a T. So as you can see, this is one way to determine the sequence of DNA. Another way has been devised, which is even faster. Instead of running the sequences all out in different polymerase chain reactions, what they do is they have some of each of the dideoxynucleotides together. But now, they fluorescently label them, such that you have a different fluorescent label on each dideoxyribonucleic acid. Now what happens is that you can run it out all on one column. And then by just looking at the colors, you can determine what the sequence is. So once again, the sequence would be GATAAGCT. And so this way, you can more efficiently, more rapidly, determine what the DNA sequence is. This has been two examples of polymerase chain reactions and their uses. This has been another help session on recombinant DNA. We hope you join us again next time. Thank you.
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https://ocw.mit.edu/courses/10-34-numerical-methods-applied-to-chemical-engineering-fall-2015/10.34-fall-2015.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. WILLIAM GREEN JR: All right, so we're going to start a new topic today about ordinary differential equations initial value problems. But before I start to talk about that, I want to remind you next week's schedule is weird. So there's no classes on Monday, but instead we'll have Monday class on Tuesday. So I'll see you next on Tuesday morning. And then we're not going to have a class on Wednesday, but we have the quiz on Wednesday night. And then we'll get back to normal on Friday. And I would expect that possibly the TAs might be interested in giving a help session or review on Wednesday the class period. AUDIENCE: We'll be here at 3:00. WILLIAM GREEN JR: They'll be here. So if you want to come at class period, and just ask questions and stuff. AUDIENCE: And I'll have your PSAT graded back, if you like. WILLIAM GREEN JR: All right, got that? PSAT be ready as well. OK, so I'll talk about today ODE-IVPs. AUDIENCE: Yeah, we'll still have office hours Monday [INAUDIBLE] WILLIAM GREEN JR: Can you guys hear this? Yes, office hours like normal Monday. And for those of you who want to get your homework started early, we might even post homework. And I'm going to focus exclusively on ODE-IVPs you can write this way. All right, and I would just comment, what do these things mean? ODE means ordinary differential equation. It means that the only differentials in the problem are with respect to one variable, so I'm going to call that t. It can be z or x or whatever you want, but I'm going to call it t here. And so this is not partial differential equations where you have differentials respect to many variables. Initial value problems mean that all the initial conditions are given in at a single t point where t is that the variable that appears in the differential. All right so this is the form we're going to do. This is first order a way of writing the ODEs. If you recall from homework zero problem number one, we showed you there how you can write higher order differential equations-- convert them into the form of first order. And I guess that's worthwhile to point down. I would also comment that I going to just talk about f of y, however the ODE solvers in Matlab expect f of the t comma y as their from. All right, so let's just explain all the details about this. So suppose I had a differential equation like this. All right, so this is like a differential equation you'd have for a body move moving-- a particle moving with some friction force on it and some time varying force. Yeah? AUDIENCE: Are you going to be posting your notes online? WILLIAM GREEN JR: No, I don't have any notes. This is it. All right, so this is a differential equation. What you can do is to convert it into the form I have above, is to write a new variable, say v, that's defined to be dx dt. That's the velocity. And has a time dependence, but I told you I don't like time dependence, so I'm not going to write it that way. So we can also define another variable. So we can define our new y vector to be x, v, and t. So this is y-- the first component of y, this is the second component of y, this is the third component of y. And then I can write down what f of y is. So the equation for dx dt is y2. It's the second component of y, dx dt is equal to v. The equation for the dv dt is given by this equation. So that's f of y3 over m minus gamma y2. And the equation dt dt is just one. So this is my f of y. So this is f1, f2, f3. Is that all right? So this is just how you can convert-- if somebody writes you an equation this way, and you want to get it into the standard form that I'm going to show you here-- dy dt equals f of y-- this is how you convert. And a skill we haven't really emphasized so far in the course, but I think a very important skill, is to be able to get like a chemical engineering problem and rearrange it so it becomes a standard form that maps up to one of the solvers you know how to use. So you've got a problem has some differential in it. Right away you actually, can I somehow rewrite it so it looks like this standard form-- or actually for Matlab, this form. So Matlab will say f of t, y. If you can do that, then you can go ahead and use those Matlab solvers to solve it. But your variables may not have a letter y appearing anywhere, right? You have to figure out how to write it into the standard form. And then once you can do that, you can use the solvers. And the same thing with the non-linear equations solvers, with the optimization software. If you can rewrite it into the form that matches the standard form, then once you're done, you're done. Then you can just go and call the canned programs that work well for those forms. All right, now what do we want as the output? So this is the problem. This is the problem as proposed. What do we really want out of this? What we're going to get out of it is a bunch of points y-- actually a matrix, right? It would be a vector of y-values. Typically y is a vector. And we want to know it a bunch of time points. And we would a lot of time points so we're getting pretty plots of how the components of y vary with time. That's what we usually want. Sometimes we just want the final value, y time value. We integrate to some point and then stop, and just run through the final value. But a lot of times, we actually want to make pretty plots, so we really want a lot of points here and making plots. Now how many points do you need to make a nice plot? Maybe, I don't know, 100 points, 50 points, you can make a nice plot. So that's the kind numbers you'd like. And so the output of these programs is going to be something like a vector of the time points, and then a matrix of the y-values corresponding to each time point. And the way it does in that lab, it's reasonable. Each row of y are the y-values that correspond to the same time point, the first time point. So the first row of y corresponds to the first number in the time vector. The second run is the second. Last one will be at time final. And then you can plot them. OK so far? And ideally, we would like these y-values that we compute, these y's we get out, ideally one that really close to what the true solution of the true differential equation is. But we're doing numerical stuff, so it's never going to be exactly the same. And so a big part of the effort is trying to figure out, how can we get the thing to compute y calculated at each time point to be as close as possible to the truth, of what the true solution of the equations are? And that turns out to be difficult, so that's a very important thing to worry about. What else can we say about this? We specify this from t naught, and you typically have to input your t final, how far you want to go away from your t naught as another input. Normally, people write it where t naught is less than t final, and so you're like integrating from left to right. But you could actually write them, put a negative sign equation which corresponds to a negative t, change your variable to of t prime that was equal to t final minus t if you wanted to, and integrate the other direction. And in fact in homework zero problem one, we did that too. So you can integrate frontwards. You can integrate backwards. If somebody tells you the initial condition or condition in the center of the range, you integrate forward for part of the range, and backwards the other way, so anything like this. So what's important for this to be what's called an initial value problem is just that all of the specifications of y are given at the same value of t. It doesn't matter exactly what value of t it is. Later, we'll talk about what happens if you don't know all the specifications at one value of time. All right, now how would you approach this? Probably, a bunch of you've done this already as undergraduates, or maybe even high school. You can write like a Taylor expansion, y of t plus delta t-- is y of t plus delta t-- times dy dt. You can write that. And then I say, wow, I know dy dt. That's actually f, so this is actually equal to-- right? And then if I truncate the Taylor expansion, now I have an update formula. So if I just forget this last bit, I can put in my current value of y, like the initial condition. The initial condition here, evaluate f, multiply by delta t, add them together, and I'll get a new value of y. And then I can repeat over and over again. And in fact, this is a very famous method. It's called the Forward Euler method. Euler was a pretty smart guy, so it's not completely ridiculous, but as I'll show you, it has some problems. So the Forward Euler, another way to write it is y new is equal to y old plus delta t times f of y old. It's called Forward Euler. And we can write a little implementation of this pretty easily. So you can write y is equal to y naught, t is equal to t naught for i equals 1 to the number of steps, y is equal to y plus delta t times f of y, t is equal to t plus delta t. Somewhere up here, I need to write delta t. OK, so there's your code for doing Forward Euler. How many of you have done this before? OK, so I can skip over this very fast. How many did this, and it didn't work for a problem? OK, so you're not doing hard enough problems. Well, we'll correct that. All right, so I noticed that this closely resembles the rectangle rule. So the rectangle rule would look a lot the same. However, if you're integrating a function-- so I have I is equal to the integral of f of t dt, then I can notice that the derivative dI dt is equal to f of t. And so if I was going to do the rectangle rule, it would look just the same as this, except this would be a t. Maybe I would pick y0 to be 0. And the final value of y I would get to would be my integral, my i value that I want. So how many of you have done rectangle rule? Yes? All right, so you did this already. Now you remember in high school, when they took the rectangle rule, they probably mentioned that it's not very accurate. And so then they also taught you some other ones. You guys, how many of you learned the trapezoid rule? How about the midpoint rule? How about Simpson's rule? All right, so at least a high school math teachers thought that this is inadequate for doing real work, so they tell you all these other things instead. This was like your first baby thing. So let's talk about why was that. So let's do the Taylor expansion a little bit further. So let's do it for the numeric [INAUDIBLE] case. So y t plus delta t is equal to y of t plus delta t times f of t plus 1/2 delta t squared times the second derivative. And so when we made the approximation to do the rectangle rule, we just threw this term away. So that's a leading term that we threw away. So we made an error order or delta t squared. Because normally the second derivative is not exactly zero. So we have an error in each step that's the order of delta t squared. But how many steps do we make? The number of steps is equal to t final minus t naught over delta t. So therefore, we're making one over delta t order steps. So the total error is going to be approximately the number of steps times how much error you making each step. So it's going to be something the second power in delta t divided by something to the first power of delta t. So the total error in the integral I is going to be older of delta t if you do the rectangle rule. And so order delta t is not very good. So you want to increase your accuracy-- if you cut your step size in half, that means you double your CPU time, because you have to do twice as many function evaluations. But you only gain-- you reduce your error by a factor of two. So if you want to, say, gain three significant figures of accuracy in your number, then you have to do 1,000 times much work as you did before. So I have some amount of some precision with some number of steps. I want to increase, get three more significant figures. I'll have to do 1,000 times as much work. That's kind of bad scaling. If I want to get six more significant figures, I've got to do a million times more work. So at some point, this going to be kind of expensive. So if you contrast it, you can do things like trapezoid rule. And as you're probably-- well let's talk about trapezoid rule for a bit. So the idea of this is you have your function, you have some point t, and t plus delta t. You're trying to integrate between them. Let's say t naught, t naught plus delta t. Try to integrate. Here's the value of the function at t. Here is the value of the function that t plus delta t. If you do the rectangle rule, what you say is I just draw a line here and integrate that rectangle. But say the real function really looks like this. Then I missing that area there, so that's the error. That's why the error's so big. If I do that trapezoid rule, then the error-- I actually overestimate the value here, but the integral of the difference between this dotted line and the solid line is less than the integral between the solid line and that dotted line, so its more accurate. So that's why people like the trapezoid rule better than the rectangle rule. If you did the midpoint rule, you'd evaluate this function halfway between these two points and then draw a dotted line here and a dotted line there, and integrate that guy. And it would overestimate for part, and underestimate for part, and they would partially cancel each other out. And so that also would be more accurate than doing the rectangle rule. So you did it before. And so what we're going to try to do is, instead of using this formula we did before, we're going to replace this with something else that I'm going to call g. And g is something that's supposed to be the average value of f over the step. Instead of just taking the value f at the starting point, I want to get sort of the average. Right, because an integral is related to an average over the delta t. So g is going to be my way to estimate the average. And there's going to be a zillion update formulas that different mathematicians have proposed over the years. They give you different g's that tell you how to do the update. And the key is try to find a g that's really accurate. You want to have one that's the most accurate you can, because then your errors will be less. And if your errors are less, than your delta t can be bigger, which means your number of steps will be less, which means your CPU time will be less. So you might be able to get more accuracy and use less CPU time both if you can find a really good formula for g, for the update. So what's the g for the trapezoid rule? It's f of t plus f of t plus delta t over 2. That was the g you used when you did the trapezoid rule when you were kids. What is the g for the midpoint rule? It's f of t plus delta t over 2. And so the different g's are called different update formulas. And these particular ones reduce the error from delta t squared for [INAUDIBLE] step to, I think, delta t to the cubed power. And then when you multiply by this 1 every delta t, it turns out you go from being order of delta t to order of delta t squared, and that's a really big change. So now if I cut the step size by a factor of 10, I gain a factor of 100 in the accuracy. If I want to get six more significant figures, I do 1,000 times more work, not a million times more work. So it's a pretty significant improvement. And then people have pushed this on further and further. And so actually very common integrators that you might use nowadays would go out to fourth and fifth order methods. And so they have complicated update formulas that are carefully designed to cancel out all the low order delta t errors and just leave very high order ones. And that's kind of the-- typical ones are like that. And then some fancy ones might go even up to eighth order, I've seen, if you're really pushing it. You have really complicated g formulas then. Now, this is all for the integration, but really our problem was that the ODEs [INAUDIBLE].. So we have to do f of y not f of t. Let's see if I can show you that. Yeah, so we did the rectangle rule, we had f of t. But the real problem is we have to do f of y. Now the problem is, we don't know what y is, y is our unknown. So we generally have a problem here. In Forward Euler, we can get away with it because we just put the y old value in. Like for example, if we wanted to do midpoint here, we'd have to know f at a different time point that