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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, time for us to do one example, a non-trivial example, which is the ionization of hydrogen. It's a fun example, and let's see how it goes. So ionization of hydrogen. Ionization of hydrogen. Very good. So we're going to think of a hydrogen atom on its ground state sitting there, and then you shine an electromagnetic field, and if the electromagnetic field is sufficiently strong, the electron is going to be kicked out by interacting, typically with the electric field of the wave. So the wave comes in and interacts. So we usually think of this in terms of photons. So we'll have a hydrogen atom in its ground state. And then you have a harmonic e field, and the electron could get ejected. So typically, you have a photon in this incident. We think of this in terms of electromagnetism, although our treatment of electromagnetism was not going to include photons in our description. But we physically think of a photon that this incident on this-- here's a proton. Here is the electron going in circles, and the photon comes in, a photon gamma, and it kicks the electron out. So the photon has energy e of the photon is h bar omega. So when we think of putting a light beam, and we're going to send many photos, we have to think of each photon, what it's doing-- whether the energy is very big, whether the energy is very small, and how does it affect our approximation. So half of the story of doing such a calculation is understanding when it could be valid, because we're going to assume a series of things in doing the calculation. And the validity still won't turn out to be for a rather wide range of values. But we have to think about it. So a couple of quantities you'll want to consider is that the energy of the electron, energy of the electron that is ejected is h bar squared k of the electron that is ejected, k squared over 2m of the electron. And it's going to be equal to the energy of the photon minus the ground state energy of the electron in the hydrogen atom. So the hydrogen atom here is energy equal 0. Ground state is below 0. So if you supply some energy, the first part of the energy has to be to get it to energy 0, and then to supply kinetic energy. So the kinetic energy is the energy of the photon minus what this called the Rydberg, or Ry. The Rydberg is 13.6 eV with a plus sign. And that's the magnitude of the depth of the well. The Rydberg is e squared over 2a0 or 2 times the Rydberg is e squared over a0. That's what was calculated, and this is actually equal to h bar c the constant alpha times h bar c over a0. remember, the constant alpha was e squared over h bar c. So those are some quantities. OK. So we need the electron to be able to go out. The energy of the photon must be bigger than a Rydberg. OK, so conditions for our approximation. One. We're going to be using our harmonic variation. We said in our harmonic variation, Hamiltonian delta h was 2h prime cosine of omega t, and we want this h prime to be simple enough. We want to think of this photon that is coming into the atom as a plane wave, something that doesn't have big spatial dependence in the atom. It hits the whole atom as with a uniform electric field. The electric field is changing in time. It's going up and down, but it's the same everywhere in the atom. So for that, we need that-- if you have a wave, it has a wavelength, if you have an atom that is this big, you would have the different parts of the atom are experiencing different values of the electric field at the same time. On the other hand, if the wavelength is very big, the atom is experiencing the same spatially independent electric field at every instant of time. It's just varying up and down, but everywhere in the atom is all the same. So what we want for this is that lambda of the photon be much greater than a0. So that-- 1. So this means that the photon has to have sufficiently long wave, and you cannot be too energetic. If you're too energetic, the photon wave length is too little. By the time it becomes smaller than a0, your approximation is not going to be good enough. You're going to have to include the spatial dependence of the wave everywhere. It's going to make it much harder. So we want to see what that means, and in the interest of time, I will tell you with a little bit of manipulation, this shows that h omega over a Rydberg must be much smaller than 4 pi over alpha, which is about 1,772. So that that's a condition for the energy. The energy of the photon cannot exceed that much. And let's write it here. So that's a good exercise for you to do it. You can see it also in the notes just manipulating the quantities. And this actually says h omega is much less than 23 keV. OK, 23 keV is roughly the energy of a photon whose wavelength is a0. That's a nice thing to know. OK. While this photon cannot be too energetic, it has to be somewhat energetic as well, because it has to kick out the electron. So at least, must be more energy than a Rydberg. But if it just has a Rydberg energy, it's just basically going to take the electron out to 0 energy, and then you're going to have a problem. People in the hydrogen atom compute the bound state spectrum, and they computer the continuous spectrum in the hydrogen atom, in which you calculate the plane waves of the hydrogen atom, how they look. They're not all that simple, because they're affected by the hydrogen atom. They're very interesting complicated solutions for approximate plane waves in the presence of the hydrogen atom. And we don't want to get into that. That's complicated. We want to consider cases where the electron, once it escapes the proton, it's basically a plane wave. So that requires that the energy of the photon is not just a little bit bigger than a Rydberg, but it's much bigger a Rydberg. And saw the electron, the free electron does not feel the Coulomb field of the proton. And it's really a plane wave. So here, it's a point where you decide, and let's be conventional and say that 1 is much less than 10. That's what we mean by much less 1/10. So with this approximation, h omega must be much bigger than 13.6. So it should be bigger than 140 eV. That's 10 times that. And it should be smaller, much smaller and 23 keV, so that's 2.3 keV. So this is a range, and we can expect our answer to make sense. If you want to do better, you have to work harder. You can do better. People have done this calculation better and better. But you have to work much harder. I want to emphasize one more thing that is maybe I can leave you this an exercise. So whenever you have a photon in this range, you can calculate the k of the electrode, and you can calculate how does the k of the electron behave. And you find that k of the electron times a0 is in between 13 and 3 for these numbers. When the energy of the photon is between those values, you can calculate the momentum of the electron, k of the electron. I sometimes put an e to remind us of the electron. But I'll erase it. I think it's not necessary. And that's the range. OK. So we're preparing the grounds. You see, this is our typical additive. We're given a problem, a complicated problem, and we take our time to get started. We just think when will it be valid, what can we do. And don't rush too much. That's not the attitude in an exam, but when you're thinking about the problem in general, yes, it is the best attitude. So let's describe what the electric field is going to do. That's the place where we connect now to an electric field that is going to produce the ionization. So remember the perturbation of the Hamiltonian, now, it's going to be the coupling of the system to an electric field. And this system is our electrons. So it's minus the electron times the potential, electric potential, scalar potential. Now, needless to say, actually, the electron that is going to be kicked out it's going to be non-relativistic. That's also kind of obvious here. You see h omega is 2.3 keV. You subtract 13.6 eV. Doesn't make any difference. So that's the energy of the emitted electron roughly, and for that energy, that's much smaller than 511 keV, which is the rest mass of an electron. So that electron is going to be non-relativistic, which is important, too, because we're not trying to do Dirac equation now. So here is our potential, and then we'll write the electric field. Let's see. The electric field. We will align it to the z-axis to begin with. 2e0 cosine omega t times z hat. These are conventions. You see, we align it along the z-axis, and we say it has a harmonic dependence. That's the frequency. That's the frequency of the photons that we've been talking about, and the intensity, again, for these convention preference will put 2e0 times cosine omega t. So some people say ep, which is the peak e field is 2e0. That's our convention. Well, when you have an electric field like that, the electric field is minus the gradient of phi, and therefore, we can take phi to be equal to minus the electric field as a function of time, times z. So if you take the gradient minus the gradient, you get the electric field. And therefore, we have to plug it all in here. So this is plus e, e of t, z, and this is e. e of t has been given 2e0 cosine omega t. And z, we can write this r cosine theta. In the usual description, we have the z-axis here, r, theta, and the electric field is going in the z direction. So z's are cosine theta. So I think I have all my constants there. So let's put it 2 e, e0, r cos theta cos omega t, and this is our perturbation that we said it's 2h prime cosine omega t. That's harmonic perturbations were defined that way. So we read the value of h prime as this one. e e0 r cosine omega t. No, r cosine theta. OK. We have our h prime. So we have the conditions of validity. We have our h prime. Two more things so that we can really get started. What is our initial state? The initial state is the wave function of the electron, which is 1 over square root of pi a0 cubed e to the minus r over a0. That's our initial state. What is our final state, our momentum eigenstates of the electron? So you could call them psi, or u sub k of the electron. And it would be 1 over l to the 3/2, e to the ik of the electron times x. These are our initial and final states. Remember, the plane waves had to be normalized in a box. So the box is back. u is the wave function of the plane wave electron. And we could use a plane wave, because the electron is energetic enough. And it has the box thing. This is perfectly well normalized. If you square it, the exponential vanishes, because it's a pure phase, and you get 1 over l cubed. The box has volume l cubed. It's perfectly normalized. This is all ready now for our computation.
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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. PROFESSOR: All right. So, so far we've recently been talking about examples in SET2, and we're going to continue to do so today. So the example that we did last time was the plan photon form factor. That did not have any soft degrees of freedom. It just had colinear and higher degrees of freedom. So it was a particularly simple example of something we could think of in SET2. We'll start with a slightly more complicated example, this decay, B to D pi, where we have both colinear and soft degrees of freedom. This was an example that we mentioned at the very beginning of our discussion of SET, and now we're going to see how factorization looks for it. And then we'll talk about something called the rapidity renormalization group, which has to do with situations in SET2 where the separation of degrees of freedom is a little more complicated than in the previous examples. And we'll see that there can be a new type of divergence that shows up. And that new type of divergence leads to a new type of renormalization group. So B to D pie. So there's going to be, in some sense, three hard scales of this problem. The mass of the B quark and the mass of the charm quark are going to be taken to be much greater than lambda QCD, so we'll have an HQET type description of the B quark and the charm quark. And also the energy of the pion, which is in some sense, related to the difference of the B quark and the charm quark mass, will also take that to be much greater than the lambda QCD. So just by kinematics this thing is proportional to MB minus MC roughly. You could say, it's the difference of the squares of the hadron masses. OK. So let's first-- we know how to treat this decay if we were integrating out the W. This is a weak decay, so B is changing to charm. Integrate of the W boson, run down to the scale MB, which is the larger scale here, that's the electroweak Hamiltonian. So that's what we'll call the QCD operators, which are the relevant description at the scale of order MB. Some pre-factor. And I'll write the operators in the following way-- a slightly different basis than we used previously, or a long time ago when we were talking about this particular case. So just a different color basis, singlet and octet. OK, so that's our description, where p left is projecting us onto, the left-handed components. So what we want to do is we want to factorize the amplitude. This is an exclusive process where we make a transition between specific states. So we'd like to separate scales in the D pi, and then we have O0 or O8 in this matrix element. So we have two matrix elements, one with O0 and one with O8. And so, what could it possibly look like? Well, we already talked about the degrees of freedom here. The D is going to be soft and the B is going to be soft. So this is soft, this is soft. This is going to be colinear. And so, if it's going to factorize, and the soft degrees of freedom are not going to talk to the colinear, the kind of thing that you would expect to show at leading order in the lambda expansion is that you have the following kind of process. Let me reclaim this space. So here's a heavy quark. Here's one of these operators. Here's the valence quarks in the pion. Another heavy quark-- this was B charm, U and D. And there's an anti-quark, and we have to address this by gluons. And if it's going to factorize, then the way that we should dress it by gluons is as follows. We would have soft gluons here, and they could interact, if you like, between things in the B and the D, because the B and the D are both soft. We can also have back and polarization diagrams. And that is going to factorize from things in the pion which are going to be colinear. So we have our colinear gluons and colinear quarks inside here, and maybe there's some Wilson lines too. So we would expect some kind of picture like that. And that's actually going to be what we do find. But exactly what happens at this vertex, what kind of convolutions there are, that we have to work out. All right. So what factorization in this context means is that there's no gluons that are directly exchanged between the B to D part and the pion part. So that it effectively factorizes into a matrix element that's like a B to D transition and a vacuum to pion transition. So you can even guess what kind of objects this would depend on. If you have something like this green thing, that's a B to D form factor. So we expect a B to D form factor. And for the pion, if you have something like this, well, we already talked about something like this when we were talking about gamma star, gamma to pi zero. So for the pion, we expect the pi zero-- the light cone distribution, which is the sort of leading order operator for the pion. So we'd expect a 5pi of x. And we'll see that we do indeed find that. So the B and D have P squared of order lambda QCD squared for their constituents. The pion is colinear, and its constituents have P squared of order lambda QCD2, but they're boosted. We can again use SET-- this is SET2, but we can use SET1 as an intermediate step, just like we did-- we talked about last time. So let's do that again. So match-- step one was to match QCD onto the SCET1. So there was some hard scale. And the harder scale in this case could be any one of these three. So collectively I just denote them by Q. And so what's going to happen in that matching is we take these operators O0 and O8, and we have to match onto SET operators. So let me call the SET operators Q0. I just there's-- because of the fact that there's a heavy quark in the way that that works, there's two possible spin structures here. But we'll see that actually only one of them has the quantum numbers in the end. So we have heavy quark fields, charm, and bottom, that we can imagine that type of operator. And then the colinear part, we can dress it with the Wilson line, as always. And let me put the flavor upstairs. And let me put in the most general Wilson coefficient that I can think of for this process. This could also depend on V dot V prime. I didn't denote that. But in general, it will. And it could depend on any of the scales Q. So it can depend on the large momentum of the colinear fields, as always. And there's one combination that's not restricted by momentum conservation. So there's one combination that's not related to these scales, and that's the, in my notation, the P bar plus operator. And then there's likewise, there's another thing with the TA. So same thing, TA, everything the same, TA. And then this has got a different coefficient like that. And so, what are the Dirac structures just to be explicit? The heavy one-- so the light one is going to be M bar slash over 4 in my notation, the 1 minus gamma 5. And for the heavy one, you could have either 1 or gamma 5. And that's because you originally have a left handed-- originally here you have a left handed guy between the charm and the B. But remember that a mass term connects left and right. So after you integrate out the mass of these quarks, you don't know whether-- you don't know the chirality anymore in this. So that's why you can have both the possibilities 1 and gamma 5. So if you put any other Dirac structure-- so here you could use chirality. And so, you know that these guys should be left handed still, and that's why that this should be this structure. You know it should be an M bar slash, because any other structure that you would stick in here between them would give you something that's power suppressed. Because you know that N slash, when CN 0, and also CN bar gamma [? per mu ?] with any kind of P left or whatever, is also 0. So you'd have to have something more complicated. All right, so there's those operators. And when you do this matching, it is non-trivial in the sense that these two operators-- it's not diagonal. It's not like O0 goes to Q0, and O8 goes to Q8. As soon as you start adding loop corrections these two mix, and then they give you some contributions to these coefficients C0 and C8. OK? So what you mean by octet operator in the electroweak Hamiltonian is different than what you mean by octet operator in the SET1 factorized result. But that's just a complication that you deal with when you are doing the matching. This guy can be proportional to the Wilson coefficient C0F and C08, and that's not really a big deal. Any questions so far? STUDENT: [INAUDIBLE] PROFESSOR: Yeah STUDENT: So is the point of matching to get one of them as being [INAUDIBLE] to get the softs? PROFESSOR: Yeah. STUDENT: So you're going to distribute them-- PROFESSOR: That's right. I'm going to do that right now. STUDENT: OK, so that's-- PROFESSOR: Yeah. So then step two, field redefinition in the SCET1. And let me not put superscript zeros, let me just make it as a replacement. So then we get RQ0 again. And OK. For this guy, it's exactly the same, because in the Q01 comma 5, all the Y's cancel. In the octet guy, it's not quite that way, because there is-- because we do have-- in that case, we do have the Wilson and lines getting trapped by the TA. So let me write out that case. So we have-- so that's what we would get after the field redefinition for two operators. OK. So the next thing to do would be to instead of calling them Y's call them S's. So, but there's one more thing I can do too. So these are, remember, these are soft fields and these are colinear fields. So this isn't factorized, because we have contractions between these Y's in the fields over here. Gluons can attach to heavy quarks. So in order to factorize we want to move those Y's from there over to here. And we can do that. So here's how that works. This is a formula that I could have told you earlier. So if you have a Y that lies around a TA, that's just actually the adjoint Wilson line. So this is a formula that relates fundamental Wilson lines and an adjoint Wilson line. So in the adjoint Wilson line, you'd build it out of matrices. If you like, they're like this. So the matrix indices would be B and C, and instead of having fundamental indices for the TA alpha beta, you have an FABC, and this is the kind of thing that you would exponentiate. But other than that it's the same thing. And there's just a color identity relating them. So because of this identity, you can also write down another identity, which is Y dagger TAY. This guy is-- so Y dagger TAY is just the other YAB-- this guy's an orthogonal matrix, TB. So if you reverse the indices then you get the opposite way. And also, this guy is-- remember this is a matrix just in the AB space. So if I use this formula in here, that allows me to take these Wilson lines here and move them over here. Right? Because I can take them, write them as a Y, and then the Y is just something that doesn't care about-- it just moves over because it's contracted with that index A. I guess I've got some problems with capitals and small letters. Let's make them all capital. So then not going to move it over here. And then I can convert it back to a Y dagger Y if I want to. OK. So we can take this guy and this thing and write it over there as HV Y TAY dagger HV prime. I was careful about that. This was the prime. And then all the soft gluons-- all the ultra soft ones are over here in this matrix element and all the colinear fields are over there. So if you like, you could say, we get this, and then we get our colinear matrix element that has TA, but it has now no ultra softs. And so we have a product of colinear and ultra soft things tied together by one index A. OK, so that's just a little color rearrangement. It's useful, because now they're really factorized. And now when you take matrix elements, the matrix elements will factorize. Oh, sorry, before we take matrix elements, let's switch to SCET 2. So this is, again, an example where it's trivial. Because what we have is, we have one type of operator that we're considering, this weak transition, and we don't have a time limited product of any type of two operators in SET2. We just have a single operator that has both types of fields, and then we have the Lagrangians So this is, again, simple. So we have one mixed operator plus L0 colinear and L0 soft. And those things are already decoupled, and so this is simple. And so, we simply replace Y's by S, renaming it soft instead of ultra soft. Really nothing is changing. And these colinears, we just put them down onto SET2 colinears from SET1 colinears. OK, so to make it look like I've done something, I'll write it out again. But there's really nothing happening except that now the fields are in SET2, with the correct SET2 scaling. So there's no Wilson coefficient that's generated by this stuff-- there's no additional Wilson coefficient because of that fact. So these are SCET2 now. And similarly for the octet. So here we would really have the-- OK. So now we can take matrix elements. STUDENT: [INAUDIBLE] PROFESSOR: Yeah? STUDENT: What do the coefficients look like when they're not just 1? PROFESSOR: So they would be functions of say, plus times minus momenta. So we could have-- if it wasn't 1, what would happen is effectively-- so yeah, we talked a little bit about this last time, but let me remind you. If it wasn't 1, that would happen in a situation where you had something like this. Some off shell field, O, say like this. So this is a T product of two things rather than just one thing that mix off the colinear. And this field here was off shell in a way that basically, this field here is an off shell field that would be a product of the plus and minus momentum of these guys. And so, you could get something that's effectively living at this hard colinear scale below the hard scale in the problem, from sort of T products of soft and colinear operators. This is getting a little sketchy, but-- since here we only have one operator, that couldn't happen, because you just start attaching softs. And colinear is been-- it's already factored. So there's no way that you could sort of get this intermediate off shell guy. STUDENT: [INAUDIBLE] PROFESSOR: So this guy would be colinear, right? And the way to think about this is like this. This is just a diagram that exists. But then you go over to the SET2, and then you have your change of colinear to the right colinear. But this guy doesn't change. He's still hard colinear. STUDENT: OK, so it's a matching when the material actually has a different scale than matching-- PROFESSOR: Yeah. This scale would show actually up. And so if it doesn't show up-- so this would be kind of a situation, an going from step one. And actually, we could have done some examples where that happens. But I'm choosing to do this rapidity renormalization group instead. Basically this happens if you look at matrix elements, and you could look at matrix elements where you have some subleading interactions. And there are examples in exclusive decays where you could have this happen. One example is if you look at B0 to D0, pi 0, just having all neutral charges, then actually this will happen. It'll be-- and it'll be more complicated than what I'm telling you. But you can derive a factorization theorem for this. It's power suppressed relative to the one we're talking about, because the one we're talking about always has a charge pion, and it turns out that that happens at leading order, whereas the neutral pion process with the neutral B and neutral D is something that's power suppressed. Hope I'm remembering that right. Yeah. I am. All right. So number four, this is an aside. Number four, take matrix elements, and here we find actually that one of the matrix elements is just 0, the one with the octet. So let me write the nonzero ones first. I wasn't too careful about the two different-- about this. So there's some guy that's just giving a convolution between the Wilson coefficient and 5 pi. And then this guy, which is some normalization factor times a form factor. These things are all mu dependent in general. And this thing here is the Isgur-Wise function, which is the HQT form factor. And W0 is kind of the kinematic variable that that form factor can depend on, which is V dot V prime, the labels on the fields, and that encodes the momentum transfer. Which here is just related to the kinematics. So W0 is some function of MB and MC, which I'm not going to bother writing down. STUDENT: [INAUDIBLE] PROFESSOR: Just some constant. I mean, it could have some kinematic factors that makes up the dimension, like some-- depends on my gamma. Yeah, it's just some number. STUDENT: [INAUDIBLE] PROFESSOR: OK so these are the singlet operators. So in the singlet case, we have initial state and final state. In all cases we have initial state and final states, which are color singlets. And these operators are color singlets. In the case of the octet, we also have color singlet states. And we factorize such that we do have a matrix element for example. So let me just write one of them, and no. And this is 0, because there's nothing that could carry the index A in this matrix element. So the octet matrix element's 0. It's important that we factorized it for that to be true, right? If we had D pi, then we'd have a color singlet operator here, B. So we wouldn't have been able to make this statement in the original operator in the electroweak Hamiltonian. This would just not be true. But once we factored it and put all the ultra soft fields here, that everything that's going to be contracted together, then we can make this statement. We couldn't even really make this statement when the Y's were on the other side. We had to move them over here to ensure that this statement is completely true. OK. So color octet operator is color singlet states. OK, so then you just put things together and we can multiply these two things to get the final result. So if we write it as a matching from the electroweak Hamiltonian, there are some normalization factors. So grouping together these factors of F pi, E pi, and M prime, which I'm not worrying so much about, there's a Isgur-Wise function, and then there's a single convolutional between the hard coefficient, which is kind of like our example of the photon pi on form factor. And then the slight [INAUDIBLE] distribution. So we see an example where it showed up in a totally different type of process from the one we were considering previously, thereby showing us kind of the universality of that function. And then there would be some power corrections to this whole thing that we're neglecting, that go like lambda QCD over those hard scales. OK, so this is the Isgur-Wise function. And actually I did write down what the W would be. So this W would be that, some-- you can write it in terms of the meson masses like that, so it's the Isgur-Wise function at max recoil. And this function is measured, for example, in a semi-leptonic transition. So you can imagine that the pion distribution function, or properties of it, were measured in the photon pion transition. Is Isgur-Wise function is measured in the B to DL new transition. And then you can make predictions for this B to D pi. OK, so that gives you an example of how you would use these factorization theorems. So this applies to basically-- this type of factorization that we just talked about, it applies to a lot of different things with charged pi minuses. Or you can make the pi minus a rho minus, and that wouldn't really change anything. So you could have B0. If you wanted to look at charges, you could have B0 to D plus pi minus. Or you could have B minus to D0 pi minus. And there's a third one, which is the one we were talking about over here, B0 to D0 pi 0. So there's three different ways-- three different B to D pi transitions depending on the charges. And what we've derived applies to charge pions or charge rhos. The neutral ones end up being power suppressed. And you can see that kind from our discussion there was kind of never a-- there was-- it just writing down the leading order operators, well, maybe you have to work a little harder. But effectively, the leading order operators don't make this transition with the two charges being the same such that you could get a pi 0, so you have to do something more in order to get that case. All right. So questions? STUDENT: Can you [INAUDIBLE]? PROFESSOR: You can if you want. Oh, MB over MC. So MB and MC, we're treating both of them as avoiding the same in what we've done. So if we wanted to some logs of MB over MC, we'd have to do something a little different than what we did. You'd have to first integrate [? O to MV, ?] treat the charm quark as a light quark. You could do that. STUDENT: Would it make the SCET [INAUDIBLE]---- PROFESSOR: It turns out that-- STUDENT: --analysis different? PROFESSOR: --yeah. So those are single logs, actually, they're not double logs. And it's related to the fact that you have massive particles. When you have massive particles you don't get the extra singularity. And people in HQT worried about something-- logs of MB over MC for a while-- and then after a flight of doing enough calculations they realized it was totally irrelevant, and you should just not bother. You should just calculate the alpha S corrections, treating MB and MC as comparable, and summing the logs-- if you sort of think of leading log as being more important than the order FS calculation, that misleads you. Sometimes the sign is even wrong. And so there's sort of a general experience that something logs of MB over MC and HQT is not even-- STUDENT: Just because they're single logs? PROFESSOR: Just because they're single logs. I mean, that's one thing that makes it different than, say, the double logs that you would resum in this process. Actually, these double logs are also single logs. So you could decide whether or not to resum of them. But either-- whether or not you do resummation, this is still useful, because you could make a prediction for this decay rate, and it works really well. All right. You can actually also make predictions for these decay rates. You can predict actually the relations between the D0 and D star 0 using the factorization-- subleading factorization theorem. OK, so let's move on to our second topic, which will take the rest of today. And that's rapidity divergences. OK. So when we were talking about SET1, and we were talking about loop calculations, we saw that there was a subtlety where when we were doing our colinear loops, that could double count the ultra soft loop, if you remember. So kind of schematically, I could say, that this true CN was sort of a CN naive minus a CN 0 bin. So you could do a calculation ignoring that, but then you have to be careful, and there's a subtraction. And that subtraction avoids the double counting with the ultra soft. So if you think about there being ultra soft amplitudes and colinear amplitudes, this avoids a double counting. Now, we never talked about whether there's something analogous to that in SET2, and we just did a lot of SET2 examples without ever even saying those words. So why it's actually-- for what we've talked about so far-- OK to ignore this issue. But in general, it's not OK. So if we go back to our picture of the degrees of freedom in SET2, have this hyperbola, and you could have softs, you could have some colinears. And then the example that we just did, it's like these were kind of the relevant modes. And in general, you might have some guy down here as well. So these are the degrees of freedom in the SET. And effectively what's happened is, if you want to think about double counting, you're sliding down this hyperbola so this hyperbola is kind of at a constant invariant mass, say, lambda QCD squared, or it could be lambda QCD. And unlike the case in SET1, where this guy lived in a different hyperbola here, to get between them you would be sliding down the hyperbola at fixed invariant mass. So that's a little different. But in general in SCET2, there are also 0 bins. So in general, you would have something like, met me denote it this way, CN minus CN soft. And what I mean by this, is this is my original amplitude, where P mu was scaling like Q lambda squared 1 lambda. This was the original. And this would be a subtraction where you take that amplitude, and you'd make it scale like in the soft regime. So it would be P mu, lambda, lambda, lambda. So that's different than the example of SET1. In The SET1 case, we would really just be scaling down the 1 in the lambda, so that they would be both of order lambda squared. Here we're actually scaling down the 1 and scaling up the lambda squared to a lambda. And that's the right thing to do to go from here to here. So we're just taking the amplitudes in this region and subtracting them in this region. And in general, we do have that. But actually that's not the real complication that shows up in the SET2. The real complication has to do with whether there's any divergences associated to that. If this amplitude here didn't have any divergences, it wasn't kind of log singular, then you wouldn't really care about doing this subtractions, because then there would be no infrared singularities that you're double counting and it would just be effectively a constant. And the constants are always ambiguous. So whatever mistake you make in constants here you just make up by changing your hard matching. So you don't have to worry if when the colinear goes down into the soft region there's no divergences. And that is actually what's happened in all the examples we've treated so far. That when the colinear goes into some region where it's not supposed to have singularities, that you just end up with no singularities. There's no log singularities. OK, so, so far there's no log singularities from the overlapped regions. But that's not in general the situation. And we'll do an example in a minute where there are singularities that overlap. And the true difficulty here is the following. If you think about what's separating these modes, you might draw lines like this. Just to draw some straight lines separating the modes. And remember that we're plotting here in the P minus, P plus plain. And that fixed P squared is like fixed product of P minus P plus. So P squared is P plus P minus, up to the P perp squared piece, which we're ignoring. So you can think of these lines as lines of constant P plus over P minus. And if I-- this is the-- so something orthogonal to P squared. All right. So that would be one way of thinking about-- so you need something that's orthogonal to P squared in order to distinguish these modes. And the real issue with that is related to the regulators. When you use dimensional regularization it turns out the dimensional regularization is not sufficient to regulate a divergence that would happen when the CN comes down on top of the S. And the reason is, because dimensional regularization regulates P squared. It regulates-- remember, it's a Lorentz invariant regulator. So it's regulating Lorentz invariant things like P squared, not something like the rapidity, which is this P plus over P minus that you would need to distinguish these modes. So invariant mass does not distinguish the low energy modes. So rapidity, you could define-- is usually defined this way. So exponent of 2Y, where Y is the rapidity, is P minus over P plus. And if you look at the scaling of that, that scaling either lambda minus 2, lambda 0, or lambda squared for the different cases for CN, S, and CN bar. So it's this variable that's really distinguishing the different modes. OK, and that's these lines-- these orange lines are just putting dividing lines between these in rapidity. All right. OK. So there's a complication that dimensional regularization doesn't suffice. So you can think of it as regulating P Euclidean squared once you do the Wick rotation, for example. So it regulates-- it separates hyperbolas, but it does not separate modes along a hyperbola. It's a way of regulating singularities between hyperbolas, but not along hyperbola. So that's one complication. We'll need an additional regulator. And we'll see that that regulator will eventually lead to a new type of a normalization group flow, which is flow along a hyperbola. It's not a flow in invariant mass, but a flow in rapidity. OK. So let's explore what can happen in an example where there are these divergences in sort of the simplest possible example. So there's enough going on that we want to make our lives as simple as possible. So what I'll talk about is something called the massive Sudakov form factor. So you should think of it set up as follows. We're going to consider a form factor, and it's going to be a space-like form factor. So it's a space-like quark-quark form factor. Q of the photon here is space-like. And we're going to think about, rather than having photons or gluons, we're going to think about massive gauge bosons. So this is going to be some kind of Z, if you like. It could be a Z boson. And I'll just call the mass M. OK. So the thing that I'm going to want to iterate is the mass-- rather than doing QCD, I'm doing electroweak corrections-- electroweak corrections from a massive gauge boson. So this is relevant if we have electroweak corrections in a situation where Q squared is much greater than at M squared. And we're not going to be having gluons. Instead we'll be talking about these massive gauge bosons-- multiple Z bosons, if you like. We could also put the W's in, but let's make it simple and just talk about Z's. OK, so let's do this example. So in the full theory, you would start with a vector current, say, and you'd want to match that onto SET. Before we do that, let's just-- let me just write down a kind of full theory object using Lorentz invariance. So you could think about the quark form factor. And four massive gauge bosons, this is just some form factor that you can calculate that's a function of Q squared and M squared, and then there's some spinners. And so, really kind of the dependence is encoded in this F, which is a function of Q squared and M squared. M squared is acting kind of like an infrared regulator. So this is Z boson, there's not a soft singularity associated to it. And so, what you'd like to do-- and in this process-- is factorize Q squared and M squared, i.e. expand this thing in Q squared and M squared, and maybe some logs of Q squared and M squared. So we want to factorize some logs, et cetera. OK, so what are the type of degrees of freedom that we could have here? So lambda is going to be M over Q, massive Z boson over the energy scale of the collision, of the gamma star. And if you just look at the Z boson, then it could be colinear or it could be soft. So it could be actually three different possibilities. And I could think about doing this effectively in a bright frame. Just like we did earlier. And then what you would have is that the colinear guy is like the quark, and then the anti-colinear guys is like the outgoing quark. [INAUDIBLE] So you'd be making a transition in this diagram from N colinear of objects to N bar colinear of objects. And you could likewise have a Z boson which could be colinear. Or you could have a Z boson that's soft. And you have a soft rather than ultra soft because of the mass. If Q times lambda is M, and so if we want a propagator that's like P squared minus M squared, P squared better be of order M squared. That happens for softs, not for ultra softs. So that's why we have softs. And the same thing for colinears, P squared of order M squared for these colinear. So it's really an SET2 type situation, where the hyperbola is just set by P squared of order M squared. We have our mode sitting on that hyperbola. OK, so it's exactly of this type over there. And the thing that's new in this example is that we're going to encounter these rapidity divergences. STUDENT: You mentioned there's the only form for vertical in your theories, is that? PROFESSOR: Yeah. Make things as simple as possible. So you could talk about mixed sort of electroweak in QCD, but yeah, we don't. Let's make it-- in some sense that's kind of just like mixing a problem that we already would know how to deal with, with this one. So let's just deal with this one. All right. So in terms of the external court momenta, we can therefore kind of treat them as follows. Let's just let them be of a large component. They're massless particles. So this is P, and this is PR. And these guys are massless-- should have said that. And if you go through the kinematics, Q squared, which is minus P, minus P prime squared. If you square that, you just find it to P minus P plus. And we're just effectively, if we pick the bright frame-- which is what we're going to do-- each one of those is separately keep going at prime. Call it bar. Each one of those is separately Q. The large-- these guys are just fixed-- both would be Q. All right. So we could factor is this current with these degrees of freedom. The quarks are colinear. So at lowest order they've just become a CN and a CN bar. And then we have to address that with Wilson lines. And we know how to do that. So let me just break down the answer. We could again follow our procedure of going through SCET1, but it's now so familiar we just know what to write down. OK. So the current would look like that. And I'm not to worry too much about the Dirac structure. So I won't worry, for example, about gamma fives. And we could put that in, it's easy. So that's the leading order current, and then we'd have leading order Lagrangians, and we need to start calculating. And if we calculate what you would expect from a major from that is, you'd expect that F of Q squared and M squared is going to split up-- given the degrees of freedom we have into some kind of hard function-- and then some kind of amplitude for the colinear parts, and then some kind of amplitude for the soft part. OK, so you'd expect some hard times colinear factorization of the form factor. And this is what we'll be after. So it's always good to sort of have an idea where you're going. And that's where we're going. So let's consider just one loop diagrams. And it suffices, in order to make the point, just to consider the most singular one. So I'm going to consider-- so there's various loop intervals that you could have to do when you're doing the diagrams if they're fermions. Let's just take the simplest, which is a scalar loop interval. And I'm going to contrast how that scalar loop interval would look if you were doing the full theory calculation with how it would look in the effective theory. And then we'll see where the divergences come from. So you can think about this as kind of-- the piece where the numerator is independent of the loop momenta, so the numerator just factors out. OK, so if I took our vertex triangle diagram over there, then a piece of it-- where the numerator is trivial and factors out-- would look like this interval. So let's just study this guy. If we did this interval in the full theory, this would be both UV and IR finite. So this is just giving us some result that involves logs of Q squared over M squared. So it does have double logs of Q squared over M squared, and single logs of Q squared over M squared. But it's perfectly-- there's no 1 over epsilons. So now let's see what-- let's think about what would happen in the effective theory. So we have a kind of analogous loop interval for colinear, where the gauge boson is colinear. There's some numerator that again I'm not going to worry about. If this numerator is constant, it doesn't-- it's effectively the same constant. In the case of the-- if you take the leading order numerator in the full theory, it'll be the leading order numerator the effective theory as well. But the denominators do change. And so, if we took the N colinear, then yeah, so this guy doesn't change. Because that's just like saying P minus and K are-- P minus and K minus are the same size. But this guy does change. OK, so this guy would be K minus P bar plus, because K minus is big and B plus is big. Both of these are big. So both of those are big in this diagram. And so that's effectively the Wilson line diagram. OK, where the propagator here was off shell, got integrated out, and just became iconal. And the K squared is smaller, so we don't keep it in a leading order term. And then analogously, for IN bar, it's the other way. So both of these are big, and this one remains. And then they're soft. And in the soft case, what happens is that both of the propagators end up being iconal. And in our SET operator, that's a diagram where we have our colinear lines, and then we have kind of a self contraction of the S. But we're taking an SN with an SN-- we have a contraction that's like this. We have two of the Wilson lines-- soft Wilson lines that are sitting in that operator. That's a non-zero contraction. That would lead to a diagram like-- that would lead to this amplitude. All right. I'm going to leave a little space here, because I'm going to add something in a minute. All right. So how do we see that there's a problem with these intervals that they're not regulated by dim reg? Well, you could look at the soft integral and you could just do the perp. The perp is only showing up in this K squared minus M squared. So you would get something by doing that. And so, if we do the perp with dim reg and IS, we would end up proportional to something that's DK plus DK minus K plus K minus, minus M squared to the some power of epsilon, divided still by the factors of K plus and K minus. So you see that the invariant mass is being regulated. We just did the perp. Perp is gone. Plus times minus is being regulated, because plus times minus-- if plus times minus grows larger, or goes small, and regulated by this epsilon-- but either one, plus going large or minus going large, or plus going small minus going small, with plus and times minus fixed is not regulated. And that's the rapidity divergence. If the invariant mass is fixed and K plus over K minus goes large or small. So let me write it as K minus over K plus going to 0, or going to infinity. Let me say, with K plus times K minus fixed, then it diverges as these things happen. And if you think about what's happening in our picture over there within these limits, it's exactly a situation where this X here would be sliding up or sliding down. So in one of these limits, this one's going towards the CN bar, and this one would be going towards CN. So it's exactly a region where you would be overlapping-- sliding down the hyperbola. And the interval has log singularities. So this is exactly a situation where we can't ignore the overlaps and we have to worry about them. OK so we need another regulator. Dim reg is not enough. We have to do something else. So what could we do? So there's lots of different things that you could do. One thing is, you could just sort of put it in some plus something in these denominators-- that's called the delta regulated, K plus, plus delta-- that's one choice. We'll do something a little bit more dim reg-like. Which makes it sort of easier to think about the renormalization group. So one choice for an additional regulator is the following. So if you think about where these divergences came from, they came from the Wilson lines. So what you really need to do is regulate the Wilson lines. And you can do that as follows. Let me write out the Wilson lines in our kind of momentum space notation. So we have some N dot P type momentum for the soft Wilson line, and an N dot AS field. And really, it's this one over this iconal denominator that's giving rise to these denominators here that are giving rise to the singularity. So if we want to regulate that singularity we need to add something, and we could do that as follows. So this is the regulator we'll pick. And I'll just write everything as kind of a momentum operator. So I've just tucked the Z momentum in and raised it to some power. So PZ is the difference between P minus and P plus. And that seems kind of arbitrary, but that'll do the job for us. You can motivate why you want to do PZ-- so this is 2PZ actually. You can motivate why you want to do PZ rather than something else in the following way. And it's a true fact, that once you have enough experience you realize it's good to use PZ, because PZ doesn't involve P0. And the softs don't really make a distinction between any of the different components. And if you put in P0, something that involved P0, that would be dangerous. So this is nice, because there is no P0. So it's the combination of P plus and P minus that you can form that doesn't have the P0, which is energy. And remember that the polls in P0 are related to things like quarks and anti-quarks. They're related to unitarity. So not messing up the structure in P0 means that you'll be fine with unitarity, fine with causality, you're not messing up a lot of nice things about the theory. So if you do put P0 in, then you have to be careful about those things. So if you just arbitrarily put in some power of P0, then you'd have more trouble. And so that's kind of why we're avoiding and just putting in PZ. For the colinears, we can do something similar. But for the colinears we also have a power counting between the minus and the plus. So for the colinear, we can still make the power counting OK by thinking about putting in PZ, but then just expanding it to be a P minus. And that's true up to power corrections. And we don't really need to worry about power corrections when we're regulating these divergences. So it's just putting in the large momentum. And so W written in a similar notation. So I'll explain what the other things in this formula are in a minute. But the important thing for regulating is that we have some factor. In this case, it would be a factor of N bar dot P. STUDENT: Dot [INAUDIBLE]? PROFESSOR: Sorry? STUDENT: Dot eta to bar dot P? PROFESSOR: No. It's supposed to be an N. Looks like eta. Too many variables. There's the eta. So there's some factor raising of the-- again the iconal propagator sort of mixing up with the iconal propagator. In this case, it's even more obvious that it's just regulating that iconal propagator. OK. So if we were to do that, and go back over here, and put the regulators into these integrals, what would happen? So here, we'd get an extra factor-- K minus to the eta. And so, that would regulate this K minus. And these integrals will also have polls from the 1 over K minus. You could think about-- well, OK. If we did those integrals, we would also have rapidity divergences that are kind of the analog ones, and the colinear SECT are still the soft ones. Here we have two propagators. And so if I have two soft Wilson lines, and so I get two factors. But I've conveniently chose it to be the square root. So it comes out kind of looking the same here. So one thing that is just a part of this regulator-- which actually I don't know a good argument for-- is kind of a priority from the symmetries of the theory, you might like to argue that that should be eta over 2, and this should be eta. But it's really just part of-- it's just a choice, a convention that we've made, as far as I know. There probably is some nice deep argument for it, but I don't know it. So what are these other factors? So nu is going to play the role of mu. We've changed the dimension of the operator. We've compensated it back with nu, just like we were doing with mu. We're going to get 1 over eta divergences, which are like our 1 over epsilon divergences. And we're going to get logs of nu, which are the analogs of logs of mu. And that's the sense in which there's kind of an analog of this M up with our usual dim reg setup. In order to have a full analog, we should think about having a coupling. And so, that's what this W factor is. You can think about it like there was some bare pseudo coupling, which is really just 1. But just imagine that you're switching from bare to renormalized, in order to set up a renormalization group equation. And then this guy here, which is in eta dimensions, if you like, would have a renormalization group which would say nu D by D nu of this W of eta and nu is minus eta over 2. Sorry, this is eta over 2. So that's the analog of saying mu by D mu of alpha is minus 2 epsilon alpha. So an analog statement. I think this is OK. So this guy here is like a dummy coupling. And the boundary condition for it after you've carried out these-- this is just to set it to back to 1. So it's identically 1, it's really just a bookkeeping device. It's just--e it's a dummy coupling once you go to the eta dimensions. But you just set it always the renormalized coupling is just identically set to 1. And identically setting it to 1 is what you need to keep gauge invariance in these Wilson lines. It turns out actually that this regulator here is gauge invariant, though it doesn't look like it. We've modified the structure of the Wilson line in some kind of way that looks like it might be drastic. But actually these factors here are gauge-- still leave a gauge invariant object. So-- STUDENT: Can you write [INAUDIBLE] space, I assume? PROFESSOR: Not that I know of. Yeah. STUDENT: I think [INAUDIBLE]. PROFESSOR: Maybe you can. Yeah. But it's not-- since it's not-- yeah, I don't know how to-- I don't know what it would look like. You could probably transform that power, and it-- STUDENT: [INAUDIBLE]. PROFESSOR: Yeah. I'm sure you can probably just try out before you [? transplant ?] that. I'm just not sure if it would look nice. Yeah. It might not look too bad. Yeah, and it might actually be a nice way of saying what I'm about to say in a less nice way, which is, if you look at the gauge symmetry, why is this not messing it up? So one way of thinking about that is just to look at general covariant gauge. So note, the 1 over eta and eta 0 terms are gauge invariant. And you can think about that by just going to a general covariant gauge and seeing the parameter dependence drop out. So for example, at 1 loop, you would take [? g mu ?] nu in the contractions and replace it in general covariant gauge by some gauge parameter-- of general covariant gauge, K mu, K nu over K squared. And you'd like to see it independent of this. But this eta to the 0 piece is kind of independent of that for the usual reasons. And the 1 over eta term is independent of that, because this guy actually doesn't deuce any rapidity singularities. What happens is that if you have an N dot K, then you have a corresponding N mu in the numerator. And so, basically what happens is, you get an extra N dot K in the numerator. So any time you have 1 over N dot K, you would get for this piece multiplied by an N dot K upstairs. And so this is cancelling, you don't have a rapidity divergence in the C-dependent part. So that's why this is invariant under the gauge symmetry. And then, because of this boundary condition, the kind of cancellation of the C-dependence in the order [? A ?] to the 0 piece, this kind of works out in the standard way. So it gives you an idea of why it's gauge invariant without giving you a kind of full proof or anything. So we have both 1 over epsilon polls and 1 over eta polls in general, and we have to understand what to do with them. So here's what we're going to do. For any fixed invariant mass, it turns out that we can have these one over eta polls. And the right procedure for dealing with them is as follows. First you take eta goes to 0 and deal with these new polls that you have introduced in your amplitude. In order to deal with them, because you can have them for any invariant mass, you actually have to add counter terms that can be a whole function of epsilon, where you have an expanded in epsilon and then divide it by eta. So let me abbreviate counterterm as CT dot. Then, after you've done that, you take epsilon goes to 0, and you find your 1 over epsilon counterterms. And this is the correct way of doing it. And we'll see how that works in practice in a minute. So let's go back to our integrals that I've now erased and just write out the answers. We're doing those integrals with this regulator. And I'll also make them fermions, so I'm putting in the numerators. We wrote them down for scalars. The scalars where the most divergent integrals actually. I can include the numerators, that doesn't really change the story. And I can include the pre-factors as well. And I'll kind of write things in a QCD type notation, even we can imagine that it's a non-abelian group, just so CF is the whatever group it is, it's the Casimir of the fundamental. Whatever group our gauge boson's in. So here's the eta poll. It has a whole function of epsilon in the numerator, and it's even divergent. So this is 2 eta. And then the rest of it I can expand. So there's going to be 1 over epsilon times the log. When the log replaces that 1 over epsilon then I can start to expand, and I get another 1 over-- when the log nu replaces the one over eta, I can expand this gamma, and it gives me 1 over epsilon. And there's also some other pieces. So over 2 epsilon there's a log mu over M. And there's a constant. And I'm never going to write the constants. So let me read all the results and then we'll talk about them. So ICN bar is the same. The only difference between this is that P minus close to P bar plus. It was really symmetric. And then IS is different. So the 1 over epsilon-- 1 over eta poll comes with the opposite sign, and it also comes to a factor of 2 different. And in this case, there's actually 1 over 2 epsilons squared term. So there's also a double log of mu or M. And then there's plus constant. OK? And so, you could think about adding them up. And what happens when you add them up is, you have 1 over 2 eta, 1 over 2 eta, minus 1 over eta, and so that 1 over eta polls cancel. And that's exactly what you'd expect, because in the full theory the eta was something we introduced in order to distinguish these effective theory modes. It wasn't something that was there needed for the full theory integral. And so you don't really-- you'd expect that it's sort of-- that there's a corresponding regulator between the two sectors. So that when you add them together, that the dependence on that parameter is canceling away, because it was just an artificial separation, if you like, or separation that we're doing. So if I add them up, 1 over eta is cancelled, and so do all the logs of nu. We have alpha-- I'm left with a log of mu over Q, 1 over epsilon poll, double log, some types of single logs, and some other type of double log. So it would look like that. All the nu dependence is canceling away. So sort of various things which we'll start talking about now, and we'll continue talking about next time. So the rapidity 1 over eta divergence, which we can call a rapidity divergence, cancels in sum. And of course, so does the log nu's. And that's as expected. And if you add an overall counterterm, for the entire thing it just involves the hard scale log mu over Q. So if you were to think about there being some Wilson coefficient, which is sort of C bare is ZC minus 1, Z bare is ZC, C renormalized, then ZC and the C renormalized only involve logs of mu over Q, which is the hard scale. OK? And that means that our hard function, which is the Wilson coefficient squared, or just the Wilson coefficient in this case, is only a function of Q and mu. OK, so integrating out the hard scale physics didn't know about the separation. The separation was really something that we needed to do in the effective theory to distinguish the CN and S modes. And you can see why we needed to do it if you look at these answers, because if you look at the types of logs that are showing up here, in this case, we have a nu over P minus. And in this case, we have a-- is it nu over mu? Just make sure I got that right. I guess it is. In this case, we have a nu over M. And we also have a mu over nu. And so, the sort of right scale to-- in order to minimize the logarithms here we're going to have to, again, as usual take different values of mu and nu. Well, it's the same value of mu. All of them are M. But it's a different value of nu, because it's the nu that would need to be of order Q here. P minus is Q. And the nu would need to be of order M here. So it's the nu that distinguishes the modes. So the logs NCN are minimized. Or mu of order M, which says being on the hyperbola, but the nu should be of order P minus, which is Q. And that's precisely actually where we put the X in our picture, if you think about it. OK, so that's saying that you have a large P minus momentum, and we have-- and we're on this hyperbola where P squared is an order M squared. So this is-- and it's likewise for the other pieces. So for the soft piece-- so for the-- say for the anti- for the other colinear piece, we need the same thing. And then for the soft we need a different value for this new parameter. So having this regulator is behaving like dim reg, where we needed different mu's, when we had different hyperbolas. Now we have different places on the hyperbola, and we're tracking that with the new parameter, and that's showing up in the logarithms. And if you think about what the logarithms are doing, you can see that when you combine terms, let's see-- if you look at the 1 over epsilon, and the mu's are canceling out, you're getting a mu over Q. Yeah, that's maybe not the best example. Look over here. You have this log of M over Q In this kind of complete decomposition. The way that that logarithm here gets made up is by having m over nu and nu over Q. All right? So in order to get this log that doesn't have any mu's in it, mu is not telling you that there's that large log. But there is a large log. So there's large logs associated to these rapidity divergences. And what we'll talk about next time is how you do the renormalization group with the diagrams like this. How you write down in almost dimension equations. There'll be an almost dimension equations in both mu and nu space. So we'll have to move around in that space and see how it works to sum of the large logarithms. But I'll postpone that to next time. So any questions? So the general idea is really, as usual, it's just that now we're dealing with a situation where there's two regulators. And they're actually independent regulators. One's, if you like, is regulating invariant mass, and the other is regulating these extra divergences. And so, we'll be able to move around in the space without any worrying about path dependence, for example. We'll talk a little bit about that next time-- in this two dimensional space of mu and nu. But you can see just from looking at the logs, as usual, you can see where you need to be. And you can see that you need to be in different places for the different modes in order to minimize the logs of these amplitudes. And if you do that, then there should be some renormalization group that would connect these guys. So there should do something RGE that goes between these guys, and it'll be an RGE in this new parameter.