we haven't computed yet, because it's forward in time. Same thing with trapezoidal rule. We have to know f at some future timepoint. So it's not so easy, actually, to go directly from these formulas to some explicit formula like this with a g in here, where the things inside g, the y's inside g, are all y-values we actually know. So this is a serious problem. But this didn't stop people from doing it. So one idea is you could go back to the Taylor expansion, and now do the Taylor expansion but in terms of f of y. So this is f of y of t. And then over here, this is d squared y dt squared, but dy dt is f, so this is really df dt. But f is actually not a function of t, it's a function of y, right? But we can do chain rule. So df dt is equal to the sum df dyn dyn dt. But dyn dt is f. And this is what we call the Jacobian [INAUDIBLE].. So this is really saying that this is equal to J times f. So that would be one possibility, is you could go and write the Taylor expansion out to the next higher order explicitly, putting in J times f here, and then just being in the cubic terms. Now, people don't do this usually. Any idea why this is not a popular thing to do? Yeah? AUDIENCE: It's computationally expensive to [INAUDIBLE].. WILLIAM GREEN JR: That's right. That's right. So how many function evaluations does it take to get the Jacobian? Right, so each function is n values, right, because it's a vector. And you have to do at least, say, forward differences would be n function evaluations. So instead of just doing one function evaluation like we were doing with the Forward Euler, now we have to do n plus 1, or something like that, function evaluations. And actually, you probably want to use center differences, so you need 2 times n plus 1. And if you do them analytically, unless it's sparse, it's still going to be really expensive to compute it. So it's a lot more effort to do it. So you need to think, well, is it really worth it? And so people have found other formulas that basically get rid of the delta t squared term that don't require so much effort, and so that's what people do instead. But this is an option. You could do it. All right, so instead, what people do is they've decided to generalize the midpoint rule and try to figure out, how can we get an estimate of the midpoint value that's cheap to evaluate? And so what they say is, well, this is g midpoint for numerical integration. Let's make an new g midpoint that works for when f is a function of y. So let's make it function of y evaluated at t plus delta t over 2. So that's what the formula looks like. But then you say, well, I don't know y is at t plus delta t over 2, because I only know it at y of t. So then they say, well, let's do Forward Euler for that. So then they say, well, let's say it's approximately f of-- what's the Forward Euler formula? It's y of t plus delta t over 2 times f of y of t. How many parenthesis there? So this is the formula that's actually used in practice a lot. And this one is called Runge-Kutta two, second order Runge-Kutta formula. And it's just a-- it's the midpoint rule where we estimate the value of y of t plus delta t using Forward Euler. And the advantage of this is it's very cheap, right? Just evaluate-- I just have to evaluate one function here, one function here, so two function calls. [INAUDIBLE] and that way I can get an update formula. And this one, turns out that it has the nice delta t scaling that you might like. So that's the kind of thing people do. And the Runge-Kutta guys went crazy, and so one of the most popular solvers is called Runge-Kutta four five, and that's the ODE four five program in Matlab. And what that does is the fourth order extension of this, and the fifth order extension of this. And it uses them both, and it compares them, and uses the difference between an estimate the error in the calculation. And then uses that to decide what size delta t you need, depending on what tolerances you demand. And the formulas for all those-- [INAUDIBLE] formulas are given in the textbook. Now, this starts to lead the problem. So we weren't able to actually evaluate the true g midpoint, which we'd like to be here. Instead, we have to use some approximation in order to extrapolate forwards to the y. So this is hinting at the whole problem. This whole problem is actually we're just extrapolating. We know the true value of y only at t naught, and all the rest we're doing is extrapolating forwards. And we're do it over and over again, so we're going to do it for n steps. Maybe we'll do 1,000 steps. So you're extrapolating. And then based on that extrapolation, you going to extrapolate again. And then based on extrapolation, you going to extrapolate again. And you can see that this is not really the most robust thing to do, right? Because if you have any experience, you're not that comfortable extrapolating at all. And then you have to extrapolate 1,000 times. You should be a little bit worried about what can happen. So let's think about how the error will propagate. So we're going to have some errors while we do these extrapolations. So we have here a t naught, t naught, we're happy. We know y naught. This is exact. We're like woohoo. We know one value of y, really good. All right, so now the question is the first step. Now we're going to get to value y1 that we compute with one of the formulas, and it's going to be good to some accuracy depending on how good or g is, our update formula. And so it's going to have an error. I should comment that this y naught, although we treat it as exact, it could have an error too. But it will probably be more like a machine precision kind of error because we don't really know the initial conditions that well. Or might be an experimental error about how well we know what the initial conditions are. But I'm not going to worry about that one for now. But be aware, this is true, we don't really know initial conditions perfectly either. But certainly, even if we treat this as being known perfectly, we're going to have some error by the time we extrapolate to y1. So now what happens at y2? So now we've added some more, so now we're at t2. And we see started from y1 true plus some error. This is y-- y naught was true. Now we're starting from y1 true with some error in it, because of a mistake we made the first step. And now we see how that step propagates further. So let's see what that's going to say. Well say y2 is going to equal y1 plus delta t times some update formula which would depend on y1. But y1 was not true anymore, so this is actually equal to what should have been the true value of y1 if we only knew what it was, but we don't know. And we have our error that we don't know exactly how big it is that we added on. And then we have another term from this guy, so it's going to be delta t times g of y1 true plus delta 1. And then we know our update formula's not right anyway, so there's actually another error, delta 2, just because of the error in the update formula. Now if we had a great update formula, these deltas are kind of small. And if somehow, by luck, our delta 1 was 0, and we had a great update formula, then we think that this error in y2 is going to be pretty small. But it's not true. So really the errors are pretty big. So let's see if we can figure out-- we're going to do a Taylor expansion here, so this'll be g of y1 true plus d dgy-- the Jacobian of g with respect to y-- times the delta 1. That's the first order Taylor extension. All right, so now let's group the terms. So y2 is equal to y1 true plus the update formula of y1 true plus the error in our update formula if we had started at the true value. So this is what we-- if somebody told us what the true value of y1, this is the size error we'd expect. It's to be that little error we calculated before. But then on top of it, we have these other terms. We have a delta 1, because y1 was not the true value. Sorry, I lost a delta t here. And then we have delta t times dg dy times delta 1. And I'll emphasize these are all vectors. I think that's it if we only keep to first order. So what is this thing? This is a matrix, right, dg dy. I could write it this way. It's the identity matrix plus delta t times the Jacobian dg dy times the vector d1. And you can see, at every iteration when I do this, I'm going to get a similar factor like this. And it's going to again multiply delta 1. I keep on multiplying the-- errors from my previous step keep getting multiplied by this kind of factor. So it's going to be really important to us about what the norm of this matrix is. What's going to happen if the normal of this matrix is big? What's going to happen? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: That's right, the error's going to get multiplied by some factor, say bigger than 1, every iteration. And after we go 1,000 iterations, how big is the error going to be? Really big, right? Even if that thing is quite close to 1, 1 plus something to the 1,000 power it's gigantic. So what was originally a pretty small error-- because we chose a good g method-- is going to get multiplied by this huge amplification factor. So the key thing for a method to be what is called numerically stable is that the norm of this matrix has got to be less than 1. Or maybe even equal, equals 1 might be OK. If this is much bigger than 1, you're doomed. Your numeric order is just going to blow up from step to step to step. And what's bad about it, is actually sometimes you might not even be aware of this. So you'll get some solution, because it still will calculate some numbers for the y's at each type-- y1, y2, y3, y4-- it's just running through the calculation. But the numbers may become increasingly meaningless further away from the y true as time goes on, as the steps go. AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: Yes, it can. Yeah, so that's what we call a stable numerical method is one that, if you make an error in an earlier step, its impact diminishes as the steps go on. That's what you want. You can't always achieve this. But if you can, that's really great. Then that's you would call a really stable, robust kind of method. Even though I give you some stupidity early on, my stupidity evaporates away, and it goes to the true solution. That's what you want. OK, so let's think of cases where it's going to be problematic for sure. So if dg dy-- say if all it's eigenvalues are positive and you add it to identity matrix, the thing is still going to have eigenvalues bigger than 1. Can you see that? And actually, what do we say when dg dy is positive, has a positive eigenvalue? You guys have a word for this already you learn in your differential equations class. What do you call these kind of differential equations? Maybe not dg dy. Back up, how about df dy? If df dy, if that Jacobian has a positive eigenvalue, what do you say? You guys remember this? So this is what we call unstable differential equations. So if it has any positive eigenvalues, then two initial conditions that differ by a little differential exponentially separate as time goes on as long as the difference as any projection in the direction of the bad eigenvector. So those are called numerically unstable. A lot of those actually exist in reality. So explosions, that's because you had some positive eigenvalue somewhere for example. Amplifiers, like you buy an amplifier to crank up the music, you get positive feedback when-- remember Professor Swan walked to some strange place, and for some reason, he got the feedback from the speakers? So he had a positive eigenvalue going there. A little tiny noise and his microphone, and it amplified to a big squeak. So there are real systems like this that amplify. And sometimes we want to model them, like explosions for examples. That's a pretty important for chemical engineers. But it's going to be really tough, because if it's got positive eigenvalues, then this is probably going to have a norm bigger than 1. And then whatever little errors we may get any step in the whole calculation, we're going to amplify and amplify and amplify. And who knows if we're going to have any reality left in our solution by the time we get to the end. All right, so that's one kind of problem. So how about we have a problem that's intrinsically stable. That means that all the eigenvalues of the Jacobian of the original problem are negative. So the thing should be perfectly stable. From whatever initial condition they started from, the whole thing should sort of come down to like an equilibrium point. So that's a pretty common situation in chemical engineering too. You start from some state, and it relaxes down to like an equilibrium state or a steady state. Well-behaved systems act like that. So what I'm going to show you is that, despite the fact that the df dy is well-behaved, it's possible that this sum matrix might still be poorly behaved and still have a norm bigger that one. And so that means you started for a problem that was pretty well-behaved, and you broke it by your choice in numerical method. So this is a very bad one. This is the kind that can make you really embarrassed if you're working in a company. Your boss gives you a problem. He says, please tell me the time is going to take our reactor to relax after a start up. the system is perfectly well-behaved. He gives these beautiful equations that your previous coworker worked out. Everything's fine. You put it in there. You use your stupid numerical solver, and you get a wild oscillation, and the thing blows up. OK, and that's going to be because of a poor choice of g so that it makes this norm of this matrix bigger than one. It's pretty easy for that to happen. So you don't need a very complicated case to show it. So let's just do a scalar case. Say dy dt is equal to negative 2y. So this is a well-behaved differential equation. What's the solution? So with a y naught y at t equals 0 is equal to 1. What's the solution? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: Yeah, OK, so it's perfectly well-behaved function. And the plot when you plot it, here's one. It goes [PLANE DIVING NOISE]. So now let's solve this with the Forward Euler method. So I'm going to say that y new is equal to y plus delta t times f. So in this case, this is f, right there, negative 2y. So for Forward Euler, this is just saying that this is negative 2y. All right, so let's choose a delta t, say delta t equals 1 to start with. So we're going to compute. We start at t equals 0. We're computing out to t equals 1. so we're out here. So we put 1 in here. Our initial condition was y was 1, so that's 1. So as 1 times negative 2 times 1 is negative 2. This is a 1. 1 minus 2 is negative 1. So at the first step, it goes from here down to there. This is not looking good. This physical variable is supposed to be gradually relaxing towards 0. Going negative is not what you want. And then I think we could go the next step, if you want. So we can put in suppose y is negative 1. And we're still doing a delta t of one step. So negative 1 times negative 2 is positive 2. Plus negative 1, so this goes back to the starting point. We'll get the sawtooth, is what you get. And it turns out, actually, if you make delta t any larger than 1, then this sawtooth amplifies, so you have a oscillating solution that explodes. Which is a little bit different than the physical solution, which gently relaxes. And so this is just a scalar equation that you guys could have done in high school. If you high school math teacher had been brave, he would have showed it you. But he don't want to ruin your faith in the Forward Euler method, so he just showed you a simple one that worked out. Now you can make this work better by making the delta t smaller. If you made the delta t small enough-- in this case, I think it's delta t is less than half-- then this actually behaves somewhat reasonably. And so what's going on there is that the identity matrix is positive. My Jacobian is negative. And if I make delta t small enough, this big positive number is bigger than this smaller negative number, and the whole thing stays positive but less than 1, and so then the normal is good. But if I make delta t too large, I make this second term much larger. If it gets much larger than 1, it overwhelms the first term. The thing's negative, but I what do the norm it's positive. And so it has an expansion. Then the negative is why we get these oscillations. So if you ever do a numerical differential equation solution and you start seeing this kind of stuff, it's almost certainly related to this numerical instability of the method. and not really a physical thing. I mean, there are physical things that do this, but not too often in chemical engineering, fortunately. All right, questions? No questions, OK. So in the textbook, they have a big discussion about how to evaluate different g methods in order to figure out if they're guaranteed to be stable, or under what conditions they'll be stable. And so that's worth a look to see. And they have detailed derivations for several of them. Questions? All right, so for the homework, homework four, you're going to have to write your own ODE solver. So you'll do update formulas and compare them. When we come back on Tuesday, I'll show you how this is usually dealt with by switching to implicit solvers and talk about that. All right, have a good weekend. Enjoy the three days off.
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https://ocw.mit.edu/courses/3-091sc-introduction-to-solid-state-chemistry-fall-2010/3.091sc-fall-2010.zip
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The following content is provided under a creative commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JOCELYN: Hi. Jocelyn here and today we're going to go over fall 2009 exam three problem number four. Starting with part A-- let's read the problem. In the 1920s, Jack Breitbart of Revlon laboratories found that acne could be treated by the use of benzoyl peroxide. The oxygen-oxygen bond in peroxide is weak and under the influence of modest heating, benzoyl peroxide readily decompose to form free radicals according to the reaction given. The rate of decomposition is measured at 92 degrees C at various concentrations and found to be-- and you're given two different concentrations and two different rates. So part A asks, determine the order of reaction for the decomposition of benzoyl peroxide. The first thing to do is, when asked for the order of the decomposition reaction, we want to think about the general rate loss, right? We're given concentrations. We're given rates. We're asked for the order. Something you might want to have written down on your equation sheet or look up is the general rate law equation. Remember that students taking these exams did have an equation sheet so they could have this ready. The important part is that you know when to use this equation, what each of the terms mean. So r is the rate. k is a rate constant. It is a function of temperature, but if you're at the same temperature, that rate constant will not change. It's constant. c is your concentration. And n is your reaction order. So this would be the place to start for this question. And now we know that we're solving for n. So what is n in this case? Looking at what we're given, we can write two different equations. The first line gives us a specific rate for a specific concentration. The second line gives us another rate for a second concentration. So now we have two equations and two unknowns. That means we can solve for one of the unknowns and then could solve for the other one if we wanted to, but here we only care about n. So how I'm going to choose to solve this and you may choose to do it differently-- I'm going to divide the first line by the second line and I get rate one over rate two equals c1 divided by c2 all to the n power. So that made the k drop out, which is one of the unknowns that I don't care about because the problem isn't asking for that. Now all we need to do is solve for n. So if we remember algebra, we're going to take the log of both sides. Because of the properties of the log, we can bring the n out front. So n equals-- and plugging in the numbers given in the problem-- they're given in the same units so everything will cancel out and that's good-- we get that the n equals one. So this is a first order reaction. Now you could've probably done that by inspection if you're familiar with the rate law and how to determine that by just seeing that as the concentration went down by a fourth, the rate also went down by a fourth. However, the question asks for you to determine the order of reaction and so we were looking for more of a derivation of the problem instead of just an inspection because that shows that you have a familiarity with the general rate law equation. So now moving to part B-- we are asked, on the plot below, sketch the variation in energy with extent of reaction for the decomposition of benzoyl peroxide. Assume that the ratio of activation energy to-- delta e, which is the energy of the reaction-- equals -2.5. Label the energy states and label the delta e of reaction, the activation energy for the forward reaction and the activation energy for the backward reaction. So that's asking us to do a lot. So first we're going to write down, what are we actually asked to do? So the main thing is that we're asked for the variation in energy with the extent of reaction. So as the reaction progresses, what's the energy of the molecule? The energy state that it's going through? In that plot, we're asked to label the energy states of the reactant and product and we're asked to label the delta e of reaction. So what's the net change in energy as well as the activation, both forward and backward? I would always write this to the side or scrap piece of paper or something because this is a lot of things to put on one plot and this will be a nice reference to help us to make sure we answered all the things the question is asking. So on the plot given, which I'll reproduce here, it has axes of extent of reaction and energy. The first thing to do would be to label the beginning and start energy states. So because we know at 92 degrees C, this-- the benzoyl peroxide-- decomposes readily, we can assume that there's a decrease in energy. So we'll have our products start at a higher energy than our-- sorry-- our reactant start at a higher energy than our products. And because the question asks us to label that, we're going to put in the actual labels. And so that's your reactant and then we have the radical product. Now that we have our beginning and end energy states, we need to think about what's happening in the middle. A clue is that he talks about the activation energy, right? And we know that most processes that we talk about have a certain activation energy. You can't just go straight from the molecule to the radical. You have to have a little bit of energy cuffed. And so we know we're going to have some type of hill that we have to go over. Furthermore, in the problem he states that the activation energy divided by the delta e is -2.5. So we know that because this is a negative delta e-- so this is our delta e here and it's negative-- our activation energy is two and a half times as large as that net change in energy. So we want to kind of just eyeball that in here. You didn't have to be exact, but relatively-- it's going to be larger than twice the distance between here. And remember that activation energy is the hill, the energy costs that you have to go through to get to your lower energy state in the product size. So now that we have all of our energy states, we want to draw a smooth curve to show the variation of energy with the reaction. So connecting all these points, we get two valleys and a hill, right? We have two stable energy states and we have an activated complex corresponding to the activation energy there. So we have our energy diagram, but we need to go back to what the question is asking and see that we've labeled the energy states, we've labeled the delta e of reaction, but we need to label both the forward and backwards activation energy. And this is one of the things people had issues with-- the backward activation. So forward activation is pretty self explanatory, right? It's the energy we need to overcome the activation of this process and so this is the Ea of the forward. Now the activation energy of the backward reaction is just the same concept, but going backwards. So if we started down here, what is the energy we need to get over this activation barrier? And so that's this full energy, this full amount of energy here. And so we can see that it's bigger than the activation energy for the forward reaction by the delta e of the reaction. So going back to our checklist here, we've answered all over the questions and so we can move to part C. So part C asks us, on the same plot above-- which we have over here to the right-- sketch the variation in energy with extent of reaction for the decomposition of benzoyl peroxide under the influence of a catalyst. So with a catalyst-- first to answer this question, we need to know what a catalyst does. And in class, we learned that a catalyst will lower the activation barrier, thus allowing for the reaction to proceed faster. You don't necessarily get more product because that's governed by a different set of laws, but you do get the product faster. And so the way we can change this is that instead of having our activation barrier up there, we can say it's going to be somewhere here. It's going to be lowered by the presence of the catalyst. So we want to make a smooth curve again and connect this lower activation barrier. And you'd want to label this-- I'll label it in white-- this is the catalyzed curve there. As long as you've showed that the activation barrier was lowered by the presence of a catalyst, that's fine. That was what the question was asking for. So we've labeled everything, answered all the questions and we're done with this problem.
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https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/8.06-spring-2018.zip
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PROFESSOR: We have to set up a little better the geometry of the calculation. And for that we have to think of various angles. We oriented there the electric field along the z direction. But that's not going to be too convenient for our calculation. So these calculations are a bit of an art to do them. They're not that trivial. Not terribly difficult. But I think if you appreciate it, next time you ever have to do one of these things it will become clear. So first your have to think physically. Is there an angle in this problem? Is there any important angle happening here? It's a question to you. I think if you figure out that you have a chance of doing a diagram that reflects this. Is there a physical angle, you think, in this process? A relevant angle? Yes, Lou? AUDIENCE: [INAUDIBLE] PROFESSOR: Perfect. Yeah. The relevant angle is you have a directional ready for the electric field. So if the electron comes off there's going to be an angle with respect to that electric field. So that's our physical angle in this question. So let's try to draw this in a way. So I will draw it this way. And I brought colored chalk for this. I'll put the z-axis here, and I'll put the electron momentum in this direction. k is the lateral momentum. Now the electric field is going to come at some angle with respect to the-- or the electron momentum is at some angle with respect to the electric field. So that angle will presumably stay there for the rest of the calculation. So let's do it with green. So here is my electric field. And it's going to have an angle theta with respect to the direction of the electric field or the photon incident direction. Now, this is actually more like the polarization of the photon. The electric field is the direction of the polarization of the photon. So one more vector, however, the position that we have to integrate over-- because we have the whole hydrogen atom and the whole of space to integrate. So x has a position. So I'll use another color here. Unfortunately, the colors are not that different. Here is r, the vector r. So now I have several angles. I have theta for the electric field. But now that I've put all this axis I not only have theta, but I also have phi for the electric field. And for r, I will have theta prime and phi prime. So r has a theta prime and a phi prime. We usually have theta and phi. But the answers, at the end of the day, are going to depend on theta. And if you have theta here, we want the theta to remain. So theta prime and phi prime are going to be our variables of integration because you integrate over r. Finally, we have one more angle that we have to define. So we have theta phi, theta prime, phi prime, and the angle that was here-- this was the angle between r and the direction of the electric field. And we called it theta. So we have to give it a new name here. And this angle, I'll call it gamma here. The angle between E and r. OK. So We have everything defined here. Maybe I should list it. E has angles theta and phi, the direction of E. r has angles theta prime and phi prime. And gamma is the angle between E and r. And k along z hat. So this is our situation. So what are we trying to calculate? Well, we want to use Fermi's golden rule. So we need to calculate final H prime initial. That's the matrix element of the Hamiltonian H prime between the final and initial states. So what is this? Well, these are all wave functions that depend all over space, and this is a function of space. So let's do it. This is integral d cubed x. Let's put the final state first because it shows up there. So it's 1 over L cubed e to the minus i k dot r. So here is our final state, u final. Then the Hamiltonian. That's simple. e E0 r cosine what? Cosine gamma. Is that right? This is what we had there. It's the angle between the electric field and r. It was theta to begin but now has become gamma. And then the final state. So this is our H prime. And then the initial state is 1 over pi a0 cubed e to the minus r over a0. OK. This is our task. This is a matrix, and here it is. An integral of a plane wave against an electron wave function and an extra r dependence here. This could range from undoable to difficult, basically. And happily, it's just a little difficult. But this is an integral-- we'll see what are the challenges on this integral. So let's take a few constants out. So e, E0, pi, a0, that all will go out. So this is e E0 over square root of pi l cubed a0 cubed. So I took the l's out, the e, E, this thing. All right. Let's write the integral more explicitly. This is r squared dr sine the volume element. Sine theta d theta d phi. But I'm integrating over x, which is integrating over r. And r is theta prime and phi prime. So these are all these ones. I'm integrating over all values of theta prime and phi prime. Then what do I have? I have this exponent, k dot r. Well, my diagram shows how k dot r is easy. It involves a cosine of theta prime. So it's e to the minus ik, the magnitude of