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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: So welcome back, everybody. This is the final exam checklist. For the single oscillator, we need to make sure that you know how to write down the Equation of Motion. We have discussed about damped, under-damped, critically damped, and over-damped. We did that. Oscillators, and we have tried to drive oscillators. We observed transient behavior in steady state solution. Resonance, right, so which we actually demonstrated that by breaking the glass. And then we moved on and tried to couple multiple objects together. And that brings us to the coupled system. What are the normal modes? And how to actually solve M minus 1 K matrix, the eigenvalue problem. What is actually the full solution for the description of coupled systems. And can we actually drive the coupled system, and we found out we can. So the system would respond as well similar to what we have seen in the single oscillator case. We see resonance as well. We can excite one of the normal modes by driving the coupled system. Then we put more and more objects until at some point, we have infinite number of coupled objects. What is actually the solution of refraction and the transmission-- refraction and the translation symmetric system. That is actually the discussion of symmetry. We go to the continuum then, and we actually found wave and wave equations. So we found that finally, we made the phase transition from single object vibration to waves, and that is actually an achievement we have done in 8.03. We have discussed about different systems, massive string, massive spring, sound wave, electromagnetic waves, and we have discussed a progressive wave and standing waves. For the bound system, we have also normal modes. We discussed about how to actually do Fourier decomposition, and what is actually the physical meaning of Fourier decomposition in 8.03. For the infinite system, we also learned about Fourier transform and uncertainty principles. And we learned to apply boundary conditions so that we constrain the possible wavelengths of the normal modes. Therefore, we also learned about how to put a system all together. Finally, how to determine the dispersion relation, which is omega as a function of K, the wave number. Until now, we discussed idealized systems, and we also moved on to discuss dispersive medium. We have learned some more, even more about dispersion relation for the dispersive medium and signal transmission, how to send signal through a highly dispersive medium. The solution we were proposing is to use an amplitude modulation radio and also the pattern of dispersion. The group velocity and phase velocity, we covered that. As I mentioned before, the uncertainty principle. A 2D/3D system. We have bound system, which we have normal modes for two-dimensional and three-dimensional systems, as well. Because we're all over the place, so just make sure that you know how to actually dewrap all those standing waves for different dimensions of systems. We showed and approved geometrical optics, which essentially is the direct consequence of waves. Wave function, a continuation of the wave function and boundary condition. We learned about the refraction rule and also Snell's law. We talked about polarized waves, linear, circularly polarized, elliptically polarized, and the polarizer and quarter-wave plate. At the end of the discussion of 2D/3D systems, we discussed about how to generate electromagnetic waves by accelerated charge. Finally, we went on and talked about how those EM waves propagate in dielectrics and again, boundary conditions, which leads to interesting phenomena, which belongs only to electromagnetic waves. For example, Brewster's angle. So the refraction amplitude-- refraction-- the wave amplitude is governed by the property of the electromagnetic waves, which is coming from the laws which governs electromagnetic waves, which is Maxwell's equations in matter. We were trying to also manipulate those waves by adding them together, and we see constructive and destructive interference and diffraction phenomena. Then we connect that to quantum mechanics by showing you a single electron interference experiment. That connects us to the beginning of the quantum mechanics, which is the probability waves, which behave very different from other waves we have been discussing. But you are going to learn a lot more in 8.04. OK, so don't worry. All right. So that is the checklist. You can see that I can write it in two pages, so it's not that bad, probably. I hope that there was nothing really sounds like new to you by now. If you find anything is new, you have to review that part. That means you missed a class. All right. So what I'm going to do now is to go through all the material faster than the speed of light. All right. So that you will get nauseated. No, you are going to get a list of the topics. You just have to feel it. If you feel good, like when you are having a cupcake, right, then you are good for the final. If you don't feel good, what is Professor Lee talking about? He's talking about nonsense now. Then you are in trouble, and you have to review that part. All right? OK. So that's what we'll do. So let's start. All right. Why 8.03? We started a discussion-- welcome, We started a discussion of 8.03 and it's vibrations and wave systems, is name of this, 8.03. And the motivation is really simple, because we cannot even recognize the universe without using waves and vibration. You cannot see me, and you cannot hear anything, and you cannot feel the vibrations-- sorry, the rotation of a black hole by your body anymore, then it's not very cool. Therefore, we study 8.03 to understand the basic ideas about waves and vibrations. And we found that waves and vibrations are interesting phenomena. Waves are connected to vibrations. Because if you look at only, for example, a single object on these waves, you see that it is actually a single object which is oscillating up and down, oscillating up and down, and this is your vibration. So there's a close connection between single particle vibration and the waves. And that is the first thing that you learned. Therefore, we need to first understand the evolution of a single particle system. And we make use of this opportunity to start the discussion of scientific matter. So using this opportunity, basically, what we have been doing for the whole class is the following. So the first step is always to translate the physical situation which we are interested into mathematics, right? Because mathematics is the only language which we know which describes the nature. If you come out with a new language, and that is going to be a super duper breakthrough, it cannot be estimated by Nobel Prize. But the problem we are facing now is that this is the only language we know which works. Therefore, we really follow this recipe, which is similar to many, many other physics classes. And we have a physical situation, we use laws of nature or models, and we have a mathematical description, which is the Equation of Motion. And this is actually the hardest part, because you need to first define a coordinate system so that we can express everything in a system in that system, then you can make use of the physical laws you have learned from the previous 8.01 and 8.02 to write down the Equation of Motion. And most of the mistakes, and also most of the problems or difficulties you are facing is always in this step. Then we can solve the Equation of Motion, which is, strictly speaking, not my problem. It's the math department's problem. Yeah, that's their problem. Then we solve the Equation of Motion, and you will be given the formula. Then we use initial conditions and then make predictions. And then we would like to compare that to experimental results. And that is the general thing which we have been doing for physics. So let's take a look at those examples. Those are examples of simple harmonics motions. And you can see that these, all these systems have one object, which is oscillating. And you can see that their Equations of Motion are really similar to each other. It's theta double dot plus omega, zero squared, theta equal to zero for those idealized simple harmonic motions. And we learned that the solution of those equations are the same, which is a cosine function. Then we went ahead and added more craziness to the system. So basically, what we tried to do is to add a drag force into the game. And we were wondering if this more realistic description can match with experimental data. So this is the Equation of Motion, and the additional turn is the one in the middle, the gamma theta dot turn. And after entering these turns, not only is this an interesting model to describe the physical system we are talking about, but the mathematical solution is far more richer than what we talked about in the single harmonic oscillator case. Basically, you see that a general solution depends on the size of the gamma compared to omega dot zero, which is the oscillation-- the natural frequency of the system. OK. And then you can see there are three distinct different kinds of solutions. They have different mathematical forms. And we call them under damped solution, critically damped solution, and over damped solution. So those equations will be given to you. And the excitement is the following. So you can see that those solutions, if you plot the solution as a function of time, they look completely different as a function of time. So in the case of no damping, the amplitude is actually the constant, it's not actually reducing as a function of time. But when the damped system, the damping is turned on, then in the under damped situation, you can see that they end up reducing as a function of time. And if you have too much damping, you put the whole oscillator into some liquid, for example, and you see that oscillator disappear. The cool thing is the following. The excitement from-- as a physicist is that all of those crazy mathematical solutions actually match with experimental results. Wow. That is really cool. Because there is nobody saying that these should match and how, naturally, I should learn that OK, when should I change the behavior of the system. So this is really a miracle that this complicated mathematical description is useful and that it is super useful to describe the nature. Once we have learned that, we can now add a driving force into ligand. From the equation here, we can see that there is a natural frequency, omega dot zero, of this system, and there is a drag force turn, which is actually to quantify how much drag we have, we have a gamma there. We are driving it at a driving frequency omega t. So what we have learned from here is that if you are driving this system, you are-- for example, I am shaking that student, shaking you. OK, in the beginning, this student is going to resist. No, don't shake me. Come on. But at some point, he knows that Professor Lee is really determined. Therefore, he is going to be shaked at the frequency I like. OK. So that is actually what is happening here. This is so-called transient behavior. So in the beginning, the system doesn't like it. So this is making use of the superposition principle. So you can solve that homogeneous solution, which is on the right-hand side. It depends on the physical situation you are talking about. You choose the corresponding homogeneous solution. And lamba and psi is the driving force from E and G, right, and that is going to win at the end of the experiment, because I'm going to shake it forever, until the end of the universe. So you can see that at the end, you-- what is left over is really the steady state solution. And it has this structure, A omega d, depends on omega. And you get resonance behavior. Don't forget to review that. So you have a delay in phase because when I shake the student, the student needs some time to respond. Therefore, the delta is non-zero if the student is damped. All right. So now we have learned all the secrets about a single object system. Then we can now go ahead and study coupled oscillators. There are a few examples here, which is coupled pendulums or coupled spring-mass systems. And we found that a very useful description of this kind of system is to make use of the matrix language. So originally, if you have n objects in a system, you have n Equations of Motion, and that looks horrible. But what is done in 8.03 is that we introduce a notation with a matrix. Basically, if you write everything in terms of matrix, then it looks really friendly, and it looks really like a single oscillator. OK? Although solving this equation is still a little bit more work. And basically, you can see that from this example, we can actually derive M minus 1 K matrix, and the whole equation won't be-- the Equation of Motion problem solving problem becomes an M minus 1 K matrix eigenvalue problem. What is an M minus 1 K matrix? This is describing how each component in the system interacts with each other. Once we have this, we can solve the eigenvalue problem, and we are going to be able to figure out the normal modes of those systems. So what is a normal mode? Normal modes is a situation where all the components in the system are oscillating at the same frequency and they are also at the same phase. So that is the definition of normal mode. And those are what is used in a deviation, also, which leads us to the eigenvalue problem. We define Z equal to X 1 H or I omega t plus 5. Everybody is oscillating at omega and also at phase 5, right? So that is what we actually learned. And what is actually the physical meaning of those normal modes? So if we plot the locus of the two coupled pendulum problem, what we see is the following. So basically, you will see that the locus looks like really complicated as a function of time if you plot X1 X2 versus time. But if we rotate this system a bit, then we find that there's a really interesting projection, which is the principal coordinate. You see that all those crazy strange phenomena we see with coupled systems are just illusions. Actually, you can understand then by really using the right projection. To one-- to the right coordinate system. Then you will see that actually the system is doing still simple harmonic motion. So that is actually the core thing which we learned from coupled systems. So we learned about how to solve the coupled system, and we also learned about going to an infinite number of coupled systems. So then this is an example here. So for example, I can have pendulum and springs, and we connect them all together, and I need to hire many, many students so that they plays it, plays until it fills up the whole universe. So this is the idea of an infinite system. You can see that that means my M minus 1 K matrix is going to be an infinite times infinite long matrix. It's two dimensional. And the A is infinitely long. And that sounds really scary. And in general, we don't know how to deal with this, really. And it can be as arbitrarily crazy as you can imagine. What we discuss 8.03 is a special case. Basically, we are discussing about systems which are having a spatial kind of symmetry, which is translation symmetry, as you can see from all those figures. And you can see that all those figures will have to all have the same normal modes because of this base translation symmetry. What we discussed about is that we introduce an S matrix, which is used to describe the kind of symmetry that this system satisfies. And if we calculate the commutator S and M minus 1 K matrix, if this commutator shows that the evaluate-- if you evaluate this commutator and you get zero, now it means they commute. And the consequence is that the S matrix and the S M minus 1 K matrix will share the same eigenvectors. So you don't really need to know how to derive this-- to arrive at this conclusion, but it is a very useful conclusion. So that means instead of solving M minus 1 K matrix eigenvalue problem, I can now go ahead to solve the S matrix eigenvalue problem. And usually, that's much easier. So for the exam, you need to know how to write down S matrix. You need to know how to solve eigenvalue problems, including M minus 1 K matrix and the S matrix. And then we can get to normal mode frequency, omega squared, and we can also solve the corresponding normal modes. And here is telling you what would be the solution for space translation of the matrix system. And basically what we will see is that making use of the S matrix should be-- brings you to the conclusion that A, j must be proportional to exponential i, j, k, a, where this A is the length scale of this system, the distance between all those little mass. And the j is a label which tells you which little mass I am talking about. And k is the-- some arbitrary constant. But by now, you should have the idea basically that's-- that's what? That, essentially, is the wave number, right? So that is really cool. So that's all planned in advance. And basically, you can see that we can also write down the A, k because we know that A, j will be proportional to exponential i, j, k, A, after solving the eigenvalue problem for S matrix. Then we actually went one step forward to make it continuous. So basically, we made the space between particles very, very small. And also, at the same time, we make sure that the string doesn't become supermassive. And we concluded that we get some kind of equation popping out from this exercise. M minus 1 K matrix becomes minus T over rho L partial square, partial x squared. You don't have to really derive this for the exam, but you would need to know the conclusion and that psi j becomes psi as a function of x and t. And the magical function appeared, which is the wave equation. Oh my god, this is the whole craziness we have been dealing with the whole 8.03. This is actually really remarkable that we can come from single object oscillation, putting it all together, making it continuous, then this equation really popped out. And this equation really describes multiple systems. Then we went ahead to actually discuss the property of the wave equation. It looks like this. Basically, I replaced the t over rho L by v, p squared. By now, you know the meaning of v, p is actually the phase velocity. And we discussed two kinds of solutions, special kinds of solutions. The first kind is normal modes. The second one is progressive wave solution, or traveling wave solution, whatever name you want to call it. Let's take a look at the normal modes, what have we learned. So if you have a bound system, a bound continuous system, the normal mode is your distending waves for the wave equation we discussed. And basically, the functional form is A, m, sine, k, m, x plus alpha, m and sine, omega, m, t plus beta, m. So what we actually learned from the previous lecture is the following. So basically, you can decide the k, m and alpha, m by just boundary conditions. So before you introduce boundary conditions, which are the conditions allow you to describe multiple nearby systems consistently. So that is the meaning of boundary condition. Before you introduce that, k, m and alpha, m are arbitrary numbers. Whatever number you choose is the-- can satisfy the wave equation. But after you introduce the boundary condition, you figure that out from the problem you are given, then k, m and alpha, m cannot be arbitrary anymore. And they usually become discrete numbers. OK. So that is what we learned from the previous lectures. And finally, we also see that omega, m is determined by the property of the system, by a so-called dispersion relation. In this case, it's linear, it's proportional to k, m, because we are talking about non-dispersive medium for the moment. And we have this beta, m, which is related to the initial condition. And the a, m, which can be determined by a Fourier decomposition. So if you are not familiar with this, you have to really review how to do Fourier decomposition. I know most of did very well on the midterm, but maybe some of you forgot how to determine a, m and it will be very, very important to review that for the preparation for the final. Now the second set of solutions is the following. So you have progress-- progressing waves. And the functional form is really interesting. So you can see this can be written as F, F is some arbitrary function, x plus-minus v, p, t. Basically that is that you're describing a wave which is traveling to the positive-- to a negative or positive direction in the x direction. Or you can actually write it down as G function k, x plus-minus omega, t. Actually, they all work for wave equations. Now we went ahead and applied approach which we learned from the general solution of wave equation to massive strings, and we discussed about sound waves. For the sound waves, it will be important to review what are the boundary conditions for the displacement of the molecules in the sound wave, compare that to the pressure deviation from the room pressure. So I think it's important to make sure that you understand the difference between these two, what are the boundary conditions and basically it should be very similar to the solution-- the boundary condition for the massive strings. And we also talked about electromagnetic waves. And that is another topic which you will really have to review. Several things which are especially interesting is that an electric field cannot be without a magnetic field. They are always together, no matter what. So if you have trouble with the electric field, then there must be trouble in the magnetic field. And that is governed by the Maxwell's equation. Before we go into the detail of those, we also discussed about dispersive medium. So in the case of dispersive medium, we used a special kind of example, which is strings with stiffness. So basically, what we found is that if you have a certain kind of wave equation, like this one, I am writing this one here. Basically, if I add the additional term to describe the stiffness, then what is going to happen is that the dispersion relation, when I ask you to plot the dispersion relation, you will be-- I am requesting you to find the relation between omega and K. And I'm going over this in more detail because I see so many similar mistakes on the midterm. So basically what I'm asking is omega versus K. And in the-- if we don't have this turn, then basically, you have a straight line. Straight line means you have a non-dispersive medium. And if you add this turn, you need to know how to evaluate the dispersion relation. The quickest way to evaluate the dispersion relation is to just simply plug in the progresssing wave solution for the G function or harmonic progressing wave solution, find omega-- K, x plus-minus omega t, into this equation, then you will be able to figure out the dispersion relation. And what we figure out is the following. If we include stiffness, then you can see that the dispersion relation is not a line anymore and is actually some kind of curve, and the slope is actually changing. And there are dramatic consequence from this thing. That means if I have a traveling wave with different wavelengths, that means the phase velocity v, p equal to omega over K is going to be different for waves with different frequencies, or different wavelengths. So that is how you clear the problem. Because if I have initially produced a signal which is a triangle and I let it propagate, what is going to happen is that the slow component will be lagging behind. Those are the slow components. And the fast components will go ahead of the nominal speed. So there will be a spread of the signal. Originally, maybe you have some kind of a square wave, and this thing will become something which is actually smeared out in space, and then you lose the information. And we are going to talk about that later. And we also learned about group velocity. So what is your group velocity? Group velocity v, p-- oh, sorry, v, g is actually partial omega, partial K, which is the slope of a tangential line here. And where the phase velocity is connecting this point to that point, and the slope of this line is the phase velocity, and the slope of the line cutting through this point, which is giving you the group velocity. And we actually learned the definition of-- the consequence of group velocity and phase velocity by introducing you a bit phenomena. Basically, we add two waves with similar wavelengths, or wave numbers. Basically, what we see is the following. So basically, you see some behavior like this. We see this-- the superposition of these two waves which produce a bit phenomena can be understood by something which is oscillating really fast modulated by a much slower more variating envelope. Basically, you can actually understand the bit phenomena by actually identifying these two interesting structures. And the speed of all those little peaks is traveling at phase velocity. And the speed of the envelope is found to be traveling at group velocity. So that is what we have learned. And we can have group velocity and the phase velocity traveling in the same direction. And we can also have a negative group velocity. So that is a technique which is really, really very difficult. And I'm still trying to practice and make sure they I can demo that in 8.03. Basically, it's like the whole system, the whole detailed structure moving in a positive direction. But the body, or say the envelope, is actually moving in the negative x direction. So that is also possible. And you can actually construct a system which has a negative group velocity. So once we have done that, we also tried to understand further the description of the solution for the dispersive medium. So basically, what we actually went over during the class is that OK, now, if the f function f of t is describing Yen-Jie's hand, and I'm holding an infinitely long string and I shake it as a function of time, and that essentially, this motion, is actually described by this f function. What we know is that this oscillation, OK, I can do one, but I won't, but all kinds of f functions can be described as superpositions of many, many, many waves with different angular frequencies. So that's a miracle which we borrowed from the math department again. And you can see that f function can be written as the sum of all kinds of different waves with different angular frequencies with population c omega. This is the weight which makes that become the f function. And we can figure out the c omega by doing a Fourier transform. And finally, what will be the resulting wave function, psi, x, t, which is the wave function generated by the oscillation of my hand. And those are governed by the wave equation, which gives you the relation between omega and the k can be returned in that functional form. So the good news is that with the help of Fourier transform, we can also describe and predict what is going to happen no matter if this system is dispersive or not dispersive using this approach. OK. So that is really cool. And you can of course can do a cross-check just to-- assuring that this is a non-dispersive medium. And you are also going to get back to what you should expect the solution to non-dispersive medium for the psi, x, t. So that is one thing which is really remarkable. And I think what is needed to know is not a deviation of all those formulas, but how the plotting and the derived c omega by using the formula you are given and how to then put together all the solutions and it becomes the resulting solution for the psi, x, t, which is really the solution we really care. So for that, you need to know how to do the integration. You need to know how to derive the dispersion relation. Then one thing left over is to put the problem into that equation, which is also given to you in a formula. And we will not ask you to do a very, very complicated integration for sure on the final. So what is the consequence? Basically, one thing which is interesting to know is that if you have a wave in a coordinate space, which is really widely spread out, and you can do a Fourier transform to get the wave population in the frequency space, what we find is that when this wave is really, really wide in the space, then what we find is that the wave population in the frequency space is very narrow by using a Fourier transform. And that just gives you the result. And on the other hand, if you have a really-- a very narrow pulse in the coordinate space, for example, I do this-- shwhew --very, very-- really quickly. I create a very narrow pulse. And then what is actually happening is that I will have to use a very wide range of frequency space to describe this very narrow pulse. So that leads to-- direct consequences of that is uncertainty principle. And this is closely connected to the uncertainty principle we talk about in quantum mechanics. Delta, p times delta, x greater or equal to h bar over 2. All right. So we have done with the one-dimensional case. And we also talked about a two-dimensional and a three-dimensional case. And this is the example of two-dimensional membranes, and they actually are constrained so that their boundary condition at-- the boundary is equal-- no, the wave function is equal to zero. And you can identify all those normal modes. And we went ahead also to talk about geometrical optics laws. Basically, how we derive that is to have a plane wave. First, you have a plane wave propagating toward the boundary of two different mediums, and we were wondering well, what is the refracted wave and the transmitting wave. By using the-- by making sure just one point, which is that the membranes don't break, the wave function is continuous at this boundary. That's the only assumption which you use. We went through the mathematics, which you don't really need to remember all of them. But you really need to remember the consequence. The consequence is the following. Basically, what we see is that if you have incident plane wave with incident angle theta 1, the refractive wave will be having an angle of theta 1 as well. So that's the first law of refraction, refraction law. And then the second one which we learned is that the transmitting wave will satisfy Snell's law, n, 1, sine, theta, 1 equal to n, 2, sine, theta, 2. And that is very interesting because this, Snell's law has also nothing to do with Maxwell's equation. You see? Right? That's actually what you can learn from here. We usually use electromagnetic waves to demonstrate Snell's law. But from 8.03, we learned that it has nothing to do with Maxwell's equation. It applies to all kinds of different systems, which you can-- which can be described by wave functions. So that is actually the very important consequence. But on the other hand, as we all discussed later, the relative amplitude of the incident wave, refracted wave, and transmitting wave, the relative amplitude is governed by Maxwell's equation. So I would like to make that really crystal clear. So the relative amplitude is governed by really the physical laws, which actually governs the propagation of those plane waves. OK. So I think we can take a five minute break to have some air. And of course, you can-- you are welcome to continue to use all this juice and coffee. And coming back at 38. OK. So welcome back, everybody, from the break. AUDIENCE: [INAUDIBLE] YEN-JIE LEE: So we are going to continue the discussion. We have learned about the two important laws for the geometrical optics. And we also went ahead to discuss the polarization that's solved in greater detail. So for example, we can have linear depolarized wave. So basically, the wave is essentially moving up and down, up and down. But the direction of the background field doesn't change. It's always, for example, initially, if it's in x direction, then it is x direction forever. And in that case, I call it linearly polarized. Of course, I can also have the case that I can have a superposition of two waves. One is having the electric field in the x direction. And the other one is in the y direction. And they are off by a phase of pi over 2. If that happens, then basically, you will see that it produces something really interesting. That direction of the electric field is going to be rotating as a function of time-- as a function of the space these waves travel. And we call it circularly polarized waves. And we can also have elliptically polarized wave. Then we learned about how to do a filtering, which is the polarizer. So suppose I have a perfect conductor here, where I have the easy axis, which is described by the green arrow there. And you can see that easy axis means that if you have electric field parallel to the easy axis, and then since that's the easy axis, so it is supposed to be easy, therefore, this electric field is going to be passing through the polarizer. On the other hand, if the electric field is perpendicular to the direction of the easy axis, that means it's taking the perfect conductor in the hard way. Therefore, when it pass through-- when it is trying to pass through with the perfect conductor, the electrons in those conductors are going to be working like crazy to deflect this wave when the direction of the electric field is perpendicular to the direction of the easy axis. So that is how this works. For example, in the first example, you can see that in this case, you have an easy axis which is perpendicular to the direction of the electric field, which is the red field, then this wave actually got refracted. There will be no transmission-- sorry, no electromagnetic field passing the perfect conductor. And on the other hand, if you have another perfect conductor, in which you have easy axis which is parallel to the electric field, then you can-- you will see that it will pass through the perfect conductor. So that is the polarizer. And also, we discussed about quarter-weight plate, which I would suggest you to have a review about the concept which we have learned about polarizer and quarter-wave plate so that you make sure that you understand how to calculate the electric field after passing through a polarizer and quarter-wave plate and how the secondary, or the elliptically depolarized waves are created using all those wave plates, et cetera. All right. So the next thing which we discussed during the class is how do we produce electromagnetic waves. I think by now, you should know that a stationary charge doesn't produce electromagnetic waves. Even a moving charge at constant speed doesn't create electromagnetic waves. So how do we create an electromagnetic wave which propagates to the edge of the universe? That is-- the trick is to create a kink in the fuel line. So you have to accelerate and stop it. Accelerate and then try to actually stop the acceleration. So then you can create a kink. And this kink is going to be propagating out of the-- as a function of time. And this kink is creating the so-called radiation from this accelerated charge. So you don't really need to remember all the deviations, but you really need to know the conclusion. So what is the conclusion is the following. The radiated electric field is equal to minus-- very important that there's a minus sign in front of it, which is a common mistake to drop it, and the q is the charge of the oscillating-- the accelerated charge, proportional to the charge. If the particle is more charged, then you have more radiation. Aperp is the acceleration projected to, which is-- the perpendicular projection of the acceleration of the particle with respect to the direction of propagation is so-called the Aperp. And only the perpendicular direction acceleration counts. The one which is parallel to the direction of propagation doesn't really count, as you can see from this equation. And the t prime what is t prime? t prime is t minus r divided by c. So t prime is the retarded time, so that is telling you that it takes some time for the information to propagate from the origin, which is the position of the moving charge to the observer, which is r, this distance, away from the moving charge. So the information takes some time to propagate, and you cannot know what is really happening, for example, 100 light years away from Earth. You have no idea about what is happening. Maybe a black hole is created there and is going to suck everybody up in a few years. But nobody knows, and we don't care because we cannot control it. All right so that is very important. And also very important to know the magnetic field must be there. You can see the relation between magnetic field and the electric field. And the Poynting vector is also its joint field. And when we went ahead, given all the knowledge we have learned, we discussed about how to take very beautiful photos using a polarizer filter. And we discussed about how to filter out the scattered light from the sun. And it would be nice to figure out why this is the case, how these polarizer lines, scatter lines are created. It's purely geometrical. And also, we discussed about Brewster's angle and also how it leads to the explanation of the filtering of the light, the refracted light from the, for example, window of a car or from the water. And this is the demonstration of-- the summary of Brewster's angle. So somebody reminded me that the amplitude should be given. So I think, this is the amplitude formula for Brewster's angle will be given to you. If not, it's asked in the final exam. So don't be worried about it, and you don't have to remember this formula. And I'm not going to ask you to derive that just in such a short time, the three hours in the final exam. But what is very important is to know how this Brewster's angle, why there's no refracted light polarizing in a way that the polarization should be-- why the refracted light is polarized, for example. And also why the transmitting wave is slightly polarized. And I think the conclusions you need to remember, and you need to know how to calculate the angle, at least. Because for this purely polarized light to be produced in a refracted light, you need to have normal angle between the direction of the refracted light and the direction of the transmitted light. And that, you should be able to remember. And you should be able to derive that also from your mind as well, because that means the direction of the oscillation of the molecule at the boundary will be in the direction of propagation of the refracted wave. Therefore, that cannot be the solution to the progressing electromagnetic wave. Therefore, the refracted waves are polarized. So if you follow this logic, then you don't really need to memorize all those formulas. All right. So finally, in the last part of the course, we focused on the superposition of many, many electromagnetic waves so you can produce constructive interference. Or that means all those waves are in phase. And you can have destructive interference when they are out of phase. And that is a very important topic, so you should review that for the preparation of the final. And you can see that there are three concrete examples which we used during the class. A laser beam. We talked about a water ripple in a demo. And we also studied how it make use of this phenomena to design a phased radar. So to detect this unknown object in the sky, what we really need to have is electromagnetic waves pointing to a specific direction. And that can be achieved by using multi-slit interference. And this is the property of the two-slit interference pattern. And you are going to have many, many peaks. They have equal height for two-slit interference If you ignore any effect coming from diffraction. So we've assume that the slit is infinitely small. The slit is super narrow. And then we can ignore the diffraction-- single-slit diffraction. In fact, then all the peaks due to this two-slit interference will have the same height. On the other hand, when we start to increase the number of slits, for example, unequal to 3, unequal to 4, unequal to 5, unequal to 6, as you can see that, the structure of the intensity as a function of delta, which is the phase difference, is actually changing. And you can see that the general structure is the following. So if you have unequal to 3, then basically, you have 2 of adult, and between them, you have 1 child. And if you have unequal to 6, then basically, you have 2 adults and somehow there are 4 children in this collection. So basically, that is what we learned from the solution of the multi-slit interference. And in this way, we can actually make the width of the principal maxima as narrow as you want. So that is why phased radar works. And then we discussed about diffraction. So that is related, again, to the explanation of laser beams. And we discussed about the design of a Star Trek ship, the gun for the ship. And we also talked about resolution. And what is actually happening here is the following. A single-slit diffraction essentially can be viewed as an infinite number of source interference. And you just need to integrate over all the point-like sources between the two walls. And all of them are acting like a spherical wave source. So basically, for every point-- continuously, every point between these two walls are a point source of spherical waves. And that is Huygens' principle. And we can see that the structures is-- of the intensity as a function of position is the following. So basically, you have a principal maxima, which is a peak in the middle. And at some angle, basically, you have destructive interference such that if you integrate over all the contributions from an infinite number of sources in this window, basically, you would see that they completely cancel each other. So that is the origin of all those deep structure minima. And then, after the minima, actually, you will see another peak, but the height of the peak is suppressed by 1 over beta squared. And it would be good to review that. And what is the consequence? So if you shoot a laser beam to the moon, the size of the laser beam will be very large. After you learn 8.03, you know that the size of the laser beam is going to be very, very large due to interference between all the point-like sources from the laser beam. And finally, we can put them all together. So the single-slit diffraction and the multi-slit interference, you can put them all together, and basically, what you get is the following. So basically, you have a multi-slit interference pattern, which is showing there. But now the intensity of the multi-slit pattern is modulated by the single-slit diffraction pattern. And of course, the full formula will be given to you. But on the other hand, you are also requested to know how to calculate, just to add the contribution from multi-slit together in case if we change the amplitude of the incident light or we change the phase, like what we did in the homework. And I think that is one important point, and you should review that. And if you are not sure about how to proceed with that, it would be good to review Lecture 22, Lecture 23. So finally, we talk about the connection to quantum mechanics. Einstein already told us that "I have said so many times, God doesn't play dice with the world." But what we actually find is that there are two very interesting things which we found. The first thing is that if we have a single photon source, and basically, if we don't play dice, we cannot explain the intensity of the-- after this single photon source passes through two polarizers. And what happens is the following. Basically, the result of a single photon source tells you that you really need to play dice so that you can get the resulting polarized light intensity. And also, the second pseudo-experiment we discussed is that if you have billiard balls, basically, you have them pass through the two-slit experiment, what you are going to get is two piles, Gaussian-like distribution. And if you have a single electron source, what it does is that it interferes with itself. An electron, a single electron, can interfere with itself and produce a pattern which is very similar to what we see in the double-slit interference pattern. So that is really remarkable. And also, we talked about a single-slit-- single electron experiment. That gives you also a diffraction pattern. We have to use the wave function to describe the position-- the probability density of the position of the electron on the screen. And know this issue closely connected to the uncertainty principle, which we discussed earlier, delta, p, delta, x is greater than or equal to h bar over 2. So if you have a very narrow window, that means you have very similar delta x, so you have very, very good confidence about the location of the electron. And then the momentum is in the x-- in the momentum in the x direction, you have large uncertainty, according to this equation. And that can be seen from this single-slit diffraction pattern and it is closely connected to what we have learned before. So where is this-- how to actually describe what this is really the dispersion relation of the probability density wave is actually coming from Schrodinger's Equation. And this is given here. We briefly talked about that. And the consequence is the following. You can describe the evolution of the wave function as a function of time by using this wave equation. And this wave equation is slightly different from what we have learned before. And we also can use what we have learned from 8.03 to solve a particle in a box problem, which is covered in lecture number 23. And I just wanted to say that you need to know the general principle, but I'm not teaching 8.04, so I'm not expecting you to solve a quantum mechanics problem. But I would like to say that OK, from this point, it's motivating you to take 8.04, right? Because there can be a lot of fun there as well. And it is closely related to what we have learned from 8.03. So I just want to say, the last point is that this is really not the end of the vibrations and waves. It's just the beginning. And that there is a path toward the peak. And it may take a long time to reach the peak. All right. And I would like to let you know that I'm really, really very happy to be your lecturer this semester. And I really enjoyed teaching this class and getting your responses when I asked questions. Thank you for the support. And I would like to say good luck with the final exam. And we have 800 contributions on Piazza, many thanks to Yinan, who is actually doing all the hard work, day and night. And thank you very much, and see you around MIT in the future. Thank you.
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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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Today is my last class with you. Awe, I'm sorry, too. You guys are a lot of fun. This has actually been the most interactive 7. 1 I've ever had. Usually there are a couple of people who perk up and say things, but you guys are great because all sorts of people are willing to contribute. So, I've had a wonderful time and it certainly seems like you guys have learned a lot. What I'd like to do for my last lecture is pick up again a little bit like I did with genomics and try to give you a sense of where things are going. I always like doing this because I get to talk about things that are in none of the textbooks that, well, I mean, it's just stuff that many people working in the field don't necessarily know. And that's what's so much fun about teaching introductory biology is because it only takes a semester for you guys to get up to the point of at least being able to understand what's getting done on the cutting-edge. Even if you might not yet be able to go off and practice it, you might need a little more experience for that, but you'd be surprised, it's not that much more. Take maybe Project Lab and you'll be able to start doing it already. It's really wonderful that it's possible to grasp what's going on. And, in many ways, you guys may have an advantage in grasping what's going on because, as I've already hinted, biology's undergoing this remarkable transformation from being a purely laboratory-based science where each individual works on his or her own project to being an information-based science that involves an integration of vast amounts of data across the whole world and trying to learn things from this tremendous dataset. And, in that sense, I think the new students coming into the field have a distinct advantage over those who have been in it. And certainly the students who know mathematical and physical and chemical and other sorts of things, and aren't scared to write computer code when they need to write computer code have a really great advantage. So, anyway, all that by way of introduction. I want to talk about two subjects today of great interest to me. One is DNA variation and one is RNA variation. The variation of DNA sequence between individuals within a population, and in particular our population, and the other is RNA variation, the variation in RNA expression between different cell types, different tissues. And the work I'm going to talk about today is work that I, and my colleagues, have all been involved in. And it's stuff I know and love. So, feel free to ask questions about it. I may know the answers, but what's reasonably fun about these lectures is if I don't know the answers it's probably the case that the answers aren't known. So, that's good fun because it's stuff I really do know well, and I love. So, anyway, here's some DNA sequence. It's pretty boring. This is a chunk of sequence from, let's say, the human genome. How much does this differ between any two individuals? If I were to sequence any two chromosomes, any two copies of the chromosome from an individual in this class or two individuals on this planet, how much would they differ? The answer is that much. That's the average amount of difference between any two people on this planet. Not a lot. If you counted up, it is on average one nucleotide difference out of 1, 00 nucleotides on average, somewhat less than one part in 1, 00 or better than 99.9% identity between any two individuals. Now, that is a very small amount, not just in absolute terms, 99.9% identity is a lot, but in comparative terms with other species. If I take two chimpanzees in Africa, on average they will differ by about twice as much as any two random humans. And if I take two orangutans in Southeast Asia, they will on average differ by about eight times as much as any two humans on this planet. You guys think the orangutans all look the same. They think you all look the same, and they're right. So, why is this? Why are humans amongst mammalian species relatively limited in the amount of variation? Well, it's a direct result of our population history. It turns out that the amount of variation that can be sustained in a population depends on two things. At equilibrium, if population has constant size N for a very long time and a certain mutation rate, Mu, you can just write a piece of arithmetic that says, well, mutations are always arising due to new mutations in the population and mutations are being lost by genetic drift, just by random sampling from generation to generation. And those two processes, the creation of new mutations and the loss of mutations just due to random sampling in each generation, sets up an equilibrium, and the equilibrium defines an equation there, Pi equals one over one plus four and Mu reciprocal which equation you have no need to memorize whatsoever and possibly even no need to write down. The important point is the concept, that if you know the number of organisms in the population and you know the mutation rate, those set up the bounds of mutation and drift, and you can write down how polymorphic, how heterozygous random individuals should be at equilibrium. That is if the population has been at size N for a very long time. Well, the expected amount of heterozygosity for the human population -- Sorry. For a population of size 10, 00 would be about one nucleotide in 1300. We have exactly the amount of heterozygosity you would expect for a population of about 10, 00 individuals. Yeah, but wait, we're not a population of 10,000 individuals. Why do we have the heterozygosity you would expect from a population of 10, 00 individuals? We're six billion. It's a reflection of our history. Because remember I said that was the statement about what the population heterozygosity should be at equilibrium? We haven't been six billion people except very recently. The human population has undergone an exponential expansion. It used to be a relatively small size, and then it very recently underwent this huge exponential expansion. If you actually write down the equations, the amount of variation in our population was determined by that constant size for a very long time. And then a rapid exponential expansion that's basically taken place in a mere 3, 00 generations, it's much too rapid to have any affect on the real variation in our population. What do I mean by that? What's the mutation rate per nucleotide in the human genome? It's on the order of two times ten to the minus eighth per generation. In a mere 3,000 generations, a tiny mutation rate like two times ten to the minus eighth is not going to be able to build up much more variation. So you might as well ignore the last 100,000 years or so. They're irrelevant to how much variation we have. The variation we have was set by our ancestral population size. Now, don't get me wrong. Eventually it will equilibrate. A couple million years from now we will have a much higher variation in the human population as a function of our size, but the population variation we have today is set by the fact that humans derive from a founding population of about 10, 00 individuals or so. And that means that the variation that you see in the human population is mostly ancestral variations, the variation that we all walked around with in Africa. And, in fact, that makes a prediction. That would say that if most of the variation in the human population is from the ancestral African founding population then if I go to any two villages around this world, in Japan or in Sweden or in Nigeria, the variance that I see will largely be identical. And that prediction has been well satisfied. Because when you go and look and you collect variation in Japan or Sweden or Africa and you compare it, 90% of the variance are common across the entire world. Most variation is common ancestral variation around the world, and only a minority of the variance are new local mutations restricted to individual populations. This is so contrary to what people think because there's a natural tendency to kind of xenophobia, to imagine that world populations are very different in their genetic background. But, in point of fact, they're extremely similar. So, anyway, there's a limited amount of variation. That's why we have such little variation in the human population. Now, that variation, humans have a low rate of genetic variation. Most of the variance that are out there are due to common genetic variance, not rare variance. If I take your genome and I find a site of genetic variation at the point of heterozygosity in your genome, what's the probability that somebody else in this class also is heterozygous for that spot? It turns out that the odds are about 95% that someone else in this class will also share that variance. So that the variance are not mostly rare, they're mostly common. And it turns out that some of this common variation, that is most of this variation is likely to be important in the risk of human genetic diseases. So human geneticists have gotten very excited about the following paradigm. If there's only a limited amount of genetic variation in the human population, actually, if you do the arithmetic, there are only about ten million sites of common variation in the human population, where common might be defined as more than about 1% in the population. There are only ten million sites. Folks are saying, well, why not enumerate them all? Let's just know them all, and then let's test each one for its risk of, say, confirming susceptibility of diabetes or heart disease or whatever? After all, ten million is not as big a number as it used to be. We now have the whole sequence of the human genome. Why not layer on the sequence of the human genome all common human genetic polymorphism? Now, that's a fairly outrageous idea but could be a very useful one. Some of these variance are important, by the way. We know that there are two nucleotides that vary in the gene apolipoprotein E on chromosome number 19. Apolipoprotein E is also an apolipoprotein like we talked about before with familiar hypercholesterolemia. But, in fact, it turns out that apolipoprotein E is expressed in the brain. And it turns out, amongst other tissues, that it comes in three variances, the spelling T-T, T-C and C-C at those two particular spots. And if you happen to be homozygous for the E4 variant, homozygous for the E4 variant, you have about a 60% to 70% lifetime risk of Alzheimer's disease. In this class 13 of you are homozygous for E4 and have a high lifetime risk of Alzheimer's. And it would be fairly trivial to go across the street to anybody's lab and test that. Now, I don't particular recommend it, and I haven't tested myself for this variant because there happens to be no particular therapy available today to delay the onset of Alzheimer's disease. And, therefore, I don't recommend finding out about that. But a number of pharmaceutical companies, knowing that this is a very important gene in the pathogenesis of Alzheimer's disease, are working on drugs to try to delay the pathogenesis using this information. And it may be the case that five or ten years from now people will begin to offer drugs that will delay the onset of Alzheimer's disease by delaying the interaction of apolipoprotein E with a target protein called towe, etc. So, this is an example of where a common variant in the population points us to the basis of a common disease and has important therapeutic implications. There are some other ones, for example. 5% of you carry a particular variant in your factor 5 gene which is the clotting cascade. It's called the leiden variant. Those 5% of you are going to account for 50% of the admissions to emergency rooms for deep venous clots, for example. The much higher risk of deep venous clots. And, in particular, there are significant issues if you have that variant and you are a woman with taking birth control pills. Some of you were at higher risk for diabetes, type 2 adult onset diabetes. There's a particular variant in the population that increased your risk for type 2 diabetes by about 30%. 85% of you have the high-risk factor, so you might as well figure you do. 15% of you have a lower risk, et cetera. And one I'm particularly interested in here, it turns out that HIV virus gets into cells with a co-receptor encoded by a gene called CCR5. Well, it turns out that if we go across the European population, 10% of all chromosomes of European ancestry have a deletion within the CCR5 gene. If 10% of all chromosomes have that deletion then 10% times 10%, 1% of all individuals are homozygous for that deletion. Those individuals are essentially immune to infection from HIV. They are not susceptible. It's not through immunity, it's through lack of a receptor. Yes? You certainly can. It's not hard. It's a specific known variant. You could test for it. Absolutely. Now, of course, that only helps the 1% of people who have that variant. But what it did do was point to the pharmaceutical industry that the interaction between the virus and that variant is essential. And now companies are developing drugs to block the interaction with that particular protein. And that tells you that it's an important protein. Yes? Over the whole world? I just specified European population for that one. That one, interestingly, is not found at as high a frequency outside of Europe, and no one knows why, whether that might have been due to an ancient selective event or a genetic drift. By contrast, the apolipoprotein E variant, at that frequency of about 3% of people being homozygous and being at risk for Alzheimer's, is about the same frequency everywhere in the world. So, there's a little bit of population variation in frequency. Now, the HIV variant is found elsewhere but at considerably lower frequencies there. And that's an interesting question as to what causes that variation. So the notion would be, I've given you a couple of interesting examples, but, look, there's only ten million variants. Just write them all down. Make one big Excel spreadsheet with ten million variants along the top and all the diseases along the rows, and let's just fill in the matrix and then we'll really, you know, this is the way people think in a post-genomic era. Now, could you do something like that? You would have to enumerate all of the single nucleotide polymorphisms, or SNPs we call them, single nucleotide polymorphisms. Now, to give you an idea of the magnitude of this problem, as recently as 1998, the number of SNPs that were known in the human genome was a couple hundred. But then a project has taken off. In 1998 an initial SNP map of the human genome was built here at MIT that had about 4, 00 of these variants. Then within the next year or so an international consortium was organized here and elsewhere to begin to collect more of these genetic variants. The goal was going to be to find 300, 00 of them within a period of two years. In fact, that goal was blown away and within three years two million of the SNPs in the human population were found. And as of today, if you go on the Web, you'll find the database with about 7.8 million of the roughly ten million SNPs in the human population already known. Now, that isn't all ten million. And it takes a while to collect the last ones, you know, collecting the last ones are hard, but we're already the hump of knowing the majority of common variation in the human population. Not just a sequence of the genome, but a database that already contains more than half of all common variation in the population. So, we could start building that Excel spreadsheet. Now, it turns out that it's even a little bit better than that because if we look at many chromosomes in the population, here are chromosomes in the population, it turns out that the common variance on each of those chromosomes tend to be correlated with each other. If I know your genotype at one variant, like over at this locus, I know your genotype at the next locus with reasonably high probability. There's a lot of local correlation. So, instead of looking like a scattered picture like that, it's more like this. If I know that you're red, red, red you're probably red, red, red over here. In other words, these variations occur in blocks that we called haplotypes. Here's real data. Across 111 kilobases of DNA there's a bunch of variants, but it turns out that the variants come in two basic flavors. 98% of all chromosomes are either this, this, this, this, this or this, this, this, this, this. Then there tends to be sites of recombination that are actually hotspots of recombination where most of the recombination of the population is concentrated. And you get a couple of possibilities here. So, the human genome can kind of be broken up into these haplotypes. Blocks that might be 20, 30, 40, sometimes 100 kilobases long in which within the block you tend to have a small number of haplotypes, or flavors as you might think of them, that define most of the chromosomes in the population. So, in fact, I don't actually need to know all the variants. If they're so well correlated within a block, if I knew this block structure I would be able to pick a small number of SNPs that would serve as a proxy for that entire block of inheritance in the population. So, what you might want to do is determine that entire haplotype block structure of hwo they're related to each other, and pick out tag snips. And it turns out that in theory, a mere 300,000 or so of them would suffice to proxy for most of the genome. So, you might want to declare an international project, and international haplotype map project to create a haplotype map of the human genome. And indeed, such a project was declared about a year and a half ago through some instigation of scientists and a number of places, including here. And this is $100 million project involving six different countries. And, it is already more than halfway done with the task, and it's very likely that by the middle of next year, we will have a pretty good haplotype map, not just knowing all the variation, but knowing the correlation between that variation, being able to break up the genome into these blocks. By the next time I teach 701, I should be able to show a haplotype map of the whole human genome already. That will allow you to start undertaking systematic studies of inheritance for different diseases across populations. And in fact, people are already doing things like that. Here's an example of a study done here at MIT like this, where to study inflammatory bowel disease, there was evidence that there might be a particular region of the genome that contained it, and haplotypes were determined across this, and blah, blah, blah, blah, blah, blah, blah. And this red haplotype here turns out to confer high risk, about a two and a half or higher risk of inflammatory bowel disease. And it sits over some genes involved in immune responses, certain cytokine genes and all that. And, things like this have been done for type 2 diabetes, schizophrenia, cardiovascular disease, just right now at the moment, a dozen or two examples. But I think we're set for an explosion in this kind of work. In addition, you can use this information to do things beyond medical genetics. You can use it for history and anthropology as well. It turns out rather interestingly, that since the human population originated in Africa and spread out from Africa all the way around the world arriving at different places in different times, you can trace those migrations by virtue of rare genetic variants that arose along the way, and let you, like a trail of break crumbs, see the migrations. So, for example, there are certain rare genetic variants that we can see in a South American Indian tribe, and we can actually see that they came along this route because we can see that residual of that. In fact, we can do things with this like take a look at Native American individuals and determine that they cluster into three distinct genetic groups that represent three distinct migrations over the land bridge. And, you can assign them to these different migrations. You can do this on the basis of mitochondrial genotype, etc. You can also, for example, determine when people talk about the out of Africa migration, there's now increasing evidence that there really were two, one that went this way over the land, and one that went this way following along the coast into southeast Asia. And, it looks like we're now beginning to get enough evidence of these two separate migrations by virtue of the genetic breadcrumbs that they have left along the way. So, it's really a very fascinating thing of how much you can reconstruct from looking at genetic variation, both the common variation that allows us to recognize medical risk, and the rare genetic variation that provides much more individual trails of things. None of this is perfect yet. There's lots to learn. But I think anthropologists are finding that the existing human population has a tremendous amount of its own history embedded in pattern of genetic variation across the world. You can do other things. I won't spend much time on this. Well, I'll take a moment on this, right? There's some very interesting work of a post-doctoral fellow here at MIT named Pardese Sebetti who has been trying to ask, can we see in the genetic variation in the population, signatures, patterns of ancient selection, or even recent selection in the human population? Now, hang onto your seats, because this will get just slightly tricky. But, hang on. It's only a couple of slides. Here was her idea. You see, when a mutation arises in the population, it usually dies out, right? Any new mutation just typically dies out. But, sometimes by chance it drifts up to a high frequency. Random events happen. But it usually takes a long time to do that. If some random mutation happens, and it happens to drift up to high frequency with no selection on it, then on average it takes a long time to do so. If you want, I could write a stochastic differential equation that would say that, but just take your gut feeling that if something has no selection on it and it's a rare event that'll drift up, when it drifts up it's kind of a slow process. It was a slow process. Then over the course of time that it took to drift to high frequency, a lot of genetic recombination would have had to have occurred many generations. And the correlation between the genotype at that spot and genotypes at other loci would break down. And there would only be short-range correlation. So, in other words, the amount of correlation between knowing the genotype here and the genotype here, maybe allele A here and a C here. That is an indication of time. It's a clock almost. It's like radioactive decay, right, that genetic recombination scrambles up the correlations. And, if something's old, the correlations go over short distances. But suppose that something happened. Some mutation happened that was very advantageous. Then, it would have risen to high frequency quickly because it was under selection. If it did so quickly, then the long-range correlations would not have had time to break down, and we'd have a smoking gun. A smoking gun would be that there would be a long-range correlation around that locus, much longer than you would expect across the genome. Things even out of this distance would show correlation with that, indicating that this was a recent event. So, we just measure across the genome, and look for this telltale sign of common variance that have very long range correlation that indicate that they're very recent. So, a plot of the allele frequency, common variance, sorry, if something has a common high frequency and long-range correlation, you wouldn't expect that by chance. So, something that was common in its frequency and had long-range correlation would be a signature of positive selection. So anyway, Pardise had this idea, and she tried it out with some interesting mutations, some mutations that confer resistance to malaria, one well-known mutation causing resistance to malaria called G6 PD and another one that she herself had proposed as a mutation causing resistance to malaria, variants in the CD4 ligand gene. And to make a long story short, both the known and her newly predicted variant showed this telltale property of having a high frequency and very long range correlation. Well that's very interesting because she was able to show that each of these mutations probably were the result of positive selection. But what you could do in principle is test every variant in the human genome this way: take any variant, look at its frequency, and compare it to the long range correlation around it, and test every single variant in the human population to see which ones might be the result of long range correlation. Now, when she proposed this, this was about a year and a half ago or two years ago, this was a pretty nutty idea because you would need all the variants in the human population, and you would need all this correlation information. But in fact, as I say, that information's almost upon us, and I believed that this experiment, this analysis to look for all strong positive selection in the human genome will in fact be done in the course of the next 12 months. So, I'm hoping by next year I can actually report on a genome-wide search for all the signatures of positive selection. Now, this doesn't detect all positive selection. It will detect sufficiently strong positive selection going back pretty much only over the 10, 00 years. When you do the arithmetic, that's how much power you have. Of course, 10,000 years has been a pretty interesting time for the human population, right? The time of civilization and population density, and infectious diseases, and all that, and I think we'll have an interesting window into the change in diet. All of that should come out of something like this. So, there's a lot of really cool information in DNA variation to be had. All right, that's one half. The other half of what I would like to talk about is totally different. It's not about inherited DNA variation. It's about somatic differences between tissues in RNA variation. So, let's shift gears. RNA variation: let me start by giving you an example here. These are cells from two different patients with acute leukemia. Can you spot the difference between these? Yep? More like bunches of grapes and all that. Yeah, it turns out that's just a reflection of the field of view you have if you move over to look like that. But I mean, that's good. It's just that it turns out that that isn't actually a distinction when you look at more fields. Anything else? Yep? White blood cells like different. They look broken. There's more of them in this field of view. But you look at 100 fields of view and it turns out that's not either. Well, the reason you're having trouble spotting any difference is that highly trained pathologists can't find any difference either. I generally agree there's no difference between these two if you look at enough fields of view. But you can convince yourself if you look that you see things there. But these actually are two very different kinds of leukemia. And, these patients have to be treated very differently. But, pathologists cannot determine which leukemia it is just by looking at the microscope, it turns out. This is the work of this man, Sydney Farber, namesake of the Dana Farber Cancer Institute here in Boston, who in the 1950s began noticing that patients with leukemias, some of them seemed different in the way they responded to a certain treatment, and he said, look, I think there's some underlying classification of these leukemias, but I can't get any reliable way to tell it in the microscope. And he put many years into working this out, first by noticing certain difference in enzymes in the cells, and then people noticed certain things in cell surface markers, and some chromosomal rearrangements. And nowadays, there are a bunch of test that can be done by a pathologist when a patient comes in with acute leukemia to determine whether they have AML or ALL. But it turns out that you can't do it by looking. You have to do some kind of immunohystochemical test of some sort in order to do that. So this is a triumph of diagnosis. After 40 years of work, we can now correctly classify patients as AML or ALL. And they get the appropriate treatment. And if they don't get the right treatment, they have a much higher chance of dying. And if they do get the right treatment, they have a much higher chance of living. So, this is great. There's only one problem with the story. It took 40 years, 40 years to sort this out. That's a long time. Couldn't we do better? Surely these cells know what they are. Surely we could just ask them if they are. Well, here's the idea. Suppose we could ask each cell, please tell us every gene that you have turned on, and the level to which you have that gene expressed. In other words, let us summarize each cell, each tumor by a description of its complete pattern of gene expression to 22,000 genes on the human genome. Let's write down the level of expression, X1 up to X22, 00 for each of the 22,000 genes of the genome. So, ever tumor becomes a point in 22, 00 dimensional space, right? Now clearly, if we had every tumor described as a point in 22, 00 dimensional space, we ought to be able to sort out which tumors are similar to each other, right? Well, it turns out you can do that now. These are gene chips, one of several technologies by which on a piece of glass are put little spots, each of which contains a piece of DNA, a unique DNA sequence. Actually, many copies of that DNA sequence are there. Each of these is a 25 base long DNA sequence, and I can design this so whatever DNA sequence you want is in each spot. The way that's done is with the same photolithographic techniques that are used to make microprocessors. People have worked out a chemistry where through a mask, you shine a light, photodeprotect certain pixels; the pixels that are photodeprotected you can chemically attach an A, then re-protect the surface. Use a light. Chemically photodeprotect certain spots. Wash on a C. And in this fashion, since you can randomly address the spots by light, and then chemically add bases to whatever spots are deprotected, you can simultaneously construct hundreds of thousands of spots each containing its own unique specified oligonucleotide sequence. And you can get them in little plastic chips. And then if you want, all you do is you take a tumor. You grind it up. You prepare RNA. You fluorescently label the RNA with some appropriate fluorescent dye. You squirt it into the chip. You wash it back and forth. You rock it back and forth, wash it out, and stick it in a laser scanner. And it'll see how much fluorescence is stuck to each spot. And bingo: you get a readout of the level of gene expression. I guess each spot, you should design it so that this spot has an oligonucleotide complementary to gene number one. And the next one, an oligonucleotide matching by Crick-Watson base pairing complementary to gene number two and gene number three. So, if I knew all the genes in the genome, I could make a detector spot for each gene in the genome. And of course we know essentially all the genes in the genome. So you can make those detector spots and you can buy them. So, you can now get a readout of all the, I mean, this is like so cool because when I started teaching 701, which wasn't that long ago because I ain't (sic) that old still, the way people did an analysis of gene expression is they used primitive technologies where they would analyze one gene at a time, certain things called northern blots and things like that, right? And, you know, you'd put in a lot of work and you get the expression level of a gene, whereas now you can get the expression of all the genes simultaneously, and it's pretty mind boggling that you can do that. How do you analyze data like that? So, we still use northern blots. It's true. So, every tumor becomes a vector, and we get a vector corresponding to each tumor. So, this line here is the first tumor, the second tumor, the third tumor, the fourth tumor. The columns here correspond to genes. There are 22, 00 columns in this matrix, and I've shown a certain subset of the columns because these genes here have the interesting property that they tend to be high red in the ALL tumors, and they tend to be low blue in the AML tumors, whereas these genes here have the opposite property. They tend to be low blue in the ALL tumors and high red in the AML tumors. These genes do a pretty good job of telling apart these tumors. So, here's a new tumor. Patient came in. We analyzed the RNA, squirted it on the chip. Can somebody classify that? Louder? AML. Next? Next? Congratulations, you're pathologists. Very good. That's right, you can do that. It works. And in fact, in the study that was done that was published about this, the computer was able to get it right 100% of the time. Not bad. So now you say, wait, wait, wait, but you're cheating. You're giving it a whole bunch of knowns. Once I have a whole bunch of knowns it's not so hard to classify a new tumor. What Sydney Farber did was he discovered in the first place that there existed two subtypes. Surely that's harder than classifying when you're given a bunch of knowns. And that's true. So, suppose instead, I didn't tell you in advance which were AML's and which were ALL's, and I just gave you vectors corresponding to a large number of tumors, do you think you would be able to sort out that they actually fell into two clusters? Could you by computer tell that there's one class and the other class? Turns out that you can. Now, I've made it a little easier by not listing most of the 22,000 columns here. But think about it. Every tumor is a point in 22, 00 dimensional space. If some of the tumors are similar, what can you say about those points in 22,000 dimensional space? They're going to be clumped together. They're near each other. So, just plot every tumor as a point in 22,000 dimensional space, and your question is, do the points tend to lie in two clumps up in 22, 00 dimensional space? And there's simple arithmetic you can learn using linear algebra to get some separating hyperplane and ask, do tumors lie on one side or the other? And, it turns out the procedures like that will quickly tell you that these tumors clump into two very clear clumps. They're not randomly distributed. And so, if you get these tumors, and you do gene expression on them and put the data into a computer, the amount of time it takes the computer to discover that there were actually two types of acute leukemia is about three seconds marked down from 40 years. That's good. So, you can reproduce the discovery of AML and ALL in three seconds. Now you know what the pathologists say about this. They say, oh, give me a break. It's shooting fish in a barrel. We know there was a distinction. Big deal that the computer can find the distinction. We knew that there was distinction there. I know the computer didn't know it and all that. Tell us something we don't know. That's a fair question. So it turns out that you can ask some more questions. You can say, suppose I take now just the ALL's. Are they a homogeneous class, or did they fall into two classes? It turns out that extending this work, folks here were able to show that we can further split that ALL class. There was a hint that you might be able to do so because there's some ALL patients who have disruptions of a gene called MLL. And this tends to be a little more common in infants, and tends to be associated with a poor prognosis. But it was really very unclear whether this was simply one of a zillion factoids about some leukemia patients, whether this was a fundamental distinction. So, what happened was folks took a lot of ALL patients, got their expression profiles, and lo and behold it turned out that ALL itself broke into two very different clusters. This is an artist's rendition of a 22,000 dimensional space. We can't afford a 22,000 dimensional projector here, so we're just using two dimensions. But, the two forms of ALL were quite distinct from each other, and so actually ALL itself should be split up into two classes, ALL plus and minus, or ALL one and two, or MLL and ALL. And it turns out that these forms are quite different. They have different outcomes and should be treated differently. It also turns out that a particularly good distinction between these two subtypes of ALL is found by looking at this particular gene called the flit-3 kinase. The flit-3 kinase gene, whatever that is, was of great interest because people know that they can make inhibitors against certain kinases. And so, it turned out that an inhibitor against flit-3 kinases, against this flit-3 kinase gene product. If you treat cells with that inhibitor, cells of this type die, and cells of this type are not affected. So in fact, there's a potential drug use of flit-3 kinases in the MLL class of these leukemias, and folks are trying some clinical trials now. So, not only did the analysis of the gene expression point to two important sub-types of leukemias, but the analysis of the gene expression even suggested potential targets for therapy. So, I'll give you a bunch more examples. I have a bunch more examples like that there. They are examples of taking lymphomas and showing that they can be split into two different categories, examples of taking breast cancers into several categories, colon cancers. Basically what's going on right now is an attempt to reclassify cancers based not on what they look like in the microscope, and based not on what organ in the body they affect, but based on, molecularly, what their description is, because the molecular description, as Bob talked to you about with CML and with Gleveck, turns out to be a tremendously powerful way of classifying cancers because you're able to see what is the molecular defect and can make a molecular targeted therapy. So, these sorts of tools are quite cool, and I've got to say, in the last year we've begun using these expression tools not just to classify cancers, but to classify drugs. We've begun an interesting and somewhat crazy project to take all the FDA approved drugs, put them onto cell types, and see what they do, that is, get a signature, a fingerprint, a gene expression description of the action of a drug. And then we hope, here's the nutty idea, that we can look up in the computer which drugs do which things and might be useful for which diseases, because we'd put the diseases and the drugs on an equal footing. All of them would be described in terms of their gene expression patterns. So, I'll tell you one interesting example, OK? This is an interesting enough example. I don't even have slides for it yet. It turns out that these patients with ALL that I've been talking about, some of the patients with ALL will respond to the drug dexamethasone. Some won't. If you take patients who respond to dexamethasone, and patients who are resistant to dexamethasone, and you get their gene expression patterns, you can ask are there some genes that explain the difference? And you can get a certain gene signature, a list of, say, a dozen or so genes that do a pretty good job of classifying who's sensitive and who's resistant. Then you can go to this database I was telling you about of the action of many drugs and say, do we see any drugs whose effect would be to produce a signature of sensitivity? If we found a drug X, which when we put it on cells turned on those genes that correlate with being sensitive to dexamethasone, you could hallucinate the following really happy possibility that when you added that drug together with dexamethasone, you might be able to treat resistant patients because that drug could make them sensitive to dexamethasone, and that you could find that drug just by looking it up in a computer database. So, we tried it and we hit a drug. There was a certain drug that came up on the screen, yes? That's very much in the idea too. We found a drug that produced the signature sensitivity, and tested it in vitro. In vitro, if you take cells that are resistant and you add dexamethasone, nothing happens because they're resistant. If you add drug X, nothing happens. But if you add both drug X plus dexamethasone, the cells drop dead. It's now going into clinical trials in human patients. It turns out drug X is already a well FDA approved drug, so it can be tested in human patients right away, so it's going to be tested. So, the gene expression pattern was able to tell us to use a drug which actually had nothing to do with cancer uses in a cancer setting because it might do something helpful. Now, what's the point of all this? We can turn up the lights because I think I'm going to stop the slides there. The point of all of this, which is what I've made again, and I will make again, because you are the generation that's going to really live this, is that biology is becoming information. Now, don't get me wrong. It's not stopping being biochemistry. It's going to be biochemistry. It's not stopping being molecular biology. It's not stopping any of the things it was before. 45:57 But it is also becoming information, that for the first time we're entering a world where we can collect vast amounts of information: all the genetic variants in a patient, all of the gene expression pattern in a cell, or all of the gene expression pattern induced by a drug, and that whatever question you're asking will be informed by being able to access that whole database. In no way does it decrease the role of the individual smart scientist working on his or her problem. To the contrary, the goal is to empower the individual smart scientist so that you have all of that information at your fingertips. There are databases scattered around the web that have sequences from different species, variations from the human population, all of these drug database, etc., etc., etc., etc. It's a time of tremendous ferment, a little bit of chaos. You talk to people in the field, they say, we're getting deluged by data. We're getting crushed by the amount of data. I don't' know what to do with all the data. There's only one solution for a field in that condition, and that is young scientists because the young scientists who come into the field are the ones who take for granted, of course we're going to have all these data. We love having all these data. This is just great, couldn't be happier to have all these data. We're not put off by it in the least. That's what's going on. That's what's so important about your generation, and that's why I think it's really important that even though it's 701 and we're supposed to be teaching you the basics, it's important that you see this stuff because this is the change that's going on, and we're counting on this very much to drive a revolution in health, a revolution in biomedical research, and we're counting on you guys very much to drive that revolution. It has been a pleasure to teach you this term. I hope many of you will stay in touch, and some of you will go into biology, and even those of you who don't will know lots about it and enjoy it. Thank you very much. [APPLAUSE]
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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: Let's get started. Settle down. Settle down. Welcome to 3.091. I'm Donald Sadoway and I'm going to be your instructor this semester. So what I want to do today is to introduce my plans. I've got plans for you. Plans for learning. I want to talk a little bit today about those plans, introduce myself, introduce my class curriculum-- the path forward. So let me begin by saying that 3.091 is the most important class you will take at MIT. It's true. But, you know, anybody who stands in front of you to lecture should say the same thing about his or her class. If they don't believe that they shouldn't be standing in front of you lecturing. The difference is when I say it, I'm right. [LAUGHTER] PROFESSOR: And it's not just an opinion, it's because of the content of 3.091. 3.091 is about chemistry. It's about the central science, but it's not just about any old chemistry. This isn't a class that you could take anywhere else in this country. This is the only place that you can take 3.091. 3.091 is solid state chemistry. Now why do we talk about solid state chemistry? Because engineering systems are made of solids. Now I know what you're thinking-- oh, he said solid state, he's going to talk about the chemistry that constitutes his laptop computer or the chemistry that constitutes this laser pointer. And that's true. We will talk about that chemistry. But we will also talk about soft matter. We as human beings are chemical machines. When this hand changes shape, it is a polymer that is changing conformality. These eyes are photodetectors, band gap of about two electron volts. They're not made of gallium nitride. They're made of organic compounds. Inside, what supports us, it's a ceramic skeleton. So solid state chemistry describes life science as well. Well, what we're going to learn in 3.091 is the rudiments. We're going to learn the rudiments. So what I'm going to do today is now go through some of the basic organization. So tomorrow you're going to meet your recitation instructors and get to know each other and get to know the recitation instructor. Today I'll say a few words about myself so you know who's standing in front of you. I graduated from the University of Toronto in chemical metallurgy, came down here in 1977 to spend a year as a postdoctoral fellow and, I guess, I lost track of time. Now what's my research? My research is in electrochemistry. Electrochemistry is the most important branch of chemistry. Do you notice some theme in my professional life? See, I have tenure. So what does that mean? It means you find your passion and pursue it. You don't waste time on trivia, all right? And that's what I urge you to do: Find your passion and pursue it. So what's my passion? My passion is electrochemistry in nonaqueous media. Anything but water. Let the rest of the world work on water. I work on molten salts, ionic liquids, and polymers. And what's the reason for this? There's an application. I'm in the School of Engineering. So I'm interested in environmentally sound technologies for metal production, all right? Right now looking at titanium, iron recycling as well. I'm also interested in electrochemistry as it applies to energy storage, energy storage for mobile power. See this gal here? She's got the cell phone. She's not even looking where she's driving. [LAUGHTER] PROFESSOR: So making safe batteries out of earth-abundant, accessible materials for portable power, ultimately to drive the car with electrons, electric fuel, to eliminate this country's dependence on imported petroleum. We can do it. How? By inventing. By inventing. And the way we're going to invent is to learn the lessons in 3.091 that will give us the chemistry we need to invent a battery that can send that car 250 miles on a single charge, and put it in a show room for the same price as a car with an internal combustion engine. The only thing that stands between that image and where we are today is invention, and the requisite material is right here in this class. We're also looking at colossal batteries. Store the grid. Enable renewables, wind, solar. And then, lastly, let's never forget about dreaming. So if we want to dream and imagine that man will return to the moon or maybe even go to Mars, we're going to have to be able to produce oxygen. We'll live like the pioneers, produce oxygen from local resources, produce structural metals, and even produce photovoltaic silicon so they can generate their own energy from local resources. And so electrochemistry is the key enabler for that as well. So now let's turn to the whole underpinning of 3.091. If we take a look at the performance of any engineering system it's a combination of the design and the construction. Now the construction is both the workmanship and the choice of materials. Now how do we choose materials? We choose them on the basis of their properties. So, for example, here's an application: a beverage container. This one I'm holding is made out of an aluminum alloy. It's a metal. When I was your age, they had steel beverage cans. You can take this same beverage and contain it in a glass bottle. You can contain it in a polymer bottle. Why do we make those various choices? Because they have the right mix of properties. Now how do you determine the properties? Properties obviously are determined by the composition. We wouldn't make this thing out of sodium chloride. It would dissolve. So composition is important, but atomic arrangement is important as well. Example is 1/2-inch-thick pine board compared to a 1/2-inch-thick piece of plywood. Both pine, but the 1/2-inch-thick pine board is one solid pine board. I can take that pine board and I can fracture it if I strike it along the grain. But the 1/2-inch-thick plywood, I can't do that because it's 1/8-inch sheet pine cross-laminated. 1/8 inch north-south, 1/8 inch east-west. If I try to cut through that board, I can't advance the fracture. Same composition, different atomic arrangement. So the thesis of 3.091 is that the electronic structure of the elements holds the key to understanding that relationship between composition, atomic arrangement, and properties. And once you have properties that's how you make your selection. And away we go. And that's how we got the syllabus of 3.091. So the syllabus has two major blocks. The first block is general principles of chemistry-- it's going to be about the first five weeks-- and that's the same stuff that you get here, 5.111, 5.112, the basics of electronic structure and bonding. And then we part company and in 3.091 we dig down into solid state chemistry. You don't have to take notes. You can relax. All this is going to be burned to PDF and posted at the website. So everything that you see in this class that goes up on the screen is archived for you. This lecture is being video recorded right now and within an hour will be posted at the website. So in the unlikely event that you can't quite come to class, it is available. [LAUGHTER] PROFESSOR: So now what I want to do is introduce some of the operational aspects of the class. And we had some handouts going around. If you didn't get them we can get you extras. But again, this is all posted at the website. So you know who I am. The key person in this operation is my administrative assistant, Hilary Sheldon, and she's just down the hall. You can roll a penny down the hall from here and get to my office. So if you tell me you tried looking for me, you couldn't find me, I know you couldn't have been trying very hard. The text is a two-volume set consisting of the text Chemistry: Principles, Patterns, and Applications by Averill and Eldredge, and then a second volume that consists of miscellaneous readings taken from other Prentice Hall texts. And it's identical in content to the blue book that was used last year and the year before. So if you get a copy of that blue book, you don't have to buy these. I don't work for Prentice Hall, I'm not trying to sell books, but we will take the homeworks out of the books to a great extent. And the page numbering is the same, so anything this year or the last two years will be fine. If you have to buy something, you'll have this. This one will come with a CD that will get you into a mastering chemistry, which is a sort of a tutorial, computerized system to help you with certain homework problems. The lectures, Monday, Wednesday, Friday, here in this room at 11 o'clock. The lecture starts at five minutes after the hour-- gives you five minutes to get in-- and then the lecture stops at five minutes to the next hour-- time for you to leave and then for the next class to arrive. And what I'm going to try to do is establish some plug flow here. So what I'm going to try is to have everybody leave by the exit to the north here to my right, and there's also an exit over the top back. So leave to your left and that way it'll be easier for the next class to come in. And maybe we can persuade the people here before us to do the same and then it'll be easy to change. I don't know why but there's this sort of-- everybody has to charge to this door. And I just stand here and watch people collide. And I don't know. Just interesting social experiment. And there are two doors. I don't understand why people do this, but I'm not in the social sciences. Recitations. Recitations will meet Tuesdays and Thursdays. So you'll go to the same hour on Tuesday and Thursday. The section should be roughly 20 students. And that's where the question and answer occurs. Here it's largely I talk, you listen. I've got time for a question like, hey, shouldn't that be a minus sign? I can take something really quick. But where you get to really interact with the instructor is in recitation. And I've given a direction to my recitation instructors not to give you a fourth and fifth lecture. You control the content of the recitation. So you have to come to the recitation prepared with questions. So they're supposed to walk in and say, good morning, good afternoon, what are your questions? And you can say, I didn't understand the last five minutes of the lecture yesterday. Would you do number five on the homework? Would you go over secondary bonding? That's the sort of thing that's supposed to happen in section. You've been assigned by the Registrar. You are forbidden to change your section on your own. You can't just squat in another section. We do this with the intention of trying to keep the enrollment roughly uniform. We don't want sections growing to 35 because then that limits your ability to interact. If you've changed classes or you've picked up a UROP, or what have you, and your situation has changed, please go down the hall and meet with my assistant Hilary, who will then arrange to move you. And that way we can keep some control over the section size. Again, you may not change simply on your own. And it has to be with cause, academic cause. You can't go in and say, I was assigned to the 9:00 am section and I don't do mornings. That's not going to work. Homework. Homework is very important here. Homework is a little bit different, though, in 3.091. You're not going to be asked to turn in homework for grading. Instead, at the beginning of each unit, you will be given homework with the model solutions right at the beginning. This is a study aid, and you can use the model solutions to help you understand the homework material. Now since we're not asking you to turn anything in and we do want to stimulate interest in the homework, we found a way to stimulate interest. And that is that once a week in section you will have a 10-minute quiz based on the homework. That will ensure that you've at least looked at the homework. And those weekly quizzes will be graded, and the aggregate of all of those quiz scores will constitute your homework grade in the subject. And you must take those quizzes. If you don't take the quiz-- you got sick or some personal emergency came up-- contact your recitation instructor and within a week of the date of the original quiz arrange to take a makeup. If you don't take the makeup, we're going to give you a zero. And there's no dropping of lowest score. Somewhere there's this in the lore here that, oh, you can drop the lowest score, the lowest two scores. People come to me with this proposal and I say no. [LAUGHTER] PROFESSOR: So let's get started. This is homework number one. It's assigned today. We're going to save paper-- just go to the website-- and you'll be tested on Tuesday, next Tuesday, September 15, at the beginning of section. At the beginning of section there'll be a 10-minute quiz based on that homework, and you can get the homework and so on. By the way, here's what the homework looks like. Chapter 1, Chapter 3 of Averill. Averill is the major text. We just refer to it by the author's last name. So taken from Averill. By the way, I don't like to use this unpleasant term "test." I like to refer to it as a celebration of learning. [LAUGHTER] PROFESSOR: So it's a celebration. We're going to start celebrating on Tuesday. There's the website by the way. You probably want to bookmark that. You're probably going to go to that a lot. We're going to keep celebrating, and so we will have some monthly celebrations, monthly tests, and they've already been scheduled. And you will write those tests during the normal class time. But on those days we will spread you out a bit. I like to see some vacancies. So you will not be sitting one next to another. There'll be at least one human vacancy next to every human, and that way you've got room to spread out, keep your eyes on your own work. So those are the dates and there'll be more of that in time. Now when you take the monthly test, I allow you to use aids. So everyone will be given in recitation tomorrow a Periodic Table of the Elements, very nice one, laminated one. So you take that with you to the test. In fact, you should take this with you everywhere. [LAUGHTER] PROFESSOR: If I run into you at Harvard Square, I want to see the Periodic Table, because every educated person has the Periodic Table. You get a table of constants. So you'll have one of these. You'll get that tomorrow as well. Paper copy, and on there are all the constants so you don't have to remember that the permittivity of vacuum is 8.85 times 10 to the minus 12 farads per meter. It's on there. I urge you to use that when you do your homework. So it's always amusing to me during the time of the first monthly celebration, somebody's looking at this as though it's today's newspaper and they're looking, where is that? Sort of revealing about the intensity with which the homework has been embraced. Calculator, something to calculate with, and I don't care if you bring in a graphing calculator or a mainframe. I don't care. [LAUGHTER] PROFESSOR: And you're allowed an aid sheet, an 8 1/2-by-11 sheet of paper. And you can write on the front, you can write on the back. You can photoreduce previous exams down to micro-dot size. I don't care what you put on it, but with this, you have then no excuse to say I really understood this stuff, but I couldn't remember a formula. Actually, many students tell me that the act of preparing the aid sheet organizes the subject matter in their minds. They bring this with them to the exam, and they never consult it, but it just soothes the nerves and makes sure that everybody does well on the monthly test. The weekly test, no. The weekly test you bring your Periodic Table and table of constants, but no aid sheet. For the weekly test, it's a very concentrated amount of material. And I'm not going to test your memories, because that doesn't prove anything to me. And then, of course, the celebration of celebrations is the final exam. [LAUGHTER] PROFESSOR: That's huge. That's a huge celebration. It's three hours. See, these are really 50 minutes. This is three full hours. The time and location will be set by the Registrar. We should know by October 1. The final exam period is December 14 to 18. And so what I urge you to do is as soon as that schedule comes out and we know when all the final exams are going to be held, then you book your passage for the holidays. Do not get that order reversed. You cannot come to me and say, I got a really good price on a ticket. I'm going Acapulco on the 15th of December, and your exam is on the 16th of December. I'll say, you just got a zero on the final. You have to be here for that. You know why? I have to be here for that, you can be here for that, too. There are about a quarter of a million students in Boston. It's a great college town, but at Christmas time, it's pandemonium at Logan Airport. So you want to book your passage early, but you can't do it until you know what your final exam obligations are. Grading. Freshman, you know, it's pass/no record. Pass/no record. And so that means that if you struggle and things don't go well, you don't have any blemishes on your record. Unfortunately, some of the upperclassmen will tell you as pass/no record, you now, barest pass imaginable/ no record. Well, I hate to let you know this, but increasingly I am being asked by medical schools, law schools, scholarship providers to reveal the scores on the freshman year. So think about that before you call Hilary in late November saying, what do I need to get a 50? I need a 32 on the final? OK. But Lord help you, you get a 31 you go down. [LAUGHTER] PROFESSOR: Upperclassmen get the luxury of the entire alphabet: A, B, C, D, F. The final grade composition. 1/6 for homework-- that's the aggregates of the weekly test scores-- and 1/6 for each of the three tests. And then the final is 2/6, or 1/3. I didn't want to get into transcendental numbers, so I made it 33 exact, 16.75 exact. I could have made it 16.77, I could have made it 16.81, I could have made it anything I want. I'm the professor. But I chose 16.75. Bottom line here is it's really dumb to fail the final. If you passed the final, you're pretty much assured that we're going to be favorably disposed to passing you. But you fail the final that says two things. It says, number one, you don't have a grasp of the overall year, but it also sort of indicates that you ran out of steam partway through the semester and stopped working. It's very bad. You get a lot of good advice from upperclassmen, but sometimes-- I'd ignore that one that says, hey, you don't have to do that well on the final. Anyway, so you have to get a C level as a freshmen or greater. And, by the way, we do not grade on a curve. I've seen it in magazines this last year that MIT grades on a curve. I don't where they get that from. I don't grade on a curve. Your success does not come at the expense of your neighbor. As far as I'm concerned everybody in this class can get an A. Again, I'm the professor, all right? So you say, well, how do you know that 50 is the right number? Why isn't it 55? Why isn't it 75? Well, I know. I know. How do I know? Because when we grade, we set up the point scheme so that if the student has the mastery of the barest level of competency of the key concept the point scheme must reflect a passing grade-- 5 out of 10, 5 out of 9-- and you propagate that through. I don't care how much is written, if it doesn't demonstrate basic mastery of the key concept the point scheme must give 4, 3, 2, 0. Maybe a 1. And so that, if you propagate it through the whole semester, means 50 is a pass. There's some wiggle room there. How do I know 49.7 is a fail and 50 is a pass? That's when I call your recitation instructor. [CELL PHONE RINGING] PROFESSOR: Let's kill that. We're going to get to that in a second. Let's call the recitation instructor. And what happens? I ask the recitation instructor, well, what can you tell me about this young lady? And, oh, she came to all my classes, she tried really hard, she came to office hours. I don't think the exam is a proper reflection of her understanding. I'll listen to that and maybe we'll promote. If, on the other hand, I get the response, I never saw her, didn't come to class. I repeatedly reached out to her, ignored my entreaties, she's going down. [LAUGHTER] PROFESSOR: OK. Website. This is what the website looks like. There's a number of tabs here. The readings are there, the videos are there archived. The schedule-- what's going to be coming up. So, for example, this is what's going on today. It says what the topic is, roughly what the readings are. I know today you didn't come to class having read. And it's OK, we'll get through it. But from now on I urge you to do the readings. A couple of topics I'm required to talk about. My management requires me to do so. Academic honesty. There's a lot of texts here, but in plain English, this says don't cheat! You know what this means! Now I don't want to hear, well, in my country, the custom is-- I don't care what they do in your country! You're in my class, and if you cheat in my class you will pay for it. Very simple. Accept information of any kind from others-- wrong. Represent somebody else's work as your own-- wrong. You know this. It was Juvenal, the Roman politician, said men need not so much be instructed as reminded. I'm not telling you anything you don't already know, but I'm going to say it so no one can say, well, nobody told us it wasn't OK to erase our answers and hand them back in for more points. I just did. All right? So if in the unlikely-- but it happens every so often. Maybe every two, three years. And people get caught. You know why they get caught? Because for the first time in your life you're being taught by people who are as smart as you are. [LAUGHTER] PROFESSOR: My TAs are really smart. And it never ceases to amaze me-- somebody succumbs and does something dishonest, and they get caught! They get busted! And what happens then? Then I get angry. You know why? Because we can't settle that in my office. You can't come and cry in my office. I have to take this episode to a committee here at MIT called the Committee on Discipline. It is staffed by faculty, by administrators, and by students. And the case is brought before the COD at a hearing and a punishment is decided. And it can be anything from suspension to expulsion. And of the three categories, faculty, administrators, students, which category do you think is most severe on infringers? AUDIENCE: Students. PROFESSOR: Your peers. You got it. You know, because people my age, old faculty, they'll be like, oh, they're just kids, whatever. The students say, no! Throw them out! [LAUGHTER] PROFESSOR: So it's not going to be me that's going to expel you, it's going to be your peers. So don't do it. And if you're ever in an exam and somebody's pressuring you, just raise your hand and ask to be reseated. For all I know, the guy next to you has got a bad case of B.O., you just want to move. We will not ask any questions, so take yourself out of the situation. Conduct yourself appropriately. Classroom behavior. Now this is the first lecture, there's a whole bunch of violations in here right now. So I'm going to say it now and we'll fix it for next time. We've got 425 seats here, we've probably got 475 people, and there's only one way we can make this system work and we have to observe certain rules of decorum. And I make the rules. So if I want to maintain a fertile learning environment I'm going to ask for these rules to be observed. No talking at all. No talking. Little conversation here, little conversation here-- it disturbs people. I don't want any food or drink. No food, no drink. One exception is the professor. [LAUGHTER] PROFESSOR: Because I do not want to have my throat get so dry that I can't finish talking my way through the end of the class. Otherwise, no food or drink and no disruptive behavior. No horseplay or anything like that. And wireless communication devices must be silenced. Cell phone goes off, you get up, you leave the room. That's it. AUDIENCE: Is water OK? PROFESSOR: If you need water because it's some kind of a health thing, fine. But I do not want to see a whole bunch of people drawing on water bottles. Don't need it. Don't need it. You can go 50 minutes without your little-- whatever. [LAUGHTER] PROFESSOR: You know why? I'll tell you why. It's not because I'm trying to be a control freak. This is a chemistry class, but it's chemistry-centered. You didn't just come to MIT to learn some geeky techno stuff. You're preparing for a professional career, and part of that is how you behave, how to act as a professional. And you cannot learn behavior by doing problem sets. How do you learn behavior? You learn behavior by observation. And how does one teach behavior? By modeling. If you're in some high-level committee meeting and all of a sudden your cell phone goes off and you're scrambling and you can't find the damn thing, it's in the bottom of your briefcase, right? I can't tell you how vulgar that is. You're in a professional setting, you commit professional suicide. And I need to tell you that. If you think it's OK you're wrong. So let's get in the habit. Disable the damn thing. What do you need? You're going to call your stockbroker? What's so important right now? And on an exam, that thing goes off I take the exam in front of everybody and-- [MAKING RIPPING NOISE] PROFESSOR: --we introduce a defect, it's called a tear. [LAUGHTER] PROFESSOR: And then I put a zero on it like this, with a circle around it. It's called a donut. [LAUGHTER] PROFESSOR: All right. Now let's talk about some lighter things, more upbeat things. Look, I want you to succeed in this class. Now how are we going to succeed? You go to the listing in the bulletin you'll see this, right? And you zoom in here it says 5-0-7. What's this mean? Well the 5 is 5 contact hours. 3 here and 2 with your recitation. The 0 is lab. There's no lab with this class. So what's the 7? The 7 is the reading, the homework, preparation, et cetera. I pledge to you, you give me 7 hours-- you go to the 5 hours contact and 7 hours-- you will not just pass this class, you will flourish in this class. How do I know this? Because I used to chair the committee on admissions and I've read applications. I know the quality of individual in this room. You should have seen, there were 11,000 applications piled up like this, each dossier like this, and they just get copied down on a four-by-six card. We get here on a Presidents' Day weekend and we got stacks and stacks of the cards. We're looking at your whole academic life is on a four-by-six card. And I pick that up and I go, yes. Literally at this speed. About 45 seconds. No. [LAUGHTER] PROFESSOR: Hell no! [LAUGHTER] PROFESSOR: That's how I spent Presidents' Day weekend. And so you got through a very grueling selection process. What's the corollary of that? Listen carefully to this: Everybody in this room has the intellectual apparatus to pass 3.091. The only people who fail 3.091 are people who choose to fail 3.091. They choose not to come to class, they chose not to go to recitation, they chose not to work the homework. I don't know why they make those choices. But I guarantee you if you give me this amount of time, you'll do well. It's straightforward. Number one problem you are going to face this fall is not secondary bonding, it's time management. So I used to call this strategies for-- what did I used to call it? I forgot. I used to have it something like, survival strategies. But I don't want you to survive, I want you to flourish. So I call it recipe for success. There are different venues for learning. OK? So lecture, that's here. That's my responsibility. Recitation, that's Tuesdays and Thursdays. That's my staff. Now reading, that's you. Homework, that'd be you. Weekly quizzes is you, monthly tests is you, final exam is you. [LAUGHTER] PROFESSOR: What we have here is a partnership. See? [LAUGHTER] PROFESSOR: So I'll do my part, you do your part. You know, one of the things beyond the basic learning of the chemistry that we're going to attempt here is a transition. You are going to change in ways you can't imagine. I want you to think about the way you are right now, and I'm going to ask you to think about how you feel about yourself on the last day of class. You will be amazed, and it's not because you know a few more chemical equations. And what one of the things that I want to see happen and to help facilitate is the transition from student, which you are as a high school graduate, to scholar. And the difference between a student and a scholar is a scholar takes ownership of his or her learning. So you're going to take ownership of it. You know, is this going to be on? Are we responsible for this? Go back to high school. Here you're a scholar. You say, how can I learn more about this? That's the difference. So I think we've covered enough. I think we don't want to just have the whole day, welcome to MIT, welcome to MIT. So what we're going to do is we'll get into the lecture, and in the very brief amount of time we have left I'm going to talk about the beginnings of chemistry. We're going to talk about taxonomy, classification, nomenclature. And to help introduce this I'm going to refer to the writings of William Shakespeare. We're going to integrate some humanities here. We're going to read from Romeo and Juliet. Maybe there are people here who were admitted on the strength of the performance in Romeo and Juliet. Maybe one of you was Romeo, one of you was Juliet. Maybe one of you did the lighting. Maybe one of you did the set design. Maybe one of you took the tickets. Somehow you were involved, right? Because we're all involved. So let's take a look. Act 2, Scene 2. Romeo: "But soft, what light through yonder window breaks? It is the east"-- that's the east. East Campus is that way, right? [LAUGHTER] PROFESSOR: "It is the east, and Juliet is the sun." That has nothing to do with nomenclature. Maybe photon emission, but-- that's a joke. We'll get to it. Now Juliet: "O Romeo, Romeo! Wherefore art thou Romeo? Deny thy father and refuse thy name. Or, if thou wilt not, be but sworn my love, and I'll no longer be a Capulet." See she's a Capulet, he's a Montague. You know, the Hatfields and the McCoys. And this is a metaphor. It's as old as literature. It's about warring factions, two communities that can't stand each other based on prejudice. And then these two youngsters fall in love, and how love triumphs over hatred, and so on. It's powerful stuff and it's written here. Romeo: "Shall I hear more, or shall I speak at this?" Remember, she's up high, he's doing "Shall I hear more, shall I speak at this?" Fellows, the answer to that question is don't speak. Don't interrupt her. All right. So now she goes, "'Tis but thy name that is my enemy. Thou art thyself, though not a Montague. What's Montague? It is nor hand, nor foot, nor arm, nor face, nor any other part belonging to a man. O, be some other name! What's in a name? That which we call a rose by any other word would smell as sweet." That's properties, right? "So Romeo would, were he not Romeo called, retain that dear perfection which he owes without that title. Romeo, doff thy name, and for that name which is no part of thee take all myself." Beautiful writing. 400 years ago and it's just fantastic. Well, that's no good. That's not the way it works in science. In science we have to agree-- [LAUGHTER] PROFESSOR: We have to agree on the name. And so this is taken from Chapter 1 of the text, and this is the classification of matter. This is about stuff and the different forms of stuff. Over here we have the simplest form of stuff, which is the element. And we're going to start here in 3.091 and we're going to work our way all the way through this table, starting with electronic structure and how electronic structure governs stuff. So let's start with a history lesson. We'll start with a history lesson. And the history lesson goes like this. What are the origins of chemistry? The ancient Egyptian hieroglyphs refer to khemeia, which was a chemical process for embalming the dead. You know the Egyptians were very fixated on the afterlife. And the chemists, the chemists were revered in that society-- not like here. They were revered in that society because they knew how to prepare the body for the afterlife. Embalming is a chemical process. A few years ago I was in London. I toured the British Museum and came upon this. This is 18 inches tall. It's the mummified cat. It's a fantastic example of mummification. It's a beautiful-- most people go zooming right past it. They're looking at the big dinosaurs and all that other nonsense. But this, this is beautiful. Mummified cat. And then khemeia expanded to other chemical processes: dying of cloth, glassmaking, and metals extraction. The chemist could take dirt and turn it into metal. Sorcery. Some things haven't changed. I'm in that tradition: producing metal from dirt. And they were always looking for an overarching theme, to unite the heavens with the simple elements. So the seven known, naked-to-the-eye, astronomical bodies were associated with the seven known chemical elements. And you could even talk in this priestly language. So if you wanted to make a bronze, which is an alloy of tin and copper, you could say, well let's have the confluence of Jupiter and Venus. It's all there. Mercury. Why do we call it Mercury? Mercury's a metal that's liquid at room temperature. It's fast. Because Mercury is the planet that moves quickest around the sun. Very nice. So this is what we knew 2,400 years ago. We had the seven metals, carbon, and silicon. And the beginning of the shift from practice to theory, or from craft to science, is with the work of Democritus. Democritus, who lived around 400 BCE. Democritus, who was Greek. He described the physical world as consisting of a combination of void plus being. Void plus being. These are lofty words. Listen to this. Void, being not some little equation. It's big, big ideas. And how did he describe void? Void is something that we would recognize, in his language, as akin to what we recognize to be vacuum. And being, he said, was comprised of an infinity of atoms. He coined the term "atom." "Atom" comes from the Greek tomoi, which is to slice. And then if you put "a" in front of it as in apolitical or amoral, cannot be sliced: indivisible. The atom cannot be sliced, and so to these atoms he attributed these properties. They're indivisible and eternal. I mean, this takes you all the way to E equals m c squared. From 400 BCE, and that's all he had to work with. This is brilliant. Absolutely brilliant. But to show you that things don't always go in a linear fashion for the better, along comes Aristotle. Aristotle, another Greek, and he decided, nah, we don't need all this carbon, sulfur, and so on. He said, we will have four essences that will describe the earthly world. Four essences. And here are the four essences. Earth, water, air, fire. So those are the Aristotelian essences. And there's actually a fifth essence that describes the heavens. That's why we say something is quintessential. It's heavenly, it's et cetera. So these are the four essences. All right? And then it gets even worse because he has compounds, a combination, so you can take fire plus earth and make dry, earth plus water make cold, air plus water make wet, air plus fire make hot. This is nuts. [LAUGHTER] PROFESSOR: And it dominated science for a long, long time. Because we know really earth is an aggregate. It's an aggregate of different minerals which are compounds. Water, as you know, is a compound H2O. Fire is the product of combustion. It doesn't even belong in this set. And air is a solution. It's a homogeneous mixture of nitrogen, oxygen, argon, rising levels of carbon dioxide, sulfur dioxide. And if you're next to an aluminum smelter, tetrafluoromethane. And so on and so forth. And finally this thing was knocked down. So we can look at this chart and go back to this one and say, now we can make the connections. All right. Here's the way the elements looked at the time of the American Revolution. The alchemists gave us arsenic, antimony, and bismuth in the 12th, 13th, 14th centuries. In 13th century India, there was zinc isolated. There's a tall zinc pillar still standing there. Platinum is an American metal. It was unknown to the Europeans. It was discovered when the Spanish came to the Americas. Actually, it's kind of ironic because plata is silver, so platina is sort of like a diminutive of silver, which in point of fact is backwards because silver melts at 962 degrees Celsius, platinum melts at 1768. It is nobler than silver. It has catalytic properties. It is even fashionable in jewelry now if you know how to work with it, because gold melts at 1063 and platinum melts at 1762. A lot of jewelers can't work with platinum. It's too high-melting. It's a fantastic metal. It's an American metal. [LAUGHTER] PROFESSOR: It really is. I mean it's an American metal. I didn't make this up. It's true. And then we see these other elements. Discovered, discovered, discovered, discovered. Now we see modern science. So what does it mean, discovered 1766? There was no hydrogen before 1766? They found it? No. It means that Cavendish isolated it and documented its properties. Hence, it was discovered. Now I've been cheating here. This is actually lined up in a way that we already know, where the story is going to end with this, with the Periodic Table. I'm just lying them on the table in a way that makes it possible to anticipate. But this was the first table of elements that is of record, and this was by John Dalton, the English chemist, who put these in order of atomic mass. They're organized in ascending order of atomic mass. And he also started a system of chemical symbols. And you can see that iron has an "I" with a circle around it, and zinc has a "Z" with a circle around it. And the Swedish chemist Berzelius said, you know I don't think the French are going to like to use "I" with a circle around it for iron, or Germans aren't going to like that. We better choose something a little bit neutral. So they chose Latin. And that's why iron is "Fe" for ferrum, and gold is "Au" for arum, and so on. But this was the first attempt. John Dalton actually was a polymath. He was also working on vision, human vision, and he suffered from an affliction that is present in about 10% of men, which is red-green color blindness. And he did the original work on red-green color blindness. In fact, in some circles it is known as Daltonism. It's the same man, John Dalton. Other classifications. Dobereiner in Jena talked about triads. And if you took the atomic mass of chlorine, added it to the atomic massive of iodine divided by two, you get something that's not too far off the atomic mass of bromine. Newlands was a musician and he talked about octaves. So if you start here, if this is a diatonic scale, so this is C, D, E, F, G, A, B, C. So potassium lies an octave away from sodium. He was ridiculed. They said, have you considered perhaps putting the elements in alphabetical order? They were cruel. Scientists can be very cruel to new ideas. And, in fact, in your book this is some of Newlands work, and you can see to what extent it helps understand things. But the first proper organization came in 1869 with Mendeleyev, and also 1870 with Lothar Meyer in Tuebingen, the Periodic table as we know it. And this is the set of elements that were known at the time. And this is a page from the paper in which Mendeleyev published the Periodic Table. And here's the smoking gun right here. It's right here where he says, in this new system, in this proposed system, there are very many missing elements. Very many missing elements-- that was the key. Because he knew that even though arsenic has the atomic mass next highest to that of zinc not to put arsenic under aluminum, not to put it under silicon. It had more in common with phosphorus. So he put it under phosphorus, and so did Meyer. But what Mendeleyev did, which was a first, he said, there are missing elements here. There is an element that lies between silicon and tin, and I predict what it's mass will be. I predict what it's chemical formulation will be, how it will react with oxygen. The predictions. And how did he get this? Because he used to travel by train and he would sit in the train station on his trunk playing solitaire. And he'd be going down one suit, and then there'd be a gap. And he'd go down the other suit and he could go farther. And so the concept of, I know there should be an eight of spades here, but I have to stop at the nine. I've got a seven but there's something in there-- triggered his imagination and led him to have the courage to say, there are elements there and I will predict their properties. And when they were discovered how close he was is shocking. Absolutely shocking. So we'll get to that next day. But before you go I want to leave you with this. This is the portrait of Mendeleyev that you typically see. This man on in years, disheveled, sort of mad scientist look. Don't remember that. Look at this. This is Mendeleyev, age 35, when he proposed the Periodic Table of the Elements. OK. We'll adjourn. We'll see you on 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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Let's now consider our rolling wheel, and we want to look at some special conditions. So at time t equals 0-- and we'll have our wheel that's rolling, here's the ground-- let's say that our point P is right up here at the top. That's cm. And we'll be in the ground frame now. And then at a later time, time t, the wheel has moved to the right. So let's draw the wheel over here. Not the greatest picture of the wheel, but we'll have the wheel over here. And now the point P has moved some angle delta theta. And we'll call this time interval delta t. Now, the center of mass of the wheel has moved a distance. Xcm is the velocity of the center of mass times delta t. And the point P on the rim, this arc change, this length here on the rim that P has moved around in the center of mass frame, is R delta theta. Now, we want to ask ourselves-- we'll call this delta x. We now have three possible conditions. We call rolling without slipping. That will be our first case 1. And that's the case when the arc length delta s is exactly equal to the distance along the ground. So we have delta Xcm is delta s. And so we get Vcm delta t equals R delta theta, or Vcm equals R delta theta over delta t. Now, in the limit as delta t goes to 0, we have that delta theta over delta t in this limit as delta t goes to 0 is d theta dt And that's what we called the angular speed. So in our limit as this wheel is rolling without slipping, we have the condition that the velocity Vcm equals R omega. So that's our first condition, and we call this the rolling without slipping. Now what is Vcm? Vcm? That's the velocity of the center of mass of the wheel, and every single point on this wheel has that same speed. And R omega, you can think of that as the tangential velocity in reference frame cm. This is just the speed in the reference frame moving with the center of mass. So this is our condition for rolling without slipping. Now, our second case is imagine that the wheel is not moving forward at all, but it's just spinning. That's what we call the wheel is slipping on the ground, for instance, if there were ice. And so what we call slipping is a little bit more general. It's whenever the wheel is spinning, and the arc length is much greater than the horizontal distance that the wheel has moved. So we have delta s representing the arc length that the point has moved in the center of mass frame is greater than how far the center of mass is moving. And so, again, we have R delta theta is greater than Vcm delta t, or in the limit R omega is greater than Vcm. You can say it's spinning faster than it's translating. And finally, the skidding condition. Skidding-- imagine that the wheel-- you're braking a wheel. The wheel is not spinning at all, but it's just sliding along horizontally. So the horizontal delta x center of mass is bigger than delta scm. And so this is the case where delta Xcm, how far it moved horizontally, is greater than the amount of arc length that the point moved. And so in the same type of argument, when we put our conditions in we get that Vcm is greater than R omega. And again, what that corresponds to in the skidding case, imagine the limit where it's not rotating at all, this would be 0, and it's just skidding along the ground, Vcm. So we have our three conditions. We have the slipping condition, where it's spinning faster than it's translating. We have the skidding condition, where it's translating faster than it's spinning. And we have the rolling without slipping condition, in which the arc length is exactly equal to the distance, horizontal distance, along the ground.
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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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CATHY DRENNAN: So I'm Cathy Drennan. I'm a Professor of Chemistry and Biology here at the Massachusetts Institute of Technology. I'm also an investigator and professor with the Howard Hughes Medical Institute. And I just wanted to give a little introduction to the training video you're about to watch, this workshop that we've developed, that is on the topic of stereotype, stereotype threat, Why Is Criticism Unconscious Bias. And so in this workshop you're about to see, this is the workshop that we run for teacher training at MIT, both in the biology department and in the chemistry department. So most of the people in this room are about to go on an adventure of teaching for the first time here at MIT. So I'm going to talk about the motivation for doing this workshop actually in the workshop. But let me just tell you the goals for a minute. So we want the people who are going to be teaching our students here to know about stereotypes, know about stereotype threat, feel comfortable with terms such as wise criticism, know some of the data behind the idea that stereotype threat can actually lead to underperformance by students, and think about how do you create an environment in your classroom where every student can reach their full potential. So that's what this training is about. And we hope the students coming out the other end are going to be really excited about their classroom experience and how they're going to create this environment for every student. So although this training is really designed for teaching assistants, we found that this material is broadly applicable. It can be used to train faculty members. It can be used to train mentors in a research laboratory. It can be used to become a better student oneself or a better mentee oneself. It's really for everyone. And it relates to everyone. So I'm hoping at the end of watching this video, that maybe you will feel comfortable enough with this material that you might start these discussions at your home institution. So I'm going to make available to you both this workshop on videotape, all of the slides that you're about to see, a booklet with more information for you to read and exercises in it, references in it. It can be used-- people can read about this before they do a workshop. We're hoping to have this discussion just sort of spread all over the place. And I know what some of you are thinking. It's like, no way, I cannot lead this discussion myself. But you know, that's what I used to think. And everyone can talk about these things. Sometimes it makes us feel a little uncomfortable to talk about things like stereotypes. But we can talk about them. We need to talk about them. And after watching this video, we hope that you'll be ready to start a discussion of your own. So it's important to do this. It's an important time to be having these discussions, because there are a lot of problems in the world. And in addition to being a professor, I'm a mom of a 10-year-old. And I can tell you, I'm really worried about the planet that she's going to be inheriting. There's global warming. There's warming of oceans. There's antibiotic resistance. We need smart people to be thinking about these problems. We need scientists to be thinking about these problems. So we need everyone who wants to be a scientist to be able to reach their full potential so that we can tackle these important problems moving forward. And that's what this training is designed to do-- start the discussion so that everyone can be in an environment where they can reach their full potential. So enjoy watching.