k, the magnitude of r cosine theta prime, because after all, k was along the z-axis. r is along the phi theta prime direction. OK. We're progressing. This, this, that term, it's r cosine gamma. And the final term is not that difficult. e to the minus r over a0. That's what we have to do. An integral over phi prime, theta prime, and r. This is our challenge. And the reason it's a challenge is the cosine gamma because this gamma is the angle. It depends on theta, depends on theta prime, depends on phi, phi prime. That's the problem. If we can solve that cosine gamma thing we can do this integral. And I think even if you were doing it numerically, that cosine gamma there is a little bit of a headache. You don't want to do an integral that you can really do like this numerically. So you really want to do it. So what we need is to calculate. So we can begin by saying, I'm going to calculate what cosine gamma is. And here is the way you can calculate cosine gamma. When you have two unit vectors, cosine of the angle between two unit vectors is just the dot product of those two unit vectors. So for gamma, we can consider a unit vector along e and a unit vector along r and take the dot product. And remember, for an arbitrary unit vector, it's sine theta cos phi sine theta sine phi cos theta. This is the theta phi decomposition of an arbitrary unit vector. So cosine gamma is the dot product of a vector n along the e times a vector n along r. And vector n along e and r are just with theta in one case and phi and theta prime and phi prime. So when I make the dot product, I get sine theta, cos phi, sine theta prime, cos phi prime. That is the product of the x components. Plus sine theta, sine phi prime times sine theta and sine phi. Sine theta prime sine phi prime, plus cos theta, cos theta prime. OK. Doesn't look much easier, but at least it's explicit. But it's actually much easier. And why is that? Because there's a lot of factors in common here. In fact, sine theta and sine theta prime are in both. So what you get here is cosine gamma is equal to sine theta sine theta prime. And then you have cos phi, cos phi prime, plus sine phi, sine phi prime. And that's cosine of phi minus phi prime plus cos theta cos theta prime. OK. So that's cosine gamma. Now suppose you were to put this whole thing in here. It's a big mess but all quantities that we know. But there's one nice thing, though-- a very nice thing happening. Think of the interval over d phi prime. This does not depend on phi prime. This does not depend on phi prime. But cosine gamma can depend on phi prime. But here it has this thing, cosine of phi minus phi prime. So when you try to do the integral of this phi prime with this term, this term will give, eventually, the integral of d phi prime times cosine of phi minus phi prime. There is a lot of messy things. But this integral is 0 because you are averaging over a full term. So happily, all this term will not contribute. And that's what makes the integral doable. So what do we have then? Our whole matrix element, f H prime i, has become e E0 over square root of pi l cube a0 cube. And now you just have to integrate. This is the only term that contributes. And when you put this term, for sure you can now do the d phi prime integral because that term is phi dependent. So that integral gives you just a factor of 2 pi from the integral of d phi. And from this thing, cosine theta is the angle between the electric field and k. So it's a constant for your integral. You're integrating over theta prime and phi prime. So cosine theta also goes out. And here is the integral that remains. r cube dr. It was r squared, but there was an extra r from the perturbation. r cubed dr into the minus a over r-- r over a0. And this thing you can pass two cosine variables. It's pretty useful. So this goes minus 1 to 1 d cos theta prime, cos theta prime e to the minus ikr cos theta prime. This cos theta prime came from here. And that's it. So this is a nice result. It looks still difficult, but we've made great progress. And in fact, we've dealt with a really difficult part of this problem, which is orienting yourself of how you're going to approach the matrix element. So to finish up I'll just give you a little more of the answer. We'll complete the discussion. We need probably 15 more minutes to finish it up. So the only thing I'm going to say now is that if you look at the notes, every integral here is easily doable. So basically, there's two integrals. And the way to do them is first do the r integral. You will have to have these terms. And then do the theta integral. And they are kind of simple, both of them. You have to keep up a lot of constants. But here is the answer for the matrix element. So that part of the integral I think you all can do. But you have to take your time. i 32 square root of pi e E0 a0. I did a lot of work here, actually, in writing it in a comprehensible way because it's pretty messy. l cube a0 cube, 1 over 1 plus k squared a0 squared cube cosine theta. By now, it starts to simplify a little. OK. That was actually plenty of work to get it to write it this way. I feel pretty happy about that writing. Why? First, OK, there is these numbers. Nothing I can do about it. But there's a multitude of constants that I have simplified and done all kinds of things. But it was not worth it. First, ka0, that is unit free. This is unit free. This is unit free. k is 1 over length. ka0 to the 4 is the length cube, and here is the length cube. No units here either. Here, nice units. This is units of energy. Why? Electric field times distance is potential, times electric charge is energy. Energy. So this is how this should be. The matrix element of an energy between normalizable states should be an energy. And that has become clear here. The next steps that we have to do, which we'll do next time, is to integrate over states, and put in the density of states, and do a little simplification. But now it's all trivial. We don't really have to integrate anymore because Fermi's golden rule did the job for us.
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https://ocw.mit.edu/courses/3-60-symmetry-structure-and-tensor-properties-of-materials-fall-2005/3.60-fall-2005.zip
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PROFESSOR: Resume by going back to our one-dimensional body that has undergone some elastic deformation. And what I would like to do now is to distinguish between displacement of an object and fractional change of length, which turned out to be measured by the same thing, that thing that we're going to name strain when properly defined. OK, here is our one-dimensional case. And we said that originally some point, P, at a location x gets mapped to a point P prime that is at x plus some displacement U. So this is the displacement vector U. Our point Q, which is originally at some location x plus delta x, where delta x is the original separation between P and Q, gets mapped to a point Q prime, which is going to be equal to a whole collection of terms. It's going to be equal to x plus delta x, the original location, plus the linear variation of U with x that has to go U times x plus delta x. And if we simplify this a little bit, Q prime is going to be at a location, factoring out x plus delta x, x plus delta x times 1 plus e. AUDIENCE: Is e represented here? PROFESSOR: e is the linear relation between displacement U and position along the body. OK, so what has happened here? The relative change of length is going to be P prime Q prime minus PQ. And if we just substitute in our two expressions, the relative change of length is going to be equal to delta x times 1 plus e minus delta x. This is the relative change of length. And I should label this such. That's going to be equal to delta x times 1 plus e minus delta x divided by delta x. And that simply going to be e, which is equal to delta U over delta x. And that is what we define as strain. And that is a dimensionless quantity. Because it has units of length over length. All right, how can we patch this one-dimensional sort of behavior up to a three-dimensional situation? What we are saying in this one-dimensional case is that the displacement of a particular point P varies linearly with position in the body. So let's generalize that into three dimensions by saying that the component of displacement U1-- and this is going to look like an algebraic identity. It's going to be the way in which U changes with x1 times the position x1. So this is saying that the displacement will vary linearly with x1. And the rate will be du1 dx1. But also U1 is going to change with the coordinate x2. And the coefficient there will be du1 dx2. And there's going to be a third term that will be the way in which U1, the x1 component of displacement, changes with x3 times the position in the body x3. So this is saying exactly the same thing that we did in one dimension, that displacement depends linearly with position, except that there's now three coordinates for position. And the displacement will change with an increment of a change in position along each of those three axes. Similarly, we can say that U2 is going to be equal to the way in which U2 changes with x1 times the position x1, the way in which U2 changes with x2 times the coordinate within the body x2, and the same for U2 dx3 and the actual positional coordinate x3. And so, in general, we're going to propose that the i-th component of displacement is given by the way in which displacement changes with x sub j times x sub j. Or you could do the same thing, not in terms of actual displacement, but differences in displacement, the same way we could define strain as the fractional change of length of our line segment on the elastic band or, alternatively, as the shift in position. So we could define it also as the difference in distances or displacements let me call them. And that really is a trivial change. We'll say the change of a length along x1 delta U1 is going to be the way in which U1 changes with x1 times delta x1 plus the way in which U2 changes with x1 times the way in which U1 changes with x2 times delta x2 plus the way in which U1 changes with x3 times delta x3 or, in general, that the fractional change in length is going to be equal to dui dxj times delta xj. All right, so what we're going to define now is dui. Dxj will be defined as something like strain. It's not exactly strain yet. And this is going to be a measure of deformation. I'm hedging my words because of something that we'll want to impose upon the proper strain tensor. But first, let's see what these three-dimensional terms mean. And to see that, let me look at a place within the body. And I'm going to look at a line segment that goes between a P and Q that is oriented along x1. And I'm picking that deliberately. Because I want to keep things simple and isolate one of these terms so we can identify its meaning. So let's say here we have two points P and Q separated by a distance delta x1 prior to deformation. And now we squish the body. And what happens is that P will move to a position P prime. Q will move to a position Q prime. And this will be the new line segment, P prime Q prime. This will be the original delta x1. To that we're going to tack on an instrument which is delta U1. But then there will also be a delta U2. And clearly what has happened is that the length of the line segment P prime Q prime has changed. But also the orientation of the line segment has changed by an angle, which I'll define as phi. So again, we have a line segment. We deform the body. P is displaced. Q is displaced. The component of that line segment along x1 has changed in length by an amount delta U1. The coordinate along x2 will change by an amount delta U2. So we not only change the length. But we rotate the line segment. We can express these in terms of our general relation that I proposed a moment ago. Delta U1 is going to be equal to du1 dx1 times delta x1. And that is going to be given by the element epsilon 1, 1 times delta x1. The value of x2 was going to be equal to du2 dx1 times delta x1. And that's going to be by definition e2, 1 times delta x1. And the reason this is so simple is that I initially picked the line segment which was parallel to x1. So not all of the four terms that would be present in the x1, x2 system have come in. So there's no contribution of delta x2 because of the fact that I've picked this special orientation. OK, the fractional change of length resolved on x1 is going to be, well, it's going to be exactly delta x plus delta U1 quantity squared, it's going to be this horizontal line segment, plus delta U2 quantity squared all to the power 1/2. And now I'm going to say that delta U2 is negligible compared to this big delta x plus delta U1. And I'll say that this is approximately equal to delta x plus delta U1, OK, just taking the whole works outside of the square root sign. So delta U1 over delta x1 is going to be equal to the term e1, 1 from this expression here. So the term e1, 1 represents the tensile strain along x1. So we can see how these derivatives are going to enter into changes of length. The P prime Q prime has also been rotated by phi. And we can say exactly what that is that the tangent of phi is going to be given exactly by delta U2 divided by the original length of the line segment delta x1 plus the little increment of displacement along x1. And this is-- I'll walk down here so I can see it. This is delta U1. This is delta U2 over times delta x1-- sorry, I can't see what I've got down here-- delta U1 plus delta x1. OK, so it's the amount of displacement along x2 over the original line segment delta x1 plus the change in displacement delta U1. And clearly delta U1 can be claimed to be small with respect to delta x1. So this is approximately