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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 open courseware at ocw.mit.edu. JOANNE STUBBE: Recitation 2 and recitation 3 are on the same paper. You only have to read one paper that Liz has been discussing in class, the Rodnina paper. The paper was published in 1999, OK? And it's still, I would say, a seminal paper. And what they propose or what you read about their model is still the working hypothesis in the field. But if you go and Google the ribosome in elongation, you will find out that in the last 10 years there are hundreds of papers now taking pot shots at this model using modern technological mechanisms, like single-molecule spectroscopy, Cryo-EM. So they're flushing things out, but so still the basic model holds. So we continue to go through this because, in my opinion, all the machines that you're going to be talking about-- and this is part of the course-- have complex behavior like this with numerous substrates and many, many steps. And so hopefully one thing you got out of this paper is that kinetics are important. OK, so today what I want to do, I'm going to ask you questions. I'm going to put some things on the board. We get you at talking points. I'm going to ask you some questions. And then the discussion will continue into the next recitation on the same exact same topic. But kinetics are important. But to do kinetics, what do you have to have? What's required to do kinetics if you look at this model? So this is the model out of your paper. If you want to do kinetics, what do you need? Not only that, you have to speak loud because I'm deaf anyhow. What do you need? AUDIENCE: To do kinetic studies? JOANNE STUBBE: Yeah, to do kinetic studies. AUDIENCE: [INAUDIBLE] a detection method in very controlled conditions. JOANNE STUBBE: Yeah, so you have to have an assay, OK? And you'll see that everything you're doing over the course of the semester requires development of an assay. And I would say the more complex you get the more complex these machines. And that's what people are studying now as opposed to if you looked at it Liz's lecture on tRNA synthetases, you saw a simple reaction, OK? That assay was developed decades ago. But when you get into these more complicated machines, you have to be really pretty creative to develop an assay. And you need to have substrates. You need to get them from somewhere. And then you need to do kinetics. And so today, what I want to do is go through the kinetics part of this, asking you questions as we go along. And I'm going to start. So kinetics, in my opinion, is a key tool. So we're using kinetics as a tool to study machines. And the machine we're studying is-- and have been studying is, is the ribosome. OK, so how many of you have had an introductory last lab course where you did kinetics? Only one? Two? OK, because steady state kinetics is where you start for everything, OK? And I find when I-- I've been teaching for many years-- that there are certain things about steady state kinetics that people don't seem to get. And furthermore, were steady state kinetics important in this paper you had to read? Can anybody tell me? Did you get anything about steady state kinetics? Did you think about it? This will tell me how closely you read the paper. No? No one? OK, so this paper is hard and this is a paper that, even though I read it probably 20 times, I still learn stuff every time I read it. So you can't read a paper once. There's huge amounts of information in this paper. And if you go back and look at it three weeks from now, you'll probably get a lot more because we're continually filling in pieces of information from you in this complex system. Yeah? AUDIENCE: The part, I think, related to the steady state kinetics they measure Kcat, and Km, and their ratio. JOANNE STUBBE: Right, so that that's where the steady state kinetics is. And so if it goes to the question of what can you learn from steady state kinetics, OK? So let me just put down a simple system, which you've all seen if you take an introductory biochemistry course. People use this system because you don't have very many rate constants. So when I write down rate constants, I don't put K's. I just put 1, 2, 3 because it becomes hard to read anything, OK? So this is a simple system for any catalyst, OK, where some substrate could be EFTU, and tRNA, and GTP binding to the ribosome, OK? You do some chemistry to form some product. OK, and then the product dissociates. So if you look at the rate of the reaction-- so this involves the assay. You have to develop an assay where you can monitor something as easily as possible. That's the key thing. So I think here is where your chemistry background plays an incredibly important role because you can be creative about your assays. And so and you look at this as a function of the concentration of your substrate. What does the spectrum look like? What does the graph look like? How would you describe the graph? This is something you've seen in 507. We're just going back. What does it look like? Right, exactly-- rectangular hyperbole. OK, and so I think what's important is that this kind of behavior has been observed over and over and over again since 1940s when this curve was first described by Michaelis and Menton with many variations on the theme. And so what you need to think about is you have two parts of the curve. What's happening up here? What is the dependence on the reaction on substrate? So we have an enzyme that's a catalyst. It doesn't matter whether you're an organic chemist, an inorganic chemist, a biochemist. All of these things can be described by this simple, simple cartoon. So what's happening up here? What's happening in this part of your graph? AUDIENCE: It's saturated. JOANNE STUBBE: Yeah, see, you're saturated. So you're zero water and substrate, OK? And then what's happening over here? Your first order N substrate, OK? So from those observations, people derived equations, a general equation. So the rate of product formation, whatever you're assay is that you're using, is equal to Vmax times a concentration of substrate over Km plus the concentration of substrate. OK, so you've all seen this before. And if you look at this one simple case, and you look at what is Vmax equal to-- can anybody tell me what are the rate constants within Vmax? So Kcat times the concentration of enzyme. OK, so Vmax, and what does that mean? Kcat we'll see in a minute, is the turnover number times the concentration of enzyme. That means all your catalysts have stuff on it. It can't go any faster. It doesn't matter if you add more, more, and more substrate. You have no catalysts. So that's what's limiting the reaction. So if you derive this equation using steady state assumptions, what are the four sets of equations you need to be able to derive this expression? Can anybody tell me? What are the conditions you need to do? So what's the goal of deriving this equation, first of all? And then what are the assumptions you make? OK, so you want to be able to describe what you see experimentally, OK? So the first thing you have to do is be able to measure it experimentally, OK? So you have to have something in terms of an experimentally measurable parameter. And if you look at e, es, ep, which one of these are going to be measurable? AUDIENCE: Going to be the substrate and the product. JOANNE STUBBE: OK, so substrate and product. Yeah, you can measure substrate. You can measure product. But I'm talking about e. OK, so we have the e, we have an es, in this case we have an ep, most of the times you have 20 more e, equilibria. So which one can you measure experimentally? AUDIENCE: [INAUDIBLE]. JOANNE STUBBE: Pardon me? AUDIENCE: [INAUDIBLE] enzyme. JOANNE STUBBE: Yeah, so you can measure the total enzyme. OK, so that's this the enzyme conservation equation. So you have-- I'm not going to draw this all out. This is a review that you've already seen, presumably. So that's the conservation equation. How do you measure the concentration of an enzyme? AUDIENCE: UV vis. JOANNE STUBBE: UV vis. What amino acids absorb in the visible? AUDIENCE: Tryptophan. JOANNE STUBBE: In the visible. AUDIENCE: Oh, in the visible? None. JOANNE STUBBE: None. So don't say UV vis. Say UV, OK? So this is key to being able to sort things out. So what are the amino acid side chains that absorb in the UV? This comes back to-- you need to-- AUDIENCE: Tryptophan, tyrosine. JOANNE STUBBE: Right. So tryptophan and tyrocine are the major ones. Then phenylalanine is much smaller. So you can measure this. But, in general, you can't measure all the other forms. OK, so you know this, and that's required to be able to get this expression that describes this rectangular hyperbole. What about the substrate concentration? Under normal assays, if you've done an assay in the lab, how is the reaction set up? How much enzyme do you have in there? How much substrate do you have in there? AUDIENCE: A lot of substrate. JOANNE STUBBE: A lot of substrate. And what conditions are you under, perhaps, if you have a lot of substrate, over here in this graph? AUDIENCE: You have to saturate your-- JOANNE STUBBE: Yeah, well, you don't have to, but you would be saturated if you had a lot of substrate. How much enzyme do you have in there? A lot or a little? AUDIENCE: A little. JOANNE STUBBE: A little, OK? So the enzyme, the concentration of the substrate is much, much greater than the concentration of the enzyme, OK? So that's a typical steady state assay when you go to determine the rate of your reaction. So because the concentration of the enzyme is much, much greater than the concentration-- the concentration of the substrate is much, much greater than the concentration of the enzyme, you don't have to worry about substrate being bound in these forms. So the second equation you routinely use is called the substrate conservation equation, because it doesn't change, because the amount of this es on the enzyme, which is tiny, you don't need to measure it. So this is the second. So these are both conservation equations. OK, so we just said we were doing steady state kinetics, OK? So now you need to be able to make the steady state assumption, which hopefully all of you know. So the rate of change of some intermediate with respect to time is equal to 0, that is we're under a set of conditions where the rate of formation is equal to the rate of disappearance of whatever this species is. And what is the fourth equation we need to be able to set up this? What is the fourth thing, which is probably the most straightforward? And, again, it needs to be in terms of experimentally measurable parameters. What are we measuring in our reaction? AUDIENCE: You said the position from vs to [INAUDIBLE] is irreversible. JOANNE STUBBE: No, you I could have done this. And what would that have done to my equation? It just would have put in more rate constants. I'm going to show you what the rate constants are in a minute. There's nothing-- in fact, almost no enzyme reactions are irreversible. If you look, you can find reversibility in almost all reactions. This is-- so why do people write equations like this? They like irreversible reactions because it makes the kinetic derivation simpler. You don't have as many rate constants, OK? So, but what do you need now? We're monitoring the reaction? What are you going to monitor? So we know how much enzyme we have. We can measure that. We know how much substrate we can have. We can measure that. We know what the steady state assumption is, and we have an equation. So we can describe that. What's the other thing you need? It's the standard thing. How do you describe the rate of product formation? That's this guy over here. So what do you need? You need some kind of an equation that expresses just appearance of substrate, formation of products. So you need a way of measuring this. And you can do this many ways, even from a simple equation we've shown over here because a description of the rate of product formation is simply the net flux through any step in the pathway. And so what you see people writing is they immediately go to an irreversible step because it makes the algebra simpler. So k3 times the concentration of ep would be the net flux through this step. But I could write the net flux through this step and I would get the same answer. So it's the net flux through any step in the pathway. OK, so why am I going through all of this? OK, and the reason I'm going through this is because of this Kcat over km, which I just described to you. So one of the questions I asked you to think about when you're thinking about steady state kinetics is what are the two important parameters you get out of Michaelis Menton analysis? And the reason I ask this is because, in my opinion, it's not correct in most textbooks. So what are the two important parameters you get that you learned about that you probably even evaluated if you did something in the lab? AUDIENCE: Kcat. JOANNE STUBBE: Kcat is one of them. OK, and what's the other one? AUDIENCE: Km. JOANNE STUBBE: OK, so this is what everybody says, is km. And that's not correct, OK? So let me put down what the-- what did I do with it-- the values for the kinetic constants are here. So in this particular simple equation, it's Kcat is 2 times 3 over 2 plus 3. OK, so this is Kcat. And Km out of this analysis is 3. The numbers really aren't important. What I want you to see is that there are a huge number of first order rate constants in each of these parameters Km and Kcat, OK? Can you measure these? Can you measure these rate constants? That's what you want to know if you want to understand how this works, you would like to understand the reaction coordinate and what the rate constants are for every step in the pathway. That's what the whole Rodnina paper is about with the long range goal of understanding fidelity. Can we come up with a model for fidelity in the translational process that's contributed by EFTU? So can we measure these guys from an assay the concentration of the enzyme, The concentration of the substrate, the steady state assumption? What do you think? Don't be afraid. This is a discussion. What do you think? Can we measure? AUDIENCE: Does it depend how fast it is? JOANNE STUBBE: No. AUDIENCE: No? JOANNE STUBBE: No. It is dependent on how fast it is, but it doesn't matter how fast it is to answer this question, OK? Anybody else got another guess? What? Your name? AUDIENCE: Rebecca. JOANNE STUBBE: Rebecca. What's your name? AUDIENCE: Nicole. JOANNE STUBBE: Nicole. OK, yeah? AUDIENCE: Yeah. [INAUDIBLE] measure them [INAUDIBLE] we measure the initial rate, [INAUDIBLE] take that [INAUDIBLE]. JOANNE STUBBE: So you get Kcat and you get Km. That's not the question I asked. I asked, can you measure all the first order rate constants that make up Kcat and Km? No. So the problem with steady state kinetics is you can't really learn very much, OK? So what can you learn from steady state kinetics, and why do we keep looking at it? OK, why is it the first thing you've seen this with the tRNA synthetases? You saw Kcat over Km values charging with valine or isoleucine, right? In this paper, if you go back and look carefully at the discussion at the end of the paper-- so hopefully after this class you'll go back and you'll read that-- a lot of the discussion is about mechanisms of fidelity where they are thinking about these initial steps. And so these initial steps are really the selection steps of these things binding, OK? And if you go back and you look at the equation that they derive, it's amazingly complicated. Why? Because we have many more equilibria in our equation, but what you can get out of all this is Kcat and Kcat over Km. So Km really is not very informative at all because it's composed of a whole bunch of first order rate constants. It's always never equal to the dissociation constant, OK? So you can't-- so what it is mathematically, it's the concentration required to reach half maximum of velocity. So it doesn't really tell you anything. It's just half maximum of velocity. OK, so the two parameters that you need to think about-- and this goes back to the way you do experiments in the steady state versus the pre-steady state, which is what we're focusing on in this paper, is that you have two extremes when you do kinetics. And kinetics is something-- how do you learn how to do kinetics? You do them yourself. And you think about-- you think about what you think is going on. And then you make guesses about what's going on. And these are one of the types of experiments, when you're doing them you change your experimental design in the middle of your experiment. So it takes a lot of practice to get good at kinetics. But what you do with all kinetics, look at the extremes, the limits. So one extreme is the concentration of s goes to infinity. OK, so if you look at that extreme, what do you have? If s goes to infinity, what happens to b? What happens to this equation? AUDIENCE: [INAUDIBLE]. JOANNE STUBBE: Yeah, so it goes to Vmax, or Kcat times the concentration of e. So you're up here, OK? And so what is Kcat? So you can get out of this Kcat. Why is Kcat an important parameter? Why do people care about Kcat? Hey, what's your name? AUDIENCE: Alex. JOANNE STUBBE: Alex. My nephew's name is Alex. I'll remember that, OK? You're stuck. What's Kcat? AUDIENCE: It's like how quickly the enzyme turns over-- JOANNE STUBBE: Per active site. So it's called the turnover number. OK, so what does it tell you. It tells you how good your catalyst is, OK? So that's pretty important. So this is the turnover number. And I would also say it's pretty-- in the age of recombinant production or proteins, where we never isolate proteins from the normal source-- we isolate them all from bacteria or from yeast-- Kcat becomes really important to know, OK? So how do you know what the real Kcat should be if you isolate your enzyme from a protein that's expressed in E. coli? Do you think you get the real Kcat? Have you ever thought about that? Most people haven't, OK? So you're not alone. What could happen if you expressed your protein in a bacteria, or in another model organism like yeast? AUDIENCE: [INAUDIBLE]. JOANNE STUBBE: Yeah, it might not have the appropriate-- it's probably not-- it could be post-translational modification. It could be co-factors. So there are examples in the literature of very, very smart scientists who have spent 25 years of their life studying an enzyme that was only 1% active. So this is, in this course-- and I think in general in biochemistry-- you've got to go back and forth between the cell and what you see in the test tube. So this Kcat, if the number is 0.00001 per second, you have to have some intuition that tells you, oh, my god. That's so slow. Something-- something is wrong. So this number of turnover is incredibly important. It gives you a feeling for how good your catalyst is. But the number we're really after is the second example and the other limit. And what happens is s goes to 0, what happens to this equation. So those are the two extremes, OK? So as s goes to 0, OK, that's the other part of this equation. What happens to the equation? The rate of product formation is equal to-- and I'll write Vmax as KT times the concentration of total enzyme, OK? I didn't write it down. Hopefully you all know that. So what you now get is Kcat over Km times the concentration of e times the concentration of s. So what is this guy, if you look at this equation? What's Kcat over Km? What are the units? Kinetics isn't that hard. These are pretty-- if you think this is hard, wait till you start getting-- we're not going to go into derivation of steady state, pre-steady state analysis. But this is pretty simple compared to pre-steady state analysis. So what's Kcat over Km? What are the units? AUDIENCE: [INAUDIBLE]. JOANNE STUBBE: Yeah. Yeah. So it's second order rate constant. So that's the key thing. So what are you looking at? You can look at that by this equation. What you're looking at is the enzyme combining with the substrate, OK? And that's what we care about. That's the specificity, specificity, or efficiency of your reaction. So if you have a tRNA loosing and an tRNA phenylalanine, they're both competing for binding to the substrate. So the important parameter to think about that selection-- and that's why that's important at the end of this paper-- relates to Kcat over Km. It's the specificity or efficiency number. And if any of you ever work in a pharmaceutical industry, you'll find out that, of course, you never-- and you're looking for inhibitors, you never look at Kcat. Why don't you look at Kcat? You always look at Kcat over Km. Why is that true? Can anybody tell me? If you were looking for a drug, if you were looking for an antibiotic, fusidic acid that we talked about today that inhibits the EFG that Liz talked about, how would you set up the experiment to look for inhibition? What would you do with your concentration of substrate? Do you want it high or do you want it low? AUDIENCE: You want it low. JOANNE STUBBE: You want it low. Why do you want it low? AUDIENCE: Because [INAUDIBLE]. JOANNE STUBBE: Yeah, so if you're inhibitor is binding to the same site, and you have a huge amount of this, no matter what you do, even if this was a great inhibitor, if you had 10,000 times the amount of this, you're never going to see any inhibition. So understanding these simple principles-- which I can tell you there are people that don't get this in drug companies-- are pretty important, OK? So Kcat and Kcat over Km, boring. But it's not really so boring. It's sort of central to everything that you'll be thinking about over the course of the semester and almost all the modules in some form, although we won't highlight it like we're highlighting it here. OK, so, again, the reason we care about Kcat over Km is this question of selectivity. And I urge you to go back and look in the methods section of your paper. Now, this paper is packed full of stuff, OK? so as I said, I read it 20 times. Every time I read it, I find out something new. And furthermore, I think the paper-- how many of you found it a tough slog to go through this paper? This is probably the hardest paper you're going to look at in my opinion? Did you think it was well written? Did you get the ideas? OK, Did you all get the ideas or not, or where you completely confused, or you didn't spend enough time on it? How much time did you have to spend on it? AUDIENCE: Probably about an hour. JOANNE STUBBE: OK, an hour. OK, so I would say-- I read the paper 45 times, and it takes me two hours to read a paper like this. OK, so again, it's a question of what level you want to look at things. And I think part of what this course is about is looking at experimental details. You're want to see that. And the problems set, you're going to see that in lecture. You're going to see that probably next time when we continue to look at the primary data that they collected, how important it is to look at the axes, and not just looking at it rapidly. You really have to think about what the data is telling you. So this paper is complicated from my point of view because it's based on-- it's based on 15 other papers. OK, so for you to really believe what they say, which is what you need to do as a scientist, how to critically evaluate somebody else's data, you need to really go back-- and we didn't ask you to do that-- and really critically evaluate the earlier experiments they've done, because some of the conclusions they've done, when we look at the primary data, I could have drawn-- without knowing all that primary data, I could have drawn a conclusion completely different. So you see something and you've got to explain it, OK? And so when you start out, you have no idea. You have a very simple model. And in general, the model's almost always get more and more complex. That's what you're going to see over and over again. You start out as simple as possible, and then things get more complex. OK, so what we want to do now is ask the question. And I've just told you, you can't evaluate these individual rate constants. We just don't have enough variables, OK? We don't have enough that we can measure, that we can change the substrate concentration we can change, which changes the rate of product formation. So those are the two variables. But we have many more unknowns. We have k1, k2, k minus 2, k2, et cetera. So we can't evaluate these things. So the question is, is there any way you can start getting the primary rate constant, the numbers to the primary rate constants, OK? And so one way that people do this nowadays-- and when this paper was done, this was not an easy task. OK, now because of molecular biology where you can get large amounts of protein, it has become much more of an easy task-- you can get a large amount of protein-- you want to turn to the pre-steady state. So what I want to do very briefly is discuss the pre-steady state. I asked you to think about-- I asked you draw this out. This is one of the talking points in the questions I handed out. But in the steady state, we're over here. And the pre-steady state is before we get to the steady state. And does anybody have any idea what timescale you are on in that region of the curve? Is it hours? AUDIENCE: Milliseconds? JOANNE STUBBE: Yes, so it's milliseconds. So, fortunately, this didn't necessarily have to be true-- most enzymatic reactions occur. the catalysis occurs in that time regime, or maybe 0.1 milliseconds to milliseconds, allowing you to be able to use this method in an effort to try to understand what these-- evaluate what the rate constants are. And when you look at the table in the Rodnina paper, we're going to talk about where all those three constants came from, OK? Are they good or are they not good? But that's what you'd like to know for every system to really understand the question of fidelity, whether it's translation fidelity, DNA fidelity in replication, transcriptional fidelity. And you'll even see in Liz's section on polyketide syntases, which make natural products, you also have fidelity issues almost everywhere. So you'd like to be able to evaluate these things. And you can get a handle on this if you're a chemist and really care about the molecular details using potentially kinetics. So this is why kinetics is one of the first places that you actually start to think about what's going on in any reaction. OK, so let's say a few things about pre-steady state. I'm going to ask you a few questions, if I can remember what I'm going to ask you. OK, so OK, so let's suppose in this simple case, which I just covered up, this step is rate limiting, OK? So what is that step? Do you think it's common that a step like this-- so we have e plus s. And eventually, the enzyme gets recycled. I'm saying this is the rate-limiting step. Where is the chemical steps? Where are the chemical steps in this reaction? AUDIENCE: 2, 2. JOANNE STUBBE: Yeah, so k2. What are these steps over here, k1, and k minus 1, and k3? AUDIENCE: It's like association of the-- JOANNE STUBBE: Yeah, so the physical steps, OK? So as a chemist, and you're trying to understand what's going on, isn't it a problem if the rate limiting step is physical? It masks all the chemistry, OK? So what you see in this paper is they have to figure out clever ways to get around these kinds of issues. And you see that over and over again when you're studying enzyme systems because enzymes have have had billions of years to evolve. They are evolved. Their catalytic transformations are amazing. They go 10 to the 15th per second, right? That's totally mind boggling. Chemists can't come close. And so what happens then issued a limited by physical steps. So what we want to do is try and then look at the first part of this transformation. And basically, what we're then doing is using the enzyme sort of as a reagent. There are numbers of ways you can do this so that you can have a way of not looking at multiple turnovers because you can only look at one turnover if this is blocked in terms of product release. OK, so I think this product release is quite often the rate-limiting step in biological transformations. And what have you seen from reading the Rodnina paper? Have you seen conformational changes in thinking about the kinetic model we had up there before, and Liz had on the slide? Have you seen conformational changes? Is that part of the mechanism? Are they fast or are they slow? AUDIENCE: Wasn't that part of their reasoning that the difference between if you have a cognitive versus if you have a mismatched? That influences the rate of the reaction based on how it can affect the conformational change? JOANNE STUBBE: Wait, so that's exactly what's going to happen. So there are multiple places. How are you going to discriminate between two amino acids? Cognate and near cognate, whatever they are, will get to the data. The question is, how do you do that? And one of the steps is-- we talked about today GTP hydrolysis. But GTP hydrolysis is limited by a conformational change. And then once that go, the hydrolysis is very fast. And so it looks like the rate constant for GTP hydrolysis is the same as the rate constant for the conformational change. Where else have we seen a confirmation change the accommodation? AUDIENCE: Peptide. JOANNE STUBBE: Yeah, so peptide confirmation. This is shown here is this little cartoon where this red ball is the amino acid. It needs to reorient itself so it can form a peptide bond. So confirmation changes are all over the place in entomology. And if you look at the ribosome, do you think it's easy to tell with those conformational changes are from a molecular point of view? What do you think? Do you think it's easy or hard? AUDIENCE: Hard. JOANNE STUBBE: Very hard. OK, so here-- one of the most amazing things about the ribosome-- you've got to think this is amazing. You have this called the anti-codon loop way down here on the [INAUDIBLE]. And the GTPase is 80 angstroms away. And somehow, twiddling-- you saw this in class today-- these guys to form the right confirmation is transferred 80 Angstroms. And that triggers the reaction, rapid and irreversible. And the reaction goes to the right. You see this over and over and over again in these machines. OK, so this is a really important observation. How does that happen? Well, I think what's mindboggling about the ribosome-- again if you Google ribosome and elongation, you'll see we have another 150 papers published where people are trying to sort out-- because we have cryoem structures, we have stagnant crystallographic structures, we have single molecule stuff now. On top of all this model we have from Rodnina, people are trying to sort all those out because they care about how it works in some detail. So who ever would have thought we could get to the stage where we-- you've seen the pictures you saw in class today. Those pictures-- when I was your age, do you know how many structures there were? Maybe three. OK, we had hemoglobin. We had chymotrypsin. There were no structures. And why was that true? Because we had no molecular biology. So I used to spend three-- I'm digressing. This happens to me all the time. You're going to hate me for this. I'm going to get hammered for this. But I used to spend three months in the cold room isolating a microgram or protein, OK? And then molecular biology came in. And it's still not easy. And Liz will tell you what the issues are with purification of protein. But you can get grams of protein now in a day, OK? So there's been a revolution. And that allowed these crystallographic-- a pure material that crystallized more readily. And then the technology on top of that has really revolutionized what you can do. I think it's a very exciting time. And I think any of you who want to be biochemists, or are thinking about drug design, you really need to learn how to look at structures. So that was the first module. It takes a lot of practice. You need to figure that all out. OK, so pre-steady state-- so we're going to look at pre-steady state. And the goal is to evaluate the individual rate constants. OK, so that's the goal. And you may or may not be able to achieve this goal. But this happens, we just said, on the millisecond timescale. And so one of the questions-- and we're doing this under single turnover. OK, so let's look at a simple-- and we've just talked about it in the steady state. The concentration of the substrate has to be much, much greater than the concentration of the enzyme. And the enzyme concentrations are really low. So let's say we have an enzyme concentration of 0.01 micromolar, OK? So that's our enzyme concentration. And that would be typical if you would be using in a steady state assay if you have done those. And let's say that we're going to monitor this reaction by some kind of absorption change, a unique absorption change. So we're looking at absorption at some wavelength, OK? And let's say the extinction coefficient for this is 10 to the 4 per molar per centimeter. It would be ATP or coA. Then you can ask yourself the question, under these conditions, the change in absorption at whatever this wavelength is, is equal to the path length of light in centimeters times 10 to the minus 8th molar times 10 to the fourth molar per centimeter. OK, so what would your change in absorption be if you were measuring this in a single turnover? It would be really, really small, 0.0001. Can you measure that? Maybe you could measure this if you took hundreds of samples and you did a statistical analysis on it. But it's really low. So what do you want to do then to do pretty steady state? What's the thing to change so that you will be able to see something? AUDIENCE: [INAUDIBLE]. JOANNE STUBBE: Yeah, so you increase. So when you have this, and you can't see something-- and, obviously, it depends on what this extinction coefficient is-- but this is a pretty high extinction coefficient. So what you do is you increase the concentration of enzyme. And if we increase it, say, 1,000 fold, then then so we're now at 10 to the minus 5. Then now what is the change in absorption? The change in absorption is 0.1, which you can measure fairly easily in any kind of current instrumentation. So the thing is you have to be able to see. So the key thing with pre-steady state, and the reason you need to have large amounts of enzyme, is you need to be able to see what you're monitoring. So it's all about sensitivity. You need to see. And this usually implies increasing the concentration of the enzyme. OK, so what's the problem if you increase the concentration of the enzyme? Say we normally are at 0.01 micromolar steady state. We now are at 1,000 times higher. What's going to happen that makes the analysis complicated? If you increase the concentration of the enzyme, what does that do? AUDIENCE: You're going to burn through-- [INTERPOSING VOICES] JOANNE STUBBE: You're going to-- yeah, it increases the rate of the reaction because the rate of the reaction is proportional to the concentration of your catalyst. If you don't remember anything else out of this course, or anything in chemistry, that's pretty important no matter what area of chemistry you're in. So now what happens is the reaction is going like a bat out of hell instead of pipetting by hand. By the time you pipetted and put this into however you're observing it in the spectrophotometer, reaction's over, OK? So that's what the issue is, OK? So the sensitivity is key. And then the second thing you need to think about-- so sensitivity is one thing. And the other thing you need to think about is, what are the limitations of this method? How fast can the rate come-- on the millisecond timescale, what are the limitations in terms of the rate constants you can actually measure? So when you're looking at these reactions, you're looking at, in general, first order reactions. So all of these take place on the enzyme. So everything is stuck on the enzyme. So it's all first order. So the half life of the reaction is, if you go back and you look at your introductory kinetics, is 0.693 divided by k observed. And so if you had, say, a rate constant of 500 per second, then that gives you a half life of 1.5 milliseconds, OK? So that means you have to be able to make your measurement faster than that, OK? So the instrumentation we're going to be using can't do that. So the instrumentation we're doing-- so this would give you a half life if you calculate this. I don't even remember what the number is. But the dead time of the instruments that you would be using to make pre-steady state kinetics is approximately 2 milliseconds. So by the time you could stop looking at the reaction in some form, you know more than 50% of the reaction is gone, OK? So the rate constant then limits also what you can measure. So we asked this question before-- how could you modify this rate constant? What could you actually do? How could you make it so you might be able to say your rate consent was 500 per second-- you missed more than half your reaction. What could you potentially do so that you could monitor more of the reaction? What's the parameter that you would change? Kinetics. Think about kinetics. What do you always control in kinetics? AUDIENCE: Substrate concentration. JOANNE STUBBE: OK, you can control substrate concentration, but that's not the one I'm looking for. AUDIENCE: Temperature? JOANNE STUBBE: Temperature. Yeah. So in our body, we're at 37 degrees. That really is where you want to be making all of your measurements. In reality, many of the measurements are right on the edge. And so if you read the papers carefully, you'll see that people do lower the temperature, and that the rate of the reaction is related to the temperature. What's the problem with lowering the temperature? These are all things just you got to think about in the back of your mind. They're all playoffs in terms of how bad you want your experimental data and what the issues are with interpretation of data. Yeah? Rebecca. AUDIENCE: It makes it difficult to compare the different values. And you might not know exactly what the relationship between temperature and rate is, like if scales linearly. JOANNE STUBBE: OK, so that's true. You could have a huge conformational change that doesn't have erroneous behavior. I think quite frequently, most enzymes, they're designed to work at 37 degrees. And when you start cooling them down, they do weird things. So you could make the measurements, but then you have this issue-- I think, which is what you were saying, of how do you extrapolate that? So a lot of times you will change the temperature because that's the only way you could make the measurement. But the caveat is, like with everything, that you need to think about what the consequences of that actually are. OK, so the methods that we're going to be using Liz already introduced you to in class, not today, but the previous time. And so what you want to do, since you can't pipette on the millisecond time scale by hand, you want to have an instrument that allows you to control the rate of reaction by-- you put two different things in your syringes. And then you have an instrument-- push the two syringes into a chamber where they're mixed. Do you know what the rate-limiting step in this process is? It's the mixing process, OK? So the mixing processes is 2-millisecond dead time. I don't know if any of you have ever mixed something viscus with something not so viscus? What do you see? Have you ever made up a solution of glycerol? No? They probably give you all this in a kit nowadays. You don't have to make up your solutions of glycerol aluminum anymore. So this goes back to experimental design. And I'm not here to teach you how to do experimental design. But if you had very high concentration of the enzyme-- because we need a lot to be able to see something-- and you're mixing it against substrate-- you have something very viscous and something not viscous-- and when you mix them it takes a lot longer to remove all the lines from the mixing process. So experimental design, you really need to do some thinking about that. Once it gets into the mixer, the liquid in the mixer pushes a third syringe back. It fills the syringe up with liquid. It hits some kind of a stop position, which then triggers detection, OK? So that's what you're looking at. And the beauty of this method is it's continuous. And so what do you have to do to be able to look at this? What did they do in the case of the Rodnina paper? What kind of method did they use? Did you think about that? They described it, but you might not have thought about it in terms of experimental design. How did they monitor their reaction, one of their reactions? AUDIENCE: [INAUDIBLE]. JOANNE STUBBE: Yeah, so they are going to be able to somehow tag-- and this is a key thing, is how do you tag something in the right place so you can see a unique fluorescent change? That's not so easy, OK? So you mix this. You can monitor this continuously by change in fluorescence. If you had something that has a visible absorption, could you use that? What would limit that? Say, if you looked at tRNA synthetases that you talked about in class two times ago, where you were looking at ATP that isolates the amino acid to form the adenylate, which then reacts with the tRNA, could you use-- ATP as an absorption at 260. Could you use that absorption change? Do you remember what that equation is? Do you remember installation of amino acid? You're going to see this again in polyketide syntases. It's used quite frequently in biology. Nobody has any idea? Nebraska, how about you? OK, so here we have amino acid plus ATP. I'm digressing. But if you learn this part, you've learned something out of all of this, that forms the acyl adenylate. OK, how did they monitor this reaction? You discussed this in class. How do they monitor this reaction? AUDIENCE: [INAUDIBLE]. JOANNE STUBBE: You need to talk louder. You need to be assertive, OK? AUDIENCE: [INAUDIBLE]. JOANNE STUBBE: Yeah. So we're going to talk about radioactivity next time. This will be one of the methods we're going to be introduced to. Why couldn't they use ATP? AUDIENCE: The absorbent's different between [INAUDIBLE].. JOANNE STUBBE: Yeah, they're the same. Yeah. So you have to have a difference in absorption to be able to measure the visible. So the total absorption is due to the adenosine moiety which is the same in both molecules. OK, so you can't do that, OK? So that's one. And then let me just do one more thing, and then we'll quit. I still have another minute according to my watch. OK, so the second method, which they also used in the Rodnina paper is rapid chemical quench, OK? So, again, you have two things. You mix them. There's some plunger that allows the mixing. And then this is a discontinuous method. So what happens is you mix. And then you have to stop the reaction, OK? So you have a third syringe where you mix in something to stop the reaction. And then you have to analyze what comes out the other side. And this is where they're going to use radioactivity. And so this is rapid chemical quench. And how can you monitor a rapid chemical quench experiment? What's the best way to stop the reaction? What did they do in this paper? How else could you stop the reaction? Anybody got any ideas? So what what's the first criteria if you're going to stop the reaction? What does it have to be able to do? You just can't throw in anything, right? What is the key criteria for successful stopping? AUDIENCE: Something that will turn off the catalytic activity? JOANNE STUBBE: Yeah. And it has to be able to do it on a millisecond timescale. So you need millisecond stopping. OK, how could you millisec? How could you stop something on the millisecond? What would you use? Anybody got any ideas? So this is not trivial, actually. AUDIENCE: You could change the pH? JOANNE STUBBE: Yeah, so changing the pH. But so you could go acid or base. Acid works. In general, base doesn't work. So if you read the handout that I've given you, it does work. But it's much, much, much slower. And every base is different. Acid works all the time. There's another thing that you can use that is quite frequently used, especially with polymerases that work on nucleic acids. And that's EDTA. So this is a chelator and EDTA chelates the metal, which is essential for viability. So that also-- the chelation can occur in the millisecond timescale. So what we're going to do next time, I've sort of introduced you to the pre-steady state. The next time we'll come in. And we're going to look at the actual experiments. We'll look at fluorescence. We'll look at radioactivity and how you measure radioactivity. We're going to look at antibiotics, like you talked about today. We're going to look at non-hydrolyzable GTP analogs. If you look carefully at this paper, it's amazing how many methods they used to come up with this model. And that's one of the take home messages that you have to use many, many methods. And then you never get an exact solution to your equations. It's numerical integration of all the data that leads you to the model that they've actually used, OK?
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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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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 view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JOANNE STUBBE: Hi, everybody. You're in 507, and if you look at your syllabus, you'll find one of the things in the front page of your syllabus is called a lexicon. And I'd like to introduce you to why we've chosen to have a lexicon for this course. So if you look at this particular slide or overhead, this is what John is going to teach you this semester-- introductory metabolism. The glycolysis pathway, fatty acid biosynthesis, fatty acid oxidation, amino acid metabolism. A complete jungle. How are we ever going to learn anything out of this mess? Well, that's actually exactly the point. So what we're going to do over the course of the semester is convince you that all of this mess can be simplified to 10 basic reactions. And those 10 basic reactions are what's in the lexicon. So it turns out all of biochemistry for primary metabolism can be described using 10 different sets of reactions and your vitamin bottle. And so if you look at your vitamin bottle, what do you see? Most of you probably take vitamins. You have vitamin B1, vitamin B2, vitamin B3, vitamin B6, vitamin B12. All of those provide enzymes the catalysts for all the reactions in this complex metabolic pathway I showed you on the previous slide. They expand the repertoire of reactions that enzymes can actually catalyze. And so within the lexicon, what we're going to show you is the chemistry of actually how these vitamins work. So again, if we come back to the lexicon, we'll talk about how carbon-carbon bonds are made and broken, fatty acid metabolism, sugar metabolism. We'll talk about oxidation reduction reactions and the vitamins that are used for that transformation. We'll talk about the energy storage and the energy currency in the cell-- ATP, etc. So what the lexicon is meant to do is be an aid when you can't remember what does oxidation and reduction? You can go back to your lexicon and look up what are the redox cofactors that are involved in transformation. And by practicing the chemistry in the first part of the semester of all these reactions, metabolism should be very straightforward in terms of all the connectivities. So hopefully what you will do is look through your lexicon tonight, see what these reactions are, and then keep it by your side during the rest of the semester. And when you're having trouble understanding some chemical transformation, you can use the lexicon as a guide to think about the chemical transformations that you'll be looking at over the course of the semester.
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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, so we started talking about transition metals, and at the end of last time we'd given an introduction to all the sort of nomenclature and things you need to know, and we had just gotten up to d orbitals, and when you're talking about transition metals you're talking about d orbitals. So, we're going to start with a little review of d orbitals, some of you have seen this before, maybe some of you have not. But here is what you're going to need to know about the d orbitals. You need to know the names of all the d orbitals, you should be able to draw the shapes of the d orbitals, and so the bar is not too high for this. You see the drawings that are in your handouts, you should be able to do about that well, it doesn't have to be super fancy. But you should be able to draw their shapes, and you should also be able to recognize which d orbital it is -- if you have a picture of different d orbitals, you should be able to name that d orbital. So let's just review what the d orbitals look like. And in this class, we're always going to have the same reference frame, so we're always going to have the z-axis up and down, the y-axis is horizontal, and the x-axis is coming out of the board and going into the board. So, we'll always use that same description. It'll often be given, but if it's not, you can assume that's what we're talking about. So, the first d orbital we'll consider is d z squared. It has its maximum amplitude along the z-axis, and it also has a little donut in the x y plane. So, d x squared minus y squared has its maximum amplitude along the x and the y axis, so directly on-axis for this particular d orbital. The next sets of d orbitals have their maximum amplitudes off-axis, so they don't correspond directly to the axes that I just mentioned, they're, in fact, 45 degrees off-axis. So we have the d y z shown here, d x z shown here, so maximum amplitude 45 degrees off z and x axes, and then z -- these pictures are a little bit complicated to see -- d x y, so it's 45 degrees off of the x and the y. And so, here I tried to draw them all in the same orientation of axes, which is a little bit difficult. So now let's look at them in terms of where they're drawn so you can kind of see them a little bit better, and so why don't you try to learn to recognize all of these. So, what is this one called? d z squared. What is this one, so its maximum amplitude is along x and y? d x squared minus y squared. I think this picture might be a little -- you might have somewhat aversion later on, but this is just good practice for you in recognizing them. So here we have one 45 degrees off-axis, which one is this? Yup. And this one over here? Yup. So, y z. And this last one here? Yup, we have the x z, so it's going up and down for the for z-axis. So, a little more practice now. To show what these look like again, you want to think in three dimensions, and on paper and in most the time in Powerpoint you're not in three dimension, so here's a little movie in three dimensions. Here you can really see that donut, so this is d x squared -- see the maximum amplitude along z-axis here, and down here, and the little donut in the x y plane. So this one is d x squared minus y squared. The maximum amplitudes are right directly along axis, so that allows you to distinguish it from d x y. So when it's on-axis here, it's the x squared minus y squared. So moving along here, so this is d x y, so you see that it's in that axis but it's not directly on-axis -- the maximum amplitude is 45 degrees off, so the orbitals are in between the axes there. Now we're looking at one that has a z in it, and it looks like it's x z, so that's where our maximum amplitude is between the x and the z-axes, 45 degrees off. And our last one, we have y and z here. Again, 45 degrees off-axis between the y-axis and z-axis. So, hopefully these little movies will help cement in your brain, what the shapes of these d orbitals are. All right, so that's d orbitals, and we're going to be mentioning d orbitals in every lecture in this, and you have to be thinking about what the shapes of the d orbitals are to talk about today's topic, which is crystal field theory. So, there are two types of theories that you may hear of and that your book mentions -- crystal field theory, and ligand field theory, and like most things that you learn about in freshman chemistry, the theories were developed to explain experimental information. So there are special properties of coordination complexes, so that's where you have a transition metal in the middle and you have ligands all around it, so you have these coordination complexes and they have special properties. And so people wanted to try to rationalize these special properties and they came up with these two theories. So, the basic idea behind these theories is that when you place a metal ion with the particular oxidation number in the center of a coordination sphere, and you have all these ligands, these donor ligands, all surrounding them, that the energy of the d orbitals is going to be altered by the position of those ligands. So it's all about the d orbitals, and the d orbitals are going to experience some influence from these ligands, these donor ligands that are surrounding the metal. So then, between these two theories that are used to explain how these d orbitals are being affected. The crystal field theory is based on an ionic description, so it considers the ligands as negative point charges. It's a very simplified model, whereas as the ligand field theory considers covalent, as well as ionic aspects of coordination. It's more powerful it's more useful, but it's also a bit more complex, and so we don't cover it in this of course, and if you go on and take the first level of inorganic chemistry, which is 503, then you'll hear about this. But for this course, we're just going to talk about crystal field theory. Even though it's very much of a simplified model, it actually works very well. You can explain quite a few properties of coordination complexes just using this simplified method. So, crystal field theory, again, very simple. It's just considering the ionic interactions, it considers the ligands as negative point charges. And so, the basic idea is that ligands, as negative point charges, are going to have repulsive effects if they get close to the d orbitals. So here is a drawing of a metal, and so this is metal abbreviated m, its oxidation number is m plus here, and it has ligands all around it. What is the geometry here? It's octahedral geometry. And so we have ligands up and down along z, ligands along y, and a ligand going back along x, and a ligand coming out along x. And so here's another picture of the same thing, the metal is in the middle, and the ligands -- in this case, you have these ammonia ligands or these little negative point charges, which are all along the axes. You have four along the equatorial, and one up and one down, so this is the octahedral geometry. And so you can just think about each of these ligands as negative point charges. And so, if the negative point charge is pointing right toward a d orbital, that'll be very repulsive. If it's not it's less repulsive. That's the whole idea behind this crystal field theory. So, here again, is just another little picture, so you can kind of get the idea that we're going to be thinking about the all the shapes of the d orbitals, and we're going to think about where the ligands are. Today we're going to talk about octahedral geometry, but we're also going to go on and talk about tetrahedral geometry later. So here in octahedral geometry, you can think about the positions of all of these negative point charges surrounding your d orbitals. And when the d orbitals are on axis, like the ligands, there's going to be more repulsion, so you can see here that would be quite repulsive -- you have a negative point charge by that d orbital. When the d orbitals are off-axis and the ligands are on-axis, that's less repulsive. And that's the basic idea. So, let's look at each one of these orbitals now in detail and think about how a ligands that's pointing directly toward it is going to be affected. So we have the ligands, l, as these point charges directed toward the d z squared, and the d x squared minus y squared orbitals, and these would result in quite a bit of repulsion. So if you had a ligand right up here along z, and so that would be a very close interaction. In this case, you're going to have ligands along x and y, again pointing directly toward the orbitals, that would be quite repulsive. And I'll just mention, we'll come back to this later, that one can think about the case where you have the octahedral geometry where the ligands are in a definite position, and you can also think about this sort of hypothetical case where you have a metal in the middle and you have the ligands, here are the little ligands, and they're everywhere, there's ligands everywhere all around. And so, in this case where you have ligands everywhere all around your metal, then all your d orbitals would have the same energy, but if you take the ligands and you isolate them in particular positions, then you can consider how the different shapes of the d orbitals will be affected. We'll come back to that in a minute. All right, so here we have a case where our ligands are on-axis, our orbitals on-axis, this is a large repulsion. So, I will tell you that d x squared and d z squared and d x squared minus y squared orbitals are destabilized, and they are destabilized by the same amount. So there's repulsion now, and so they're destabilized, and they're destabilized by the same amount of energy. So what's it called when orbitals are of the same energy? Yup. So, d z squared and d x squared minus y squared are degenerate. So, d z squared and d x squared minus y squared orbitals are destabilized more than the other three orbitals, and let's consider now why that is true. So here are our other sets of orbitals, and remember, here the maximum amplitude of these orbitals are 45 degrees off-axis, whereas our ligands are all on-axis. So, the ligand negative charges are directed in between these orbitals, not directly toward them. So that is stabilized compared to this hypothetical case where the ligands are everywhere, so some of them will be pointing toward them, and also stabilized compared to the other sets of orbitals where the ligands are now pointing directly at them. So these three sets of orbitals are stabilized relative to the d z squared and the d x squared minus y squared orbitals, and they're stabilized by the same amount. So these three orbitals are also degenerate with respect to each other. So then to sort of summarize this set of orbitals, we have for d z squared and d x squared minus y squared, we have large repulsions by those negative point charges, they're pointing directly at the orbitals, and so they're destabilized, higher in energy than the other -- the d x y, d y z and d x z. For the d y z, d x z and d x y, they're smaller repulsion, because these orbitals are off-axis, and so the negative point charges aren't pointing directly at them. So they're stabilized relative to these guys up here. So that's the whole idea behind an octahedral case of crystal field theory. And we can look at this just one other way, if pictures help you. Here it's a little clearer that those negative point charges are pointing directly toward the orbitals, here I think you can see that the negative point charges are not directly pointing toward any of the orbitals. So I'll show you a bunch of different figures, this all shows you the same thing, but some might help you see this relationship better. OK, so now we're going to draw some diagrams. I'm going to start over here. So we're going to draw what's called a crystal field splitting diagram, and this is for an octahedral case. And the diagrams are going to look different depending on what the geometry is. So when we draw this diagram, energy is going up, and we're going to start with our 5 d orbitals, and so this is going to be the average energy, the average energy of our d orbitals. And so, this is then with a spherical crystal field. So that's where the ligands are distributed around uniformly. So it's all spherical, they aren't set up in the octahedral case yet, our octahedral diagram's going to be over here, but this is the case that this represents. If you have all your ligands spherically distributed around your metal, then the energy of all the d orbitals are identical, because every d orbital has the same amount of ligands, it's uniform, it's symmetrical all around the metal. And I just want to tell you that this is was very exciting to me when I saw this. I've been teaching this class for a while, and I never had a real spherical crystal field around my metal before. And then I walked into Walgreens one day, and I was very excited to see that Walgreens sold spherical crystal fields. I mean you never know what you're going to get. I'm a big fan of Walgreens, I've found a lot of good stuff, toys for my dog, etcetera, but this was really amazing. So I asked the cashier on the way out whether they knew they were selling spherical crystal fields, and they did not actually. So, you just never know what you're going to get. OK, so in that case, where the ligands are uniform all around, the energy is the same. But now, if we have an octahedral crystal field over here, so we have our octahedral crystal field, then we get some splitting. So some of our orbitals are going to be destabilized, and they'll be higher in energy here. So we have the d x squared minus y squared, and d z squared over here are going to be higher in energy. And we're going to have three that lower in energy, so we'll have our d x y, our d y z, and our d x z over here, will be lower in energy. This difference is called the octahedral field splitting energy, because it's the amount of energy that the octahedral field is split. So over here, we can put this is for the octahedral case, crystal field splitting energy. And again, some of the orbitals go up in energy, some of the orbitals go down in energy, and the overall energy needs to be conserved. So, if two orbitals go up in energy, and three go down in energy, then to have everything add up, you can say that three go up in energy by 3/5, and two, these three orbitals are going to go down by 2/5. So overall, the energy of the system is maintained. OK, so that's a crystal field splitting diagram for an octahedral case, and now let's look at some examples of this. So let's look at an example, and we're going to have a chromium system that has three n h 3 ligands and three b r ligands. Now, you tell me what the d count of that is. Let's just take 10 more seconds. They're not as high overall, but still more people got the right answer. So, let's take a look at this. What's the oxidation number of bromium? What is it? Bromium? What's 1 b r minus? What's its oxidation number. Minus 1. There are three of them. What about ammonia? 0. So, 3 times 0. And the overall charge of this is 0, so there's nothing up there. So what does that mean about chromium? Plus 3. All right, so now we have to figure out the d count. So the d count is going to equal the -- and the periodic table, the group number, so if you switch to my slides, we can see what that is. So what is that for chromium? 6. And then we have 6 minus 3, because our oxidation number is 3, and so we have a d 3 system. So, did some of you get this wrong because you stopped too early? Here, the answer, if you had 3, you could have stopped with oxidation number and still gotten it correct. So that's a d 3 system. So, we're going to worry about three d electrons, and we're going to put three d electrons into our splitting diagram. OK, so if you had a hypothetical spherical crystal field, you would have your one's in here, but now let's consider what happens in the octahedral case. So we can come down here. Am I going to put my first electron down here or up here? Down. Oh, I just realized that I didn't put two things on this diagram, so these are in your notes, but these diagrams have little abbreviations in them for the orbital levels. So we have an e g and t 2 g, and that's an abbreviation for the names of the ts of orbitals, which you'll see later is very convenient in terms of writing things out. All right, so we're going to put them down here. Am I going to put two of them together with the spins up in the first orbital? No. So you know that that is not good, give you the same four quantum numbers, you don't want to do that. So, we can put them in. What about just putting in a paired set over here yet? No. They have the same energies here, so we're going to put them in all singly, and then we're done. So we put three electrons in these three orbitals. Now we can introduce a couple other terms, which is where I realized I forgot to put the labels on. So, you'll often be asked for -- OK, you'll often be asked for something called the d n electron configuration. And so here you can use the abbreviations for the orbitals, so we point three electrons in to the t 2 g orbitals, so we can just say that's t 2 g raised to the three. So that let's someone know that you have three electron in the set of orbitals that are stabilized in an octahedral crystal field. Then we can consider something else that's called c f s e, and I think I should have -- OK, so I'll write that out. So, this is the crystal field stabilization energy. So it's not the crystal field splitting energy, it's the stabilization energy, which indicates how much those electrons are stabilized by being in an octahedral field, rather than this hypothetical spherical crystal field. And so what we can do there is you see that you have three electrons in those lower sets of orbitals, and those orbitals are stabilized by 2/5 times the octahedral crystal field splitting energy. And so that gives an answer of minus 6/5 times the octahedral crystal field splitting energy. So that's how much those electrons are stabilized. So they are lower in energy -- see, the average energy is much, much higher in this hypothetical case, but because of having this octahedral geometry, and there are only three electrons to consider, they all go into the stabilized energy, and so they're stabilized by minus 6/5 times whatever the octahedral crystal field splitting energy is for this particular case. So, now let's look at another example. So let's look at an example of a coordination complex -- you have manganese and you have six water ligands and some chlorides hanging around. So why don't you tell me what the oxidation number is for this now -- not the d count, but the oxidation number? OK, so let's just take 10 more seconds. OK, so let's just look at that one for a minute, people did very well on that. So, what's the overall charge in this coordination complex? So we can write this out here. Our six ligands and we say that here it's plus 3 overall, because we have three chloride ions hanging around with a negative charge. So this tells us that the overall charge on the coordination complex had to be plus 3. And this again is 0, so that means that is plus 3. So, people did very well on that. All right, so then what is the d count? What is it? 4, right. So we have 7 minus 3 is 4, so we have a d 4 system. All right, so now, if you look up there, we have to make a decision about that fourth electron. three electrons were easy, four makes it complicated. Do we put the fourth one down in the lower set or do we go up? So there are two possibilities here. So I have two diagrams drawn over here. And you might be in a case where you have a small crystal field splitting energy, or you might be in a case where you have a large crystal field splitting energy. And so, there are two different ways the electrons can go. So over here where you have a small crystal field splitting energy, that's called a weak field. And when you have a big splitting energy, that's a strong field. So, with the weak field there's not that much of an energy difference between them. And so, when you're putting in, you do your first three electrons, that's always going to be the same. But then the fourth electron, in this case, if there's not a big difference in the energy, if it's pretty small, if it's a weak field, you can put that fourth one up there. Because it takes energy to pair the electrons up, and so you're asking the question, does it take more energy to pair them, or does it take more energy to put one in the upper set? And for a weak field you say that the crystal field splitting energy is smaller than the pairing energy or p e, so p e is the pairing energy. And so it takes more energy to pair than it does to bump one electron up to the higher level. So that's what a weak field situation would look like. Now in a strong field situation, boy, there's a big splitting difference, a big energy difference here. So in this case, the crystal field splitting is much larger than p e, that pairing energy for the electron. So it's better to pair, then to put one up. I can't even reach those, that's really far up. So I'm not going to do that, I'm just going to put them in here. I'll put the first three in, and then the fourth one is going to go down where I can reach it. I don't have the energy to put it up there. So that's a strong field. Weak field I can handle, strong field I'm going to try pair them all up. So, now we can write the different d n electron configurations for these two. So, in this case, if we have our d n electron configuration, so we have three in the t 2 g, we have three. And in the e g set we have one, so that is our electron configuration for this weak field case. And for the strong field case, we didn't put any up in the e g, so we just have t 2 g, four electrons in that set of orbitals. OK, so let's just put up what we've done here, and introduce another term, which are high spin and low spin. So we have these two cases here, and again, we're considering how big is this octahedral crystal field splitting energy compared to a pairing energy. The energy involved in pairing electrons together. In a weak field, the splitting energy is small, so electrons are placed singly with spins parallel to the fullest extent in all the sets of orbitals. In this other case with the strong field, the pairing energy is smaller than the splitting energy -- strong field you have a big splitting energy. And so, in that case, with the splitting energy is large, you're going to put all your electrons in and pair them up in t 2 g and don't put any electrons in the e g sets of orbitals until you completely filled your t 2 g set. So, the net result of this is for a weak field, you have the maximum number of unpaired electrons, so see, you have a maximum number, you have four electrons, all four are unpaired. So that's the maximum number of unpaired electrons, and this is referred to as high spin. And in the other case, you have the minimum number of unpaired electrons that you can have, and so you have this one set that's paired here, so that would be considered a low spin case. We can also talk about the stabilization energy of these two cases, and so we have the crystal field stabilization energy. So why don't you go ahead and tell me for a weak field case, what is our stabilization energy? Let's just take 10 more seconds. OK, pretty good. So let's work this out. So, first we consider how many electrons we have in the lower set of orbitals, so we have three. Those three are stabilized by 2/5. And then we have one in the upper set, so we have one at 3/5 times the octahedral crystal field splitting energy. So this ends up with minus 3/5 times the octahedral crystal field splitting energy. And notice why a is not correct -- you don't have the symbol for the octahedral crystal field splitting energy. On a test, you need to make sure that you remember to write this term, so this was a little test so that you hopefully emphasize that you want to have that there. All right, so that's for this one. Let's look at the low spin case, the strong field case, and do our crystal field splitting energy for that. So in this case we're going to have four electrons times minus 2/5 times the octahedral crystal field splitting energy, so that's equal to minus 8/5 times the octahedral crystal field splitting energy, and some books will also indicate, and they'll say plus p e to indicate that there's pairing energy that's associated there. So it's not quite as beneficial as one might think. You do have a lot of electrons in these in lower energy orbitals, but you did have to pair some of them, so you had one, and if two of them are paired, you can see sometimes 2 p e. So, some books will do that, some books will not. So I just wanted to mention that both are OK. And whether you're asked to write that or not, and the question will say include the pairing energy, so you know whether you're supposed to do that or not, that there is that energy associated with pairing in this strong field case. OK, so let's look at another example, and then you should be all set to do these problems for an octahedral field. So we'll take our electrons down. All right, so let's at a case of cobalt plus 2. So let's consider how many we're going to put in, the oxidation number is? Plus 2. And what is the d count then? What group number? 9 minus 2 is 7, so we're doing a d 7 system. All right, so now for the weak field case here, why don't you tell me what is the correct electron distribution. OK, let's just take 10 more seconds. OK, very good. This is the weak field case, so in this case, we are going to fill up singly all of the orbitals, because the splitting energy is less than the pairing energy, so we're going to put them to the fullest extent possible before we have to pair. So we have 7, so we're going to do 1, 2, 3, 4, 5, and now we have to pair. We have no -- we've used up all our orbitals, so we're going to start pairing down here in the lower energy, so that would be 6, 7. So we had no choice, we had to pair them, but because it was a weak field, we filled them all up singly first before we paired. So now let's consider what we're going to do down here. Now remember, this is a strong field, so there's a big splitting energy, so it takes less energy to pair them than to reach this higher level. So first we put them in the same way -- 1, 2, 3, but now that we have the 3 in, it's better to pair than to bring them up here. So we'll do 1, 2, 3, we'll pair them all up. Now it's all filled, we have no other choice but to go up here. So, we have these very different cases depending on the splitting energy. So what controls the splitting energy? Well, what controls the splitting energy is nature of the ligands, so we're going to be talking about that next time, and you'll recognize for certain kinds of ligands you're going to have a strong field, and other kinds you'll have a weak field. But right now we're not talking about that, we're just showing those two possibilities. So, for this system, is this going to be a high spin or a low spin system? So, this will be high spin, because we have the maximum amount of unpaired electrons. And over here we're going to have a low spin system. We have the minimum number possible of the unpaired electrons. All right, so let's finish this up now, and do our d n electron configurations and our crystal field splitting energies. So, for this case, we're going to have in our t 2 g system, how many? 5. And in e g? 2. And for our splitting energy, we have 5 times minus 2/5 times the octahedral crystal field splitting energy plus 2 times plus 3/5, and what does that end up equaling? Minus 4/5 times the octahedral crystal field splitting energy. And we could also optionally put 2 p e because we have two sets paired. All right, so let's look at our strong field system. How many do we have in our e 2 g set? 6. What about e g? 1. And for our splitting energy then, we have 6 times minus 2/5 times the octahedral crystal field splitting energy plus 1 times 3/5, and what is that going to equal? Minus what? 9/5. And how many pairing energies? three pairing energies, great. I think you have that part down. Next time we get more complicated, we're going to talk about types of ligands, we're going to talk about tetrahedral, we're going to talk about square planar, and that's, of course, after the exam on Wednesday. So, good luck, everyone, with the exam on Wednesday.
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https://ocw.mit.edu/courses/20-020-introduction-to-biological-engineering-design-spring-2009/20.020-spring-2009.zip
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Sally: This is Systems Sally. Dude: Hey Sally. It's Dude. Sally: Dude, are you OK? You sound out of breath. Dude: Well I started growing some bacterial cells with Izzy last night, and today I had to get to practice but coach kept us late and now I'm worried that the cells. Sally: Dude? You still there? The cells what? Dude Well I guess I don't really know. Izzy said dog phase but that seems wrong. Sally: Izzy probably said log phase which is when the cells are actively growing and dividing. If you have some bacteria growing, I'm happy to put the cultures on ice. That will slow down their growth, and you can measure their density when you get here. Dude: Thanks Sally! Last night we put two flasks at room temperature and two in the 37 degrees incubator. I'll be there as fast as I can. Izzy: Hi Sally. Have you seen Dude? Sally: He just called to say he's on his way. I'm putting these cultures on ice, but look: the ones that were at room temperature don't seem to have grown much. Izzy: I was a little afraid of that. They carry a high-copy protein generator that Dude and I made, but the growth of the cells takes a real hit when they carry this plasmid. We've seen them spend 12 hours in lag phase and then take two days to reach stationary phase. Without the plasmid, it's more like 4 hours to go from lag to log and they're saturated overnight. Sally: When did you start these growing? Izzy: About 8:00 last night we diluted the strain 1:1000. And I don't think these have grown much since then, but I'll measure their density anyway. Sally: The cells that were growing at 37 degrees look a little more dense. Maybe they're doubling at an almost normal rate. Izzy: Yup. Their density has definitely changed since yesterday and if Dude gets here fast, we might be able to catch them in log phase and transform in a control device. Sally: You might also want to count the cells since some of the light scattering you're measuring at the spectrophotometer could be coming from cell debris. Izzy: I hadn't thought about that, but you're right. I'll teach Dude to count cells when he gets here Sally: One other thing that I've seen happen is for the cells to lose a plasmid if it's making them sick. You and Dude might want to check that the cells are still resistant to whatever antibiotic marker you have on the plasmid. Dude: Hi Sally! Hi Izzy! Sorry to be late. I hope you've left me some things to do. Izzy: In lab there's always plenty to do. Let's get started figuring out what's going on Dude: Getting here, I saw the most amazing pack of dogs getting walked over the bridge.
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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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PRESENTER: 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 we begin with the harmonic oscillator. And before we get into the harmonic oscillator, I want to touch on a few concepts that have been mentioned in class and just elaborate on them. It is the issue of nodes, and how solutions look at, and why solutions have more and more nodes, why the ground state has no nodes. This kind of stuff. So these are just a collection of remarks and an argument for you to understand a little more intuitively why these properties hold. So one first thing I want to mention is, if you have a Schrodinger equation for an energy eigenstate. Schrodinger equation for an energy eigenstate. You have an equation of the from minus h squared over 2m. d second dx squared psi of x plus v of x psi of x equal e times psi of x. Now the issue of this equation is that you're trying to solve for two things at the same time. If you're looking at what we call bound states, now what is a bound state? A bound state is something that is not extended all that much. So a bound state will be a wave function that goes to 0 as the absolute value of x goes to infinity. So it's a probability function that certainly doesn't extend all the way to infinity. It just collapses. It's normalizable. So these are bound states, and you're looking for bound states of this equation. And your difficulty is that you don't know psi, and you don't know E either. So you have to solve a problem in which, if you were thinking oh this is just a plain differential equation, give me the value of E. We know the potential, just calculate it. That's not the way it works in quantum mechanics, because you need to have normalizable solutions. So at the end of the day, as will be very clear today, this E gets fixed. You cannot get arbitrary values of E's. So I want to make a couple of remarks about this equation. Is that there's this thing that can't happen. Certainly, if v of x is a smooth potential, then if you observer that the wave function vanishes at some point, and the derivative of the wave function vanishes at that same point, these two things imply that psi of x is identically 0. And therefore it means that you really are not interested in that. That's not a solution of the Schrodinger equation. Psi equals 0 is obviously solves this, but it's not interesting. It doesn't represent the particle. So what I claim here is that, if it happens to be that you're solving the Schrodinger problem with some potential that is smooth, you can take derivatives of it. And then you encounter that the wave function vanishes at some point, and its slope vanishes at that same point. Then the wave function vanishes completely. So you cannot have a wave function, a psi of x that does the following. Comes down here. It becomes an inflection point and goes down. This is not allowed. If the wave function vanishes at some point, then the wave function is going to do this. It's going to hit at an angle, because you cannot have that the wave function is 0 and needs the derivative 0 at the same point. And the reason is simple. I'm not going to prove it now here. It is that you have a second order differential equation, and a second order differential equation is completely determined by knowing the function at the point and the the derivative at the point. And if both are 0s, like the most trivial kind of initial condition, the only solution consistent with this is psi equals 0 everywhere. So this can't happen. And it's something good for you to remember. If you have to do a plot of a wave function, you should never have this. So this is what we call the node in a wave function. It's a place where the wave function vanishes, and the derivative of the wave function better not vanish at that point. So this is one claim that it's not hard to prove, but we just don't try to do it. And there's another claim that I want you to be aware of. That for bound states in one dimension, in the kind of thing that we're doing now in one dimension, no degeneracy is possible. What do I mean by that? You will never find two bound states of a potential that are different that have the same energy. It's just absolutely impossible. It's a very wonderful result, but also I'm not going to prove it here. Maybe it will be given as an exercise later in the course. And it's discussed in 805 as well. But that's another statement that is very important. There's no degeneracy. Now you've looked at this simple potential, the square well infinite one. And how does it look? You have x going from 0 to a. And the potential is 0. From 0 to a is infinite otherwise. The particle is bound to stay inside the two walls of this potential. So in here we've plotted the potential as a function of x. Here is x. And the wave functions are things that you know already. Yes? AUDIENCE: Is it true if you have two wells next to each other that there's still no degeneracy if it's an infinite barrier? PROFESSOR: If there's two wells and there's an infinite barrier between them, it's like having two universes. So it's not really a one dimensional problem. If you have an infinite barrier like two worlds that can talk to each other. So yes, then you would have degeneracy. It's like saying you can have here one atom of hydrogen or something in one energy level, here another one. They don't talk to each other. They're degenerate states. But in general, we're talking about normal potentials that are preferably smooth. Most of these things are true even if they're not smooth. But it's a little more delicate. But certainly two potentials that are separated by an infinite barrier is not part of what we really want to consider. OK so these wave functions start with m equals 0, 1, 2, 3, and psi ends of x are square root of 2 over a sin n plus 1 pi x over a. Things that you've seen already. And En, the energies are ever growing as a function of the integer n that characterizes them. 2ma squared. And the thing that you notice is that psi 0 has technically no nodes. That is to say these wave functions have to varnish at the end, because the potential becomes infinite, which means a particle really can go through. The wave function has to be continuous. There cannot be any wave function to the left. So it has to vanish here. These things don't count as nodes. It is like a bound state has to vanish at infinity. And that's not what we count the node. A node is somewhere in the middle of the range of x where the wave function vanishes. So this is the ground state. This is psi zero has no nodes. Psi one would be something like that, has one node. And try the next ones. They have more and more nodes. So psi n has n nodes. And the interesting thing is that this result actually is true for extremely general potentials. You don't have to just do the square well to see that the ground state has no nodes. That first excited state was one node, and so on and so forth. It's true in general. This is actually a very nice result, but its difficult to prove. In fact, it's pretty hard to prove. So there is a nice argument. Not 100% rigorous, but thoroughly nice and really physical that I'm going to present to you why this result is true. So let's try to do that. So here is the general case. So I'm going to take a smooth v of x. That will be of this kind. The potential, here is x, and this potential is going to be like that. Smooth all over. And I will actually have it that it actually goes to infinity as x goes to infinity. Many of these things are really not necessary, but it simplifies our life. OK, so here is a result that it's known to be true. If this thing grows to infinity, the potential never stops growing, you get infinite number of bound states at fixed energies. One energy, two energy, three energy, infinite number of bound states. Number of bound states. That's a fact. I will not try to prove it. We'll do it for the harmonic oscillator. We'll see those infinite number of states, but here we can't prove it easily. Nevertheless, what I want to argue for you is that these states, as the first one, will have no nodes. The second state, the first excited state, will have one node. Next will have two nodes, three nodes, four nodes. We want to understand what is this issue of the nodes. OK. You're not trying to prove everything. But we're trying to prove or understand something that is very important. So how do we prove this? Or how do we understand that the nodes-- so there will be an infinite number of bound states, psi 0, psi 1, psi 2, up to psi n, and it goes on. And psi n has n nodes. All right. So what I'm going to do, in order to understand this, is I'm going to produce what we will call screened potentials. Screened potentials. I'm going to select the lowest point of the potential here for convenience. And I'm going to call it x equals 0 is the lowest point. And the screen potential will have a parameter a. It's a potential which is equal to v of x if the absolute value of x is less than a. And its infinity if the absolute value of x is greater than a. So I come here, and I want to see this is this potential, v of x, what is the screened potential for sum a? Well, I wanted colored chalk, but I don't have it. I go mark a here minus a. Here are the points between x and a. Absolute value of x less than a. Throughout this region the screened potential is the potential that you have. Nevertheless, for the rest, its infinite. So the screened potential is this thing. Is infinite there, and it's here this thing. So it's just some potential. You take it a screen and you just see one part of the potential, and let it go to infinity. So that's a screen potential. So now what I'm going to do is that I'm going to try to argue that you could try to find the bound state of the screen potential. Unless you remove the screen, you will find, as you let a go to infinity, you will find the bound states of the original potential. It's reasonable that that's true, because as you remove the screen, you're letting more of the potential be exposed, and more of the potential be exposed. And the wave functions eventually die, so as the time that you're very far away, you affect the wave functions less and less. So that's the argument. We're going to try to argue that we're going to look at the bound states of the screened potentials and see what happened, whether they tell us about the bound states the original potential. So for this, I'm going to begin with a screen potential in which a goes to 0 and say that a is equal to epsilon, very small. So what potential so do I have? A very tiny potential here from epsilon to minus epsilon. Now I chose the original point down here to be the minimum. So actually, the bottom part of the potential is really flat. And if you take epsilon going to 0, well, the potential might do this, but really at the bottom for sufficiently small epsilon, this is an infinite square well with psis to epsilon. I chose the minimum so that you don't get something like this. If it would be a point with a slope, this would be an ugly thing. So let's choose the minimum. And we have the screen potential here, and that's it. Now look what we do. We say all right, here there is a ground state. Very tiny. Goes like that. Vanishes here. Vanishes there. And has no nodes. Very tiny. You know the two 0s are very close to each other. And now I'm going to try to increase the value of the screen a. So suppose we've increased the screen, and now the potential is here. And now we have a finite screen. Here is the potential. And I look at the wave function. How it looks. Here is psi 0. This ground state psi 0. Well, since this thing in here, the potential becomes infinite, the wave function still must vanish here and still must vanish here. Now just for your imagination, think of this. At this stage, it still more or less looks like this. Maybe. Now I'm going to ask, as I increase, can I produce a node? And look what's going to happen. So suppose it might happen that, as you increase, suddenly you produce a node. So here's what I'm saying here. I'm going to show it here. Suppose up to this point, there is no node. But then when I double it, when I increase it to twice the size, when I go to screen potential like that, suddenly there is a node in the middle. So if there is a node in the middle, one thing that could have happened is that you have this. And now look what must have happened then. As I stretch this, this slope must have been going down, and down, and down, until it flips to the other side to produce a node here. It could have happened on this side, but it's the same, so the argument is just done with this side. To produce a node you could have done somehow the slope here must have changed sine. But for that to happen continuously, at some point the this slope must have been 0. But you cannot have a 0 and 0 slope. So this thing can't flip, can't do this. Another thing that could have happened is that when we are here already, maybe the wave function looks like that. It doesn't flip at the edges, but produces something like that. But the only way this can happen continuously, and this potential is changing continuously, is for this thing at some intermediate stage, as you keep stretching the screen, this sort of starts to produce a depression here. And at some point, to get here it has to do this. But it can't do this either. It cannot vanish and have derivative like that. So actually, as you stretch the screen, there's no way to produce a node. That property forbids it. So by the time you go and take the screen to infinity, this wave function has no nodes. So that proves it that the ground state has no nodes. You could call this a physicist proof, which means-- not in the pejorative way. It means that it's reasonable, it's intuitive, and a mathematician working hard could make it rigorous. A bad physicist proof is one that is a little sloppy and no mathematician could fix it and make it work. So I think this is a good physics proof in that sense. Probably you can construct a real proof, or based on this, a very precise proof. Now look at excited states. Suppose you take now here this screen very little, and now consider the third excited state, psi three. I'm sorry, we'll call this psi 2 because it has two nodes. Well, maybe I should do psi 1. Psi 1. One node. Same thing. As you increase it, there's no way to create another node continuously. Because again, you have to flip at the edges, or you have to depress in the middle. So this one will evolve to a wave function that will have one node in the whole big potential. Now stayed does that state have more energy than the ground state? Well, it certainly begins with a small screen with more energy, because in the square well psi 1 has more energy. And that energy should be clear that it's not going to go below the energy of the ground state. Why? Because if it went below the energy of the ground state slowly, at some point for some value of the screen, it would have the same energy as the ground state. But no degeneracy is possible in one dimensional problems. So that can't happen. Cannot have that. So it will always stay a little higher. And therefore with one node you will be a little higher energy. With two nodes will be higher and higher. And that's it. That's the argument. Now, we've argued by this continuous deformation process that this potential not only has these bound states, but this is n nodes and En is greater than En prime for n greater than n prime. So the more nodes, the more energy. Pretty nice result, and that's really all I wanted to say about this problem. Are there any questions? Any? OK. So what we do now is the harmonic oscillator. That's going to keep us busy for the rest of today's lecture. It's a very interesting problem. And it's a most famous quantum mechanics problem in a sense, because it happens to be useful in many, many applications. If you have any potential-- so what is the characteristic of the harmonic oscillator? Harmonic oscillator. Oscillator. Well, the energy operator is p squared over 2m plus, we write, one half m omega squared x squared where omega is this omega that you always think of angular velocity, or angular frequency. It's more like angular frequency. Omega has units of 1 over time. It's actually put 2pi over the period of an oscillation. And this you know from classical mechanics. If you have a harmonic oscillator of this form, yeah, it actually oscillates with this frequency. And E is the energy operator, and this is the energy of the oscillator. So what defines an oscillator? It's something in which the potential energy, this term is v of x. v of x is quadratic in x. That is a harmonic oscillator. Then you arrange the constants to make sense. This has units of energy, because this has units of length squared. 1 over time squared. Length over time is velocity squared times mass is kinetic energy. So this term has the units of energy. And you good with that. And why is this useful? Because actually in any sort of arbitrary potential, rather general potential at least, whenever you have a minimum where the derivative vanishes, then the second derivative need not vanish. Then it's a good approximation to think of the potential at the minimum as a quadratic potential. It fits the potential nicely over a good region. And therefore when you have two molecules with a bound or something oscillating, there is a potential. It has a minimum at the equilibrium position. And the oscillations are governed by some harmonic oscillator. When you have photons in space time traveling, there is a set of harmonic oscillators that correspond to photons. Many, many applications. Endless amount of applications for the harmonic oscillator. So we really want to understand this system quantum mechanically. And what does that mean? Is that we really want to calculate and solve the Schrodinger equation. This is our first step in understanding the system. There's going to be a lot of work to be done even once we have the solutions of the Schrodinger equation. But the first thing is to figure out what are the energy eigenstates or the solutions of the Schrodinger equation for this problem. So notice that here in this problem there's an energy quantity. Remember, when you have a harmontonian like that, and people say so what is the ground state energy? Well, have to find the ground state wave function. Have to do things. Give me an hour, I'll find it. And all that. But if you want an approximate value, dimensional analysis will do it, roughly what is it going to be. Well, with this constant how do you produce an energy? Well, you remember what Einstein did, and you know that h bar omega has units of energy. So that's an energy associated with Lagrangian energy like quantity. And we expect that that energy is going to be the relevant energy. And in fact, we'll find that the ground state energy is just one half of that. There's another quantity that may be interesting. How about the length? How do you construct a length from these quantities? Well, you can start doing m omega h bar and put powers and struggle. I hate doing that. I always try to find some way of doing it and avoiding that thing. So I know that energies go like h over h squared over m length squared. So I'm going to call the length a quantity a. So ma squared. That has units of energy. And you should remember that because energy is b squared over 2m, and b by De Broglie is h bar over sub lamda. So h bar squared, lambda squared, and m here, that's units of energy. So that's a length. On the other hand, we have another way to construct an energy is with this thing, m omega squared length squared. So that's also m omega squared a squared. That's another energy. So from this equation I find that a to the fourth is h squared over m squared omega squared. And it's a little complicated, so a squared is h bar over m omega. So that's a length. Length squared. I don't want to take the square root. We can leave it for a moment there. But that's important because of that's a length scale. And if somebody would ask you in the ground state, how far is this particle oscillating, you would say probably about a square root of this. Would be a natural answer and probably about right. So OK, energy and units is very important to begin your analysis. So what is the Schrodinger equation? The Schrodinger equation for this thing is going to be minus h squared over 2m, d second psi, dx squared plus the potential, one half m omega squared x squared psi is equal E psi. And the big problem is I don't know psi and I don't know E. Now there's so many elegant ways of solving the harmonic oscillator. You will see those next lecture. Allan Adams will be back here. But we all have to go through once in your life through the direct, uninspired method of solving it. Because most of the times when you have a new problem, you will not come up with a beautiful, elegant method to avoid solving the differential equation. You will have to struggle with the differential equation. So today we struggle with the differential equation. We're going to just do it. And I'm going to do it slow enough and in detail enough that I hope you follow everything. I'll just keep a couple of things, but it will be one line computations that I will skip. So this equation is some sort of fairly difficult thing. And it's complicated and made fairly unpleasant by the presence of all these constants. What kind of equation is that with all these constants? They shouldn't be there, all this constants, in fact. So this is the first step, cleaning up the equation. We have to clean it up. Why? Because the nice functions in life like y double prime is equal to minus y have no units. The derivatives create no units. y has the same units of that, and the solution is sine of x, where x must have no units, because you cannot find the sine of one centimeter. So this thing, we should have the same thing here. No units anywhere. So how can we do that? This is an absolutely necessary first step. If you're going to be carrying all these constants, you'll get nowhere. So we have to clean it up. So what I'm going to try to see is that look, here is psi, psi, and psi. So suppose I do the following thing, that I will clean up the right hand side by dividing by something with units of energy. So I'm going to do the following way. I'm going to divide all by 1 over h bar omega. And this 2 I'm going to multiply by 2. So multiply by 2 over h bar omega. So what do I achieve with that first step? I achieve that these 2s disappear. Well, that's not too bad. Not that great either, I think. But in the right hand side, this has units of energy. And the right hand side will not have units of energy. So what do we get here? So we get minus. The h becomes an h alone over-- the m disappears-- so m omega. The second psi the x squared. The 1/2 disappeared, so m omega over h bar x squared psi equals 2 E over h bar omega psi. It looks actually quite better already. Should agree with that. It looks a lot nicer Now I can use a name for this. I want to call this the dimensionless value of the energy. So a calligraphic e. It has no units. It's telling me if I find some energy, that that energy really is this number, this pure number is how many times bigger is e with respect to h omega over 2. So I'll write this now as e psi. And look what I have. I have no units here. And I have a psi. And I have a psi. But things have worked out already. Look, the same factor here, h over m omega is upside down here. And this factor has units of length squared. Length squared times d d length squared has no units. And here's 1 over length squared. 1 over length squared times length squared. So things have worked out. And we can now simply say x is going to be equal to au, a new variable. This is going to be your new variable for your differential equation in which is this thing. And then this differential equation really has cleaned up perfectly well. So how does it look now? Well, it's all gone actually, because if you have x equals au, d dx by chain rule is 1 over a d du. And to derivatives this with respect to x it's 1 over a squared times the d second du squared. And this thing is a squared. So actually you cancel this factor. And when I write x equals to au, you get an a squared times this. And a squared times this is 1. So your differential equations has become minus the second psi du squared, where u is a dimensionless quantity, because this has units of length, this has units of length. No units here. You have no units. So minus d second du squared plus u squared psi is equal to e psi. Much nicer. This is an equation we can think about without being distracted by this endless amount of little trivialities. But still we haven't solved it, and how are we going to solve this equation? So let's again think of what should happen. Somehow it should happen that these e's get fixed. And there is some solution just for some values of e's. It's not obvious at this stage how that is going to happen. Yes? AUDIENCE: [INAUDIBLE]. PROFESSOR: Here for example, let me do this term. h bar over m omega is minus, from that equation, a squared. But dx squared is 1 over a squared d du squared. So a squared cancels. And here the x is equal a squared times u, so again cancels. OK so what is the problem here? The problem is that most likely what is going to go wrong is that this solution for arbitrary values of e's is going to diverge at infinity, and you're never going to be able to normalize it. So let's try to understand how the solution looks as we go to infinity. So this is the first thing you should do with an equation like that. How does this solution look as u goes to infinity? Now we may not be able to solve it exactly in that case either, but we're going to gain insight into what's happening. So here it is. When u goes to infinity, this term, whatever psi is, this term is much bigger than that, because we're presumably working with some fixed energy that we still don't know what it is, but it's a fixed number and, for you, sufficiently large. This is going to dominate. So the equation that we're trying to solve as u goes to infinity, the equation sort of becomes psi double prime-- prime is for two derivatives-- is equal to u squared psi. OK, so how do we get an idea what solves this is not all that obvious. It's certainly not a power of u, because when you differentiate the power of u, you lower the degree rather than increase the degree. So what function increases degree as you differentiate? It's not the trivial function. Cannot be a polynomial. If it could be even a polynomial, if you take two derivatives, it kind cannot be equal to x squared times a polynomial. It's sort of upside down. So if you think about it for a little while, you don't have an exact solution, but you would imagine that something like this would do it, an e to the u squared. Because an e to the u squared, when you differentiate with respect to us, you produce a u down. When you one derivative. When you take another derivative, well, it's more complicated, but one term you will produce another u down. So that probably is quite good. So let's try that. Let's try to see if we have something like that. So I will try something. I'll try psi equals 2. I'm going to try the following thing. e to the alpha u squared over 2 where alpha is a number. I don't know how much it is. Alpha is some number. Now could try this alone, but I actually want to emphasize to you that if this is the behavior near infinity, it won't make any difference if you put here, for example, something like u to the power k. It will also be roughly a solution. So let's see that. So for that I have to differentiate. And let's see what we get. So we're trying to see how the function behaves far, far away. You might say well look, probably that alpha should be negative. But let's see what the equation tells us before we put anything in there. So if I do psi prime, you would get what? You would get one term that would be alpha u times this u to the k into the alpha u squared over 2. I differentiated the exponential. I differentiated the exponential. And then you would get a term where you differentiate the power. So you get ku to the k minus 1 into the alpha u squared over 2. If you take a second derivative, well, I can differentiate the exponential again, so I will get alpha u now squared, because each derivative of this exponent produces a factor of alpha u. u to the k into the alpha u squared over 2. And a couple more terms that they all have less powers of u, because this term has u to the k plus-- already has u to the k plus 1. And this has u to the k minus 1. They differ by two powers of u. So for illustration, please, if you want, do it. Three lines, you should skip three lines in your notebook if you're taking notes and get the following. No point in me doing this algebra here. Alpha u squared over 2. Because actually it's not all that important. Over alpha 1 over u squared plus k minus 1 over alpha squared 1 over u to the fourth. That's all you get. Look, this is alpha squared u squared times psi times these things. 1 plus 2 k plus 1 over alpha 1 over u squared. So when u goes to infinity, your solution works, because these thing's are negligible. So you get a number times u squared. That is the equation you are trying to solve up there. And therefore, you get that the equation if alpha squared is equal to 1. And that means and really that alpha can be plus minus 1. And roughly this solution near infinity, probably there's two solutions. This is a second order differential equation, so even near infinity there should be two solutions. So we expect as u goes to infinity psi of u will be some constant A times u to the k times e to the minus u squared over 2. That's where alpha equal minus 1. Plus Bu to the k into the plus u squared over 2. And what is k? Well, we don't know what is k. It seems to work for all k. That may seem a little confusing now, but don't worry. We'll see other things happening here very soon. So look at what has happened. We've identified that most likely your wave function is going to look like this at infinity. So we're going to want to this part not to be present. So presumably we're going to want a solution that just has this, because this is normalizable. The integral of any power times a Gaussian is convergence. So this can be normalized. The Gaussian falls so fast that any power can be integrated against a Gaussian. Any power however big doesn't grow big enough to compensate a Gaussian. It's impossible to compensate a Gaussian. So we hope for this. But we want to translate what we've learned into some technical advantage in solving the differential equation, because, after all, we wanted be insight how it looks far way, but we wanted to solve the differential equation. So how can we use this insight we now have to simplify the solution of the differential equation? The idea is to change variables a little bit. So write psi of u to be equal to h of u times e to the minus u squared over 2. Now you're going to say wait, what are you doing? Are you making an approximation now that this is what is going to look far away? Or what are you putting there? I'm not making any approximation. I'm just saying whatever pis is, it can always be written in this way. Why? Because if you have a psi of u, you can write it as psi of u over e to the minus u squared over 2 times e minus u squared over 2. Very trivially this can always be done. As long as we say that h is arbitrary, there's nothing, no constraint here. I have not assume anything, nothing. I'm just hoping that I have a differential equation for psi. That because this is a very clever factor, the differential equation for h will be simpler. Because part of the dependence has been taken over. So maybe h, for example, could be now a polynomial solution, because this product has been taken care. So the hope is that by writing this equation it will become an equation for h of u, and that equation will be simpler. So will it be simpler? Well, here again this is not difficult. You're supposed to plug into equation one-- this is the equation one-- plug into one. I won't do it. It's three lines of algebra to plug into one and calculate the equation for h of u. You should do it. It's the kind of thing that one should do at least once. So please do it. It's three, four lines. It's not long. But I'll just write the answer. So by the time you substitute, of course, the e to the minus u squared over 2 is going to cancel from everywhere. It's very here. You just need to take two derivatives, so it becomes a second order differential equation. And indeed, it becomes a tractable differential equation. The second h, du squared minus 2u dh du plus e minus 1 h equals 0. OK, that is our equation now. So now we face the problem finally solving this equation. So before we start, maybe there's some questions of what we've done so far. Let's see. Any questions? Yes? AUDIENCE: Do you have right there in the middle would be-- this equation is linear, so can we just [INAUDIBLE] minus u squared over 2 and you stuck it to that u to the k. PROFESSOR: It's here? This thing? AUDIENCE: Yeah. Could you then just power series what's going on at 0 with those u to the k terms [INAUDIBLE]? PROFESSOR: No. This is the behavior as u goes to infinity. So I actually don't know that the function near 0 is going to behave like u to the k. We really don't know. It suggest to you that maybe the solution is going to be near 0 u to the k times some polynomial or something like that. But it's not that, because this analysis was just done at infinity. So we really have no information still what's going on near 0. Other questions? Yes? AUDIENCE: So is k some arbitrary number or is it an integer? PROFESSOR: At this moment, actually, it doesn't matter. Is that right? Doesn't matter. The analysis that we did here suggests it could be anything. That's why I just didn't put it into h or u. I didn't put it because would be strange to put here a u to the k. I wouldn't know what to make of it. So at this moment, the best thing to say is we don't know what it is, and maybe we'll understand it. And we will. In a few seconds, we'll sort of see what's going on. OK, so how does one solve this equation? Well, it's not a trivial equation, again. But it can be solved by polynomials, and we'll see that. But the way we solve this equation is by a power series expansion. Now you could do it by hand first, and I did it when I was preparing the lecture yesterday. I said I'm going to just write h of u equal a constant a0 plus a1u plus a2u squared plus a3u cubed. And I plugged it in here. And I just did the first few terms and start to see what happened. And I found after a little thinking that a2 is determined by a0, and a3 is determined by a1 once you substitute. It's not the obvious when you look at this, but that happens. So when you face a problem like that, don't go high power to begin with. Just try a simple series and see what happens. And you see a little pattern. And then you can do a more sophisticated analysis. So what would be a more sophisticated analysis? To write h of u equal the sum from j equals 0 to infinity aju to the j. Then if you take a derivative, because we're going to need the derivative, dh du would be the sum from j equals 0 to infinity. j times aju to the j minus 1. You would say that doesn't look very good because for j equals 0 you have 1 over u. That's crazy. But indeed for j equals 0, the j here multiplies it and makes it 0. So this is OK. Now the term that we actually need is minus 2u dh du. So here minus 2u dh du would be equal to the sum from j equals 0 to infinity, and I will have minus 2jaju to the j. The u makes this j minus 1 j, and the constant went there. So here is so far h. Here is this other term that we're going to need for the differential equation. And then there's the last term that we're going to need for the differential equation, so I'm going to go here. So what do we get for this last term. We'll have to take a second derivative. So we'll take-- h prime was there, so d second h du squared will be the sum from j equals 0 of j times j minus 1 aju to the j minus 2. Now you have to rewrite this in order to make it tractable. You want everything to have u to the j's. You don't want actually to have u to the j minus 2. So the first thing that you notice is that this sum actually begins with 2, because for 0 and 1 it vanishes. So I can write j times j minus 1 aj u to j minus 2. Like that. And then I can say let j be equal to j prime plus 2. Look, j begins with 2 in this sum. So if j is j prime plus 2, j prime will begin with 0. So we've shifted the sum so it's j prime equals 0 to infinity. And whenever I have a j I must put j prime plus 2. So j prime plus 2. j prime plus 1 aj prime plus 2 u to the j prime. Wherever I had j, I put j prime plus 2. And finally you say j or j prime is the same name, so let's call it j. j equals 0. j plus 2. j plus 1. aj plus 2 uj. So we got the series expansion of everything, so we just plug into the differential equation. So where is the differential equation? It's here. So I'll plug it in. Let's see what we get. We'll get some from j equals 0 to infinity. Let's see the second derivative is here. j plus 2 times j plus 1 aj plus 2 uj, so I'll put it here. So that's this second derivative term. Now this one. It's easy. Minus 2j aj and the uj is there. So minus 2jaj. Last term is just e minus 1, because it's the function this times aj as well. That's h. And look, this whole thing must be 0. So what you learn is that this coefficient must be 0 for every value of j. Now it's possible to-- here is aj and aj, so it's actually one single thing. Let me write it here. j plus 2 times j plus 1 aj plus 2 minus 2j plus 1 minus e aj uj. I think I got it right. Yes. And this is the same sum. And now, OK, it's a lot of work, but we're getting there. This must be 0. So actually that solves for aj plus 2 in terms of aj. What I had told you that you can notice in two minutes if you try it a little. That a2 seems to be determined by a0. And a3 seems to be determined by a2. So this is saying that aj plus 2 is given by 2j plus 1 minus e over j plus 2 j plus 1 aj. A very nice recursive relation. So indeed, if you put the value of a0, it will determine for you a2, a4, a6, a8, all the even ones. If you put the value of a1, it will determine for you a3, a5. So a solution is determined by you telling me how much is a0, and telling me how much is a1. Two constants, two numbers. That's what you expect from a second order differential equation. The value of the function at the point, the derivative at a point. In fact, you are looking at a0 and a1 as the two constants that will determine a solution. And this is the value of h at 0. This is the derivative of h at 0. So we can now write the following facts about the solution that we have found. So what do we know? That solutions fixed by giving a0 and a1. That correspond to the value of the function at 0 and the derivative of the function at 0. And this gives one solution. Once you fix a0, you get a2, a4. And this is an even solution, because it has only even powers. And then from a1, you fixed a3, a5, all the other ones with an odd solution. OK. Well, we solve the differential equation, which is really, in a sense, bad, because we were expecting that we can only solve it for some values of the energy. Moreover, you have a0, you get a2, a4, a6, a8. This will go on forever and not terminate. And then it will be an infinite polynomial that grows up and doesn't ever decline, which is sort of contradictory with the idea that we had before that near infinity the function was going to be some power, some fixed power, times this exponential. So this is what we're looking for, this h function now. It doesn't look like a fixed power. It looks like it goes forever. So let's see what happens eventually when the coefficient, the value of the j index is large. For large j. aj plus 2 is roughly equal to, for large a, whatever the energy is, sufficiently large, the most important here is the 2j here, the j and the j. So you get 2 over j aj. So roughly for large j, it behaves like that. And now you have to ask yourself the question, if you have a power series expansion whose coefficients behave like that, how badly is it at infinity? How about is it? You know it's the power series expansion because your h was all these coefficients. And suppose they behave like that. They grow in that way or decay in this way, because they're decaying. Is this a solution that's going to blow up? Or is it not going to blow up? And here comes an important thing. This is pretty bad behavior, actually. It's pretty awful behavior. So let's see that. That's pretty bad. How do we see that? Well you could do it in different ways, depending on whether you want to derive that this is a bad behavior or guess it. I'm going to guess something. I'm going to look at how does e to the u squared behave as a power series. Well, you know as a power series exponential is 1 over n u squared to the n. Here's n factorial. n equals 0 to infinity. Now these two n's, u to the 2n, these are all even powers. So I'm going to change letters here, and I'm going to work with j from 0, 2, 4, over the evens. So I will write u to the j here. And that this correct, because you produce u to the 0, u to the 2, u to the fourth, these things. And j is really 2n, so here you will have one over j over 2 factorial. Now you might say, j over 2, isn't that a fraction? No, it's not a fraction, because j is even. So this is a nice factorial. Now this is the coefficient, cj u to the j. And let's see how this coefficients vary. So this cj is 1 over j over 2 factorial. What is cj plus 2 over cj? Which is the analogue of this thing. Well, this would be 1 over j plus 2 over 2 factorial. And here is up there, so j over 2 factorial. Well, this has one more factor in the denominator than the numerator. So this is roughly one over j over 2 plus 1, the last value of this. This integer is just one bigger than that. Now if j is large, this is roughly 1 over j over 2, which is 2 over j. Oh, exactly that stuff. So it's pretty bad. If this series goes on forever, it will diverge like e to the u squared. And your h will be like e to the u squared with e to the minus u squared over 2 is going to be like e to the plus. u squared over 2 is going to go and behave this one. So it's going to do exactly the wrong thing. If this series doesn't terminate, we have not succeeded. But happily, the series may terminate, because the j's are integers. So maybe for some energies that are integers, it terminates, and that's a solution. The only way to get a solution is if the series terminates. The only way it can terminate is that the e is some odd number over here. And that will solve the thing. So we actually need to do this. This shows the energy. You found why it's quantized. So let's do it then. We're really done with this in a sense. This is the most important point of the lecture, is that the series must terminate, otherwise it will blow up horrendously. If it terminates as a polynomial, then everything is good. So to terminate you can choose 2j plus 1 minus e to be 0. This will make aj plus 2 equal to 0. And your solution, your h of u, will begin. aj will be the last one that is non-zero, so it will be aj times u to the j, and it will go down like aj minus 2 u to the j minus 2. It will go down in steps of 2, because this recursion is always by steps of two. So that's it. That's going to be the solution where these coefficients are going to be fixed by the recursive relation, and we have this. Now most people here call j equal n. So let's call it n. And then we have 2n plus 1 minus e equals 0. And h of u would be an u to the n plus all these things. That's the h. The full solution is h times e to the minus u squared over 2 as we will see. But recall what e was. e here is 2n plus 1. But he was the true energy divided by h omega over two. That was long ago. It's gone. Long gone. So what have you found therefore? That the energy, that' we'll call en, the energy of the nth solution is going to be h omega over 2 2n plus 1. So it's actually h omega, and people write it n plus 1/2. Very famous result. The nth level of the harmonic oscillator has this energy. And moreover, these objects, people choose these-- you see the constants are related by steps of two. So just like you could start with a0, or a1 and go up, you can go down. People call these functions Hermite functions. And they fix the notation so that this an is 2 to the n. They like it. It's a nice normalization. So actually h of n is what we call the Hermite function of u sub n. And it goes like 2 to the n u to the n plus order u to the n minus 2 plus n minus 4, and it goes on and on like that. OK, a couple things and we're done. Just for reference, the Hermite polynomial, if you're interested in it, is the one that solves this equation. And the Hermite sub n corresponds to e sub n, which is 2n plus 1. So the Hermite solution from that the equation is that the Hermite polynomial satisfies this minus 2u d Hn du plus 2n. Because en is 2n plus 1. So it's 2n Hn equals 0. That's the equation for the Hermite polynomial, and interesting thing to know. Actually, if you want to generate the efficiently the Hermite polynomials, there's something called the generating function. e to the minus z squared plus 2zu. If you expand it in a power series of z, it actually gives you n equals 0 to infinity. If it's a power series of z, it will be some z to the n's. You can put a factor here n, and here is Hn of u. So you can use your mathematic program and expand this in powers of z. Collect the various powers of u that appear with z to the n, and that's Hn It's the most efficient way of generating Hn And moreover, if you want to play in mathematics, you can show that such definition of Hn satisfies this equation. So it produces the solution. So what have we found? Our end result is the following. Let me finish with that here. We had this potential, and the first energy level is called E0 and has energy h omega over 2. The next energy is E1. It has 3/2 h omega. Next one is E2 5/2 h omega. This polynomial is nth degree polynomial. So it has n zeros, therefore n nodes. So these wave functions will have the right number of nodes. E0, the psi 0, will have no nodes. When you have psi 0, the Hn becomes a number for n equals zero. And the whole solution is the exponential of u squared over 2. The whole solution, in fact, is, as we wrote, psi n Hn of u e to the minus u squared over 2. In plain English, if you use an x, it will be Hn u with x over that constant a we had. And you have minus x squared over 2a squared. Those are your eigenfunctions. These are the solutions. Discrete spectrum, evenly spaced, the nicest spectrum possible. All the nodes are there. You will solve this in a more clever way next time. [APPLAUSE]
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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. PROFESSOR: Today, we're going to talk about a very exciting subject regarding molecules. And this subject has to do with metals in biology. Metals in biology, by the way, is actually the name of one of the Gordon research conferences. So if you get excited about this topic, you can go to the Gordon research conference web page and apply to attend the conference next summer and learn about the latest developments in metals in biology. One of the things I want to impart to you today has to do with where we usually find metals inside of biomolecules. So that leads me to talk a little bit about nature's ligands. In biological chemistry, we find that molecules sometimes are quite a bit larger than we're used to looking at in synthetic small-molecule chemistry. So I'm going to talk about nature's ligands today at sort of two levels of size. Going to talk first about the porphyrin ligand. And then next, we're going to talk about proteins. OK. First this porphyrin ligand is one of the most pervasive small-molecule hosts for metal ions within biomolecules. And it's really quite a remarkable ligand type. And I'd like you to appreciate it at the level of being able to draw the structure of it and to see just how this system can provide a host for metal atoms. It is a system that has four nitrogens that it can donate to a metal center. So it's a tetradentate ligand. We talked about bidentate ligands earlier in the semester. This is a tetradentate ligand that nature uses to hold metal ions very tightly where it wants them in a biomolecule. And first you draw the four nitrogens at the corners of a square. And now we're drawing in the carbon parts of the molecule. You can draw four vertical lines like that and four horizontal lines like this. And then what we're going to do is, at each of the four corners of this square-shaped molecule, we're going to go ahead and complete five-membered rings. Like that. And then finally, we have four more carbons to add to this structure. And those four carbons are at the top, bottom, and left, and right, like this. OK. So you can see that the porphyrin ligand is a large ring-shaped macrocycle. And where can a metal ion go? It can go right in the center. If you are going to figure oxidation states that involved a porphyrin molecule, then you would need to know the charge like we know that one chloride is in a metal complex. It carries a one minus charge. And you need to know how many charges the porphyrin ligand would carry if a metal were in there, so that you could figure the oxidation state and the d-electron count for the metal that would be inserted into the porphyrin. And to do that, we first recognize that if you isolate a porphyrin without the metal in there, then two of the four nitrogens have hydrogens attached. OK? And also this is a system-- you'll remember the graphite structure that we looked at last time that had all these p orbitals perpendicular to the plane of the molecule. And similarly, this is an unsaturated molecule that has lots of pi bonding going on with the carbon p orbitals that are perpendicular to the plane of the blackboard here. So it can be a quite flat molecule, but as you'll see it can ruffle as well when it carries out its biological function. And so we'll go ahead and draw in the double bonds. And this is the part where people sometimes get messed up because one thing we don't want to do is draw five bonds to a carbon atom. OK? And if you see a structure that's drawn that way, you know that someone has made a mistake because they violated the octet rule. All right? So let's draw in, starting here at the upper left, we'll put in a double bond here, and here, and here. And then this is where it gets interesting because I could put a double bond here now or here because this carbon's going to need a double bond. I'm going to choose to put it here, and then I'm going to use symmetry across the way to put one there. And having done that, I can now go here, here, here, here, and-- so I didn't use symmetry across the way. This is much easier to do and it's small. OK. So now you can see that these two nitrogens here would have lone pairs that point into the center of the porphyrin ligand and same across the way by symmetry there. And then when we remove these two hydrogens as H plus and insert a metal ion, you'll see that the porphyrin is actually behaving as a dianion. When a metal ion and-- most typically, we're going to see that the metal that appears at the center of a porphyrin ligand is an iron center. And so the nitrogens provide four lone pairs in the plane of the molecule that are directed at the metal center. So that's four out of the six positions for an octahedron all in the same plane. And because of the rigid delocalized structure of the porphyrin ligand, what we find is that a metal center that is in a porphyrin is very difficult to take out of the porphyrin because these lone pairs are forced at all times to converge on the same point which is the metal ion. OK. You know that I enjoy giving lectures using shock. But sometimes when the molecules are really big, it's easier to use a computer to portray them quickly. And that's what we're going to do today, assuming my computer participates and cooperates with me today. And what I really want to do is to go through and show you a tour of six really important metalloproteins. And all of these, you'll see the heme plays an important role, the iron in a porphyrin. This unit together, when an iron is located inside of a porphyrin, that is referred to as heme. So you can have other metals in there, and then it's not called a heme anymore, but that would be very similar structurally. OK. So I'm going to take advantage of some really nice summaries that have been put together on the protein data bank website. So since the first crystal structure was determined for a protein, the structural information has been collected and put together in one site so that researchers all over the world can take advantage of this information to advance science and medicine. And since 2000, the protein data bank has been collecting information about certain biomolecules and putting out one new little short story or vignette each month. So you have their Molecule of the Month portion of the protein data bank website. So each month, you can go and look and see which new molecule has been added and learn more about it. And I want to start out here with myoglobin. OK. And I'm going to give you this website information online associated with today's notes. So you're going to see that myoglobin was the first protein structure. There are many different ways to represent protein molecules. Protein chains are composed of linear polimers of the naturally occurring amino acids with polypeptide chains. And they can sometimes be very, very large in length, and then they fold up and make three-dimensional shapes. And these are big ligands for metal ions. And what you're going to see is that in many of the most important proteins, the place where the reaction's actually happened is that the metal center that's embedded somewhere in the protein chain as it has folded up around it to modulate function. OK. So this is a pretty small protein-- myoglobin is. It's an oxygen storage protein. So as you know, diatomic molecules like dioxygen are important for life on Earth, and O2 in particular-- not only do we have hemoglobin that we'll talk about next, but there are also oxygen storage molecules like this myoglobin molecule. And this was the first structure that was determined, and that happened in 1960. So we've been getting information at the molecular level about how biomolecules work really since 1960 using x-ray diffraction techniques. So first, a worker in this area has to obtain a large quantity of the protein in question, get it purified, and then find conditions under which it will grow single crystals so that you can impinge upon it with an x-ray beam and then looked at the intensities of the diffracted x-ray beams and then work back to solve the crystal structure and find out what arrangement of atoms in 3D space could have given rise to that pattern of diffraction intensities. And that was first done for a protein in 1960. You might find it interesting that this protein was isolated from sperm whales. OK? Sperm whale muscles. And in order to get a protein pure, as I said, you need a lot of it. And the muscles of sperm whales are very large indeed, and so these workers were able to collect large amounts of this red protein myoglobin from sperm whale muscles in order to finally get enough in pure form to crystallize and solve the crystal structure. And since this is involved in oxygen storage and you know that whales and other marine mammals can dive below the ocean surface for long periods of time, they need to store lots of oxygen so that they can use their muscles during the length of a long dive. And this protein is very abundant in muscles. It gives meat its red color and so forth. The color is coming from the heme units that are embedded in the myoglobin. And I'll step through a little bit of this, but I would like to actually show you some more of the features of this. You'll see as you go through this website, there are both static and interactive ways to look at metalloproteins. And sometimes the polypeptide chain composed of the amino acids is just represented here as sort of a tube, so you can see the polymer has wrapped up. And here they've chosen to represent the heme unit as spheres to highlight it and accentuate it. And in this first structure of myoglobin, there was not an O2 molecule bound to the heme unit. Instead, there was a water molecule bound to the iron center of the heme unit. And then down here, they rotate the structure perpendicular so you can appreciate the square nature of the heme unit together with when viewed edge on, its flat nature. So that's the structure here. And you can see how this myoglobin molecule, this big protein polymer, has folded up like a clam to grasp this heme unit that inside of it will bind O2 and store it when oxygen comes to it. And so later structures of myoglobin have been carried out that do show you the position of the O2 molecule. And in these two representations that we are given here, we first see, once again, the heme unit as spheres and the protein side chain just as tubes. And then this representation in the bottom, all the protein side chain atoms themselves are represented as spheres. And you can see that when that's the case, you can tell that the atoms of the protein ligand really fill up space very well. And what they're showing you with this circle here is the location beneath that side chain of the dioxygen molecule inside there. And so they're trying to give you the idea that when O2 is on the heme unit inside the myoglobin, the myoglobin protein matrix actually covers that up. And so in order for the O2 molecule to get into the heme and also to exit from the metalloprotein, this thing has to be flexible enough to open up and allow the dioxygen molecule to exit from its chamber inside the metalloprotein. OK. Now next, I would like to go over to the site where we're actually able to look at things interactively. And you need to be able to enter the code for the protein in question. And so this is a structure of the myoglobin molecule that was obtained at molecular or atomic resolution. So one of the parameters that you'll see when people discuss protein structures is the resolution to which the data have permitted the structure to be determined. And the best quality structures, the ones where we know most precisely where the atoms are, are the ones that have very low resolution numbers. In other words, this one, if you looked at the introductory web page there so that the resolution was at the level of one angstrom. OK? So we're talking on the order of carbon-carbon bond distances so that you can actually see all the individual atoms quite clearly. And this has to do with the quality of the crystal and how large was the angle at which you could collect spots from the diffracted x-rays as they came off of the crystal. And so this is nice. This one requires Java. That's why I was not able to use Athena today. But actually, I come in here, and I see that the Athena terminal is down anyway, so this was a good choice as it turns out. And this is nice because you can actually zoom on these structures. One thing you notice is that this was crystallized with sulfate present in the medium, so there was some buffer present. And it had sulfate, and the sulfates crystallized together with the protein chains. And as we look at representations of proteins, you'll appreciate that this representation shows the polypeptide backbone as these ribbons. And that emphasizes that when proteins fold and go into a three-dimensional shape that they need for their function, they can usually do so either by making coils like this that look like springs-- these are called helices. Or they can just form random coils as you can see out here. And oftentimes you have these helices that are like long springy tubes that are then connected at their termini by coils of the protein side chain. And then, of course, here embedded within this structure is our heme unit. And one thing if you look at this one pretty closely, you'll be able to see that this heme is a little more interesting, a little more complicated, than the one that I drew on the board here because the one I drew on the board was assumed to have just hydrogens at each of these carbon positions at the periphery of the molecule. But when we look at the one that's actually been found in myoglobin at atomic resolutions-- so you see this one is actually with the O2 molecule present. For some reason, no bond drawn there, but that's OK. We've talked a lot about molecular representations, and we understand what's going on here. It's that people have decided that we would represent the atoms over here without any detail at all just showing them as ribbons. And then here we're going to look at the atom positions. And you can see the five-membered ring here, here, here, and here at each nitrogen corner of the heme unit. There's the iron center. And then, notice that the substituents, instead of being just hydrogen or organic, there's a carbon here. This looks like a methyl group perhaps. And this one looks like an ethyl group or maybe a vinyl group here if this is one of the etioporphyrin ligand types. And then coming out