equal to delta U2 over delta x1. And tangent of phi is going to be tangent of a very small angle. So this will be delta U2 over delta x1. And so this angle phi is, for small strains, going to be equal to delta U2 over delta x1. And that is the definition of our element of strain e1, 2. So e1, 2 corresponds to a rotation of a line segment that was originally parallel to x1 in the direction of x2-- AUDIENCE: [INAUDIBLE]. PROFESSOR: e2, 1, I'm sorry, e2, 1, yeah, along x1 in the direction of x2. And that is counterintuitive as I have just demonstrated. e2, 1 is a rotation of a line segment along x1 in the direction of x2, which is just the reverse of the subscript. So eij, a general off-diagonal element of the array eij, is going to be a rotation of a line segment initially along x sub j in the direction of x sub i. And for very small strains, numerically that term eij will give an angle in radiants. AUDIENCE: For the equation you have over there, should it be delta x plus delta x2? PROFESSOR: No, I would say this-- AUDIENCE: Because aren't you saying delta U1? PROFESSOR: Oh, OK, you're right. Yeah, I didn't take the difference here. AUDIENCE: Delta U1 is basically negligible? PROFESSOR: Yep, yep, so I'm saying that that [INAUDIBLE] should be delta x1-- AUDIENCE: Plus delta U2? PROFESSOR: --plus delta U2. You're right. No, I'm throwing this out. And that's right. So I'm saying that this is essentially delta x1 plus delta U1. I'm saying that this is negligible. So strictly speaking the distance between P prime and Q prime is the square root of this squared plus this squared. But if this thing is tiny, P prime Q prime is essentially going to be this distance, delta x1 plus delta U1. So that's right. OK, so we have something that looks like it measures deformation. But I would like to ask if this is a suitable measure of deformation. And I would like to show that, unless the tensor eij is symmetric, that we have included in our definition of strain rigid body rotation as well as true deformation. So in order to demonstrate that, what I'm going to look at is a case where we have actually, by the way in which we apply a stress to the material, actually done nothing more than rotate x1 to x1 prime and rotate x2 to x2 prime. And in general, for a real amount of deformation, that is something that is going to happen. Let's say I decide to deform this eraser in shear before your very eyes. And I try shearing it. Nothing much has happened. And I squeeze a little harder. And finally, I've got it wrestled down into an orientation like this. And when I finished, this was the original position of the eraser. Here's x1. Here's x2. And by the time I've wrestled it around to a deformed state, it sits down like this, maybe deformed a little bit. But would you say that this angle here is a measure of deformation? No, that's clearly a rigid body rotation. And what I would intuitively do, if I wanted to measure true deformation, if this were the original body and after deformation it went to a location with some deformation to be sure but this has been rotated to here. This has been rotated down. And I wouldn't want to say that this is a measure of deformation. This would be e1, 2. What I would want to do intuitively would be to position the body symmetrically between the axes and then say that this is a measure of shear strain. So let me show you now in a more rigorous treatment that, if there is a component of rigid body rotation, let me show you what a rotation of the coordinate system to a new orientation x1, x prime would do. So let's say that this is an original point Q1. And after deformation, it moves by a rotation phi that's equal to e2, 1 to a new location Q1 prime. Let's suppose that this is a position Q2. And after what we think is the deformation, this moves to a location Q2 prime. And this would be e1, 2. That's equal to minus phi. This is e2, 1. And that's equal to plus phi. So what we would do is like to get rid of that rigid body rotation. And what I'll do is to show that, if we have a point that's out at the end of a radial vector, R, which has coordinates xi, x1, x2, x3, and if we have a displacement, which is absolutely perpendicular to that radial vector, that this would be characteristic of rigid body rotation. And I will say that, if Ui is perpendicular to the radial vector xi, then U.R should be equal to 0 for every position xi. So if we just multiply this out, this is saying that Ui Ri should be equal to 0 for rigid body rotation. And let's simply carry out this expansion. And I will have then eij xi xj forming this dot product. And that is going to be 0 for a rigid body rotation. And if I expand this, this will contain terms like e1, 1 times x1 x1 plus e2, 2 times x2 x2 plus e3, 3 times x3 squared. And then they'll be cross-terms e1, 3 plus e3, 1 times x1 x3 plus e1, 2 plus e2, 1 times x1 x2 plus e2, 3 plus e3, 2 times x2, x3. And my claim now is that, if this represents rigid body rotation, then that should be 0. That is to say the displacement is absolutely perpendicular to the radius vector for small displacements. This must be 0 for all xi. And the only way that's going to be possible for any value of x1, x2, x3 is that each of these six terms vanish individual. It's the only way I'm going to be able to get the whole thing to disappear for any value of coordinate x1, x2, x3. So we're going to have to then have e1, 1 equals e2, 2 equals e3, 3 equals 0. All the diagonal terms are going to have to be 0. In order to get the fourth term to vanish, I'm going to have to have e1, 3 equal to the negative of e3, 1 and e1, 2 the negative of e2, 1 and e2, 3 the negative of e3, 2. So what is this going to look like for the part of the e tensor that corresponds to rigid body rotation? The diagonal terms are all going to be 0. And the off-diagonal terms are going to be the negative of one another. So this is e1, 2. e2, 1 would have to be equal to minus e1, 2. e3, 1 would have to be equal to the negative of e1, 3. And so we will have something like this. So this suggests that our definition of the true deformation, which will be a tensor epsilon ij-- I bet you kind of guessed I was going to call it epsilon and not E. I can write this as a sum of two parts. I'll turn that around and say I'll write epsilon ij plus another three by three array omega ij is equal to the tensor eij. That's a novel idea. This is addition of two tensors element by element. And if I do that and define epsilon ij equal to 1/2 of eij plus eji, so for the diagonal elements that is going to take 1/2 of e1, 1 plus e1, 1. And that's just going to give me e 1, 1 back again. And I'm going to define the terms omega ij as 1/2 of eij minus eji. And if I do that, the resulting tensor will be symmetric. OK, so from tensor eij we can split it up into two parts and a part omega ij. And I won't bother to write it out. But you can see that omega ij plus eij is going to be equal to eij minus 1/2 of 2eij minus 1/2 of 0. And that's just going to be eij. So this plus this is indeed going to give me the tensor eij. So given a tensor eij, which is not symmetric, I will define eij as the sum of two parts, a tensor epsilon ij, which is true strain, and a part omega ij, which is rigid body rotation. And now at last we have a satisfactory measure of true deformation in terms of the displacement of points within a deformed body where that displacement can arise from either rigid body rotation and or deformation. So finally, we have a tensor epsilon ij, which is the strain tensor. And now I can finally make my claim that, if this really is true deformation, that for cubic crystals-- you can see the same [INAUDIBLE] window coming again-- since second ranked tensors have to be symmetric for cubic crystals, the form of strain for a cubic crystal can only be this. It has to be diagonal if the tensor is to conform to cubic symmetry. And you say, ah, ah, you tried to pull that on us with stress, too. And I can squish a crystal anyway I want. But you can only develop a strain if you deform a crystal. And the symmetry of the crystal is going to determine how the crystal deforms. So therefore, a cubic crystal should be able to deform only in a way that stays invariant to the transformations of a cubic symmetry, right? So cubic crystals can only deform isotropically. So if I wanted to design springs for an automobile so that elastic energy could be stored most efficiently in a solid, I would want to make these automobile springs out of a triclinic metal. Because then the tensor could be one of general deformation. OK, well, this, obviously, is a swindle like my assertion that stress could only be an isotropic compressional stress for a cubic crystal. But the argument is a little more convoluted. And strain, indeed, is also a field tensor. Because, even though I can only create the deformation by application of a stress or perhaps an electric field or something of that sort in that the link between the deformation and what I do to the crystal is coupled by a property of the crystal, which has to conform to the symmetry of the crystal, nevertheless, given a tensor that describes deformation, I can always, independent of the symmetry of the crystal, devise a particular set of stresses that would produce any state of strain I want. I would have to design, though, a particular state of stress to produce a desired state of strain. And that coupling has to conform to the symmetry of the crystal. But there's no reason why strain itself has to conform to the symmetry of the crystal. I can create any state of strain I like by choosing the appropriate stress. OK, so that is a swindle. But everything now that I can say about the behavior of a second ranked tensor applies to the strain tensor. I can take a strain tensor epsilon ij. And I can perform the surface epsilon ij xi xj equals 1. And that will be the strain quadric. The strain quadric will have the property like any quadric constructed from a tensor that the value of the radius in a particular direction is going to be such that the strain in that direction is equal to 1 over the radius squared. So what does that mean? Well, we have to look at what the strain tensor is relating. The direction, remember now that the definition of the strain tensor is that U sub i equals epsilon ij times x sub j. So the quote "applied vector" is the direction in a solid. And we will get, in general, a displacement, U, which is not parallel to the direction that we're considering. And the epsilon in a particular direction is going to be equal to the part of U that's parallel to the direction of interest divided by distance in that direction. And what this is going to give us is the tensile component of deformation in the direction that we're examining. Now, the radius normal property works in this case. And what the radius normal property says, if you look at a certain direction in the solid and then look at the normal to the surface at that particular point, this will be the direction of what happened. So I've not drawn this correctly. This should be the direction of U. And this is the radio vector. So the normal to the surface gives us the direction of the displacement that's going to occur. And the reason I can say that is by definition that that property holds only for a symmetric tensor. And by definition, we have defined the strained tensor such that it is symmetric. So the radius normal property holds. If we want to change from one coordinate system to another, we do that by the all for transformation of a second ranked tensor that, if we change coordinate system, the elements of strain change to new values epsilon ij prime that are given by cil cjm times epsilon lm, the same law for transformation of a second ranked tensor that we have seen earlier. Finally, we can, because there is a quadric, we can change that by finding eigenvectors and eigenvalues, take a general form of the tensor epsilon 1, 1, epsilon 1, 2, epsilon 1, 3, and so on, and by solving the eigenvalue problem, convert this to a value epsilon 1, 1 prime 0, 0, 0, epsilon 2, 2 prime 0, 0, 0, epsilon 3, 3 prime to find out what the extreme values of tensile deformation are and the orientations within the original description of the body in which these extreme values occur. OK, I'm just about out of time. I'll save until next time another neat feature of the strain tensor. And that is contained within these elements is information on the volume change that the body undergoes. And this is something called the dilation. And we'll see that this is related in a very simple way to the elements of the strain tensor. OK, we've spent a lot of time developing the formalism of stress and strain. And next thing we'll do is to move on to some interesting properties of crystals, which are defined in terms of tensors of rank higher than two. So we'll look at some third ranked tensor properties. That will include things like piezoelectricity, the stress optical effect, and things of that sort. These are going to be really anisotropic. Second ranked tensors were anisotropic enough. But the variation of property with direction was a quasi-ellipsoidal variation When as 1 over the square of the radius of the quadric. For higher ranked tensor properties, we will encounter some absolutely weird surfaces with dimples and lumps and lobes, not this smooth quasi-ellipsoidal variation, a property with direction, which was the case for second ranked tensor properties. So we will get into some exotic anisotropies very shortly to finish up the semester. So I will probably see some of you at the MRS meeting next week. I won't see you on Thursday for reasons that I need not elaborate upon. So I hope you have a happy holiday. Relax, suck in air, and come back refreshed next Tuesday for some really wild anisotropy of physical properties.