here, we have CH2, CH2. We have something that looks like it might be a ketone residue since we have a single oxygen on a carbon there, or alternatively, it could be some other oxygenated residue. But over here, you see that there's a link to this five-membered ring of the heme. You have a couple of methylenes and then a carboxylate residue. And so when you investigate these structures, these metal complexes that are embedded in the protein usually are held in position, both by the way the protein folds up around the heme unit, but also because, in this case, the heme unit is connected to residues like this carboxylic acid residue that can hydrogen bond to specific residues on the protein side chain. So if we were to look at the atoms involved in the side chain right over here that's next to this carboxylate residue, this probably has some hydrogen bond acceptor that is interacting with this carboxylate residue-- and same over here-- to hold this heme unit in its desired position or in its required position for function. So please keep that in mind as you go ahead and look at different structures in the database. And now let's go to the 2003 description of hemoglobin. This is rather nice. OK. So myoglobin was a pretty small protein having a single heme unit. And hemoglobin is much more interesting. It actually has four big pieces that come together, and it has four heme units in it. OK. And as you go through this website, you'll find out about why blood is red versus blue. It's actually red in the oxygenated form and blue in the deoxygenated form. And there are many crystal structures of different hemoglobins and mutants of hemoglobins in the database and both with and without the oxygen molecules. There's a discussion here of artificial blood. And as you'll see what the premise is here for the design of artificial blood motivated by a desire to have blood available for transfusions and so forth that would not be contaminated, would be just synthetic-- would be just pure hemoglobin in fact. But what happens is that without the protective casing of the red blood cell, the four parts of the hemoglobin molecule fall apart from each other, and they don't do their function properly anymore. And we'll learn more about the function in a moment. So people have actually made mutants wherein you make covalent bonds between the four different subunits of the hemoglobin so that they can't fall apart. And that's one of the approaches to the synthesis of artificial blood. And then this is a rather nice little video which comes from two different hemoglobin crystal structures. If you imagine different crystal structures, like crystal structure of hemoglobin when it has O2 bound verses when it does not have O2 bound, as snapshots of the molecule in motion, then you can generate a little movie like this one shown here from the experimental data on the crystal structure. And so the really remarkable thing here is that-- see when the O2 molecule is bound to the heme, then the protein itself, this large unit that is comprised of four parts, as one confirmation, one structure, and it changes quite a bit. You see how much it is moving around when the dioxygen molecule comes off of the heme. And that is related very much to its function. In fact, it turns out that, in hemoglobin, when the deoxyhemoglobin arrives at a place in your body where there is a lot of oxygen, the first O2 molecule binding event is very difficult to achieve because it involves a big change in the structure of the entire protein molecule. But once that first O2 molecule has bound, the whole protein molecule has changed its structure such that the next three hemes take up their O2 molecules rapidly in rapid fire succession after the first one has done it. So the heme goes into a place of high-oxygen concentration and it rapidly loads up with four O2's. And then it moves on. And then when it gets to a location that needs oxygen, the first one comes off, and then the next three fire off in rapid succession because of the change in the protein shape. And so hemoglobin is, therefore, very good at collecting four O2 molecules at a time and then delivering them to where they're needed. And this is just a beautiful example of the way protein crystallography can teach us about mechanisms of biomolecules. OK. And then also you can learn about disease mechanisms in the data bank here because this is from a different crystal structure where a mutation has occurred in the protein side chain of the hemoglobin. So this is one hemoglobin molecule. And it turns out that when that mutation is present, unfortunately, it causes hemoglobin molecules to aggregate and stick together and clump together like that. And that phenomenon is associated with diseases like sickle cell anemia. OK? So the structure of the protein is changed when one residue of the amino acid side chain is changed to a different amino acid. And then these things all aggregate together instead of forming nice disks as normal red blood cells do. They stick together and form these different shapes that are not good at functioning the way they're supposed to. And then here is a close-up on where the action takes place in the hemoglobin molecule. And this, again, is from the two crystal structures that we were talking about-- the one that has the oxygen molecule bound to the iron of the hemoglobin and the one that does not. And one thing that you'll notice is that when the O2 molecules is absent, this iron in yellow represented here as this yellow sphere is sort of dipping down on one side of the hemoglobin-- the heme plane toward the nitrogen of this histidine. So this is part of the amino acid protein side chain. And a histidine residue has this Lewis space group here that is coordinated to the iron. And so in that form, this is a five-coordinate metal center with five ligands of square pyramidal geometry. It's an octahedron missing one of those six ligands. And then when the O2 ligand comes in and binds and becomes the sixth ligand in the coordination sphere of the iron, the iron responds to that by moving up to the other side of the porphyrin plane, and it draws with it the histidine ligand. So the iron moving up pulls the histidine up. And that structural change at the iron center is then propagated throughout the amino acid side chain to which this is connected and then, ultimately, throughout the entire protein to give you that huge conformational change that we saw when O2 is bound versus when it's not. So the coordination chemistry of the iron here is really controlling the peptide confirmation and the function of this protein. OK. So these bio molecules actually accumulate quite a lot of the concepts that we've talked about all semester. And that phenomenon is going to continue as I take you through a tour of a couple more of these important molecules. Now the first one was a small one that we looked at, myoglobin, and I'm going to continue that sort of progression by now talking about another small protein. This one may even remind you in its appearance of myoglobin. But instead of serving to store O2, like myoglobin, cytochrome c, which has a single heme unit in it here, serves as a shuttle for electrons. OK? The pathway that electrons take in the whole process of respiration in organisms is really quite fascinating. We require reducing agents. We eat food. Those are reducing agents. And then we go ahead and burn that fuel using the oxygen molecule. And that's where we get our energy. We don't burn the fuel literally by having us burst into flames, but we do it in a controlled manner, in a stepwise manner. And we take advantage of those steps to carry out the processes of life. And so there are proteins like these small cytochrome c proteins that are developed expressly for the purpose of moving electrons from one place in the body to another. So it's like a little electron shuttle protein. And these are just two views of it. And if you read this part of the website, you'll also learn that this one has remained virtually unchanged over eons. It's a very ancient protein. OK. And this picture shows you cytochrome c protein and where it picks up its electrons and then where it goes to drop them off. So here's an enormous protein relative to cytochrome-- cytochrome c is the little teeny one here with just a single heme unit. OK? Look at this thing. This thing is-- I don't know what this looks like, but it looks like a bizarre-shaped entity, shall we say. And this yellow stripe that goes across the page here is actually a membrane. So proteins are often classified as to whether they are membrane proteins-- meaning that they don't move around. But they're embedded or fixed in a membrane with one part of their molecule on one side of the membrane and the other part of the molecule up on the other side of the membrane and then the part of the molecule that's actually located within the membrane, and then non-membrane proteins like cytochrome c that are actually mobile and can move about within a cell. And cytochrome c goes over to this large protein which has lots of different cofactors shown here in red which incidentally are hemes themselves. And so this molecule-- the big protein here embedded in the matrix is called cytochrome bc1. And it is a protein involved in this process of respiration. It is producing electrons. And cytochrome c comes over here, and as you can see, it fits really well into this spot right here on cytochrome bc1. And that's a feature of protein-protein interactions that they often contain residues on their surface that are hydrogen bonding residues, for example, that make their surfaces complimentary. And it means that when cytochrome c comes in into cytochrome bc1, it actually locks in in one particular way. And this complex between cytochrome c and cytochrome bc1 has been crystallized. And the crystal structure of it's in the database, and that's where we get this representation shown here. So you can actually go through-- if you're a crystallographer-- and look at all the different complimentary interactions that holds cytochrome c into place on cytochrome bc1 when it docks there. And what I'm going to tell you in a moment is that as electrons are originating within cytochrome bc1 here, they have a pathway through which they move and they get up into this heme unit. And then when cytochrome c docs, the heme unit within it can get close enough to this heme unit that their wave functions can overlap, and the electron can tunnel right across into the cytochrome c which accepts it. And then it goes off, and it's looking for this protein next which has also been crystallized and characterized in the database. And we'll talk about it next. This is called cytochrome C oxidase because it oxidizes the reduced form of cytochrome c. And let me just say that probably one thing we won't have on the final will be working the molecular orbital diagram of this molecule. As you can see, we've got up to a position here where the number of orbitals is kind of prohibitive to do that sort of thing. Now you can understand why the computational chemists have quite a big challenge as they try to understand at a molecular level all of the processes of life and, really, a long way from doing that actually. When people attempt to do that these days, they're usually making some big approximations with the protein side chain because that's the electronically uninteresting part of the system. Where the metals are is where everything is happening. Let's go down here and see what else we can learn. OK. Here's a close-up with a different representation of when cytochrome c molecule docs on to the cytochrome bc1 complex. And here we've got the polypeptide chain of cytochrome c represented as these pink tubes, and then as these yellow tubes down here, this little part of the cytochrome bc1 molecule that is reaching up to interact with cytochrome c. And here's the heme unit of cytochrome bc1 that is right at the surface of that protein so that it can overlap it's wave function with the heme unit of cytochrome c to permit an electron to transfer and reduce cytochrome c. OK. And so now we'll go on to cytochrome c oxidase. OK. So in reading about cytochrome c oxidase, you're going to learn more about oxygen. And this piece of today's story, together with the one that I'm going to go to next, which will be with respect to photosystem one and photosystem two, are sort of the two ends of the chain of respiration. Because with photosynthesis, that we'll talk about next, we use light energy to create O2. And most organisms on the planet have these cytochrome so that they can use O2. In fact, what they do is they take electrons that are derived from food and get energy by combining them with oxygen, reducing them-- reducing the oxygen molecule-- to water, and at the same time pushing protons across a membrane which stores energy that can be used for other processes like building up ATP. And so, the cytochrome c oxidase is getting its electrons from cytochrome c and it's using those electrons to reduce the O2 molecule. And for that reason, cytochrome c oxidase should have somewhere in it a place for the dioxygen molecule to bind. Just like is the case for hemoglobin. We found out where oxygen binds in hemoglobin. And notice that as you go through they'll make an analogy here to charging of a battery or actually of a capacitor as you're reducing the O2 molecules with these electrons coming in from the cytochrome c shuttle, and making water and pushing protons across a membrane that is being likened to charging a battery or charging a capacitor. And this is a membrane-bound protein just like the bc1 complexes. And here's a nice graphic that shows a number of the important cofactors. A cofactor is just something that's an integral part of a protein but isn't part of the amino acid backbone of the polypeptide chain. And usually it's something like a heme, but here not only do we have hemes, but here's a heme that is thought to be involved in the electron transport chain within the cytochrome C oxidase molecule. And over here is a heme that is thought to be important in binding the O2 molecule. And also this one has copper. So these 3D transition elements are turning out to be pretty important in biology. The iron is frequently the site for the binding of a diatomic molecule, but look at this. This structure that's in the Cambridge database here, its PDB entry 1OCO, actually has a carbon monoxide ligand bonded to the iron in the active site where O2 is supposed to bind when this enzyme functions. So this is a poisoned form of the enzyme. CO comes in and binds and shuts this enzyme down, and that's what they were able to purify and crystallize and get the 3D structure of. And then here you have a heme iron which interacts with the carbon end of carbon monoxide. And then positioned over here, and also ligated by residues from the protein side chain, is copper, referred to as copper b. So this is a copper ion. And that means that when a diatomic molecule goes in and binds to the heme at one end, it's other end is binding simultaneously to copper b. And that diatomic molecule is serving as a bridge between two metal ions-- the iron heme and the copper b up here. And then also up here is a copper a site, where you have a pair of coppers. And this site is referred to and thought to be the port of entry of the O2 molecule into cytochrome c oxidase. That O2 somehow comes in and may bind here first, but then go over here and electrons are coming in through this other heme unit into the one that binds the O2 and ends up reducing it to generate water. So that is a really pretty picture of how complicated the machinery of an enzyme can be to take advantage of all this lovely transition element chemistry and redox chemistry to carry out life function. And also-- let's see-- OK, so what they're doing here is analyzing this. This cytochrome c oxidase is composed of a number of different protein chains that are all packed together into the overall complex, and they're coloring each of these residues a little differently here. And then when you get down to the bottom of this page, they're comparing a subunit of cytochrome c oxidase to an actual cytochrome c oxidase that is found in bacteria molecules. Bacteria organisms. So bacteria have their own cytochrome c oxidases and this looks like the central core of our cytochrome c oxidases. And the idea is that mitochondria in cells may, at some point way, way back in evolution, have been the result of bacterial invaders into cells. And then these bacteria became integral parts of ourselves, and we added new proteins on and elaborated to change the function as we deemed necessary through evolution. But these ways of comparing structures of simpler or ancient proteins to ones that are more modern is an interesting way to learn about how life has evolved on this planet. OK, and-- so cytochrome c oxidase is a pretty remarkable protein. And I want to see if I can get this thing up interactively because the structure is quite impressive. And I'm using this carbon monoxide substituted variant. When these structures have many thousands of atoms, sometimes these things take a little while to load. So hopefully this won't take too long. The viewer that-- there are actually several viewers that you can choose from on this protein data bank web page, and they possess features that allow you to turn on or turn off different parts of the molecule. You can subtract away the cofactors, or subtract away some of the loops or the helices, and that allows you to explore the structure in quite some detail. Now, each of these are different protein residues, different residues of this enormous protein. So let's come down here, zoom in a little bit. OK, if I notice that-- there I can subtract away the coils that hold together the various loops, OK? Or you can take away, in fact, all of those things. And notice the cofactors themselves are these heme units and they have some interesting and complicated side chains. But they're the same sort of heme units in general that we see in myoglobin or in cytochrome c. OK? Let's see . Actually-- there are orientations of this molecule that you can access that shows you how this thing fits nicely into a membrane. This is one of them. You see all these different helices that are aligned with one another. This is the part of the cytochrome c oxidase that resides in the membrane. And all of these residues are lipophilic, so they interact with the non-polar interior of the membrane very well allowing this protein to just stay in as part of the membrane. And then these loops that extend on one side of the membrane or onto the other contain polar residues that would prefer to interact with water or with charged things in solution. And so that's an important aspect of the way a membrane protein like this is the structured. If I could just take away the loops here, the coil, that becomes quite clear. So now, with the few minutes we have left, I'm going to go ahead and talk about photosystem I and II at the other end of the chain of respiration where organisms are actually-- you know, they realize that eventually you would run out of all the food on the planet and everything would die. And so they learned to absorb light and to use light energy to carry out processes of life instead of just using the energy that you get when you add electrons to water, to two oxygen and make water. And so let's go look first at photosystem I. Photosystems I and II are different parts of the photosynthetic pathway and of the electron utilization chain, electron transfer chain. And what you see is that photosystem I and II are both going to be membrane proteins. These, incidentally, are usually much harder to obtain in large quantities and much harder to crystallize, and their structures much harder to characterize. So there are a lot of important membrane proteins whose structures are not very well known yet. In photosystem I it's a trimer. There's one piece right here, it's shaped like a big disk. Here's another piece and here's another piece. And then if you flip it by 90 degrees to look at it edge-on you can see here it is sitting in the membrane. And these have colorful cofactors. These now have porphyrins, many porphyrins, that contain instead of iron, contain magnesium. And these are the chlorophylls that are present in here as light gathering entities. And then-- OK. OK, with this picture I can tell you quite a bit about photosystem I. Here, you can see, is the part of the molecule that is embedded in the membrane. And what we have is the ability to extract electrons from a molecule that is hard to oxidize. OK, so in photosystem II you'll see that the source of electrons is water. And so photosystem II is capable of pulling electrons out of water, generating dioxygen, the most important photosynthetic reaction on our planet. In the case of photosystem I, the source of the electrons, instead of being water, is a little protein called plastocyanin. So plastocyanin comes up here and it is oxidized. And it's a little bit like the LED we talked about last time, because what happens is there's a pair of heme units here, they're not hemes of chlorophylls, the special pair, that when light energy is absorbed and excitation occurs, it sends an electron to high energy. And normally, after such an absorption of light occurs, that electron would just drop back down and the molecule would go back to its ground state. But what this molecule has been built to do is to take that high energy electron and descend it through a series of redox active molecules. And it actually goes all the way through here, and these are little iron four s four cubes. So these are beautiful little inorganic molecules, Fe4 S4 cubes. The electron pathway goes through these redox active units and jumps from one of these Fe S4 cubes to another, and finally jumps out to a protein that docks up here to accept it, which is a ferredoxin protein. So after light comes in and you get a high energy electron, you get a hole, the hole is filled through oxidation in the case of photosystem I and plastocyanin in the case of photosystem II of water, and so the chain goes. Now, having run out of time here, I will talk about photosystem II at the beginning of next hour.
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https://ocw.mit.edu/courses/8-821-string-theory-and-holographic-duality-fall-2014/8.821-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. HONG LIU: OK, let us start. So first, let me remind you what we did at the end of the last lecture. So we can see there's a large N matrix field theory. And then we saw that when you write down your Feynman diagrams to calculate things, then there's a difference now between the order of contractions. Which is now refracted whether you have a Planar diagram or non-planar diagram, et cetera. And that in turn, also affects your end counting. So at the end, we discussed two observations. One observation we said, is for the example to be considered is the non-planar diagram, even though it cannot be drawn on the plane without crossing lines, can actually be drawn on the torus without crossing lines. So the diagram can actually be straightened out on the torus. And another observation, is that the power of N is related to the number of faces you have in your diagram after you have straightened it out. So now, I'm going to generalize these two observations. So first, I will tell you a fact. Many of you may already know this. So I'll first tell you a fact that any orientable two dimensional surface is classified topologically by integer, which I will call h. So this h is called a genus. So heuristically, this genus is equal to the number of holes a surface has. So for example, if you have a plane, the plane topologically is actually equivalent to a sphere, so when I draw a sphere, that means a plane, because if you identify the point on the plane of the infinity, then it becomes a sphere. So topologically, a plane is no different from a sphere. So this genus 0. So this is plane. So h is equal to 0. And then for torus, then there's one hole, and this is just genus 1. So this is torus. You can also draw surfaces with as many holes as you want. So this is surface with two holes. A surface with two holes, and so this is genius 2. So the remarkable thing is that this actually classifies the topology of all two-dimensional surfaces. And this so-called topological invariant, there's a topological number a so-called Euler number, which is defined to be Chi is equal to 2 minus 2h. So if h labels the topology, and so the Chi is related to h in this way. So any two surfaces with the same topology will have the same Chi, because if they have the same h, they have the same Chi. So Chi is what we call the topological invariants. So this is just a mathematical fact. So now I'm going to make two claims. So these two claims are in some sense, still evident after you have studied a little bit. I'm not going to prove it here, will just make the claim, and then I will leave the task to familiarize these two claims to yourself. But as I said, these two claims are actually self-evident, if you just do a little bit of studying. So claim one-- so this is regarding the structural Feynman diagrams. So remember, the example theory we conceded last time was something like this, which phi is [INAUDIBLE] matrix. So if you can look at the Feynman diagrams of the theories, the claim is the following. for any non-planar diagram, there exists integer h so that the diagram can be straightened out. Straightened out just means non-crossing on a genus h surface, but not on a surface with a smaller genus. So actually, you can classify all the non-planar diagrams by drawing it on a genus h surface on two-dimensional surfaces with non-trivial topology. And then you should be able to find the number h, which is the lowest genus you need to make this diagram to be non-crossing, instead of crossing. Last time, we saw an example which you can stretch it out on the torus, but it was a more complicated diagram and torus would not be enough. You would need, say, a genus 2 surface or higher genus surface. So you can easily convince yourself that this is doable just by some practice. Also, this is very reasonable. Do you have any questions regarding this claim? Good. So this was claim one. The claim two-- and this is a generalization of the second observation. It's a claim of the generalization of the first observation of last time, and claim two will be the generalization of the second observation. For any diagram, the power of n coming from contracting propagators is given by the number of faces on such a genus h surface. As we explained last time, the number of faces just means the number of these connected regions in the diagram. But if you are stretching it out, then there's an ambiguous way you can count the number of disconnected regions In the diagram. And that number is the power of n coming from the contracting propagators. So this claim is also self-evident because each power of n comes from a single connected line. And essentially, a single connected line will be circles some face and so the number of independents inside the face of course, are just the number of power n. Any questions regarding the second claim? Good. So based on these two claims, now immediately write down the independents. So from this, we can find a vacuum diagram. Again, we have been only talking about the vacuum diagram. Vacuum diagram means the diagram has no external lags, has the full independence. So from here, we immediately conclude that the vacuum diagram has the following, g squared and dependence. So g squared is just the carpeting constant. So remember, each propagator is proportional to g squared and that each vertex is one over g squared. Just from there. So let's just take an arbitrary vacuum diagram. Let me call this amplitude to be A. And A should be proportional to g squared to the power E. E should be the number of propagators, because each propagator gives you a factor of g squared. And also minus V. And V is the number of vertices below to the number of vertices. And each vertex gives you one over g squared, so that should give you E minus V. And then we just multiply N to the power of F, and F is the number of faces. So without doing any calculation, so this has essentially characterized the N and the g and the accompanying dependents of any diagrams. Now, if you look at this expression, you say we are doomed, because this kind of expression does not have a sensible [INAUDIBLE] So remember our goal, the original goal of the 't Hooft is that you treat this N as a parameter and then you want to take it N to be large and then to it's ranking, one over N. Doing expansion in one over N means you are expanding around the including infinity. You want to expand around the including infinity. But if you look at this expression and then go to infinity, it's not a well-defined limit, because I can draw a [INAUDIBLE] Feynman diagram, which F can be as large as possible. So you just have sufficiently many vertices, and sufficient to be mainly propogators, F is unbounded. So then, there's another well-defined angle to infinity limits. So this expression is not a sensible N of infinity limit. If you don't have a sensible N infinity limit, then you cannot talk about doing a longer expansion, and you cannot even define the limit we just finished. Yes? AUDIENCE: I remember people do partial sum to-- can they do the same thing here? Put the large in the denominator pattern from series of singular bubbles and-- HONG LIU: Yeah. AUDIENCE: Like a random face approximation-- HONG LIU: Yes. AUDIENCE: Somehow they can infinite a sum over an infinite number of A, and put the margin into like 1 minus N something for the denominator-- HONG LIU: Yeah. So people certainly have been talking about expansion. And certainly, if this just failed here, we will not be talking about this. I'm just setting up a target, which I'm going to shoot it down. Right. Yeah, I would just say if you do the margin expansion, this probably would be the place which would turn you back. You say, ah, there's no margin limits, and that's from another problem. But hopefully, that's no ordinary person. And so remember, there are several nice mathematical tricks you have to go through. First, you have to invent the stop line notation so that you can count the end very easily. Yeah, first you have to come up with this margin idea. I think maybe not due to him, maybe other people already considered similar things like right here. But first you have to think about the double line notation to make your computation easier, and then even after you reach the staff and then you need to know this kind of topology, et cetera. After you reach this staff, still, there is a roadblock here. But 't Hooft found a very simple way to go around it. Because when I say this limit it is not well-defined, I made the assumption that when I go to infinity limit, the g is kept fixed. So that's the reason that this will blow up. But then 't Hooft came up with a different image. He said when you can see that N goes infinity, but at the same time, you can see that g squared goes to zero. At the same time, you can see that g squared goes to zero. Because the problem with this, is that if you have a diagram with lots of faces and if you multiply by a finite number, if you take g equal to 0, then you have infinity multiply something potentially goes to 0, maybe you will get something finite. One second. So you said let's consider this limit, so that the product of them is kept finite. Let's consider this limit. So in this limit, the end counting will be different. So this A will be g squared N, E minus l, because we want to keep this fixed. So we put an N factor here, then we take the N factor out. N plus F plus V minus E. And then let me just write this slightly differently. So this is Lambda. If I use this notation, Lambda now is finite. So this is E minus V. So let me just write it as Lambda L minus 1 N to the power of Chi. Now, let me explain [INAUDIBLE]. First, L is equal to E minus V plus 1, is the number of loops in the diagram. So do you guys remember this formula, why this is true? The reason this is true is very easy to understand. So if you look at the Feynman diagram, the number of loops is exactly the same as the number of undetermined momentum. And each propagator will carry momentum, and at each vertex, you have momentum conservation. But then this overall momentum conservation, which is guaranteed, so the number of independent momentum is E minus V and then plus 1. Yeah, it's just E minus V minus 1, because there's only V minus 1 independent momentum constraints. So this is the number of independent momentums, and so this is the number of loops. And the Chi is just defined to be the number of faces plus the number of vertices minus the number of edge. Yes? AUDIENCE: Is there any-- in general, is there a relation between E, V, and F, like are they completely independent to each other? HONG LIU: E, V, and F? AUDIENCE: Yeah. HONG LIU: Yeah, that I'm going to explain. Any other questions? AUDIENCE: So the reason that you can take the F to infinity limits without taking it to 0, is that why we can't apply this to QCD to get a-- HONG LIU: No, because we apply this. AUDIENCE: But if g is not small? HONG LIU: It doesn't matter. So this is an expansion scheme. And you can apply it to QCD, then you can ask whether this particular expansion scheme is a good approximation or not. That's a separate question. And the focus is pretty good, it's not that bad. Yeah, focus of each squaring is finite. AUDIENCE: But it's [INAUDIBLE]. HONG LIU: No, here it does not tell you if lambda has to be small. Lambda can be very big here. It just tells you the lambda in this limit has to be finite. Lambda can be convening and it can be very big. And actually, in the future, we will take lambda to go to infinity. So now, in order to have a well-defined limit still depends on this chi. So now, you ask why this works because we still have this chi, but now again, we need another piece of mathematics. First, when you draw, let me remind you some diagram we had last time, so this is the simplest diagram. And there's a lot of non-planar diagrams which you can draw on the torus, which is the non-planar version of this, et cetera. And you can also have more complicated diagrams. So suppose you are on the torus, I can consider more complicated diagrams, like that. For example, such diagrams. So if you think about such diagrams, then in a sense, each Feynman diagram can be considered as a partition of the surface. So if you draw a diagram of the surface into polygons. Yeah, it's very clear here. I drew this now with wavy lines, just make them straight. And this one topologically, I can just draw like this, and then I have one, one, and the other part, similarly here. So each Feynman diagram can be concealed if you just partition whatever surface you draw the diagram. Is this clear to you? So this is a very important point. Yes? AUDIENCE: Can you repeat that? HONG LIU: Yes. Look at this diagram. So this is a Feynman diagram drawn on the torus. Does this look like a partition of torus into polygons? Yes, so that's what I'm talking about. And the statement that this would apply for any Feynman diagrams. Yes, so this is like a partition of a sphere. You separate a sphere into three regions, one, two, three. Any questions about this thing? Yes, do you have a question? AUDIENCE: Yeah, in that case the third one isn't abandoned, is it? The third-- HONG LIU: So this one I am going to show you, you should view the plane as a sphere. AUDIENCE: Oh, OK. HONG LIU: Topologically, it's the same as a sphere. So now, once you recognize this, now you can use a famous theorem due to Euler. Some of you could have learned it in junior in high school, because the-- so this theorem, given a surface composed of polygons-- so if you're not familiar with this, you can go to Wikipedia-- with F faces, E edges, and V vertices. Suppose you have a surface like this, and then this particular combination chi, which is defined to be F plus V minus E is precisely equal to 2 minus h. So this is the-- this is why this is called Euler's number. And they only depend on the topology. So this combination only depends on the topology of the surface and nothing else. So we can imagine 't Hooft a very good high school student. So he already knew this. So now, we can rewrite this thing as A, reading script A lambda to the power L minus 1 N to the power 2 minus 2h. So remember h is greater than zero. So now this expression has a well-defined limit. So now this has a well-defined analogy and limit. So to the leading order, and in particular, this N dependence only depends on the topology of the diagram. Only depend on the topology of the diagram. For example, to leading order in large N, then the leading order will be given by the planar diagrams. Because that should be the diagrams with h equal zero because h is not negative. So the leading term is given by h zero. So all of the planar diagrams in this limit, in this so-called 't Hooft limit, will have N dependence which is in N squared. N squared. So now if you go back to your loads, above the fourth diagram we started last time, above the two planar diagrams we started last time, you can immediately tell that indeed it is N squared in this limit. Then you still haven't explained in terms of lambda. So you can do Feynman diagrams, et cetera. Say lambda just depends on the number of loops. So if you have one loop, then start it with a lambda equal to zero, start it with some constant. If it's two loops, then lambda, three loops, lambda squared, et cetera. So the sum of four-- sum of all planar diagrams, we have this structure. So you can imagine you can sum all this, if you're powerful enough. And then you can write it as N squared, then some function of lambda. So the planar diagram would be just N squared time some function of lambda. So if you are powerful enough to compute this f0 lambda exactly, then you can say you have solved the planar. You have solved the large N limit of this series. Unfortunately for lambda being in gauge theory we cannot do that. We don't know how to compute this. We can on only compute perturbatively which actually does not work for QCD. Yes? AUDIENCE: But N going to be infinite. It just seems like-- why does this overcome the same problem that we had before? HONG LIU: Because this is a specific power. Here there's no specific power. That F can be as large as you want. AUDIENCE: But wasn't the other problem also that N was going to infinity? HONG LIU: No that's another problem. Because here, there is a specification limit, when you take N going to infinity. In there, there's no specific limit. There's no limit when you take N go to infinity. No, the limit is N squared. I can say-- this is your essentially, your vacuum energy, right? I can say E0, divided by N squared has a well defined limit. So the key is that this is a specific dependence on N. But there it's unbounded. AUDIENCE: I guess another question is, we didn't assume that lambda is a small prohibiter. HONG LIU: No we did not assume lambda. AUDIENCE: So isn't this kind of perturbative analysis, saying that that series converges to some function, is that OK? HONG LIU: It's a very good question, and it is OK. But the reasoning is more complicated. The reasoning is more complicated. The keys are falling. What I'm writing here you can understand using two perspectives. Just one is heuristic. Just one say, suppose you can only do perturbation series, and that would be the thing you compute. And then say, if you are powerful enough to solve the theory with perturbity, then you're guaranteed just to find some general functions. Yeah but whether this series actually converges is an important mathematical question, of course, one has to answer. It turns out, actually this is convergent. You can mathematically prove this is convergent, for very simple reasons. Yeah this is a side remark, but let me mention it because this is a cool fact. So if you do just lambda of g 5, 4 series, just do the standard Feynman diagram calculation, input the basic series, the series is divergent. The series is divergent, no matter how g is small, just as the radius of convergence is 0 for any non-zero value. Yeah the radius of convergence is 0 in terms of g, no matter how small g is. The reason that it's divergent is because the number of diagrams. So when you go to Nth order, it increases exponentially in N. So the coefficient of g-- say to the power N-- the coefficient can become huge. Because of the number of diagrams to increase exponentially in N. So you think, I have g to the power N times N factorial, something like that. And that series is not convergent. But the thing that is convergent here, because can see the planar diagram, the number of planar diagram is very, very small. It only increases with N as a power of N, rather than as a factorial of N. And so you can make lambda small enough, then this can be convergent. Yeah. AUDIENCE: Didn't we already see at the end of the last lecture that only planar diagrams keep the biggest contribution, so why is that equal to-- HONG LIU: Sorry? AUDIENCE: So at the end of the last lecture, we saw that-- HONG LIU: No not necessarily. That is only for the two diagrams. If you look at this thing, I can draw arbitrary complicated non-planar diagrams, with a very large F. They only depend on F. It does not depend on anything else, if you do this. OK so in general, then of course if you include the higher non-planar diagram, et cetera-- so in general the vacuum energy, which you would normally call log z. OK log z is, essentially, the sum of all vacuum diagrams. So this is the partition functions, so it's a path integral. So log z, then we'll have the expansion from h equal to 0 to infinity, N two the power 2 minus 2h, F h lambda. OK so at each genus level, you will have some function of lambda. Yeah the leading, order we just showed is f0. And then if you add the Taurus now-- we'll add to the torus now our on non-planar diagram. It's order one. And then, for the two genus, it's f over n squared, et cetera. So let me just write down z explicitly. Z is the partition function, i S phi. So if you compute this, with the right boundary condition-- if you compute this path integral, with the proper boundary conditions, then that gives you the vacuum diagrams. The log z is the sum of all connected vacuum diagrams. I should say, the sum of all connected vacuum diagrams. Any questions regarding this? Good. Now there is a heuristic way we can understand this expansion. So it's actually a heuristic way to understand this one-way expansion. So let's just look at this path integral. So let's look at the Lagrangian. So the Lagrangian I wrote there. In this 't Hooft limit, I can write it as minus N divided by lambda. So I want to write things in terms of lambda. So I multiply the pre-factor N downstairs, and then upstairs. So g squared N give me lambda. And then I have a trace, et cetera. So now it's easy to see that you're leading order, this things should give you order N squared. Because there is already a factor of N here, and the trace is the sum of N things. So generically, this should be of the order N squared. So now we-- a little bit leap of faith-- say supposing the large N limit, since the leading order is order N squared, you can argue that actually all the N squared is the expansion parameter, if you do scatter point approximation. And then naturally, you will see the power will be given by 1 over N squared. any questions regarding this? Yes? AUDIENCE: What does N over lambda factor to? HONG LIU: It's-- AUDIENCE: OK. HONG LIU: So clearly this discussion actually does not-- so when I say clearly here, it requires a little bit of practice. But clearly, our discussion only depend on the matrix nature of the Lagrangian and the fields. So I'm going to make a claim. It says for any Lagrangian, of matrix valued fields of the form L, which is N divided by some coupling constant, times a trace of something. Doesn't matter what you put inside the trace here, as far as you have a single trace. Such a series will always have the expansion like this. It will always have an expansion like this. So let me just summarize. In the 't Hooft limit-- by 't Hooft limit, I always mean this form-- the coupling constant is defined such that you have some coupling concept here, and then you have over N factor. And then of course, you can also have some coupling constant inside here. It doesn't matter. As far as those coupling constants are independent of N. And as far as things inside the trace are independent of N. So for such a kind of series, the 1 over N expansion is equal to the topological expansion. It's the expansion in terms of topology of Feynman diagrams. So this is a very, very beautiful-- and as I said, it's [INAUDIBLE], because in principle, it puts a very simple structure into something that's, in principle, very complicated. Yeah, any questions regarding this? Yes? AUDIENCE: Maybe it's not can I just understand it in such a way that he kind of treats the Feynman diagram as a triangulation of different spaces. HONG LIU: Yeah for example, you can think of from that point. Yeah, so we use that to derive, to use this formula. Yeah and that picture will be very useful in a little bit from now on. Keep that picture in mind. The Feynman diagram is like the partition of some surfaces. And that will be very useful later. AUDIENCE: Does there ever arise a situation in which you care not about two surfaces, but Feynman diagrams on three surfaces or something like that? Because this asks the question, you don't necessarily have to consider the topology of two surfaces. Are there any situations in which it's more complicated? HONG LIU: Yeah but we always draw Feynman diagrams on the paper, which is two dimensional. Yeah it's enough. Two dimensions is enough. You don't need to go to three dimensions. Yeah and this structure only comes when you go to two dimensions, because if you go to three dimension, of course, they don't cross. In three dimension, you can no longer distinguish planar or non-planar diagram. AUDIENCE: Well if you did-- OK. HONG LIU: Good any other questions? AUDIENCE: Why is it always orientable surfaces? HONG LIU: That's a good question. It's because those lines are orientable because when we draw the double roation, so you have this two, double rotation. So essentially those lines are orientable. They have a direction. And essentially, this is, of course, one and two. Yeah so those Feynman diagrams actually have a direction, have a sense of orientation. So I'm going to mention, by passing in nature, but let me just mention also now. So if it's not [INAUDIBLE] matrix, say if it's a real symmetric matrix, then the then there's no difference between two index. And then there's no orientation. And that would be related to m orientable surfaces. And then you need to slightly generalize this. Any other questions? AUDIENCE: Now you mentioned that it has something to do with string theory, but does that it has anything to do with scatter particles-- HONG LIU: Yeah, so we're going to talk about that. No, we're going to talk about that. Good? So now let me talk about general observables. I think we're a little bit short on time, if we want to reach the punchline today. So right now we only have looked at the vacuum diagrams. So now let's look at the general observables. Before talking about general observables, let me just again make a side remark, Which is the gauge versus global symmetries. So in the example we talk about here-- let me call this a, equation a. Then let me write down another equation b, which is a Yang-Mills theory. So the difference between a and b-- so a is this guy. So the difference between a and b, is that a, as we discussed last time, is invariant on the global symmetry, is a U(N) global symmetry. It's that phi is invariant under the acting of a unitary matrix. But this U must be constant. Only for a constant U is this a symmetry. But if you have studied a little bit of gauge theory, or if you have not studied gauge theory, just take my word for it, the b is invariant under a local symmetry, a local U(N) symmetry. He said A miu-- so A mu is what make up the F-- U x A mu x, U dagger x minus i partial mu U x. It doesn't matter. The only thing I want to say is that this U x is arbitrary. It can have arbitrary space time dependence. Just like of the generalization of the QED, it's the gauge symmetry. So the key difference between the two. the key difference between this local and the global symmetry, are manifested in what kind of observables we can see there. For example, for a-- AUDIENCE: I think that should be U dagger, partial mu and then U dagger. HONG LIU: HONG LIU: So this difference, you can say what's the big deal? in one case, this is constant, and this is dependent on space time. So the key difference between them is that in the case for a, operator like this phi squared, phi is a matrix. So phi squared is a matrix. Phi squared x is an allowed operator. So this operator is not invariant under this global U(N) symmetry. But it doesn't matter because this is a global symmetry. So this is an allowed operator. But if you have gauge symmetry, all operators must be gauge invariant. That means that all operator must be invariant under this kind of transformations. So the analog of this is not allows operators. So observables in the gauge theory are much more limited. So we will be interested gauge theories. We will be interested in gauge theories. So that means we are always interested in observables, which are invariant under the symmetry. So we're interested in gauge theories. So that means we're always interested in gauge invariant operators. So the kind of Lagrangian does not matter. So you can have the gauge fields. You can also have some other field, say some matrix phi, et cetera. As phi is the Lagrangian of this form, it's OK. We always only consider the Lagrangian of that form. But it doesn't have an arbitrary number of fields, and with arbitrary kind of interactions. So you start your theory. So let's for simplicity, this can see the local operators. In this kind of theory, then allowed, say local operators, must always have some kind of trace in it. Say you must have some form trace, F mu U, F mu U, a trace phi squared, et cetera. A trace phi to some power F N, phi to some power k, et cetera. You can also have operators with more than one trace, say trace phi squared, trace F mu U, F mu U. So we are going to make a distinction because the operator with only a single trace, we call them single trace operator. And the operator with more than one trace, we call them multiple trace operators. OK so the reason for this distinction will be clear soon. So multiple trace operators-- it's self-evident again-- that multiple trace operators can always be written as products of single trace operators. AUDIENCE: I have a question. Is it a possible case to have a local gauge invariant operator, the F mu U times F mu U? HONG LIU: Yeah I always can see the local gauge invariant operators. We can see the gauge theories. AUDIENCE: So this combination, F mu U times F mu U, it's only-- HONG LIU: No this is gauge invariant. AUDIENCE: Is that the only gauge invariant component? HONG LIU: Sorry? AUDIENCE: Is this is the only gauge invariant component that we can use to construct the gauge invariant local operators? HONG LIU: Sorry, I don't quite understand what you mean. No this is just one specific operator. No, you can take F to the power N, an arbitrary number of-- as far as they're inside the trace, it's always gauge invariant. AUDIENCE: I see. HONG LIU: I'm just writing down a particular example. So just for notational simplicity, I will just write-- so from now on I will just denote the single trace operators collectively just as O with some script n, which denotes the different operators. so n denotes different operators. And then for the multiple trace operator, then you have O1, O2. That would be a double trace operator. And O1, O2, O3, say times O3 would be a triple trace operator. n just labels different operators. I'm just using abstract notation. And the reason for this-- a distinction will be clear soon. Then for such gauge theories, the general observables, in the quantum field theory is just correlation functions of gauge invariant operators. So by gauge invariant operators, you can have local operators, non-local operators, et cetera. So for simplicity, I will restrict my discussion only on the local operators. But local operators means that the fields evaluated at a single point. So the typical correlation functions, then the typical observables will have this form, will be just a product of some correlation functions, a product of some operators, and I say their correlation functions. By c I mean the connected correlation functions. So you can see immediately, these multi trace operators are just the product of a single trace operator. And the correlation function of a multiple trace operator can be obtained from those of a single trace one. You just identify some of the acts. Then that will be enough. So we only need to talk about the correlation function of single trace operators. So now the question-- let me call this equation one. So now the question follows what we discussed before, is how do we decide the N dependence of the guy? Previously we determined the N dependence of this vacuum diagrams of this petition function. But now want should be the N dependence of general correlation functions? One way to do it, you just start immediately calculating. And then you can find some root, et cetera. But actually there's a very simple trick. There's a very simple trick to determine the N dependence of this. So now I will explain. Now I will tell you. So I will not give you any examples because this trick is so nice, and it just works very easy. And so the question, what is N dependence of one? So this is the question we want to address. so here is a very simple, beautiful, trick. So let's consider the following generating functional. So in quantum field theory, when we talk about correlation functions, it's always convenient to talk about the generating functional. So whatever is your field, you do the path integral of all fields. And then you look at the action. So you have your regional action, which I call S0. And then that's add those operators. Ji x, O i x. Yes so this is a standard story. When you take the derivative over Ji, then you will bring down a factor of O i. Then that essentially give you the correlation function. You have, for example, a correlation function, the connected correlation function. O n, the connected correlation function would be just you take the derivative of log z. And then delta J1 x delta Jn xn. And then you set J equal to zero. You set all the J equal to zero. And that gives you the end point function. I should write i here, so i to the power n. So now here is the beautiful trick. You can determine this in a single shot, the N dependence of this guy. And this simple trick is just to add N here. You add N here. In order to get the correlation function, we need to divide it by N to the power N. Now you wanted to get O i, you need to divide-- take the root of N times Ji, so you also need to divide it by 1 over N. But why does this help? Why does this help? It helps for the following reason. Let me call this whole thing iS effective. iS effective, so the key thing is that this O i, single trace operators, then this S effective then has the form N times the trace something. Because you already know S0, which is our starting point, has this form. And now the term you added in, precisely, also has this form, is the N factor times something single trace. So that means the whole thing still has this form. Then now we can immediately conclude this log z J1, Jn, must have this expansion. h from 0 to infinity, and to the 2 minus 2h, f h lambda J1, et cetera. So adding that N is a powerful, powerful trick. So now you can immediately, just from here, we can immediately find out that for endpoint function, connected endpoint function, the leading order is 2 minus n, because the leading order is n squared. Yeah it's 2 minus n. And then suppressed by 1 over N squared, et cetera. Good? So for example, if you look at a 0 point function, which is essentially the partition function, so this is order N squared. So this is what we found before, to leading order. And if you look at the one point function, some operator, then, will be order N. If you look at the two point function, connected two point function of some operator, so it would be order one. And the three point function of some operator will be order 1 over N, as the leading order. And then all higher order just down by 1 over N squared, compared to the leading order. And again, the leading order is given by the planar diagrams. Because of the leading order contribution to here, in terms of this S effective is planar diagram. And then they must be, under those things, just obtained by through [INAUDIBLE], so they must be planar diagrams. so again this comes from planar diagrams. Good? So now let's talk about the physical implications of this. So what does this mean? So we have found out, this is our N dependence for our gauge invariant operators. So what does this mean? Now let me talk about physical implications. AUDIENCE: What did one of those [INAUDIBLE] have with-- HONG LIU: Yeah, I defined them without chi. AUDIENCE: Then something like-- HONG LIU: You define something. So when you write down your theory, you define this 't Hooft limit. Then everything is already in terms of lambda or some other order one number. So that can depend on those numbers in an arbitrary way. It doesn't matter can depend on coupling constants in an arbitrary, but it cannot have N dependence defined inside the operator. Once we introduce this 't Hooft limit, then the operator can depend on the coupling in the 't Hooft limit in an arbitrary way. Because they're all just all the one constant. Let's talk about physical implications of this. It turns out, these simple and [INAUDIBLE] behavior actually has a very simple physical picture behind it. So first he said, in the large N limit-- so let's just look at leading order behavior. So in a large N limit, if we consider this state of O i x acting on the vacuum. So some single trace operator acting on the vacuum. So i, again, just labels different state, different operators. These can be interpreted as creating a "single particle" state. I'm first describing the conclusion. Then I will explain why. So you can see that the state obtained by adding a single trace operator on the vacuum, then this can be interpreted as a single particle state. So if you add on the double trace operator on the vacuum, then this can be interpreted as two particle state. Similarly, say O n acting on the vacuum would be N particle states. So why is this so? Why this is so. So I will support this statement using three arguments. First remember O i O j, the connecting Green function of any two operators of order one. So we can actually just diagonalize them. If you can just diagonalize them, so that O i O j are proportional to delta i j. So in some sense, a two point function, those operators can be considered as independent. And now the second statement. So if you want to call this single particle, this two particle, multi particle, then they should be that they don't overlap. Because a single particle cannot overlap with two particle, et cetera. And then to see the overlap between a single particle with a multi particle, you look at these correlation functions. So if you look at the overlap of a single particle state, with a two particle state, say some double trace operator. Let me just, to avoid confusion, let me just use the inside-- use this notation to see this as a single operator. So you can see that the overlap we saw of the single trace operator with this double trace operator. And they start off-- this whole thing is like a three point function. Just put these two over the same point. So from our discussion here, is how you connect the Green functions of order 1 over N. So this goes to 0, compared to this overlap with itself. So that means in the N goes to infinity limit, there is no mixing between what we called single particle state, single trace, and the multiple trace states. So the third thing is that now let's look at the two point function of two double trace operator. So O1 O2 x, say O1 O2 and y. OK double trace operators. So let's look at the two point function with double trace operators. So there's an even contribution to this. So let's include all diagrams. Also these can all be connected So leading order is a disconnected diagram, which is essentially O1 x O1 y and O2 x and O2 y. Because I have a diagonal next to them. have And then plus the connected Green functions, which is order 1 over N squared. So if you see this leading order contribution, it's just like two independent particle propagating. Just like the product of two independent propagators. So it's sensible to interpret this as just the propagating of two particles. So it's sensible to interpret this two particle state, this double trace operator as just creating some two particle state. Yeah, again this goes to zero, in the large N limit. So I should emphasize when we call this a single particle state, it's not necessary they really exist on a shell particle corresponding to this state. We're just saying that the behavior of these states can be interpreted as some kind of single particles. The behavior you can just interpret as single particles. It's not really necessary they exist as stable on shell particles. In certain cases, there might be. There might be actual particles, actual stable, on shell particles associated with these kind of states. But for this interpretation to be true, it does not have to be. So in QCD actually sometimes-- so I just say they exist. So we call them "glueball" state. For example, in QCD, the analog of this kind of operator can create some state, which they are short-lived. They are not long-lived. They quickly decay. And so they're typically called the glueball state. So from now on we just call them glueballs. Call them glueballs. So this is one of the first implication-- the first indication is that they're just a single particles. A single trace operator can be interpreted as creating single particle states, and the multiple trace operator can be interpreted as creating multi particle states. Another second, he said if this glueball operators, so fluctuations of glueballs suppressed. So let me explain what this means. So let me suppose some single trace operator, O, which has a non-zero expectation value, suppose some state has a non-zero expectation value. And then let's look at the variance of this operator, the variance of the expectation value, which is given by O squared, minus O squared for the fluctuation. And this is, by definition, just a collected Green function of O squared. The is the full O squared. This is the disconnected one. So this is just a collected part of the O squared. And this one, N dependence, we know this is order one, order N to the power 0. So that means that these to below this order N. So the variance of this, compared to the expectation value of this operator itself is surprising, the large N limit. So essentially, in the large N limit, so that means the variance provided by the operator itself is 1 over N because of the 0 in the large N limit. So assuming that if you have a two point function-- so suppose each two point function, so each O2 have a non-zero expectation value, then you can factorize this O1 O2, plus O1 O2. So this is disconnected part. And again this will be of order one. But this part is of order N squared. So essentially the disconnected the part is always factorized. So the disconnected part is always factorized. Yeah this connected part is always small, compared to the disconnected part. So this is like a classical theory. AUDIENCE: We have just five minutes. We can proceed before it's 5:00 HONG LIU: Yeah but it may go to 10 minutes. It's just hard to say. It's just hard to say. Yeah, let me do it next time.
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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: Let me demonstrate now with plain doing the integral that, really, the shape of this wave is moving with that velocity. So in order to do that, I basically have to do the integral. And of course, if it's a general integral, I cannot do it. So I have to figure out enough about the integral. So here it is. We have psi of x and t. It's integral dk phi of k e to the ikx minus omega of kt. OK. It's useful for us to look at this wave at time equals 0 so that we later compare it with the result of the integral. So phi psi at time equals 0 is just dk phi of k e to the ikx. Only thing you know is that phi has peaked around k0. You don't know more than that. But that's psi of x and time equals 0. Let's look at it later. So we have this thing here. And I cannot do the integral unless I do some approximations. And I will approximate omega. Omega of k, since we're anyway going to integrate around k0, let's do a Taylor series. It's omega of k0 plus k minus k0, the derivative of omega with respect to k at k0 plus order k minus k0 squared. So let's-- do this here. So if I've expanded omega as a function of k, which is the only reasonable thing to do. k's near k0 are the only ones that contribute. So omega of k may be an arbitrary function, but it has a Taylor expansion. And certainly, you've noted that you get back derivative that somehow is part of the answer, so that's certainly a bonus. So now we have to plug this into the integral. And this requires a little bit of vision because it suddenly seems it's going to get very messy. But if you look at it for a few seconds, you can see what's going on. So psi of x and t, so far, dk phi of k e to the ikx. So far so good. I'll split the exponential so as to have this thing separate. Let's do this. e to the minus i. I should put omega of k times t. So I'll begin. Omega of k0 times t. That's the first factor. e to the minus i, the second factor. k-- k d omega dk. k0 times t. And the third factor is this one with the k0. e to the minus-- it should be e to the plus. i k not d omega dk. k0 t. Plus order-- higher up. So e to the negligible-- negligible until you need to figure out distortion of wave patterns. We're going to see the wave pattern move. If you want to see the distortion, you have to keep that [INAUDIBLE]. We'll do that in a week from now. This is the integral. And then, you probably need to think a second. And you say, look. There's lots of things making it look like a difficult integral, but it's not as difficult as it looks. First, I would say, this factor-- doesn't depend on k. It's omega evaluated at k0. So this factor is just confusing. It's not-- doesn't belong in the integral. This factor, too. k0 is not a function of k. d omega dk evaluated at k0 is not a function of k. So this is not really in the integral. This is negligible. This is in the integral because it has a k. And this is in the integral. So let me put here, e to the minus i omega of k0 t e to the minus-- to the plus i k0 d omega dk-- at k0 t. Looks messy. Not bad. dk. And now I can put phi of k. e to the i k x minus these two exponentials, d omega dk at k0 times t. And I ignore this. So far so good. For this kind of wave, we already get a very nice result because look at this thing. This quantity can be written in terms of the wave function at time equals 0. It's of the same form at 5k integrated with ik and some number that you call x, which has been changed to this. So to bring in this and to make it a little clearer-- and many times it's useful. If you have a complex number, it's a little hard to see the bump. Because maybe the bump is in the real part and not in the imaginary part, or in the imaginary part and not in the real part. So take the absolute value, psi of x and t, absolute value. And now you say, ah, that's why. This is a pure phase. The absolute value of a pure phase is that. So it's just the absolute value of this one quantity, which is the absolute value of psi at x minus d omega dk k0 t comma 0. So look what you've proven. The wave function-- the norm of the wave function-- or the wave. The new norm of the wave at any time t looks like the wave looked at time equals 0 but just displaced a distance. If there was a peak at x equals 0, at time equals 0. If at time equals 0, psi had a peak when x is equal to zero, it will have a peak-- This function, which is the wave function at time equals 0, will have a peak when this thing is 0, the argument. And that corresponds to x equals to d omega dk times t, showing again that the wave has moved to the right by d omega dk times t. So I've given two presentations, basically, of this very important result about wave packets that we need to understand.
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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: Last Thursday, we began the work of moving from special relativity to general relativity, and we spent a lot of time unpacking two formulations of the principle of equivalence. So one, which goes under the name "weak equivalence principle"-- a simpler way of saying that is that, at least over a sufficiently small region, if there is nothing but gravity acting, I cannot distinguish between freefall under the influence of gravity or a uniform acceleration. The two things are equivalent to one another. Basically, this is a reflection of the fact that the gravitational charge and the inertial mass are the same thing. That is the main thing that really underlies the weak equivalence principle. When I gave my tenure talk a number of years ago here, I pointed out that there was this wonderful program called the Apollo program that was put together to test this. And the way it was done was that they put astronauts on the moon, and you actually show that, if you drop a hammer and a feather on the moon, they fall at the same rate. Of course, the Apollo program probably did a few other things as well. But I'm a general relativity theorist, so for me, that was the outcome of the Apollo program, was test of the equivalence principle. We also have a different variation of this we called the Einstein equivalence principle, which leads us to a calculation that we went through last time, which states that we can find a representation over a sufficiently small region of spacetime such that the laws of physics are reduced to those of special relativity. And we did a calculation to examine this, where we showed, given an arbitrary spacetime metric, I can find a coordinate system such that this can be written in the form metric of flat spacetime plus terms that are of order so we have coordinate distance squared. Let's put it this way-- so this, in the vicinity of a point pl-- I'll make that clear in just a moment. It ends up looking like of order of coordinate displacement squared with corrections that scale as 1 over second derivative of the metric. That sets the scale for what these end up looking like, or maybe it's actually times that. Sorry. It's times that. Why did I divide? I don't know. Oh, I know why, because I wanted to point out that that is what the scale of 1 over this thing-- for God's sake, Scott! Stop putting your square roots in there! So it looks like that. And so this is what I was saying. You have a curvature scale that is on the order of square root of 1 over the second derivative of the metric. Apologies for botching that as I was writing it up there quickly. OK, so we did a calculation that shows that. And indeed, what we did is we went through and we showed that a general coordinate transformation has more than enough degrees of freedom to make the metric flat to get the flat spacetime metric at a particular point. And in fact, there are six degrees of freedom left over, corresponding to six rotations and six boosts that are allowable at that particular point or event. Bear in mind we're working in spacetime. We have exactly enough degrees of freedom to cancel out the first order term, but we cannot cancel out the second order term. And in fact, we find there are 20 degrees of freedom left over. And in a future lecture, we will derive a geometric object that characterizes the curvature that has indeed 20 degrees of freedom in it or 20 independent components that come out of it. So this is the foundation of where we're moving forward. And so what this basically tells us is that, in a spacetime like this, we have what we call curvature. Trajectories that start out parallel to one another are not going to remain parallel as they move forward. And where we concluded last time, we were dealing with the problem that, if I want to take derivatives-- and we're going to start with vector field. If I want to differentiate a vector field on a curved manifold, it doesn't work. If I do the naive thing of just taking a partial derivative, it does not work. So I'll remind you where we left off. So we found partial derivatives of vectors. Let me put that in there. Partial derivatives of vectors, and it won't take much work to show it's true for one forms or any tensor, so let's say, partial derivatives of tensors do not yield tensors. We're familiar with this to some extent because we already encountered this when we began thinking about the behavior of, even in flat spacetime, flat spacetime and curvilinear coordinates. So that's the fact that the basis objects themselves have some functional dependence associated with them. We're interpreting it a little bit differently now. And so what we're doing is we're going to say that what's going on here is that, on my curved manifold-- and I want you to visualize something that's like a bumpy surface or, if you like, maybe a sphere. Think about if you are a two-dimensional being confined to the surface of a sphere. There's only two directions at any given point. You can go up, or you can go on the left-right axis or the north-south axis, right? And you would define unit vectors pointing along there. But as you move around that sphere, those of us who have a three-dimensional life, and can step back and see this, we see that these basis objects point in different directions at different locations on the sphere. The two-dimensional beings aren't aware of that. They just know that they're on a surface that's curved. And so they would say that the tangent space is different for all these objects. They've imagined that every one of these basis objects lives in a plane that is tangent to the sphere at any given point, and that plane is different at every point along the sphere. So we interpret this by saying that all of our basis objects live in this tangent space, and the tangent space is different at every point on the surface. So let me just write out one equation here. So when we looked at the transformation of a partial derivative of a vector, if we looked at just the components-- so this calculation is in the notes, so I would just write down the result. What we found was that, if I'm taking, say, the beta derivative of component A alpha, what I found when I want to go into a coordinate transformation is that there's an extra term that ruins the tensoriality of this. So this goes over to something that looks like-- So the first term is what we'd expect if this were a tensor relationship. That's exactly the matrix of the Jacobian between two different coordinate representations that we expect to describe how components change if they indeed obey a tensorial relationship. This extra thing here I wrote on the second line-- that is spoiling it for us. So I began to give you the physical notion of what we were going to do to fix this. So let me just reset that up again. So let's imagine I have a particular curve that goes along my manifold. I have a point P here on the curve, and I have a point Q over here. And let's say that P is at event x alpha. Q is at x alpha plus dx alpha. Here's my vector A at the event Q. And here's my vector A at the event P. So what we discussed last time is that, to get this thing, I'm just doing the normal partial derivative. I'm basically imagining that these are close to one another, that I can just subtract A at Q from A at P, divide by dx and take the limit. That's the definition of a derivative. And what this is telling us is, mathematically, yeah, it's a derivative, but it's not a derivative that yields a tensor quantity. And so we are beginning to discuss the fact that, to compare things that have different tangent spaces, that live in different points in my curve manifold, I need a notion of transport to take one from the other in order to compare them. So there are two notions of transport that we're going to talk about here. The first one is called parallel transport. So transport notion one-- we call this parallel transport. I'm going to actually focus a little bit on the math first and then come back to what is parallel about this afterwards. So what I essentially need to do is say, what I want to do is find some kind of a way of imagining that I take the vector at P and transport it over to the point Q, and I will compare the transported object rather than the object originally at point P. Abstractly, what I'm essentially going to do is I'm going to do what we always do in this. I'm going to imagine that there is some kind of an operation that is linear in the separation between the two of them that allows me to define this transport. So let's do the following. So I'm going to assume that we can define an object, which I will call Pi, capital Pi, alpha, beta, mu, which is going to do the following. So what I'm going to do is say that A alpha transported-- and to make it even clearer, how it's being transported. Let's say it's being transported from P to Q. This is given by alpha at P, and what I'm going to do is say that, whatever this object is, it is linear in both the coordinate separation of those two events and the vector field. So far, I've said nothing about physics, by the way. I'm just laying out some mathematical definitions. I'm going to bring it all together in a few moments. I'm then going to say, OK, I know that I had trouble with my standard partial derivative. Let's define a derivative operator in the following way. Define a derivative by comparing the transported vector to the field at Q. So what I'm going to do for the moment is just denote this notion of a new kind of derivative with a capital D. Right now, it's just another symbol that we would pronounce with a "duh" sound so that it sounds like derivative. So don't read too much into that for a moment. So I'm going to define this as A at Q minus A transported from P to Q and then divide by the separation. Take the limit-- usual thing. And when you do this, you're going to get something that looks like the partial derivative plus an additional term on here. It looks like this. So I've said nothing about what properties I'm going to demand of this thing. And in fact, there are many ways that one could define a transport of an object like this. In general, when you do this, this thing I'm calling Pi here is known as the connection. It is the object that connects point P to point Q. So let's make a couple demands on its properties. So now I'll start to put a bit of physics in this. So demand one is I'm going to require that, when I change my coordinate representation, when I evaluate this in my new coordinate system, that I get something that looks like this. If I do this, I'm going to find that, when I change coordinates and apply it to the entire derivative I've defined over here, that little extra bit of schmutz that's on the second line there is exactly what you need to cancel out this annoying bugger so that you have a nice tensor relationship left. So I'm going to demand that a key part of whatever this guy turns out to be is something that cancels out the irritating garbage that came along with the partial derivative. Combine that with what I'm about to say in just a moment-- that pins things down significantly. I'm going to make one more demand, and this demand is going to then be connected to the physical picture that I'm going to introduce in about five minutes. My final demand is I'm going to require that whatever this derivative is-- when I apply it to the metric, I get 0. If I do that, then it turns out that the connection is exactly the Christoffel symbol that we worked out earlier. If I put in this demand, the connection is the Christoffel, and this derivative is nothing more than the covariant derivative. Der-iv-a-tive. This is just the covariant derivative we worked out earlier. So much for mathematics. The key thing which I want to emphasize at this point in the conversation-- I hope we can see the logic behind the first demand. That's just something which I'm going to introduce in order to clean this guy up. This is my choice. Not just my choice-- it's the choice of a lot of people who have helped to develop this subject. But it's a good one, because, when I do this, there is a particularly good physical interpretation to what this notion of transport means. And bear with me for a second while I gather my notes together. So let's do the top one first. OK, so let's make that curve again. And actually, let's just go ahead and redraw my vector field. I do want to have a different copy of this. Let me introduce one other object. So I'm going to give this curve a name. I'm going to call this curve gamma. And this is a notion that I'm going to make much more precise in a future lecture, but imagine that there is some kind of a tape measure that reads out along the curve gamma that is uniformly ticked in a way that we will make precise in a future lecture. I will call the parameter that is uniform that denotes these uniform tick marks-- I will call that lambda. If you want to read a little bit about this, this is what's known as an affine parameter. We will make this a little bit more precise very soon. With this in mind, you should be able to convince yourself that this field U defines the tangent vector to this curve. So let's say that my original point here, that this is at lambda equals 2-- and let's say this is at lambda equals 7. And what I want to do is transport the vector from 2 to 7. Well, a way that I can do this is by saying, OK, I now know that the derivative that goes over here-- this is just the covariant derivative. So I forgot to write this down but from now on, this derivative I wrote down-- I'm going to go back to the gradient symbol we used for the covariant derivative. What I can do is take the covariant derivative of my vector field, contract it with this tangent vector. And what this does-- I'm going to define this as capital D alpha D lambda. This is a covariant derivative with respect to the parameter lambda as I move along this constrained trajectory. What this does is this tells me how A changes as it is transported along the curve. I'm now going to argue-- what parallel transport comes down to is when you require that, as you slide this thing along here, the covariant total derivative of this thing as you move along the curve, that it's equal to 0. OK, now let me motivate where that's coming from. Why is that? So let's put all of our definitions back in. Let's apply the definition of covariant derivative that we discussed in a previous lecture and we've all come to know and love. Now, this is the bit where we begin to introduce a little bit of physics. Let's imagine that points P and Q are sufficiently close to each other that they fit within a single, freely falling frame. They can go into the same local Lorentz frame. So remember, when I go into my local Lorentz frame, the metric becomes the metric of flat spacetime. G goes over to eta. There is a second order correction to that. So the second derivatives-- I cannot find a [INAUDIBLE] that gets rid of them. So there'll be a little bit of that there, which tells me how large this Lorentz frame can be. But I can get rid of all the first derivatives. And if you get rid of all the first derivatives, you zero out the Christoffel symbols in that frame. So in that frame, there are no Christoffel symbols. So what this means is that, in this frame, this just becomes the idea that a simple total derivative of this object-- you don't need to include all the garbage that comes along with the covariant derivative-- this is equal to 0. And that's equivalent to saying that, as I take this vector and transport it along, I hold all of its components constant as I slide it from one step to the other. So I start out with my A at point P here. And then I slide it over a little bit, holding all the components constant just like this-- slide over again, slide it over again. Da, da, da, da da. Da, da, da, da, da. Till finally, I get it over to there. What this is doing is that-- any two vectors in the middle of this transport process-- I am holding them as parallel as it is possible to hold them, given that, to be blunt, you can't even really define a notion of two objects being parallel on a curved surface if there's a macroscopic separation between them. But if you think about just a little region that's sufficiently small, that it's flat up to quadratic corrections, then a notion of these things being parallel to each other makes sense. And this idea, that I'm going to demand that, upon transport, the derivative of the metric equals 0, thereby yielding my connection being the Christoffel and this derivative being the covariant derivative-- it tells us that this notion of transport is one in which objects are just kept as parallel as possible as they slide along here. And that is why this is called "parallel transport." It's as parallel as it's possible to be, given the curvature. So as I switch gears, I just want to emphasize I went through the mathematics of that with a fair amount of care because it is important to keep this stuff as rigorous as possible here. This notion of parallel transport gets used a lot when we start talking about things moving around in a curved spacetime. In particular, if you think about an object that is freely falling, and it is experiencing no forces other than gravity, which we are going to no longer regard as a force before too long-- if you just go back to Newtonian intuition, what does it do? Well, you give it an initial velocity or initial momentum, and it maintains it. It just continues going in a straight line. In spacetime, going in a straight line basically means, at every step, I move and I take the tangent to my world line, my four-velocity, and I move it parallel to itself. So this notion of parallel transport is going to be the key thing that we use to actually define the kinetics of bodies in curved spacetime. There's a tremendous amount of work being done by all sorts of things these days that's based on studies of orbits in curved spacetime, and they all come back to this notion of parallel transport. All right. Bear with me a second. I just want to take a sip of water. And then I'm going to talk about the other notion of transport, which we are going to discuss-- I always say, briefly, and then I spend three pages on it, so we're going to discuss. All right. So parallel transport is extremely important, and there's a huge amount of physics that is tied up in this, but one thing which I really want to emphasize is that it is not unique. And there is one other one which we are going to really use to define one particularly important notion, instead of quantities, for our class. So suppose I've got my curve gamma, and I'm going to, again, take advantage of the fact that I can define a set of tick marks along it and make that vector U be the tangent to this curve. And I'm going to, again, have my favorite points, x alpha at point P plus dx alpha at point Q. There's another notion of transport that is-- basically what you do is you cheat, and you imagine that moving from point x alpha to x alpha plus dx alpha is a kind of coordinate transformation. So let's do the following. Let's say that x alpha plus dx alpha-- we're going to take advantage of the fact that, since we have this tangent notion built into the symbols we've defined, we'll just say that it's going to be the tangent times the interval of lambda. And what I'm going to do is define this as a new coordinate system, x prime. So that's the alpha component of coordinate system x prime. It's a little bit weird because your x prime has a differential built into it. Just bear with me. So what we're going to do is regard the shift, or the transport, if you prefer, from P to Q as a coordinate transformation. It's the best eraser, so I'll just keep using it. So what I mean by that is I'm going to regard x alpha, and I'm going to use a slightly different symbol. I will define what the L is. So this is transported, but I'm going to put an L in here for reasons that I will define in just a moment. This, from P to Q, is what I get if I regard the change from point P to point Q as a simple coordinate transformation and do my usual rule for changing coordinate representation. So expand what the definition of x prime is there, and what you'll see is that you get a term that's just basically dx alpha dx beta. Then you're going to get something that looks like the partial derivative of that tangent vector. And remember, this is being acted on. I should've said this is this thing evaluated at P. Great. So we fill this out. OK. So that's what I get when I use this notion to transport the field from P to Q. Let's think about it in another way. Now, these fields are all just functions. So I can also express the field at Q in terms of the field at P using a Taylor expansion. I'm assuming that these are close enough that everything is accurate to first order in small quantities, so nothing controversial about this. I'm assuming dx is small enough that I can do this. But now I'm going to get rid of my dx beta using the tangent field U. Now, before I move on, I just want to emphasize-- these two boards here, over the way the left-- we're talking about two rather different quantities. The one I just moved to the top-- that actually is the field-- if you were some kind of a gadget that managed your field A-- that would tell you what the value is that you measure at point Q. This would tell you-- what do we get if you picked P up and, via this transport mechanism, moved it over to Q? They are two potentially different things. So this motivates another kind of derivative. So suppose I look at A-- value it at a Q-- minus A transported-- whoops, that's supposed to be transported from P to Q-- defined by D lambda. I will expand this out in just a moment. Now I will, at last, give this a name. This is written with a script L. This is known as the Lie derivative of the vector A along U. Anyone heard of the Lie derivative before? Yeah. So at least in the context where we're going to be using it, this is a good way to understand what's going on with it. We'll see how it is used, at least in 8.962 in just a few moments. Filling in the details-- so plug in these definitions, subtract, take limits, blah, blah, blah. What you find is that this turns out to be U contracted on the partial derivative of A minus A contracted on the partial derivative of U. Exercise for the reader-- it is actually really easy to show that you can promote these partial derivatives to covariant derivatives. And what this means is that, when you evaluate the Lie derivative-- so notice, nowhere in here did I introduce anything with a covariant derivative. There was no connection, nothing going on there. If you just go ahead and work it out, basically, when you expand this guy out, you'll find you have connection coefficients or Christoffel symbols that are equal and opposite and so they cancel each other. So you can just go from partials to covariants. Give me just a second, Trey. And this is telling us that the Lie derivative is perfectly tensorial. So the Lie derivative of the vector field is also a tensor quantity. You were asking a question, Trey. AUDIENCE: In the second term, did you miss the D lambda? SCOTT HUGHES: I did. Yes, I did. Thank you. There should be a D lambda right here. Thank you. Yes. Yeah. If you don't have that, then you get what is technically called "crap." So thank you for pointing that out. For reasons that I hope you have probably seen before, you always compute the Lie derivative of some kind of an object along a vector field. So when you're computing the Lie derivative of a vector field along a vector, sometimes this is written using a commutator. I just throw that out there because you may encounter this in some of your readings. It looks like this. So let me just do a few more things that are essentially fleshing out the definition of this. So I'm not going to go through and apply this definition very carefully to higher order objects. What I will just say is that, if I repeat this exercise and, instead of having a vector field that I'm transporting from point to point, suppose I do it for a scalar field-- well, what you actually get-- pardon me for a second-- is this on the partial, but the partial derivative of a scalar is the covariant derivative because there's no Christoffel that couples in. If you do this for a one-form, where it's a 1 indexed object in the downstairs position, you get something that looks like this. And again, when you expand out your covariant derivatives, you find that your Christoffel symbols cancel each other out. And so, if you like, you can just go ahead and replace these with partials. And likewise, let me just write one more out for completeness. Apply this to a tensor. So it's a very similar kind of structure to what you saw when we did the covariant derivative in which every index essentially gets corrected by a factor that looks like the covariant derivative of the field that you are differentiating along. The signs are a little bit different. So it's a similar tune, but it's in a different key. OK, so that's great. And if you get your jollies just understanding different mathematical transport operations, maybe this is already fun enough. But we're in a physics class, and so the question that should be to your mind is, is there a point to all this analysis? So in fact, the most important application of the Lie derivative for our purposes-- in probably the last lecture or two, I will describe some stuff related to modern research that uses it quite heavily. But to begin with in our class, the most important application will be when we consider cases where, when I compute the Lie derivative of some tensor along a vector U and I get 0. I'm just going to leave it schematic like that. So L U of the tensor is equal to 0. If this is the case, we say that the tensor is Lie transported. This is, incidentally, just a brief aside. It shows up a lot in fluid dynamics. In that case, U often defines the flow lines associated with the velocity field of some kind of a fluid that is flowing through your physical situation. And you would be interested in the behavior of all sorts of quantities that are embedded in that fluid. And as we're going to see, when you find that those quantities are Lie transported in this way, there is a powerful physical outcome associated with that, which we are going to derive in just a moment. So suppose I, in fact, have a tensor that is Lie transported. So suppose I have some tensor that is Lie transported. If that's the case, what I can do is define a particular coordinate system centered on the curve for which U is the tangent. So what I'm going to do is I'm going to define this curve such that x0 is equal to lambda, that parameter that defines my length along the curve in a way that, I will admit I've not made very precise yet but will soon. And then I'm going to require that my other three coordinates are all constant on that curve. So if I do that, then my tangent vector is simply delta x0. In other words, it's only got one non-trivial component, and its value of that component is 1. And this is the constant. So the derivatives of the tangent field are all equal to 0. And when you trace this through all of our various definitions, what you find is that it boils down to just looking at how the tensor field varies with respect to that parameter along the curve itself. If it's Lie transported, then this is equal to 0. And so this means that, whatever x0 represents, it's going to be a constant along that curve with respect to this tensor field. Oh, excuse me. Screwed that up. The tensor does not vary with this parameter along the curve. This was a lot, so let's just step back and think about what this is saying. One of the most important things that we do in physics when we're trying to analyze systems is we try to identify quantities that are constants of the motion. This is really tricky in a curved spacetime because much of our intuition gets garbled by all of the facts that different points have different tangent spaces. You worry about whether something being true, and is it just a function of the coordinate system that I wrote this out in? What the hell is going on here? The Lie derivative is giving us a covariant, frame-independent way of identifying things that are constants in our spacetime. So we're going to wrap up this discussion. Let's suppose that the tensor that I'm looking at here is called the metric. Suppose there exists a vector C such that the metric is Lie transported along this thing. What does this tell us? So first, it means there exists some coordinate such that the metric does not vary. The metric is constant with respect to that coordinate. Essentially, if you go through what I sketched a moment ago, this is telling us that the existence of this kind of a vector, which I'm going to give a name to in just a moment-- the existence of this thing demands that my metric is constant with respect to some coordinate. I am not going to prove the following statement. I will just state it, because, in some ways, the converse of that statement is even more powerful. If there is a coordinate, such that dgd, whatever that coordinate is, is equal to 0, then a vector field of this type exists. So the second thing I want to do is expand the Lie derivative. So if I require that my metric be transported along the vector C, well, insert my definition of the Lie derivative. Now, what is the main defining characteristic of the covariant derivative? How did I get my connection in the first place? In other words, what is this going to be? OK, students who took undergraduate classes with me, I'll remind you of one of the key bits of wisdom I always tell people. If the professor asks you a question, 90% of the time, if you just shout out, "0," you are likely to be right. [LAUGHING] Usually, there's some kind of a symmetry that we want you to understand, which allows you to go, oh, it's equal to 0. By the way, whenever I point that out to a class, I then work really hard to make a non-zero answer for the next time I ask it. So the covariant derivative of g is 0, so this term dies. Because the covariant derivative of g is 0, I can always commute the metric with covariant derivatives. So I can take this, move it inside the derivative. I can take this, move it inside the derivative. So what this means is this Lie derivative equation, after all the smoke clears, can be written like this. Or, if I recall, there's this notation for symmetry of indices, which I introduced in a previous lecture. The symmetric covariant derivative of this C is equal to 0. This equation is known as Killing's equation, and C is a Killing vector. Now, this was a fair amount of formalism. I was really laying out a lot of the details to get this right. So to give you some context as to why this matters, there's a bit more that needs to come out of this, but we're going to get to it very soon. Suppose I have a body that is freely falling through some spacetime. And you know what? I'm going to leave this here. So this is a slightly advanced tangent, so I'll start a new board. So if I have some body that is freely falling, what we are going to show in, probably, Thursday's lecture is that the equation of motion that governs it is-- you can argue this on physical grounds, and that's all I will do for now-- it's a trajectory that parallel transports its own tangent factor. For intuition, go into the freely falling frame where it's just the trajectory from special relativity. It's a straight line in that frame, and parallel transporting its own tangent vector basically means it just moves on whatever course it is going. So this is a trajectory for which I demand that the four-velocity governing it parallel transports along itself. Now, suppose you are moving in a spacetime that has a Killing vector. So this will be Thursday's lecture. Suppose the spacetime has a Killing vector. Well, so there will be some goofy C that you know exists, and you know C obeys this equation. By combining these things, you can show that there is some quantity, C, which is given by taking the inner product of the four-velocity of this freely falling thing and the Killing vector. And you can prove that this is a constant of the motion. So let's think about where this goes with some of the physics that you presumably all know and love already. Suppose you look at a spacetime. So you climb a really high mountain. You discover that there's a spacetime metric carved into the stone into the top of it. You think, OK, this probably matters. You look at it and you notice it depends on, say, time, radius, and two angles. Suppose you have a metric that is time-independent. Hey, if it's time-independent, then I know that the derivative of that thing with respect to time is 0. There must exist a Killing factor that is related to the fact that there is no time dependence in this metric. So you go and you calculate it. So this thing, that C is a constant of the motion-- I believe that's P set 4. It's not hard. You combine that equation that we're going to derive, called the geodesic equation, with Killing's equation. Math happens. You got it. So suppose you've got a metric that's time-dependent and you know you've got this thing. So you know what? Let's work it out and look at it. It becomes clear, after studying this for a little bit, that the C for this Killing vector is energy. In the same way that, if you have a time-independent Lagrangian, your system has a conserved energy, if you have a time-independent metric, there is a Killing vector, which-- the language we like to use is-- we say the motion of that spacetime emits a conserved energy. Suppose you find that the metric is independent of some angle. We'll call it phi. Three guesses what's going to happen. In this case-- just one guess, actually. Conserved-- AUDIENCE: Angular momentum. SCOTT HUGHES: Angular momentum pops out in that case. So this ends up being the way in which we, essentially, make very rigorous and geometric the idea that conservation laws are put into general relativity, OK? So I realize there's a lot of abstraction here. So I want to go on a bit of an aside just to tie down where we are going with this and why this actually matters. OK. Let's see. So we got about 10 minutes left. So what we're going to do at the very end of today-- and we'll pick this up beginning of next time. So for the people who walked in a few minutes late, the stuff that I'm actually about to start talking about we need to get through before you can do one of the problems on the P set. So I'm probably going to take that problem and move it on to P set 4, but I'm going to start talking about it right now. So we've really focused a lot, so far, on tensors. We're going to now start talking about a related quantity called tensor densities. There's really only two that matter for our purposes, but I want to go through them carefully. So I will set up with one, and then we'll conclude the other one at the beginning of Thursday's lecture. So let me define this first. I'm going to give a definition that I like but that's actually kind of stupid. So these are quantities that transform almost like tensors-- a little bit lame, but, as you'll see in a moment, it's kind of accurate. What you'll find is that the transformation law is off by a factor that is the determinant of the coordinate transformation matrix. Take it to some power. So there's is an infinite number of tensor densities that one could define. Two are important for this class. So the two that are most important for us are the Levi-Civita symbol and the determinant of the metric. So we use Levi-Civita already to talk about volumes. And it was a tensor when we were working in rectilinear coordinates, where the underlying coordinate system was essentially Cartesian plus time. It's not in general, OK? And we'll go through why that is. That'll probably be the last thing we can fit in today. So let me remind you-- Levi-Civita-- I'm going to write it with a tilde on it to emphasize that it is not tensorial. So this is equal to plus 1 if the indices are 0, 1, 2, 3 and even permutations of that equals minus 1 for odd permutations of that. And it's 0 for any index repeated. Now, this symbol has a really nice property when you apply it to any matrix. In fact, this is a theorem. So I'm working in four-dimensional space. So let's say I've got a 4-by-4 matrix, which I will call m. Write a new [INAUDIBLE] notation, m alpha mu. If I evaluate Levi-Civita, contract it on these guys, I get Levi-Civita back, multiply it by the determinant of the matrix m. Now, suppose what I choose for my matrix m is my coordinate transformation matrix. So I'm just going to write down this result, and I'll leave it since we're running a little short on time. You can just double check that I've moved things from one side of the equation to the other, and you can just double-check I did that correctly. What that tells me is that Levi-Civita and a new set of prime coordinates is equal to this guy in the old, unprimed coordinates with all my usual factors of transformation matrices and then an extra bit that is the determinant of the coordinate transformation matrix. If it were just the top line, this is exactly what you would need for Levi-Civita to be a tensor in the way that we have defined tensors. It's not. So the extra factor pushes away from a tensor relationship. And so what we would say is, because this is off by a factor of what's sometimes called the Jacobian, we call this a tensor density of weight 1. So in order to do this properly-- I don't want to rush-- at the beginning of the next lecture, we're going to look at how the determinant of the metric behaves. And what we'll see is that, although the metric is a tensor, its determinant is a tensor density of weight negative 2. And so what that tells us is that I can actually put together a combination of the Levi-Civita and the determinant of the metric in such a way that their product is tensorial. And that turns out to be real useful because I can use this to define, in a curved spacetime, covariant volume elements, OK? With this as written, my volume elements-- if I just use this like I did when we're taught about special relativity, my volume elements won't be elements of a tensor, and a lot of the framework that we've developed goes to hell. So an extra factor of the determining of the metric will allow us to correct this. And this seems kind of abstract. So let me just, as a really brief aside, before we conclude today's class-- suppose I'm just in Euclidean three-space and I'm working in spherical coordinates. So here's my line element. My metric is the diagonal of 1r squared r squared sine squared theta. The determinant of the metric, which I will write g-- it's r to the fourth sine squared theta. What we're going to learn when we do this is that the metric is a tensor density of weight 2. And so to correct it to get something of weight 1, we take a square root. If you're working in circle coordinates, does that look familiar? This is, in fact, exactly what allows us to convert differentials of our coordinates. Remember, we're working in a coordinate basis. And so we think of our little element of just the coordinates. It's just dr, d theta, d phi. This ends up being the quantity that allows us to convert the little triple of our coordinates into something that has the proper dimensions and form to actually be a real volume element. And so dr, de theta, d phi-- that ain't enough volume. But r squared sine theta, dr, d theta, d phi-- that's a volume element, OK? So basically, that's all that we're doing right now, is we're making that precise and careful. And that's where I will pick things up. We'll finish that up on Thursday.