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https://ocw.mit.edu/courses/8-422-atomic-and-optical-physics-ii-spring-2013/8.422-spring-2013.zip
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The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Good afternoon. So it has been a little bit more than a week when we met the last time. Before I continue our discussion of metrology and interferometry. I just want to share something I saw on my visit to the Netherlands. When I visited the University of Delft and the Kavli Institute, they had just accomplished the entanglement of two NV centers. And we had just talked in class about the entanglement of two ions. So I'm sort of excited to show you that people have just taken the next step. And what often the next step means is that in atomic physics, we use pristine systems-- ions, neutral atoms in [INAUDIBLE] vacuum chamber. And we create new forms of entanglement. Or with quantum gases, new forms of quantum matter. But ultimately, we hope that those concepts, those methods, and this knowledge translates to some room temperature materials or solid state materials which can be handled more easily, and are therefore much closer to applications. So NV centers are kind of nature's natural ion trap, or nature's natural neutral atom trap. Let's not discuss whether this is neutral or ionized. It's a vacancy in nitrogen. And it has a spectrum which looks like an atom. So you can say once you have such a defect in nitrogen, you have an atom, a single atom in an atom trap or an ion in an ion trap. And you don't have to create the vacuum. It's there. Every time you look at it, it's there. And you can excite it with the laser. It has a spectrum similar to atoms. It has spin structure So you have these vacancies in diamond. So these are little quantum dots. But now you have two problems. One is you want to collect the light emitted by them. And what is best is to mill a lens right into the material. So this is the diamond material. A lens is milled. And that already gives some [? culmination ?] of the light emitted by it. But the bit problem until recently has been when you create those some people call it artificial atoms. Every atom is a little bit different, because it experience a slightly different environment. A crystal has strain, so if you have seemingly two identical defects in a diamond crystal, the two defects will have a resonance line, which is a few gigahertz difference. You would say, well, maybe it's just [? apart ?] in 10 to the 5, but it means the photons are distinguishable. So if you want to do entanglement by having two such artificial atoms emitting a photon onto a beam splitter, and then by performing a measurement we project the atoms into Bell state-- I hope you all remember what we discussed for the trapped ions-- then you have to make sure that fundamentally, those photons cannot be distinguished. And the trick here is that they put on some electrodes. And by adding an electric field, they can change the relative frequency. And therefore, within the frequency uncertainty given by Heisenberg's Uncertainty Relation, they can make the two photons identical. And when, let's say, experiment. Well, you have two NV centers, defects in diamond. You can manipulate with microwaves coherently the spin. You need lasers for initialization and readout. But then closer to what we want to discuss, you need laser beams which excite the NV center. And then the two NV centers emit photons. And what you see now is exactly what we discussed schematically and in context of the ions, that the two NV centers now emit photons. And by using polarization tricks and the beam splitter, you do a measurement after the beam splitter. And based on the outcome of the measurement, you have successfully projected the two NV centers into a [INAUDIBLE] state. OK, good. All right. Let me just summarize what our current discussion is about. This section is called Quantum Metrology. It is a section where we want to apply the concepts we have learned to precision measurements. It's actually a chapter which I find nice. We're not really introducing new concepts. We're using concepts previously introduced. And now you see how powerful those concepts are, what they can be used for. So we want to discuss the precision we can obtain in quantum measurement. We apply it here to an atom in the interferometer, to an optical interferometer. We could also discuss the precision of spectroscopic measurements. A lot of precision measurements have many things in common. So what we discuss here is as a generic example for precision measurement that we send light from our center interferometer. Here is a phase shift. And the question is, how accurately can we measure the phase shift when we use n photons as a resource? And of course, you are all familiar with the fundamental limit of standard measurements, which is short noise. And sort of as a warm up in our last class Wednesday, week ago, I showed you that when we use coherent states of light at the input of the interferometer, we obtain the short noise. Well, it may not be surprising, because coherent light is as close as possible, has similar to classical light. But then we discussed single mode, single photon input. And by using the formalism of the Mach-Zehnder interferometer, we've found that the phase uncertainty is again 1 over square root n. So then the question is, how can we go to the Heisenberg limit, where we have an uncertainty in the phase of 1 over n? And just as a reminder, I find it very helpful. I told you that you can always envision if you have n photons and you multiply it, put the n photons together by multiplying the frequency by n, then you have one photon which [INAUDIBLE] with n times the frequency. And it's clear if you do a measurement at n times the frequency, your precision in phase is n times better. So what we have to do is we have to sort of put the n photons together. And then we can get a precision of the measurement, which is not square root n, but n times better. So this is what we want to continue today. This is the outline I gave you in the last class. If you have this optical interferometer, this Mach-Zehnder interferometer. And if you use coherent sets or single photons as the input state, we obtain the short noise. Now we have to change something. And we can change the input state, we can change the beam splitter, or we can change the readout. So we have to change something where we entangle the n photons, make sure in some sense they act as one giant photons, with either n times the frequency, but definitely with n times the sensitivity to phase shift. Any questions? Good. The first example, which we pretty much covered in the last class, was an entangled state interferometer. So instead of having the Mach-Zehnder interferometer as we had before, we have sort of a Bell state creation device. Then we provide the phase shift. And then we have a Bell analysis device. And I want to use here the formalism and the symbols we had introduced earlier. So just as a reminder, what we need is I need two gates. One is the Hadamard gate, which in matrix representation has this matrix form. And that means if you have a qubit which is either up or down and you apply the matrix to it, you put it in a superposition state of up and down. It's a single qubit rotation. You can say for the bloch on the Bloch sphere, it's a 90 degree rotation. The second gate we need is the controlled NOT. And the controlled NOT we discussed previously can be implemented by having an interferometer and using a non-linear Kerr medium coupling another photon to the interferometer in such a way that if you have a photon in mode C, it creates a phase shift with the interferometer. If there is no photon in mode C, it does not create an additional phase shift. And as we have discussed, this can implement the controlled NOT. So these are the ingredients. And at the end of the last class, I just showed you what those quantum gates can do for us. If you have two qubits at the input, we apply the-- and I will assume they are both in logical 0-- the Hadamard gate makes the coherent superposition. And then, we have a controlled NOT where this is the target gate. Well, if this is 0, if the control beat is 0, the target beat stays 0. So we get 0, 0. If the control beat is 1, the target beat is flipped to 1. So we get 1, 1. So the result is that we have now created a state 0, 0 plus 1, 1. And if we apply a phase shift to all the photons coming out on the right hand side, we get a phase shift which is too fine. So we already get the idea. If we take advantage of a state 1, 1, it has twice the [? face ?] sensitivity as a single photon. And we may therefore get the full benefit of the factor of 2, and not just square root 2. And this is what this discussion is about. Questions? OK, so that's where we ended. We can now use another controlled NOT and bring in the third qubit. So what we create here. Just make a reference, this is where we start today. So we create here the state, which is either all beats are 0, all beats are 1, and then the phase shift gives us three times the phase shift phi. And therefore, by bringing in more and more qubits, I've shown you n equals 1, n equals 2, n equals 3. So now you can go to n. We obtain states which have n times the phase sensitivity. Let me just mention in passing that for n equals 3, the superposition of 0, 0, 0 and 1, 1, 1 goes by the name not gigahertz, GHZ. Greenberger. This third one is Zeilinger. The second one is-- AUDIENCE: Horne. PROFESSOR: Horne. Thank you. Greenberger-Horne-Zeilinger state. And those states play an important role in test of Bell's inequality. Or more generally, states which are macroscopically distinct are also regarded cat states or Schrodinger cat states. If we apply a phase shift and go now through the entangler in reverse, just a reverse sequence of CNOT gates and Hadamard gates, then we have all the other n minus, have the n minus 1 qubits. All the qubits except for the first reset to 0. But the first qubit is now in a superposition state where we have the phase into the power n. And now, we can make a measurement. And P is now the probability to find a single photon in the first qubit. So in other words, our measurement is exactly the same as we had before where we had the normal Mach-Zehnder interferometer. We put in one photon at a time. And we determined the probability. What is the photon at one of the outputs of the interferometer? But the only difference is that we have now a factor of n in the exponent for the phase shift. And I want to show you what is caused by this factor of n. Let me first remind you how we analyzed the sensitivity of an interferometer for single photon input before. So for a moment, set n equals 1 now. Then you get what we discussed two weeks ago. The probability is [INAUDIBLE] as an interferometer. Sine also, cosine also infringes with cosine phi. If you do measurements with a probability P, that's a binomial distribution. Then the deviation or the square root of the variance in the binomial distribution is P times 1 minus P. And by inserting p here, we get sine phi over 2. The derivative of P, which is our signal, with a phase is given by that. And then just using error propagation, the uncertainty in the phase is the uncertainty in P divided by dp dt. So then we repeat the whole experiment n times, and if your binomial distribution and we do m [INAUDIBLE] of our [INAUDIBLE] with probability P, we get an m under the square root. And then we get 1 over square root m. And this was what I showed you two weeks ago, the standard short noise limit. But now we have this additional factor of n here, which means that our probability has a cosine which goes with n phi. When we take the derivative, we get a factor of n here. And then dp d phi has the derivative. But then we have to use a chain rule, which has another factor of n. And therefore, we obtain now that the phase sensitivity goes as 1 over n. And therefore, we reach the Heisenberg limit. So if you have now your n photon entangler, the one I just-- the entanglement operation with these n cubits is special beam spreader replaced by the entangler, we have now a sensitivity, which goes 1 over n. And then of course, if you want, you can repeat the experiment n times. And whenever you repeat an experiment, you gain an addition with the square root of m. Just give me one second. We have this. OK, so this is an example where we had n qubits entangled, like here. And the sensitivity of this interferometer scales now as the Heisenberg limit with 1 over n. OK, I want to give you, because they're all nice and they're all special, I want to give you three more example how you can reach the Heisenberg limit. The next one goes by the name super beam splitter method. It's actually a very fancy beam splitter where you have two prods of your beam splitter. One has 0 and one has n photons. And behind the beam splitter, you have two options. Either all the n photons are in one state, or the n photons are in the other state. You never have any other mixture. Of course, you know your standard half silvered mirror will not do that for you, because it will split the n photon state with some Poissonian statistics or whatnot into the two arms and such. And you can calculate exactly what a normal beam splitter does to that. It's very, very different. This here is a very, very special beam splitter. Just to demystify this beam splitter a little bit, I want to show you that, at least conceptually, there is an easy implementation with atoms. I really like in this section to grab an example from atoms and to grab an example for photons, because it really brings out that the language may develop equally applies to atoms and photon. So if I take now an easy example, if I take an example of the Bose-Einstein condensate with n atoms, let's assume we have a double well. But now we have attractive interactions. This is not your standard BEC. Your standard Bose-Einstein condensate has repulsive interaction, with attractive interaction. If you go beyond a certain size, certain a atom number, the condensate will collapse. So let's assume we have a Bose-Einstein condensate with n atoms. There are some attractive interactions, but we stay within the stability diagram that those atoms do not-- that those atoms do not collapse. So now, if the atoms are attractive, they all want to be together. Because then they lower the energy of each other. So if you have something with attractive interaction, you want to have n atoms together. But if your double well potential is absolutely symmetric, there is no symmetry breaking, and you have an equal amplitude for all the atoms to be in the other well. So therefore, under the very idealized situation I described here, you will actually create a superposition state of n atoms in one well with n atoms in the other well. And this is exactly what we tried to accomplish with this beam splitter. OK, so if we have this special beam splitter, which creates that state, then when we add a phase shift to one arm of the interferometer, we obtain-- which looks very promising-- a phase shift phi multiplied by n. And if we read it out, we have now if you pass it through the other super beam splitter, we create again a superposition of n0 and 0n. But now, because of the phase shift with cosine and sine factors, which involve n times the phase shift. For obvious reasons, those states also go by the name noon states. That's the name which everybody uses for those states, because if you just read n00n, it gives noon. So this is the famous noon state. So it seems it smells already, right, because it has a phase shift of n phi. Let me just convince you that this is indeed the case. So the probability of reading out 0 photons in one arm of the interferometer scales now with the cosine square. If you do one single measurement, the binomial distribution is the same as before, but with a factor of n. The derivative has those factors of n. And that means that the variance in the phase measurement is now 1 over n. If you want a reference, I put it down here. As far as I know, the experiment has been done with three photons, but not with larger photon number. Questions? Collin. AUDIENCE: This picture with the double well, isn't there going to be-- all right, some people would say you spontaneously break the symmetry and [INAUDIBLE] either ends up in one side or the other. Like isn't this [INAUDIBLE]? PROFESSOR: Well, Bose-Einstein condensation with attractive interactions was observed in 1995. And now, 18 years later, nobody has done this seemingly simple experiment. And what happens is really that you have to be very, very careful against any experimental imperfection. If the two double well potentials are not exactly identical, the bosons always want to go to the lowest quantum state. Well, that's their job, so to speak. That's their job description. And if you have a minuscule difference between the two, the dates of the two wells, you will not get a superposition state. You will simply populate one state. And how to make it experimentally [INAUDIBLE], this is a really big challenge. [? Tino ?]. AUDIENCE: I have a question. Let's say we're somehow able to make the double well potential perfect. But if we didn't have attractive interactions, then wouldn't he just get a big product state of each atom being in either well? PROFESSOR: OK, if you're a non-interacting system, the ground state of a double well potential is just your metric state 1 plus 2 over square root 2. And for non-interactive BEC, you figure out what is the ground state for one particle and then take it to the power n. This is the non-interactive BEC. STUDENT: So then entanglement is because of the interaction, right? PROFESSOR: If you have strong repulsive interaction, you have something which should remind you of the [INAUDIBLE] insulator. You have n atoms. And n over 2 atoms go to one well. n over 2 go to the other well. Because any form of number fluctuations would be costly. It will cost you additional repulsive interaction. So therefore, the condensate wants to break up into two equal parts. So that's actually a way how you would create another non-classical state, the dual flux state of n over 2, n over 2 particle. And for attractive interactions, well you should create the known state. OK, so this was now a state n0 and 0n. There is another state, which you have encountered in your homework. And this is a superposition not of n0 and 0n and n0. It's a superposition of n minus 1n and n n minus 1. This state goes by the person I think who invented it, the [? Yerkey ?] state. And you showed in your homework that with that, you also reach Heisenberg Limited Interferometry, where the phase scales is 1 over n. What I want to add here to it is how one can create such a [? Yerkey ?] state. And again, I want to use the example with an atomic Bose-Einstein condensate. And here's the reference where this was very nicely discussed. So let's assume we can create two Bose-Einstein condensates. And they have exactly n atoms. And I would actually refer to [? Tino's ?] questions how to make them have two n atoms in a trap and then make a double well potential, deform the harmonic oscillator potential to a double well potential. Then for strong repulsive interactions, the condensate will symmetrically split into two flux states, each of which has n atoms. So now how can we create the [? Yerkey ?] state from that? Well, we simply have-- we leak out atoms. We leak atoms out of the trap. My group demonstrated an RF beam splitter, how you can just split in very controlled way start rotating the cloud on non-trapped state. And then atoms slowly leak out. Well, when you can measure, of course, you can take an atom detector and measure that an atom has been out-coupled, that an atom has leaked out of the trap. If you don't like RF rotation, you can also think that there is a tunneling barrier, and atoms slowly leak out by whatever mechanism. And the moment you detect that an atom, you then project the state in the trap to n minus 1 atom, because you've measured that one atom has come out. But now you use a beam splitter. And a beam splitter could simply be a focused laser beam. And the atoms have a 50% probability of being reflected or end up tunneling through. So therefore, if you have now a detector which measures the atoms on one side and the atoms on the other side, then you don't know anymore when the detector makes click form which atom trap the atom came. Or more formally, a beam splitter transforms the two input modes ab into a plus b and a minus b normalized by square root 2. So therefore, if this detector clicks, when you project the remaining atoms into the symmetric state, here you detect one atom in the mode a plus b. And that means that the remaining atoms have been projected into the [? Yerkey ?] state. If the lower atom detector would click, well, you get a minus sign here. So that's one way how at least in a conceptually simple situation, you can prepare this highly non-classical state by starting with a dual flux state of Bose-Einstein condensates, and then using-- and this is an ongoing theme here-- by using a measurement, and then the post-measurement state is the non-classical state you wanted to prepare. Question? Yes. AUDIENCE: How can you ensure that only one atom leaves the [INAUDIBLE]? PROFESSOR: The idea is that we leak atoms out very, very slowly. And then we have a detector, which we assume has 100% quantum efficiency. So therefore, we simply wait until the detector tells us that one atom has leaked out. And in an idealized experiment, we know the atoms either have been measured by the detector, or they are still in the trap. AUDIENCE: And also, is there a property that holds the [INAUDIBLE] you detect atoms [INAUDIBLE]? PROFESSOR: In principle yes, but the idea here is if you have a very slow leakage process, the probability that you detect two atoms at the same time is really zero. You leak them continuously and slowly, but then quantum mechanically, that means for most of the time, you measure nothing. That means the leakage hasn't taken place. The quantum mechanical system has not developed yet. But the moment you perform a measurement, you project-- it's really the same if you say you have n radioactive atoms. You have a detector. And when the detector makes click, you know you have n minus 1 radioactive atoms left. It's just applied here to two atom types. Other questions? AUDIENCE: [INAUDIBLE]? PROFESSOR: No. And maybe I'll tell you now why not. We have discussed the noon state. And we have discussed here a highly non-classical superposition state. Let's just go back to the noon state. We have n atoms here, zero here, or n atoms, or the reverse. But now, assume a single atom is lost, is lost from your trap by some background gas collisions. And you have surrounded the trap with a detector. So if you have the noon state, the symmetric superposition state, all atoms here and all atoms there. But by your background process, by an inelastic collision, you lose one atom. And you detect it. You could set up your detectors that you know from which that you figure out from which trap was the particle lost. So therefore, a single particle lost if you localize from which trap the particle is lost would immediately project the noon state into a state where you know I have n or n minus 1 atoms in one well, and 0 in the other well. So I've already told you, with the attenuator, you can never assume an attenuator is just attenuating a beam. An attenuator can always regard it as a beam splitter. And you can do measurements at both arms of the beam splitter. Or you have to consider the vacuum noise which enters through the other part of the beam splitter. And if you are now add that those atoms, n atoms, in a trap have some natural loss by inelastic collision or background gas scattering, the loss is actually like a beam splitter that particles don't stay in the trap, but go out through the other part. And then you can measure it. So in other words, every loss process should be regarded as a possible measurement. And it doesn't matter whether you perform the measurement or not. And I think it's just obvious that the noon state, the moment one particle is lost and you reduce this particle to figure out if n particles are here or n particles are there, the whole superposition state is lost if you just lose one single atom out of n. So the lifetime of a noon state is then not your usual trap lifetime, where you lose half of the atoms, 1 over e of the atoms. It is n times faster because it is the first atom lost, which is already completely removing the entanglement of your state. So more quantitatively, to say it more specifically, the limitation is loss. When you have a fully entangled state, maximally entangled state, usually a loss of one particle immediately removes the entanglement. We had the situa-- no, that's not a good example. But for the most entangled state, usually-- and for the noon state, it's trivial to see-- a single particle lost allows you to measure on which side of the potential barrier all the atoms are, and all the beauty of the non-classical state is lost. So if you assume that in a time window, you have an infinitesimal loss, a loss of epsilon, what usually happens is if you have an entanglement of n particles, then you lose your entanglement. If you expand 1 minus epsilon to n, it becomes 1 minus n times epsilon. So that's one reason why people have not scaled up those schemes to a large number of photons, or a large number of atoms. Because the larger n is, the more sensitive you are to even very, very small losses. Questions? AUDIENCE: [INAUDIBLE] PROFESSOR: Yes. AUDIENCE: Or is that physically? PROFESSOR: The super beam splitter would create the noon state. I've not explained to you what it physically would be for photons, but I gave you the example for atoms to BECs in a double well potential with attractive interaction. So to start with the condensate in-- AUDIENCE: Start with a bigger [INAUDIBLE]. PROFESSOR: If you start with a double well potential, and you put in n atoms initially in, but then you switch on tunnel coupling, then you would create the noon state if the interactions are attractive, and if everything is idealized, that you have a completely symmetric double well potential. OK, the last example is the squeeze light Interferometer. I just want to mention it briefly, because we've talked so much about squeezed light. So now I want to show you that squeezed light can also be used to realize Heisenberg Limited Interferometery. So the idea is, when we plot the electric field versus time, and if you do squeezing in one quadrature, then for certain times, the electric field has lower noise and higher noise. We discussed that. And the idea is that if there is no noise at the zero crossing, that this means we can determine the zero crossing of the light, and therefore the phase shift with higher accuracy. You may also argue if you have this quasi-probabilities, and with squeezing, we have squeezed the coherent light into an ellipse. And things propagate with e to the i omega t. Then you can determine a phase shift, which is sort of an angular sector in this diagram, with higher precision if you have squeezed this circle into an ellipse like that. So that's the idea. And well, it's fairly clear that squeezing, if done correctly, can provide a better phase measurement. And what I want to show you here in a few minutes is how you can think about it. So we have discussed at length the optical Interferometer, where we have just a coherence state at the input. This is your standard laser Interferometer. But of course, very importantly, the second input port has the vacuum state. And we discussed the importance of that. So the one difference we want to do now is that we replace the vacuum at the second input by the squeezed vacuum where r is the parameter of the squeezed vacuum. OK, so that's pretty much what we do. We take this state, we plug it into our equation, we use exactly the same formalism we have used for coherence states. And the question, what is the result? Well, the result will be that the squeezing factor appears. Just as a reminder, for our Interferometer, we derived the sensitivity of the Interferometer. We had the quantities x and y. And the noise is delta in this y operator squared divided by x. This was the result when we operate the Interferometer at the phase shift of 90 degrees. Just a reminder. That's what we have done. That's how we analyze the situation with a coherent state. The signal x is now-- the signal x is the number measurement a dagger a. And we have now the input of the coherent state. And b dagger b. We have an input mode a and a mode b. They get split. And then we measure it at the output. And we can now at the output have photons-- a dagger a, which come from the coherent state, and b dagger b, which come from the squeeze vacuum. So this is now using the beam splitter formalism applied to the Interferometer. So this is now the result we obtain. And in the strong local oscillator approximation, it is only the first part which contributes. And this is simply the number or photons in the coherent beam. The expectation value of y is 0, because it involves a b and b dagger operator. And the squeezed vacuum has only, if you write it down, in the n basis, in the flux basis, has only even n. So if you change n by one, you lose overlap with the squeezed vacuum. So therefore, this expectation value is 0. For the operator y squared, you take this and square it. And you get many, many terms, which I don't want to discuss. I use the strong local oscillator. Limited a dagger and a can be replaced by the eigenvalue alpha of the coherent state. So therefore, I factor out alpha squared in the strong local oscillator limit. And then what is left is b plus b dagger squared. And since we have squeezed the vacuum, this gives us a factor into the minus r. So if we put all those results together, we find that the phase uncertainty is now what we obtained when we had a coherent state with the ordinary vacuum. And in the strong local oscillator limit, the only difference to the ordinary vacuum is that in this term, we've got the exponential factor e to the minus r. And since we have taken a square root, it's e to the minus r over 2. So that result would actually suggest that the more we squeeze, that delta phi should go to 0. So it seems even better than the Heisenberg limit. However-- well, this is too good to be true-- what I've neglected here is just the following. When you squeeze more and more, the more you squeeze the vacuum, the more photons are in the squeezed vacuum, because this ellipse stitches further and further out and has overlap with flux states at higher and higher photon number. So therefore, when you go to the limit of infinite squeezing, you squeeze out of the limit where you can regard the local oscillator as strong, because the squeezed vacuum has more photons than your local oscillator. And then you have to consider additional terms. So let me just write that down. However, the squeezed vacuum has non-zero average photon number. And the photon number of the squeezed vacuum is, of course, apply b dagger b to the squeezed vacuum. This gives us a sinc function. And we can call this the number of photons in the squeezed vacuum. So we have to consider now this contribution to y square. So we have to consider the quadrator of the ellipse, the long part of the ellipse, the non-squeezed quadrator component. And we have to consider that when we calculate the expectation value of y square. And then we find additional terms, which I don't want to derive here. And the question is then, if you squeeze too much, you lose. So there's an optimal amount of squeezing. And for this optimal amount of squeezing, the phase uncertainty becomes approximately one over the number of photons in the coherent state, plus the number of photons in your squeeze vacuum. So this is, again, very close to the Heisenberg limit. So the situation with squeezed light is less elegant, because if you squeeze too much, you have to consider additional terms. This is why I gave you the example of the squeezed light and the squeezed vacuum as the last. But again, the Heisenberg limit is very fundamental as we discussed. And for an optimum arrangement of the squeezing, you can also use a squeezed vacuum input to the Interferometer to realize the Heisenberg limit. Any questions? Why is squeezing important? Well, squeezing caught the attention of the physics community when it was suggested in connection with gravitational-- with the detection of gravitational waves. As you know, the laser Interferometer, the most advanced one is LIGO, has a monumental task in detecting a very small signal. And pretty much everything which precision metrology can provide is being implemented for that purpose. So you can see, this is like precision measurement. Like maybe the trip to the moon was for aviation in several decades ago. So everything is really-- a lot of things pushing the frontier of precision measurement is motivated by the precision needed for gravitational wave detection. And what I want to show you here is a diagram for what is called advanced LIGO. LIGO is currently operating, but there is an upgrade to LIGO called advanced LIGO. And what you recognize here is we have a laser which goes into a Michelson Interferometer. And this is how you want to detect gravitational waves. But now you realize that the addition here is a squeezed source. And what you are squeezing is not, while it should be clear to you now, we're not squeezing the laser beam. This would be much, much harder, because many, many photons are involved. But it is sufficient to squeeze the vacuum and couple in squeezed vacuum into your cavitational wave detector. If you wonder, it's a little bit more complicated because people want to recycle light and have put in other bells and whistles. But in essence, a squeezed vacuum source is a major addition to advanced LIGO. Yes. AUDIENCE: Where is the squeeze actually coming into the system? I see where it's drawn, but where is it actually entering the interferometer? At that first beam splitter? PROFESSOR: OK. We have to now-- there are more things added here. Ideally, you would think you have a beam splitter here, the laser comes in here, and you simply want to enter the squeezed vacuum here. And this is how we have explained it. We have one beam splitter in our Interferometer. There is an input port and an open port. But what is important here is also that the measurement-- here you have a detector for reading out the Interferometer. And what is important is that the phase is balanced close to the point where no light is coming out. So you're measuring the zero crossing of a fringe. But that would mean most of the light would then exit the Interferometer at the other port. But high power lasers are very important for keeping the classical short noise down. So you want to work with the highest power possible. And therefore, you can't allow the light to exit. You want to recycle it. You want to use enhancement cavities. And what I can tell you is that this set up here integrates, I think, the signal recycling, the measurement at the zero fringe. And you see that kind of those different parts are copied in a way which I didn't prepare to explain it to you. All I wanted you to do is pretty much recognize that a squeeze light generator is important. And this enters the interferometer as a squeezed vacuum. What I find very interesting, and this is what I want to discuss now is, that when you have an interferometer like LIGO, cavitational wave interferometer, and now you want to squeeze. Does it really help to squeeze? Does it always help to squeeze? Or what is the situation? And this is what I want to discuss with you. So let's forget about signal recycling and enhancement cavities and things like this. Let's just discuss the basic cavitation wave detector, where we have an input, we have the two arms of the Michelson interferometer. And to have more sensitivity, the light bounces back and forth in an enhancement cavity. You can say if the light bounces back and forth 100 times, it is as if you had an arm length which is 100 times larger. And now we put in squeezed vacuum at the open part of the interferometer. And here we have our photo-diodes to perform the measurement. So the goal is to measure a small length scale. If a cavitation wave comes by, cavitational waves have quadrupolar character. So the metric will be such that there's a quadrupolar perturbation in the metric of space. And that means that, in essence, one of the mirrors is slightly moving out. The other mirror is slightly moving in. So therefore, the interferometer needs a very, very high sensitivity to displacement of one of the mirrors by an amount delta z. And if you normalize delta z to the arm length, or the arm length times the number of bounces, the task is to measure sensitivity in a length displacement of 10 to the minus 21. That's one of the smallest numbers which have ever been measured. And therefore, it is clear that this interferometer should operate as close as possible to the quantum limit of measurement. So what you want to measure here is, with the highest accuracy possible, the displacement of an object delta z. However, your object fulfills and uncertainty relationship, that if you want to measure the position very accurately, you also have to consider that it has a momentum uncertainty. And this fulfills Heisenberg's uncertainty relation. You will say, well, why should I care about the momentum uncertainty if all I want to measure is the position. Well, you should care because momentum uncertainty after a time tau turns into position uncertainty, because position uncertainty is uncertainty in velocity. And if I multiply it by the time tau it takes you to perform the measurement, you have now an uncertainty in position, which comes from the original uncertainty in momentum. So if I use the expression for Heisenberg's Uncertainty Relation, I find this. And now, what we have to minimize to get the highest precision is the total uncertainty in position, which is the original uncertainty, plus the uncertainty due to the motion of the mirror during the measurement process. So what we have here is we have delta z. We have a contribution which scores as 1 over delta z. And by just finding out what is the optimum choice of delta z, you find the result above. Or if you want to say you want that this delta z tau is comparable to delta z, just set this equal to delta z, solve for delta z, and you find the quantum limit for the interferometer, which is given up there. So this has nothing to do with squeezing. And you cannot improve on this quantum limit by squeezing. This is what you got. It only depends on the duration of the measurement. And it depends on the mass of the mirror. Now-- just get my notes ready-- there is a very influential and seminal paper by Caves-- the reference is given here-- which was really laying out the concepts and the theory for quantum limited measurements with such an interferometer, and the use of squeezed light. Let me just summarize the most important findings. So this paper explains that you have two contributions to the noise, which depend on the laser power you use for your measurement. The first one is the photon counting noise. If you use more and more laser power, you have a better and better signal, and your short noise is reduced. So therefore, you have a better read out of the interferometer. And this is given here. Alpha is the eigenvalue of the coherent state. But there is a second aspect which you may not have thought about it, and this is the following. If you split a laser beam into two parts, you have fluctuations. The number of photons left and right are not identical. You have a coherent bean and you split it into two coherent beams, and then you have Poissonian fluctuations on either side. But if you have now Poissonian fluctuations in the photon number, if those photons are reflected off a mirror, they transfer photon recoil to the mirror. And the mirror is pushed by the radiation pressure. And it's pushed, and it has-- there is a differential motion of the two mirrors relative to each other due to the fluctuations in the photon number in the two arms of the interferometer. So therefore, what happens is you have a delta z deviation or variance in the measurement of the mirror, which comes from radiation pressure. It's a differential radiation pressure between the two arms. And what Caves showed in this paper is that the two effects which contribute to the precision of the measurement come from two different quadrature component. For the photon counting, we always want to squeeze the light in such a way that we have the narrow part of the ellipse in the quadrature component of our coherent beam. We've discussed it several times. So therefore, you want to squeeze it by e to the minus r. However, when what has a good effect for the photon counting has a bad effect for the fluctuations due to radiation pressure. So therefore, what happens is-- let's forget squeezing for a moment. If you have two contributions, one goes to the noise, one goes with alpha squared of the number of photons. One goes inverse with the square root of the number of photons. You will find out that even in the interferometer without squeezing, there is an optimum laser power, which you want to use. Because if you use two lower power, you lose in photon counting. If you use two higher power, you lose in the fluctuations of the radiation pressure. So even without squeezing, there is an optimum laser power. And for typical parameters, so there is an optimum power. And what we're shown in this paper is whenever you choose the optimum power which keeps a balance between photon counting and radiation pressure, then you reach the standard quantum limit of your interferometer. But it turns out that for typical parameters, this optimum is 8,000 watt. So that's why people at LIGO work harder and harder to develop more and more powerful lasers, because more laser power brings them closer and closer to the optimum power. But once they had the optimum power, additional squeezing will not help them because they are already at the fundamental quantum limit. So the one thing which squeezing does for you, it changes the optimum power in your input beam by a squeezing factor. So therefore, if you have lasers which have maybe 100 watt and not 8,000 watt, then squeezing helps you to reach the fundamental quantum limit of your interferometer. So that's pretty much all I wanted to say about precision measurements. I hope the last example-- it's too complex to go through the whole analysis-- but it gives you at least a feel that you have to keep your eye on both quadrature components. You can squeeze, you can get an improvement, in one physical effect, but you have to be careful to consider what happens in the other quadrature component. And in the end, you have to keep the two of them balanced. Any questions? Oops. OK. Well, we can get started with a very short chapter, which is about g2. The g2 measurement for light and atoms. I don't think you will find the discussion I want to present to you in any textbook. It is about whether g2 is 1 or 2, whether we have fluctuations or not. And the discussion will be whether g2 of 2 and g2 of 1 are quantum effect or classical effects. So I want to give you here in this discussion four different derivations of whether g2 is 1 or g2 is 2. And they look very, very different. Some are based on classical physics. Some are based on the concept of interference. And some are based on the quantum indistinguishability of particles. And once you see you all those four different explanations, I think you'll see the whole picture. And I hope you understand something. So again, it's a long story about factors of 2. But there are some factors of 2 which are purely calculational, and there are other factors of 2 which involve a hell of a lot of physics. I mean, there are people who say the g2 factor of 2 is really the difference between ordinary light and laser light. For light from a light bulb, g2 is 2. For light from a laser, g2 is 1. And this is the only fundamental difference between laser light and ordinary light. So this factor of 2 is important. And I want to therefore have this additional discussion of the g2 function. So let me remind you that g2 of 0 is the normalized probability to detect two photons or two particles simultaneously. And so far, we have discussed it for light. And the result we obtained by using our quantum formulation of light with creation annihilation operator, we found that g2 is 0. In the situation that we had black body radiation, which we can call thermal light, it's sometimes goes by the name chaotic light. Sometimes it's called classical light, but that may be a misnomer, because I regard the laser beam as a very classical form of light. This is sometimes called bunching because 2 is larger than 1. So pairs of photons appear bunched up. You have a higher probability than you would naively expect of detecting two photons simultaneously. And then we had the situation of laser light and coherent light where the g2 function was 1. And I want to shed some light on those two cases. We have discussed the extreme case of a single photon where the probability of detecting two photons is 0 for trivial reasons. So you have a g2 function of 0. But this is not what I want to discuss here. I want to shed some light on when do we get a g2 function of 1. When do we get a g2 function of 2. And one question we want to address, when we have a g2 function of 2, is this a classical or quantum effect? Do you have an opinion? Who thinks-- question. AUDIENCE: When we did the homework problem where we had the linear's position, like alpha minus alpha [INAUDIBLE] plus alpha, we found that one can have a g2 greater than 1, and one can have a g2 less than 1. So maybe g2 isn't a great discriminator, whether it's very quantum or very classic. PROFESSOR: Let's hold this thought, yeah. You may be right. Let's come back to that. I think that's one opinion. The g2 function may not be a discriminator, because we can have g2 of 1 and g2 of 2 purely classical. But why classical light behaves classically, maybe that's what we can understand there. And maybe what I want to tell you is that a lot of classical properties of classical light can be traced down to the indistinguishability of bosons, which are photons. So in other words, we shouldn't be surprised that something which seems purely classical is deeply rooted in quantum physics. But I'm ahead of my agenda. So let me start now. I want to offer you four different views. And the first one is that we have random intensity fluctuations. Think of a classical light source. And we assume that if things are really random, they are described by a Gaussian distribution. And if you switch on a light bulb, what you get if you measure the intensity when you measure what is the probability that the momentary intensity is I, well, you have to normalize it to the average intensity. But what you get is pretty much an exponential distribution. And this exponential distribution has a maximum at I equals 0. So the most probable intensity of all intensities when you switch on a light bulb is that you have 0 intensity at a given moment. But the average intensity is I average. So you can easily, for such a distribution, for such an exponential distribution, figure out what is the average of I to the power n. It is related to I average to the power n, but it has an n factorial. And what is important is the case for our discussion of n equals 2, where the square of the intensity averaged is two times the average intensity squared. And classically, g2 is the probability of detecting two photons simultaneously, which is proportional to I square. We have to normalize it. And we normalize it by I average square. And this gives 2. So simply light with Gaussian fluctuations would give rise to a g2 function of 2. Since it's random fluctuations, it's also called chaotic light. And the physical pictures is the following. If you detect a photon, the light is fluctuating. But whenever you detect the first photon, it is more probable that you detect take the first photon when the intensity happened to be high. And then since the intensity is high, the probability for the next photon is higher than the average probability. So therefore, you get necessarily a g2 function of 2. So this is the physics of it. So let me just write that down. The first photon is more likely to be detected when the intensity fluctuation-- when intensity fluctuations give high intensity. And then we get this result. Yes. Let us discuss a second classical view, which I can call wave interference. This is really important. A lot of people get confused about it. If you have light in only one mode, this would be the laser of, for atoms, the Bose-Einstein condensate. One mode means a single wave. So if we have plain waves, we can describe all the photons or all the atoms by this wave function. So what is the g2 function for an object like this? AUDIENCE: 1. PROFESSOR: Trivially 1, because if something is a clean wave, a single wave, all correlation functions factorize. You have the situation that I to the power n average is I average to the power n. And that means that gn is 1 for all n. OK, but let's now assume that we have two. We can also use more, but I want to restrict to two. That we have only two modes. And two modes can interfere. So let me apply to those two modes a simple model. And whether it's simple or not is relative. So it goes like as follows. If you have two modes, both of Unity intensity, then the average intensity is 2. But if you have interference, then the normalized intensity will vary between 0 and 4. Constructive interference means you get twice as much as average. Destructive means you get nothing. So therefore, the [? nt ?] squared will vary between 0 and 16. So if I just use the two extremes, it works out well. You have an I square, which is 8, the average of constructive interference and destructive interference. And this is two times the average intensity, which is 2 squared. So therefore, if you simply allow fluctuations due to the interference of two modes, we find that the g2 function is 2. So this demonstrates that g2 of 2 has its deep origin in wave interference. And indeed, if you take a light bulb which emits photons, you have many, many atoms in your tungsten filament which can emit, re-emit waves. And since they have different positions, the waves arrive at your detector with random phases. And if you really write that down in a model-- this is nicely done in the book by Loudon-- you realize that random interference between waves results in an exponential distribution of intensity. So most people wouldn't make the connection. But there is a deep and fundamental relationship between random interference and the most random distribution, the exponential distribution, intensity which characterises thermal light or chaotic light. So let me just write that down. So the Gaussian intensity distribution-- actually it's an exponential intensity distribution. But if you write the intensity distribution as a distribution in the electric field, intensity becomes e square. Then it becomes a Gaussian [INAUDIBLE]. So Gaussian or exponential intensity distribution in view number one is the result, is indeed the result, of interference. Any questions? So these are the two classical views. I think we should stop now. But on Wednesday, I will present you alternative derivations, which are completely focused on quantum operators and quantum counting statistics. Just a reminder, we have class today and Friday. And we have class this week on Friday.
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