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https://ocw.mit.edu/courses/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2013/8.333-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. PROFESSOR: OK. Let's start. So last lecture, what we talked about was limitations of classical statistical mechanics, and what I will contrast with what I will talk about today, which is new version. The old version of quantum mechanics, which was based on the observation originally from Planck, and then expanded by Einstein, that for a harmonic oscillator, a frequency omega, the energies cannot take all values. But values that are multiples of the frequency of the oscillator and then some integer n. What we did with this observation to extract thermal properties was to simply say that we will construct a partition function for the harmonic oscillator by summing over all of the states of e to the minus beta e n according to the formula given above. Just thinking that these are allowed states of this oscillator-- and this you can very easily do. It starts with the first term and then it's a algebraic series, which will give you that formula. Now, if you are sitting at some temperature t, you say that the average energy that you have in your system, well, the formula that you have is minus the log z by the beta, which if I apply to the z that I have about, essentially weighs each one of these by these Boltzmann weights by the corresponding energy and sums them. What we do is we get, essentially, the contribution of the ground state. Actually, for all intents and purposes, we can ignore this, and, hence, this. But for completion, let's have them around. And then from what is in the denominator, if you take derivative with respect to beta, you will get a factor of h bar omega. And then this factor of 1 minus e to the minus beta h bar. Actually, we then additional e to the minus beta h bar omega in here. Now, the thing that we really compare to was what happens if we were to take one more derivative to see how much the heat capacity is that we have in the harmonic oscillator. So basically taking an average of the raw formula with respect to temperature, realizing that these betas are inverse temperatures. So derivatives with respect to t will be related to derivative with respect to beta, except that I've will get an additional factor of 1 over k bt squared. So the whole thing I could write that as kb and then I had the h bar omega over kt squared. And then from these factors, I had something like e to the minus e to the h bar omega over kt. E to the h bar omega over kt minus 1 squared. So if we were to plug this function, the heat capacity in its natural unit that are this kb, then as a function of temperature, we get behavior that we can actually express correctly in terms of the combination. You can see always we get temperature in units of kb over h bar omega. So I can really plug this in the form of, say, kt over h bar omega, which we call t to some characteristic temperature. And the behavior that we have is that close to 0 temperatures, you go to 0 exponentially, because of essentially the ratio of these exponentials. We leave one exponential in the denominator. So the gaps that you have between n equals to 0 and n equals to 1 translates to behavior of that at low temperatures is exponentially decaying to leading order. Then, eventually, at high temperatures, you get the classical result where you saturate to 1. And so you will have a curve that has a shift from one behavior to another behavior. And the place where this transition occurs is when this combination is of the order of 1. I'm not saying it's precisely 1, but it's of the order of 1. So, basically, you have this kind of behavior. OK? So we use this curve to explain the heat capacity of diatomic gas, such as the gas in this room, and why at room temperature, we see a heat capacity in which it appears that the vibrational degrees of freedom are frozen, are not contributing anything. While at temperatures above the characteristic vibrational frequency, which for a gas is of the order of 10 to the 3 degrees k, you really get energy in the harmonic oscillator also in the vibrations. And the heat capacity jumps, because you have another way of storing energy. So the next thing that we asked was whether this describes also heat capacity of a solid. So basically, for the diatomic gas, you have two atoms that are bonded together into a molecule. And you consider the vibrations of that. You can regard the solid as a huge molecule with lots of atoms joined together. And they have vibrations. And if you think about all of those vibrations giving you something that is similar to this, you would conclude that the heat capacity of a solid should also have this kind of behavior. Whereas we noted that in actuality, the heat capacity of a solid vanishes much more slowly at low temperatures. And the dependence at low temperatures is proportional to t cubed. So at the end of last lecture, we gave an explanation for this, which I will repeat. Again, the picture is that the solid, like a huge molecule, has vibrational modes. But these vibrational modes cover a whole range of different frequencies. And so if you ask, what are the frequencies omega alpha of vibrations of a solid, the most natural way to characterize them is, in fact, in terms of a wave vector k that indicates a direction for the oscillatory wave that you set up in the material. And depending on k, you'll have different frequencies. And I said that, essentially, the longest wave length corresponding to k equals to 0 is taking the whole solid and translating it. Again, thinking back about the oxygen molecule, the oxygen molecule, you have two coordinates. It's the relative coordinate that has the vibration. And you have a center of mass coordinate that has no energy. If you make a molecule more and more complicated, you will have more modes, but you will always have the 0 mode that corresponds to the translation. And that carries over all the way to the solid. So there is a mode that corresponds to translations-- and, in fact, rotations-- that would carry no energy, and corresponds, therefore, to 0 frequency. And then if you start to make long wavelength oscillations, the frequency is going to be small. And, indeed, what we know is that we tap on the solid and you create sound waves, which means that the low-frequency long wavelength modes have a dispersion relation in which omega is proportional to k. We can write that as omega is v times k, where v is the velocity of the sound in the solid. Now, of course, the shortest wave length that you can have is related to the separation between the atoms in the solid. And so, basically, there's a limit to the range of k's that you can put in your system. And this linear behavior is going to get modified once you get towards the age of the solid. And the reason I have alpha here is because you can have different polarizations. There are three different possible polarizations. So in principle, you will have three of these curves in the system hard. And these curves could be very complicated when you get to the edge of [INAUDIBLE] zone and you have to solve a big dynamical matrix in order to extract what the frequencies are, if you want to have the complete spectrum. So the solid is a collection of these harmonic oscillators that are, in principle, very complicated. But we have the following. So I say, OK, I have all of these. And I want to calculate at a given temperature how much energy I have put in the solid. So this energy that I have put in the vibrations at some temperature t, assuming that these vibrations are really a collection of these oscillators. Well, what I have to do is to add up all of these terms. There's going to be adding up all of the h bar omega over 2s for all of these. OK? That will give me something that I will simply call e 0, because it doesn't depend on temperature. Presumably will exist at 0 temperature. And I can even fold into that whatever the value of the potential energy of the interactions between the particles is at 0 temperature. What I'm interested in really is the temperature dependence. So I basically take the formula that I have over there, and sum over all of these oscillators. These oscillators are characterized by polarization and by the wave vector k. And then I have, essentially, h bar omega alpha of k divided by e to the beta h bar omega alpha of k minus 1. So I have to apply that formula to this potentially very complicated set of frequencies. The thing is, that according to the picture that I have over here, to a 0 order approximation, you would say that the heat capacity is 1 if you are on this side, 0 if you're on that side. What distinguishes those two sides is whether the frequency in combination with temperature is less than or larger than 1. Basically, low frequencies would end up being here. High frequencies would end up being here. And would not contribute. So for a given temperature, there is some borderline. That borderline would correspond to kt over h bar. So let me draw where that borderline is. Kt over h bar. For a particular temperature, all of these modes are not really contributing. All of these modes are contributing. If my temperature is high enough, everything is contributing. And the total number of oscillators is 3 n. It's the number of atoms. So essentially, I will get 3 n times whatever formula I have over there. As a come further and further down, there's some kind of complicated behavior as I go through this spaghetti of modes. But when I get to low enough structures, then, again, things become simple, because I will only be sensitive to the modes that are described by this a omega being vk. OK? So if I'm interested in t going to 0, means less than some characteristic temperature that we have to define shortly. So let's say, replace this with t less than some theta d that I have to get for you shortly, then I will replace this with e 0 plus sum over alpha and k of h bar v alpha k e to the beta h bar e alpha k minus 1. OK? Now, for simplicity, essentially I have to do three different sums. All of them are the same up to having to use different values of v. Let's just for simplicity assume that all of the v alphas are the same v, so that I really have only one velocity. There's really no difficulty in generalizing this. So let's do this for simplicity of algebra. So if I do that, then the sum over alpha will simply give me a factor of 3. There are three possible polarizations, so I put a 3 there. And then I have to do the summation over k. Well, what does the summation over k mean? When I have a small molecule for the, let's say, three or four atoms, then I can enumerate what the different vibrational states are. As I go to a large solid, I essentially have modes that are at each value of k, but, in reality, they are discrete. They are very, very, very, very finely separated by a separation that is of the order of 2 pi over the size of the system. So to ensure that eventually when you count all of the modes that you have here, you, again, end up to have of the order of n states. So if that's the case, this sum, really I can replace with an integral, because going from one point to the next point does not make much difference. So I will have an integral over k. But I have to know how densely these things are. And in one direction it is 2 pi over l. So the density would be l over 2 pi. If I look at all three directions, I have to multiply all of them. So I will get v divided by 2 pi cubed. So this is the usual density of states. And you go to description in terms of wave numbers, or, later on, in terms of momentums. And what we have here is this integral h bar v k e to the beta h bar v k minus 1. OK? So let's simplify this a little bit further. I have e 0. I have 3v. The integrand only depends on the magnitude of k, so I can take advantage of that spherical symmetry and write this as 4 pi k squared v k divided by this 8 pi cubed. What I can do is I can also introduce a factor of beta here, multiplied by k t. Beta k t is 1. And if I call this combination to be x, then what I have is k b t x e to the x minus 1. Of course, k is simply related to x by k being x kt over h bar v. And so at the next level of approximation, this k squared v k I will write in terms of x squared v x. And so what do I have? I have e 0. I have 3v divided by 2 pi squared. Because of this factor of kt that I will take outside I have a kt. I have a k squared vk that I want to replace with x squared v x. And that will give me an additional factor of kv over h bar v cubed. And then I have an integral 0 to e 0 v x x cubed e to the x minus 1. Now, in principle, when I start with this integration, I have a finite range for k, which presumably would translate into a finite range for x. But in reality none of these modes is contributing, so I could extend the range of integration all the way to infinity, and make very small error at low temperatures. And the advantage of that is that then this becomes a definite integral. Something that you can look up in tables. And its value is in fact pi to the fourth over 15. So substituting that over there, what do we have? We have that the energy is e 0 plus 3 divided by 15, will give me 5, which turns the 2 into a 10. I have pi to the fourth divided by pi squared, so there's a pi squared that will survive out here. I have a kt. I have kt over h bar v cubed. And then I have a factor of volume. But volume is proportional to the number of particles that I have in the system times the size of my unit cell. Let's call that a cubed. So this I can write this as l a cubed. Why do I do that is because when I then take the derivative, I'd like to write the heat capacity per particle. So, indeed, if I now take the derivative, which is de by dt, the answer will be proportional to n and kv. The number of particles and this k v, which is the function, the unit of heat capacities. The overall dependence is t to the fourth. So when I take derivatives, I will get 4t cubed. That 4 will change the 1 over 10 to 2 pi squared over 5. And then I have the combination kvt h bar v, and that factor of a raised to the third power. So the whole thing is proportional to t cubed. And the coefficient I will call theta d for [INAUDIBLE]. And theta d I have calculated to be h bar v over a h bar v a over k t. No, h bar v over a k t. So the heat capacity of the solid is going to be proportional, of course, to n k b. But most importantly, is proportional to t cubed. And t cubed just came from this argument that I need low omegas. And how many things I have at the omega. How many frequencies do I have that are vibrating? The number of those frequencies is essentially the size of a cube in k space. So it goes like this-- maximum k cubed in three dimensions. In two dimensions, it will be squared and all of that. So it's very easy to figure out from this dispersion relation what the low temperature behavior of the heat capacity has to be. And you will see that this is, in fact, predictive, in that later on in the course, we will come an example of where the heat capacity of a liquid, which was helium, was observed to have this t cubed behavior based on that Landau immediately postulated that there should be a phonon-like dispersion inside that superfluid. OK. So that's the story of the heat capacity of the solid. So we started with a molecule. We went from a molecule into an entire solid. The next step that what I'm going to do is I'm going to remove the solid and just keep the box. So essentially, they calculation that I did, if you think about it, corresponded to having some kind of a box, and having vibrational modes inside the box. But let's imagine that it is an empty box. But we know that even in empty space we have light. So within an empty box, we can still have modes of the electromagnetic field. Modes of electromagnetic field, just like the modes of the solid, we can characterize by the direction along which oscillations travel. And whereas for the atoms in the solid, they have displacement and the corresponding momentum for the electromagnetic field, you have the electric field. And its conjugate is the magnetic field. And these things will be oscillating to create for you a wave. Except that, whereas for the solid, for each atom we had three possible directions, and therefore we had three branches, for this, since e and b have to be orthogonal to k, you really have only two polarizations. But apart from that, the frequency spectrum is exactly the same as we would have for the solids at low temperature replacing to v that we have with the speed of light. And so you would say, OK. If I were to calculate the energy content that is inside the box, what I have to do is to sum over all of the modes and polarizations. Regarding each one of these as a harmonic oscillator, going through the system of quantizing according to this old quantum mechanics, the harmonic oscillators, I have to add up the energy content of each oscillator. And so what I have is this h bar omega of k. And then I have 1/2 plus 1 over e to the beta h bar omega of k minus 1. And then I can do exactly the kinds of things that I had before, replacing the sum over k with a v times an integral. So the whole thing would be, first of all, proportional to v, going from the sum over k to the integration over k. I would have to add all of these h bar omega over 2s. Has no temperature dependence, so let me just, again, call it some e 0. Actually, let's call it epsilon 0, because it's more like an energy density. And then I have the sum over all of the other modes. There's two polarisations. So as opposed to the three that I had before, I have two. I have, again, the integral over k of 4 pi k squared v k divided by 8 pi cubed, which is part of this density of state calculation. I have, again, a factor of h bar omega. Now, I realize that my omega is ck. So I simply write it as h bar ck. And then I have e to the beta h bar ck minus 1. So we will again allow this to go from 0 to infinity. And what do we get? We will get v epsilon 0 plus, well, the 8 and 8 cancel. I have pi over pi squared. So pi over pi cubed. So it will give me 1 over pi squared. I have one factor of kt. Again, when I introduce here a beta and then multiply by kt, so that this dimension, this combination appears. Then I have, if I were to change variable and call this the new variable, I have factor of k squared dk, which gives me, just as before over there, a factor of kt over h bar c cubed. And then I have this integral left, which is the 0 to infinity v x x cubed e to the x minus 1, which we stated is pi to the fourth over 15. So the part that is dependent on temperature, the energy content, just as in this case, scales as t to the fourth. There is one part that we have over here from all of the 0s, which is, in fact, an infinity. And maybe there is some degree of worry about that. We didn't have to worry about that infinity in this case, because the number of modes that we had was, in reality, finite. So once we were to add up properly all of these 0 point energies for this, we would have gotten a finite number. It would have been large, but it would have been finite. Whereas here, the difference is that there is really no upper cut-off. So this k here, for a solid, you have a minimum wavelength. You can't do things shorter than the separation of particles. But for light, you can have arbitrarily short wavelength, and that gives you this infinity over here. So typically, we ignore that. Maybe it is related to the cosmological constant, et cetera. But for our purposes, we are not going to focus on that at all. And the interesting part is this part, that proportional to t to the fourth. There are two SOP calculations to this that I will just give part of the answer, because another part of the answer is something that you do in problem sets. One of them is that what we have here is an energy density. It's proportional to volume. And we have seen that energy densities are related to pressures. So indeed, there is a corresponding pressure. That is, if you're at the temperature t, this collection of vibrating electromagnetic fields exerts a pressure on the walls of the container. This pressure is related to energy density. The factor of 1/3 comes because of the dispersion relation. And you can show that in one of the problem sets. You know that already. So that would say that you would have, essentially, something like some kind of p 0. And then something that is proportional to t to the fourth. So I guess the correspondent coefficient here would be p squared divided by 45 kt kt over h bar c cubed. So there is radiation pressure that is proportional to temperature. The hotter you make this box, the more pressure it will get exerted from it. There is, of course, again this infinity that you may worry about. But here the problem is less serious, because you would say that in reality, if I have the wall of the box, it is going to get pressure from both sides. And if there's an infinite pressure from both sides, they will cancel. So you don't have to worry about that. But it turns out that, actually, you can measure the consequences of this pressure. And that occurs when rather than having one plate, you have two plates that there are some small separation apart. Then the modes of radiation that you can fit in here because of the quantizations that you have, are different from the modes that you can have out here. So that difference, even from the 0 point fluctuations-- the h bar omega over 2s-- will give you a pressure that pushes these plates together. That's called a Casimir force, or Casimir pressure. And that was predicted by Casimir in 1950s, and was measured experimentally roughly 10 years ago to high precision, matching the formula that we had. So sometimes, these infinities have consequences that you have to worry about. But that's also to indicate that there's kind of modern physics to this. But really it was the origin of quantum mechanics, because of the other aspect of the physics, which is imagine that again you have this box. I draw it now as an irregular box. And I open a hole of size a inside the box. And then the radiation that was inside at temperatures t will start to go out. So you have a hot box. You open a hole in it. And then the radiation starts to come out. And so what you will have is a flux of radiation. Flux means that this it energy that is escaping per unit area and per unit time. So there's a flux, which is per area per time. It turns out that that flux-- and this is another factor, this factor of 1/3 that I mentioned-- is related the energy density with a factor of 1 c over 4. Essentially, clearly the velocity with which energy escaping is proportional to c. So you will get more radiation flux the larger c. The answer has to be proportional to c. And it is what is inside that is escaping, so it has to be proportional to the energy density that you have inside, some kind of energy per unit volume. And the factor of 1/4 is one of these geometric factors. Essentially, there's two factors of cosine of theta. And you have to do an average of cosine squared theta. And that will give you the additional 1/4. OK? But rather than looking-- so this would tell you that there is an energy that is streaming out. That is, the net value is proportional to t to the fourth. But more interestingly, we can ask what is the flux per wavelength? And so for that, I can just go back to the formula before I integrated over k, and ask what is the energy density in each interval of k? And so what I have to do is to just go and look at the formula that I have prior to doing the integration over k. Multiply it by c over 4. What do I have? I have 8 pi divided by 8 pi cubed. I have a factor of k squared from the density of states. I have this factor of h bar c k divided by e to the beta h bar c k minus 1. So there's no analogue of this, because I am not doing the integration over k. So we can simplify some of these factors up front. But really, the story is how does this quantity look as the function of wave number, which is the inverse of wave length, if you like. And what we see is that when k goes to 0, essentially, this factor into the beta h bar ck I have to expand to lowest order. I will get beta h bar c k, because the 1 disappears. H bar ck is cancelled, so the answer is going to be proportional to inverse beta. It's going to be proportional to kt and k squared. So, essentially, the low k behavior part of this is proportional to k squared c, of course, and kt. However, when I go to the high k numbers, the exponential will kill things off. So the large k part of this is going to be exponentially small. And, actually, the curve will look something like this, therefore. It will have a maximum around the k, which presumably is of the order of kt over h bar c. So basically, the hotter you have, this will move to the right. The wavelengths will become shorter. And, essentially, that's the origin of the fact that can you heat some kind of material, it will start to emit radiation. And the radiation will be peaked at some frequency that is related to its temperature. Now, if we didn't have this quantization effect, if h bar went to 0, then what would happen is that this k squared kt would continue forever. OK? Essentially, you would have in each one of these modes of radiation, classically, you would put a kt of energy. And since you could have arbitrarily short wavelengths, you would have infinite energy at shorter and shorter wavelengths. And you would have this ultraviolet catastrophe. Of course, the shape of this curve was experimentally known towards the end of the 19th century. And so that was the basis of thinking about it, and fitting an exponential to the end, and eventually deducing that this quantization of the oscillators would potentially give you the reason for this to happen. Now, the way that I have described it, I focused on having a cavity and opening the cavity, and having the energy go out. Of course, the experiments for black body are not done on cavities. They're done on some piece of metal or some other thing that you heat up. And then you can look at the spectrum of the radiation. And so, again, there is some universality in this, that it is not so sensitive to the properties of the material, although there are some emissivity and other factors that multiply the final result. So the final result, in fact, would say that if I were to integrate over frequencies, the total radiation flux, which would be c over 4 times the energy density total, is going to be proportional to temperature to the fourth power. And this constant in front is the Stefan-Boltzmann, which has some particular value that you can look up, units of watts per area per degrees Kelvin. So this perspective is rather macroscopic. The radiated energy is proportional to the surface area. If you make things that are small, and the wavelengths that you're looking at over here become compatible to the size of the object, these formulas break down. And again, go forward about 150 years or so, there is ongoing research-- I guess more 200 years-- ongoing research on-- no, 100 and something-- ongoing research on how these classical laws of radiation are modified when you're dealing with objects that are small compared to the wavelengths that are emitted, etc. Any questions? So the next part of the story is why did you do all of this? It works, but what is the justification? In that I said there was the old quantum mechanics. But really, we want to have statements about quantum systems that are not harmonic oscillators. And we want to be able to understand actually what the underlying basis is in the same way that they understand how we were doing things for classical statistical mechanics. And so really, we want to look at how to make the transition from classical to quantum statistical mechanics. So for that, let's go and remind us. Actually, so basically the question is something like this-- what does this partition function mean? I'm calculating things as if I have these states that are the energy levels. And the probabilities are e to the minus beta epsilon n. What does that mean? Classically, we knew the Boltzmann rates had something to do with the probability of finding a particle with a particular position and momentum. So what is the analogous thing here? And you know that in new quantum mechanics, the interpretation of many things is probabilistic. And in statistical mechanics, even classically we had a probabilistic interpretation. So presumably, we want to build a probabilistic theory on top of another probabilistic theory. So how do we go about understanding precisely what is happening over here? So let's kind of remind ourselves of what we were doing in the original classical statistical mechanics, and try to see how we can make the corresponding calculations when things are quantum mechanical. So, essentially, the probabilistic sense that we had in classical statistical mechanics was to assign probabilities for micro states, given that we had some knowledge of the macro state. So the classical microstate mu was a point which was a collection of p's and q's in phase space. So what is a quantum microstate? OK. So here, I'm just going to jump several decades ahead, and just write the answer. And I'm going to do it in somewhat of a more axiomatic way, because it's not up to me to introduce quantum mechanics. I assume that you know it already. Just a perspective that I'm going to take. So the quantum microstate is a complex unit vector in Hilbert space. OK? So for any vector space, we can choose a set of unit vectors that form an orthonormal basis. And I'm going to use this bra-ket notation. And so our psi, which is a vector in this space, can be written in terms of its components by pointing to the different directions in this space, and components that I will indicate by this psi n. And these are complex. And I will use the notation that psi n is the complex conjugate of n psi. And the norm of this I'm going to indicate by psi psi, which is obtained by summing over all n psi psi n star, which is essentially the magnitude of n psi squared. And these are unit vectors. So all of these states are normalized such that psi psi is equal to 1. Yes. AUDIENCE: You're not allowing particle numbers to vary, are you? PROFESSOR: At this stage, no. Later on, when we do the grand canonical, we will change our Hilbert space. OK? So that's one concept. The other concept, classically, we measure things. So we have classical observable. And these are functions all of which depend on this p and q in phase space. So basically, there's the phase space. We can have some particular function, such as the kinetic energy-- sum over i pi squared over 2 n-- that's an example of an observable. Kinetic energy, potential energy, anything that we like, you can classically write a sum function that you want to evaluate in phase space, given that you are at some particular point in phase space, the state of your system, you can evaluate what that is. Now in quantum mechanics, observables are operators, or matrices, if you like, in this vector space. OK? So among the various observables, certainly, are things like the position and the momentum of the particle. So there are presumably matrices that correspond to position and momentum. And for that, we look at some other properties that this classical systems have. We had defined classically a Poisson bracket, which was a sum over all alphas d a by d q alpha d b by d p alpha minus the a by the p alpha b d by the q f. OK? And this is an operation that you would like to, and happens to, carry over in some sense into quantum mechanics. But one of the consequence of this is you can check if I pick a particular momentum, key, and a particular coordinate, q, and put it over here, most of the time I will get 0, unless the alphas match exactly the p q's that I have up there. And if you go through this whole thing, I will get something like that is like a delta i j. OK. So this structure somehow continues in quantum mechanics, in this sense that the matrices that correspond to p and q satisfy the condition that p i and q j, thinking of two matrices, and this is the commutator, so this is p i q j minus q j p i is h bar over i delta h. So once you have the matrices that correspond to p and q, you can take any function of p and q that you had over here, and then replace the p's and q's that appear in, let's say, a series expansion, or an expansion of this o in powers of p and q, with corresponding matrices p hat and q hat. And that way, you will construct a corresponding operator. There is one subtlety that you've probably encountered, in that there is some symmetrization that you have to do before you can make this replacement. OK. So what does it mean? In classical theory, if something is observable the answer that you get is a number. Right? You can calculate what the kinetic energy is. In quantum mechanics, what does it mean that observable is a matrix? The statement is that observables don't have definite values, but the expectation value of a particular observable o in some state pi is given by psi o psi. Essentially, you take the vector that correspond to the state, multiply the matrix on it, and then sandwich it with the conjugate of the vector, and that will give you your state. So in terms of elements of some particular basis, you would write this as a sum over n and m. Psi n n o m m psi. And in that particular basis, your operator would have these matrix elements. Now again, another property, if you're measuring something that is observable, is presumably you will get a number that is real. That is, you expect this to be the same thing as its complex conjugate. And if you follow this condition, you will see that that reality implies that n o m should m o n complex conjugate, which is typically written as the matrix being its Hermitian conjugate, or being Hermitian. So all observables in quantum mechanics would correspond to Hermitian operators or matrices. OK. There's one other piece, and then we can forget about axioms. We have a classical time evolution. We know that the particular point in the classical phase space changes as a function of time, such that q i dot is plus d h by d p i. P i dot is minus d h by d q i. By the way, both of these can be written as q i and h Poisson bracket and p i and h Poisson bracket. But there is a particular function observable, h, the Hamiltonian, that derives the classical evolution. And when we go to quantum evolution, this vector that we have in Hilbert space evolves according to i h bar d by dt of the vector psi is the matrix that we have acting on psi. OK. Fine. So these are the basics that we need. Now we can go and do statistical descriptions. So the main element that we had in constructing statistical descriptions was deal with a macrostate. We said that if I'm interested in thinking about the properties of one cubic meter of gas at standard temperature and pressure, I'm not thinking about a particular point in phase space, because different gases that have exactly the same macroscopic properties would correspond to many, many different possible points in this phase space that are changing as a function of time. So rather than thinking about a single microstate, we talked about an ensemble. And this ensemble had a whole bunch of possible microstates. In the simplest prescription, maybe we said they were all equally likely. But, actually, we could even assign some kind of probability to them. And we want to know what to do with this, because then what happened was from this description, we then constructed a density which was, again, some kind of a probability in phase space. And we looked at its time evolution. We looked at the averages and all kinds of things in terms of this density. So the question is, what happens to all of this when we go to quantum descriptions? OK. You can follow a lot of that. We can, again, take the example of the one cubic meter of gas at standard temperature and pressure. But the rather than describing the state of the system classically, I can try to describe it quantum mechanically. Presumably the quantum mechanical description at some limit becomes equivalent to the classical description. So I will have an ensemble of states. I don't know which one of them I am in. I have lots of boxes. They would correspond to different microstates, presumably. And this has, actually, a word that is used more in the quantum context. I guess one could use it in the classical context. It's called a mixed state. A pure state is one you know exactly. Mixed state is, well, like the gas I tell you. I tell you only the macroscopic information, you don't know much about microscopically what it is. If these are possibilities, and not knowing those possibilities, you can say that it's a mixture of all these states. OK. Now, what would I use, classically, a density for? What I could do is I could calculate the average of some observable, classically, in this ensemble. And what I would do is I would integrate over the entirety of the six n-dimensional phase space the o at some particular point in phase space and the density at that point in phase space. And this average I will indicate by a bar. So my bars stand for ensemble average, to make them distinct from these quantum averages that I will indicate with the Bra-Kets. OK? So let's try to do the analogue of that in quantum states. I would say that, OK, for a particular one of the members of this ensemble, I can calculate what the expectation value is. This is the expectation value that corresponds to this observable o, if I was in a pure state psi alpha. But I don't know that I am there. I have a probability, so I do a summation over all of these states. And I will call that expectation value ensemble average. So that's how things are defined. Let's look at this in some particular basis. I would write this as a sum over alpha m and n p alpha psi alpha m m o n n psi alpha. So, essentially, writing all of these psis in terms of their components, just as I had done above. OK. Now what I want to do is to reorder this. Do the summation over n and m first, the summation over alpha last. So what do I have? I have m o n. And then I have a sum over alpha of p alpha n psi alpha psi alpha n. So what I will do, this quantity that alpha is summed over-- so it depends on the two indices n and m. I can give it a name. I can call it n rho m. If I do that, then this o bar average becomes simply-- let's see. This summation over n gives me the matrix product o rho. And then summation over m gives me the trace of the product. So this is the trace of rho o. OK? So I constructed something that is kind of analogous to the classical use of the density in phase space. So you would multiply the density and the thing that you wanted to calculate the observable. And the ensemble average is obtained as a kind of summing over all possible what values of that product in phase space. So here, I'm doing something similar. I'm multiplying this o by some matrix row. So, again, this I can think of as having introduced a new matrix for an operator. And this is the density matrix. And if I basically ignore, or write it in basis-independent form, it is obtained by summing over all alphas the alphas, and, essentially, cutting off the n and m. I have the matrix that I would form out of state alpha by, essentially, taking the vector and its conjugate and multiplying rows and columns together to make a matrix. And then multiplying or averaging that matrix over all possible values of the ensemble-- elements of the ensemble-- would give me this density. So in the same way that any observable in classical mechanics goes over to an operator in quantum mechanics, we find that we have another function in phase space-- this density. This density goes over to the matrix or an operator that is given by this formula here. It is useful to enumerate some properties of the density matrix. First of all, the density matrix is positive definite. What does that mean? It means that if you take the density matrix, multiply it by any state on the right and the left to construct a number, this number will be positive, because if I apply it to the formula that I have over there, this is simply sum over alpha p alpha. Then I have phi psi psi alpha, and them psi alpha psi, which is its complex conjugate. So I get the norm of that product, which is positive. All of the p alphas are positive probabilities. So this is certainly something that is positive. We said that anything that makes sense in quantum mechanics should be Hermitian. And it is easy to check. That if I take this operator rho and do the complex conjugate, essentially what happens is that I have to take sum over alpha. Complex conjugate of p alpha is p alpha itself. Probabilities are real numbers. If I take psi alpha psi alpha and conjugate it, essentially I take this and put it here. And I take that and put it over there. And I get the same thing. So it's the same thing as rho. And, finally, there's a normalization. If, for my o over here in the last formula, I choose 1, then I get the expectation value of 1 has to be the trace of rho. And we can check that the trace of rho, essentially, is obtained by summing over all alpha p alpha, and the dot product of the two psi alphas. Since any state in quantum mechanics corresponds to a unit vector, this is 1. So I get a sum over alpha of p alphas. And these are probabilities assigned to the members of the ensemble. They have to add up to 1. And so this is like this. So the quantity that we were looking at, and built, essentially, all of our later classical statistical mechanics, on is this density. Density was a probability in phase space. Now, when you go to quantum mechanics, we don't have phase space. We have Hilbert space. We already have a probabilistic theory. Turns out that this function, which was the probability in phase space classically, gets promoted to this matrix, the density matrix, that has, once you take traces and do all kinds of things, the kinds of properties that you would expect the probability to have classically. But it's not really probability in the usual sense. It's a matrix. OK. There is one other thing element of this to go through, which is that classically, we said that, OK, I pick a set of states. They correspond to some density. But the microstates are changing as a function of time. So the density was changing as a function of time. And we had Liouville's theorem, which stated that d rho by d t was the Poisson bracket of the Hamiltonian with rho. So we can quantum mechanically ask, what happens to our density matrix? So we have a matrix rho. I can ask, what is the time derivative of that matrix? And, actually, I will insert the i h bar here, because I anticipate that, essentially, rho having that form, what I will have is sum over alpha. And then I have i h bar d by dt acting on these p alpha psi alpha psi f. So there rho is sum over alpha p alpha psi alpha psi alpha. Sum over alpha p alpha I can take outside. I h bar d by dt acts on these two psis that are appearing a complex conjugates. So it can either, d by dt, act on one or the other. So I can write this as sum over alpha p alpha i h bar d by dt acting on psi alpha psi alpha, or i, or psi alpha, and then i h bar d by dt acting on this psi alpha. Now, i h bar d by dt psi alpha, we said that, essentially, the quantum rule for time evolution is i h bar d by dt of the state will give you is governed by h times acting on psi alpha. If I were to take the complex conjugate of this expression, what I would get is minus i h bar d by dt acting on psi that is pointing the other way of our complex conjugation is h acting on the psi in the opposite way. So this thing is minus psi alpha with h f acting on it. OK. So then I can write the whole thing as h-- for the first term take the h out front. I have a sum over alpha p alpha psi alpha psi alpha, minus, from this complex conjugation-- here, h is completely to the right-- I have a sum over alpha p alpha psi alpha psi alpha. And then we have h. Now, these are again getting rho back. So what I have established is that i h bar, the time derivative of this density matrix, is simply the commutator of the operators h and o. So what we had up here was the classical Liouville theorem, relating the time derivative of the density in phase space to the Poisson bracket with h. What we have here is the quantum version, where the time derivative of this density matrix is the commutator of rho with h. Now we are done, because what did we use this Liouville for? We used it to deduce that if I have things that are not changing as a function of time, I have equilibrium systems, where the density is invariant. It's the same. Then rho of equilibrium not changing as a function of time can be achieved by simply making it a function of h. And, more precisely, h and conserved quantities that have 0 Poisson bracket with h. How can I make the quantum density matrix to be invariant of time? All I need to do is to ensure that the Poisson bracket of that density with the Hamiltonian is 0. Not the Poisson bracket, the commutator. Clearly, the commutator of h with itself is 0. Hh minus hh is 0. So this I can make a function of h, and any other kind of quantity also that has 0 commutator with h. So, essentially, the quantum version also applies. That is, the quantum version of rho equilibrium, I can make it by constructing something that depends on the Hilbert space through the dependence of the Hamiltonian and on the Hilbert space and any other conserved quantities that have 0 commutator also with the Hamiltonian. So now, what we will do next time is we can pick and choose whatever rho equilibrium we had before. Canonical e to the minus beta h. We make this matrix to be e to the minus beta h. Uniform anything, we can just carry over whatever functional dependence we had here to here. And we are ensure to have something that is quantum mechanically invariant. And we will then interpret what the various quantities calculated through that density matrix and the formulas that we described actually. OK?
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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'd like to now show you two mathematical facts about how we integrate quantities of motion. Suppose we have an object, let's call this the i hat direction, and it's moving, and it has an x component of velocity. And suppose it starts at some initial position and goes at some final position. And in the initial position, it was at time ti, and the final position was at t final. Now, we have two very important integrals that we want to look at which govern how we describe, how we integrate acceleration with respect to time, and also how we'll integrate acceleration with respect to space. Now, recall that our acceleration, which may be a function of time, is the derivative of the x component of velocity with respect to time. So, let's first look at a simple integration of acceleration with respect to time and see what we get. So, if we integrate acceleration, now I'm going to introduce an integration variable, dt prime. And our integration variable goes from our initial time to some final time. And if we use our fact that acceleration is the derivative of the velocity, then we can write this is dvx, dt prime, dt prime again. After a while I'll drop the dummy variables and the endpoints of the integral. And this simply becomes the integral from t prime, t initial. t prime equals t final of dvx prime. Now notice we've done a change of variables in the integration. So instead of now talking about the endpoints of the integral from t prime, t initial to t final, now what we're doing is we changed our integration variable. And so what we have is, we have the velocity integration variable is going from some initial value. And that integration variable is going to some final value. So again, our initial conditions may have some initial velocity and some final velocity-- three different ways of describing the initial and final states. This integral is a very straightforward integral, the x final minus vx initial, which is the change in the x component of the velocity. And that's our classic result that we've done that the integration of acceleration with respect to time is a change in velocity. Now, let's see what happens as a comparison when we integrate our acceleration not with respect to time, but suppose that it's a function of space. And so now we're integrating again. We have a dummy variable. We have to be careful, because this is the x component of acceleration, but x prime is our integration variable, and that's going from some initial position to some final position. Now, we can write this as again, make the substitution, dx dt, dx prime, going from the initial to the final. But now notice we're going to rewrite this integrand as dvx times dx prime, dt prime, dt. And when we do that, we have this result that dx prime, dt prime is precisely what we mean by vx. Now, I'll introduce some dummy variables there as well. So what our integrand becomes is dvx prime, vx prime. That's not a function, it's a product of a differential times vx prime. And now our new integration variable is going from some initial value to some final value. Once again, this is a straightforward integral. For those who haven't seen integrals like this, it's just something like x dx is x squared over 2, but our integration variable is vx prime. So what we get is 1/2 vx final squared minus vx initial squared. And so what we see here is two fundamentally different facts that if you integrate acceleration with respect to time, you get the change in velocity. But if you integrate acceleration with respect to space, you get 1/2 times the change not of velocity, but of the x component of the velocity squared. And both of these facts are central to how we'll analyze the concept of work and how we applied Newton's second law.
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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: So today is going to be our last pass at bound states. So starting next week or actually starting next lecture, we're going to look at scattering. Scattering's going to be great. But we need to close out bound states. So today's topic is the finite well, finite the potential well. We've sort of sketched this when we looked at qualitative structure of wave functions of energy eigenstates. But we're going to solve the system today. So good. So the system we're interested in is going to be-- The system we're interested in is going to be a system with a finite depth and a finite width. And I'll go into detail and give you parameters in a bit. But first I want to just think about how do we find energy eigenfunctions of a potential of this form, v of x, which is piecewise constant. So first off, is this is a terribly realistic potential? Will you ever in the real world find a system that has a potential which is piecewise constant? Probably not. It's discontinuous. Right? So it's rather unphysical. But it's a very useful toy model. So for example, if you take a couple of capacitor plates, then you can induce a situation where the electric field is nonzero in between the capacitor plates and zero outside of the capacitor plates. Right. So at a superficial level, this looks discontinuous. It looks like the electric field is-- But actually, you know that microscopically there are a bunch of charges, and everything is nice and continuous except for the behavior right at the charges. So but it's reasonable to model this as a step function for an electric field. So this is going to be an idealization, but it's going to be a very useful idealization, the constant potential. OK. So what's the equation? What are we trying to do? We want to find the energy eigenstates for this because we want to study the time evolution. And the easiest way to solve the Schrodinger equation, the time evolution equation, is to expand an energy eigenstates. So the equation we want to solve is energy eigenvalue times phi e of x is equal to minus h bar squared on 2m phi e phi prime plus v of x. And I'm going to put it in the form we've been using. Phi prime prime e of x is equal to 2m upon h bar squared v of x minus e. OK. So this is the form that I'm going to use today to solve for the energy eigenvalue equations. e is some constant. Do you expect the allowed energies to be arbitrary? No. They should be discrete. Yeah, exactly. So we expect that there should be discrete lowest energy state, some number of bound states. And then, eventually, if the energy is greater than the potential everywhere, the energies will be continuous. Any energy will be allowed above the potential. So we'll have a continuum of states above the potential. And we'll have a discrete set of bound states-- Probably, it's reasonable to expect some finite number of bound states just by intuition. --from the infinite well. So we expect to have a finite number of discrete energies and then a continuous set of energies above zero. So if this is the asymptotic value potential of zero. OK. And this is intuition gained from our study of qualitative structure of energy eigenfunctions. So we are going to talk today about the bound states. And in recitation, leaders should discuss the continuum above zero energy. OK. So to solve for the actual energy eigenfunctions and the energy eigenvalues, what we need to do is we need to solve this equation subject to some boundary conditions. And the boundary conditions we're going to want to solve are going to be finite. So it's normalizable infinity. The solution should be vanishing far away. And the wave function should be everywhere smooth. Well, at least it should be continuous. So let's talk about what exactly boundary conditions we want to impose. And so in particular, we're going to want to solve for the energy eigenfunctions in the regions where the potential is constant and then patch together solutions at these boundaries. We know how to solve for the energy eigenfunctions when the potential is constant. What are the energy eigenfunctions? Yeah. Suppose I have a potential, which is constant. v is equal to 0. What are the energy eigenfunctions of this potential? AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah, e to the ikx. Yeah. Now, what if I happen to tell you-- So we're at h bar squared k squared upon 2m is equal to the energy. Suppose I happen to tell you that here's the potential. And I want to find the solution in this region where the energy is here less than the potential. What are the solutions? That's a second order differential equation. There should be two solutions. What are the solutions to that differential equation when the energy is less than the potential? AUDIENCE: Decaying and growing. PROFESSOR: Decaying and growing exponentials. Exactly. e to the plus alpha x and e to the minus alpha x. And the reason is these are sinusoidal, and these have the opposite concavity. They are growing and dying exponentials. Cool? OK. So we've studied that. Yeah? AUDIENCE: Shouldn't that be a phi e of x? PROFESSOR: Sorry. Oh, oh, yes, indeed. Sorry. Thank you. Phi e of x. It's early, and I'm still working on the coffee. It won't take long. Good. So we know how to solve the energy eigenvalue equation in all these regions where the potential is constant. So our job is going to be to find a solution where we patch them together at these interfaces. We patch them together. And what condition should we impose? So the basic condition is going to be continuity of the wave functions phi e of x. And so what are the conditions that we need? Well, if v of x-- Here's the way I like to think about this. Suppose v of x is continuous. So if the potential is continuous, then what can you say? You can say that phi prime prime is a continuous function times phi of x. So very roughly, if we look at a region where phi isn't varying very much, so if we have a potential that's varying in some way, then phi prime prime, in a region where it's not varying much compared to its value as a function of x, does something smooth because it's varying with the potential. And so phi prime, which is just going to be the integral of this-- The integral of a smooth function is again a smooth function. --and phi, the integral of that function is also going to be a nice, smooth function. OK? I've drawn it badly, but-- So the key thing here is that if the potential is continuous, then the energy eigenfunction has a second derivative, which is also continuous. That means its first derivative is continuous. So that means the function itself is continuous. Everyone agree with this? Questions? So in regions where the potential's continuous, the wave function and its first two derivatives all have to be continuous. On the other hand, suppose the potential has a step. v of x has a step discontinuity. OK. So the potential does one of those. So what does that tell you about phi prime prime? It's a function of x. So for example, let's look at that first step. Suppose the potential is that first step down by some amount. Then phi prime prime is going to decrease precipitously at some point. And the actual amount that it decreases depends on the value of phi because the change in the potential time is the actual value of phi. And if that's true of phi prime prime what can you say of phi prime? So this is discontinuous. Phi prime of x, however, is the integral of this discontinuous function. And what does it do? Well, it's linearly increasing in this region because its derivative is constant. And here, it's linearly increasing less. So it's not differentiably smooth, but it's continuous. And then let's look at the actual function phi of x. OK. What is it doing? Well, it's quadratic. And then here it's quadratic a little less afterwards. But that still continuous because it's the integral of a continuous function. Everyone we cool with that? So even when we have a step discontinuity in our potential, we still have that our derivative and the value of the function are continuous. Yeah? However, imagine the potential has this delta function. Let's just really push it. What happens if our potential has a delta function singularity? Really badly discontinuous. Then what can you say about phi prime prime as a function of x? So it has a delta function, right? So the phi prime prime has to look like something relatively slowly varying, and then a step delta function. So what does that tell you about the derivative of the wave function? It's got a step function. Exactly. Because it's the integral of this, and the integral is 0, 1. Whoops. I missed. So this is a delta. This is a step. And then the wave function itself is, well, it's the integral of a step, so it's continuous. Sorry. It's certainly not differentiable. Its derivative is discontinuous. Well, it's differentiable but its derivative's not supposed to be. It is not continuous, indeed. So this is continuous. So we've learned something nice, that unless our potential is so stupid as to have delta functions, which sounds fairly unphysical-- We'll come back to that later in today's lecture. --unless our potential has delta functions, the wave function and its first derivative must be smooth. Yeah. This is just from the energy eigenvalue equation. Now, we actually argued this from the well definedness and the finiteness of the expectation value of the momentum earlier in the semester. But I wanted to give you this argument for it because it's going to play a useful role. And it also tells us that if we do have a delta function singularity in the potential, then the upshot is that the wave function is going to be continuous, but its first derivative will not. Its first derivative will jump at the wave function. And that means that the wave function-- Let me draw this slightly differently. --the wave function will have a kink. Its derivative will not be continuous. OK. So anywhere where we have a delta function in the potential, we will have a kink in the wave function where the first derivative is discontinuous. Cool? Yeah. AUDIENCE: What does that mean as far as the expectation value of the momentum? PROFESSOR: Ah, that's an excellent question. So what do you expect to happen at such a point? AUDIENCE: Well, your momentum blows up. PROFESSOR: Yeah, exactly. So we're going to have some pathologies with expectation values in the momentum. Let's come back to that when we talk about the delta function potential, which should be at the end of today. Hold that question in the back of your head. It's a good question. Other questions? OK. So what we're going to do now is we're going to use, you know, math to find the energy eigenfunctions and eigenvalues for the finite well potential, i.e. we're going to solve this equation contingent on the boundary conditions that the wave function and its derivative are smooth everywhere. And in particular, they must be smooth here and here. I mean they've got to be smooth everywhere. We know what the solution is inside here. We know what the solution is in here. All we have to worry about is what happens at the interface. And we're going to use smoothness of the wave function and its derivative to impose conditions that allow us to match across that step. Everyone cool with that? OK. So let's do it. By the way, just a quick side note. Let me give you a definition. I've used this phrase many times, but I haven't given you a definition of it. So I've used the phrase, bound state. And its opposite is called a scattering state. So here's what I mean by a bound state. Intuitively, a bound state, if you think about this classically, imagine I have a potential well. And I have a marble, and I let go from here. This marble is bound. Right? It's never getting out of the well. It's stuck. Yeah? And so I would call that a bound marble. On the other hand, a marble that I give a huge kick to, big velocity so that it can get out, well then it's not bound to this potential well. So I'll call that a scattering state just to give it a name. And next lecture we'll see why we call it a scattering state. The important thing is that bound configurations of a classical potential are things basically in a well that are stuck. OK. So the quantum version of that statement is the following. Suppose I take this potential, and I treat it quantum mechanically, and I consider a state with total energy like this. Well, among other things, the total energy is less than the value of the potential asymptotically far away. So what is the form of a wave function in this region with this energy? Exponential, right? It's e to the minus some alpha x, where alpha squared is roughly the difference, alpha x. And out here it's going to be e to the plus alpha x. And that's for normalizability. We want to have a single particle in this state. So what that tells us is that the wave function falls off in these classically disallowed regions exponentially. And so the probability of measuring the particle at an arbitrarily far position goes to zero. And it goes to zero exponentially rapidly. Cool? So I will call a bound state, a quantum bound state, an energy eigenstate, such that the probability falls off exponentially as we go far away from wherever we think is an interesting point, like the bottom of the well. Cool? A bound state is just a state which is exponentially localized. If you put it there, it will stay there. Yeah. And it's important that when I say a bound state I'm talking about energy eigenstates. And the reason is this. Bound state equals energy eigenstate. The reason is that, consider by contrast a free particle, so a free particle with constant potential. What are the wave functions? What are the energy eigenfunctions? Well, they're plane waves. Right. So are these bound states? No. Good. OK. On the other hand, I claimed that I can build a wave packet, a perfectly reasonable wave packet, which is a Gaussian. OK. This is some wave function, psi of x times 0. It's a Gaussian. It's nice and narrow. Is that a bound state? Well, it's localized at this moment in time. But will it remain, and in particular its probability distribution, which is this thing norm squared is localized in space-- Sorry. This was zero. --it's localized in space. The probability of it being out here is not just exponentially small, it's Gaussian so it's e to the minus x squared. It's really not out here. But what happens if I let go? What happens if I look at the system at time t? It's going to spread out. It's going to disperse. We're going to talk about that in more detail later. So it's going to spread out. And eventually, it will get out arbitrarily far away with whatever probably you like. So the probability distribution is not time invariant. That is to say it's not a stationary state. It's not an energy eigenstate. Saying something is bound means that it never gets away. So bound states are specifically energy eigenstates that are strictly localized, that fall off at least exponentially as we go away from the origin. Cool? It's just terminology, what I mean by a bound state. Questions? OK. So let's talk about the finite well. So I need to give you definitions of the parameters. Let's draw this more precisely. Here's my well. Asymptotically, the potential is zero. The potential depth, I'm going to call minus v naught. OK. And I'm going to center the well around zero. And I'll call the sides minus l and l. And I want to find bound states of this potential, just like we found bound states of the harmonic oscillator, i.e. states with energy e, which is less than zero. So these are going to give us bound states because we're going to have exponential fall offs far away. So a couple of things to note. The first is on, I think, problem set three or four you showed that if you have a potential, which is symmetric, which is even, under x goes to minus x, then every energy eigenfunction, or at least every bound state energy eigenfunction, every energy eigenfunction can be written as phi e symmetric or phi e anti-symmetric. So it's either even or odd. It's either even or odd under the exchange of x to minus x. So when our potential is symmetric, the wave function or the energy eigenfunctions are either symmetric or anti-symmetric. OK. So we want to solve for the actual eigenfunctions. So we want to solve that equation. And we have this nice simple fact that we know the solutions in this region. We know the general solution in this region. We know the general solution in this region. So I'm going to call these regions one, two, and three, just to give them a name. So in region one-- That's actually sort of stupid. Let's call this inside, left, and right. Good. OK. So let's look at this equation. We have two cases. If the energy is greater than the potential in some region, then this is of the form phi prime prime is equal to energy greater than potential. This is a negative number. And so this is a minus k squared phi. And we get exponentials. And if, on the other hand, e is less than v of x, then phi prime prime is equal to plus alpha squared phi. I should say oscillator. And in particular here, I want the k squared is equal to 2m upon h bar squared. It's just the coefficient v minus e. And alpha squared is equal to 2m over h bar squared e minus v. OK. So let's apply that here. So in this region, we're going to get oscillations because we're in a classically allowed region, where the energy is greater than the potential. So we'll get oscillatory solutions. And the salient value of k inside, is k is equal to the square root of 2m over h bar squared times v minus e. So that's minus v naught. What did I do? I did. It's e minus v. I thank you. Yes. And I want the other one to be l so it'd be minus v. Good. Good. Excellent. So root 2 over h bar squared. And now we have e minus v naught, which is the actual value of e, which is negative. Right? But minus v naught or plus v naught is positive and greater in magnitude. So this is a nice positive number, and k is the square root of it. This is controlling how rapidly the wave function oscillates in this region. Similarly, out here we have alpha is equal to-- Well, here the potential is zero. So it's particularly easy. Alpha is equal to the square root of 2m upon h bar squared of-- Now z minus e is zero minus e, which is a negative number. So we can just write e absolute value. So we can write the general solution of this eigenfunction, of this eigenvalue equation, as phi e of x is equal to-- Let's break it up into inside and outside. Well, inside we know, since it's constant with this value of k, we get superpositions of oscillatory solutions. It' a second-order difference equation. There are two solutions and two integration constants. So first we have a cosine of kx plus b sine kx. This is inside. And then on the left we have a combination of exponentially growing and exponentially decreasing. So the exponentially growing is e to the alpha x plus de the minus alpha x. And on the right, similarly by symmetry, we have some combination of e-- But I don't want to call it the energy, so I'll call it the curly e. --to the alpha x plus fe to the minus alpha x. OK. So that's the general solution. We solve the problem as a superposition of the two oscillatory solutions or a superposition of the two exponentially growing and damped solutions or exponentially growing and collapsing functions on the left and right. Questions? So a couple of things to note at this point. So the first is we have boundary conditions to impose. We have boundary conditions at these two interfaces. But we also have boundary conditions off in infinity. What are the boundary conditions at infinity? Yeah, exactly. It should vanish. So we want the system to be normalizable. So normalizable is going to say that phi goes to zero at minus infinity. Phi of x goes to minus infinity goes to zero, which it equals. And phi of x goes to plus infinity should also be zero. OK. And then we're also going to have the conditions at the left boundary, and we're going to have conditions at the right boundary. [LAUGHTER] All right. So what are the boundary condition at the left boundary condition? So first off, what are the boundary conditions we want to impose at the left and right boundaries? AUDIENCE: Continuous. PROFESSOR: Continuous, and the derivative should be continuous. Exactly. So we have that phi is continuous, and phi prime-- Good god. --phi prime is continuous. Similarly, phi continuous, phi prime continuous. OK. Do we have enough boundary conditions to specify our function? So we have now for our solution, we have six undetermined coefficients. And we have six boundary conditions. So that looks good. Are they all independent? Ponder that one. So in particular, let's start with the normalizable. So in order for phi to go to zero at minus infinity deep out on the left, what should be true? Yeah, d goes to zero. Oops, equals zero. And on the right? Yeah, that curly e equals 0, which is nice so I don't ever have to write it again. So that's zero. And that's zero. OK. That's good. We can take advantage though of something nice. We know that the wave function has to be either symmetric or anti-symmetric. Right? So we can exploit that and say, look, the wave function is going to be different from our boundary conditions, but it's a true fact, and we can take advantage of it. We can use the parody of the well. I can never-- So we can use the parody of the potential to say that the system is either symmetric or anti-symmetric. And these are often said as even or odd because the function will be, as a function of x, either even or odd. You either pick up a minus sign or a plus sign under taking x to minus x. So if the system is even, what can we say about these coefficients? What must be true of b, for example? AUDIENCE: It's zero. PROFESSOR: Yeah. So b equals 0. And what else? AUDIENCE: It equals f. PROFESSOR: This equals f. Yeah, good. OK. This equals f. And if the system is anti-symmetric, then a equal to 0. And we see that c is equal to minus f. Yeah? So that's a useful simplification. So it's easy to see that we could do this either way. We could do either symmetric or anti-symmetric. I'm going to, for simplicity in lecture, focus on the even case. b is equal to 0, and c is equal to minus f. Sorry, c is equal to f. So plus c. So now we're specifically working with the even solutions. And on your problem set, you'll repeat this calculation for the odd functions. So we're going to focus on the even solutions. And now what we have to do is we have to impose the boundary conditions for phi and phi prime. So that's easy enough. We have the function. All we have to do is impose that the values are the same. So for example, let's focus on the left boundary conditions. Sorry. Let's focus on the right because I don't want to deal with that minus sign. So let's focus on the right boundary conditions. So this is x is equal to plus l. So when x is equal to plus l, what must be true? Phi and phi prime must be continuous. So what's phi? So phi is equal to-- Well, from inside, it's a cosine of kl. Yeah. And b is gone because we're only looking at the even functions. On the right, however, it's equal to c e to the minus alpha l because we're evaluating at the right boundary. Yeah? OK. So this is cool. It allows us to determine c in terms of a. And if we solve that equation for c in terms of a, we'll get an eigenfunction, phi even, with one overall normalization coefficient, a. And then we can fix that to whatever it has to be so that everything integrates to one. Yeah? So that seems fine. It seems like we can solve for c in terms of a. c is equal to-- This is weight. Well, OK. So c is equal to a cosine e to the plus alpha l. On the other hand, we also have a condition on the derivative. And the condition on the derivative is that phi prime is continuous. And the derivative of this, well that's easy. It's the derivative of cosine. So this is going to be minus sine. But we pull out a factor of k, because we're taking derivative with respect to x. Minus k is sine of kx. Evaluate it out, sine of kl, l. And this is going to be equal to minus alpha c e to the minus alpha l. AUDIENCE: You forgot an a. PROFESSOR: Oh, yes. There should be an a. Thank you. OK. So but now we've got a problem because this says that c is equal to minus a times k over alpha sine of kl times e to the alpha l. And that's bad because c can't be equal to two different numbers at the same time. There's a certain monogamy of mathematical equations. It just doesn't work. So how do we deal with this? Well, let's think about what these equations would have meant. Forget this one for the moment, and just focus on that first expression. I'm going to rewrite this slightly. a cosine of kl. OK. So what does this expression say? Well, it seems like it's just saying if we fix c to be equal to this value, for fixed value of kappa l and alpha, then there's a solution. However, what is k? What are k and alpha. k and alpha are functions of the energy. So it would seem from this point of view, like for any value of the energy, we get a solution to this equation. Everyone see that? But we know that can't possibly be right because we expect the solutions to be discrete. We don't expect any value of energy to lead to a solution of the energy eigenvalue equation. There should be only discrete set of energies. Yeah? AUDIENCE: Did you pick up an extra minus sign in the expression for c? PROFESSOR: You do. Thank you. The sign's cancel. Yes, excellent. I've never written this equation in my life. So thank you. Yes, extra minus sign. So what's going on here? Well, what we see is that we've written down the general form of the solution. Here were imposing that we've already imposed the condition that we're normalizable at infinity. Here, we're imposing the continuity condition on the right. And if we impose just the continuity condition for the wave function, we can find a solution. Similarly, if we impose only the continuity condition for the derivative, we can find a solution for arbitrary values of the energy. But in order to find a solution where the wave function and its derivative are both continuous, it can't be true that the energy takes just any value because it would tell you that c takes two different values. Right? So there's a consistency condition. For what values of energy or equivalently, for what values of k and alpha are these two expressions equal to the same thing? Cool? So we can get that by saying, look we want both of these equations to be true. And this is easy. I can take this equation and divide it by this equation. And I will lose my coefficients c. I will lose my coefficients a. What do we get? On the right hand side, we get-- And I'm going to put a minus sign on everything. So minus, minus, minus. So if we take this equation and we divide it by this equation, on the right hand side, we get alpha, because the c exponential drops off. And on this side, we lose the a. We get a k. And then we get sine over cosine of kl, also known as tangent of kl. Here we have a kl. Here we have a k. These are all dimensionful things. Let's multiply everything by an l. And this is nice and dimensionless. Both sides are dimensionless. So we get this condition. This is the consistency condition, such that both the wave function and its derivative can be continuous at the right boundary. OK? And this is a pretty nontrivial condition. It says, given a value of k, you can always determine the value of alpha, such as this equation as true. But remember that k and alpha are both known functions of the energy. So this is really an equation, a complicated, nonlinear equation for the energy. So this is equal to a horrible expression, a condition, badly nonlinear, in fact, transcendental condition on the energy. And where's it coming from? It's coming from normalizability and continuity everywhere. And a useful thing to check, and I invite you to do this on your own, is to check that the boundary conditions at the left wall give the same expression. Yeah. AUDIENCE: For our final form of that equation, is there a reason that we prefer to multiply both sides by l than divide both sides by k? PROFESSOR: Yeah. And it'll be little more obvious in a second. But here's the reason. So let's divide through by l. This is the form that we got. What are the units? What are the dimensions of this expression? k is a wave number, so it has units of 1 upon length. Right? And that's good because that's 1 upon length times the length, and you'd better have something dimensionless inside a tangent. But it seems there are two things to say about this. The first is it seems like l is playing an independent role from k in this equation. But this is dimensionless. These are both dimensionful units of 1 over length. So we can make the entire expression dimensionless and make it clear that k and l don't have an independent life. The dimensionless quantity, kl, times the tangent of that dimensionless quantity is equal to this dimensionless quantity. So the reason that this is preferable is twofold. First off, it makes it sort of obvious that k and l, you can't vary them independently in this sense. But the second is that it makes it nice and dimensionless. And you'll always, whenever possible, want to put things in dimensionless form. I mean it's just multiplying by l. So it's obviously not all that deep. But it's a convenient bit of multiplication by l. Other questions? OK. So where are we? So I'd like to find the solutions of this equation. So again, just to-- Let me write this slightly differently where k squared is equal to 2m upon h bar squared v0 plus e. And alpha squared is equal to 2m upon h bar squared e, the positive value of e. So this is a really complicated expression as a function of e. So I'd like to solve for the actual energy eigenvalues. I want to know what are the energy eigenvalues of the bound states of the finite potential well, as a function of l, for example. Sadly, I can't solve this equation. It's a transcendental equation. It's a sort of canonically hard problem to solve. You can't write down a closed from expression for it. However, there are a bunch of ways to easily solve it. One is take your convenient nearby laptop. Open up Mathematica, and ask it to numerically find solutions to this. And you can do this. It's a good exercise. I will encourage you to do so on your problem set. And in fact, on the problem set, it asks you to do a calculation. And it encourages you do it using Mathematica. Let me rephrase the statement in the problem set. It would be crazy for you to try to do it only by hand. You should do it by hand and on computer because they're both easy. And you can check against each other. They make different things obvious. This should be your default is to also check on Mathematica. The second thing we can do is we can get a qualitative solution of this equation just graphically. And since this is such a useful technique, not just here, but throughout physics to graphically solve transcendental equations, I'm going to walk through it a little bit. So this is going to be the graphical solution. And we can extract, it turns out, an awful lot of the physics of these energy eigenstates and their energies through this graphical technique. So the first thing is I write this in nice, dimensionless form. And let me give those dimensionless variables a name. Let me call kl is equal to z, just define a parameter z. And alpha l is a parameter y. And I want to note that z squared plus y squared is equal to a constant, which is, I will call if you just plug these guys out, kl squared plus al squared. That's easy. Kl squared is this guy times l squared. Al squared is this guy times l squared. And so the e and the minus e cancel when we add them together. So we just get 2mv0 over h bar squared times l squared. So 2m upon h bar squared l squared v0. And I'm going to call this r naught in something of a pathological abusive notation. OK. So this is our expression. And I actually want to call this r0 squared. I know. I know. It's awful. But the reason I want to do this is that this is the equation for a circle. Yeah? And a circle has a radius. The thing that goes over here is r squared. OK. So at this point, you're thinking like, come on, circle. So let's plot it. So how are we going to solve this equation? Here's what I want to solve. I have now two equations relating z and y. We have that from this equation z tangent z is equal to y. And from this equation we have that z squared plus y squared is a constant r0 squared. Where r0 squared depends on the potential and the width in a very specific way, on the depth of the potential and the width in a very specific way. So we want to find-- Bless you. --simultaneously, we want to find simultaneous solutions of these two equations. Yeah? So that's relatively easy. So here's y, and here's z. So this equation has solutions. Any time that y plus z squared is equal to r0 squared, that means any time we have a circle. So solutions for fixed values of r0 lie on circles. Oh, I really should have drawn this under here. Sorry. y and z. So those are the circles. Notice that I'm only focusing on y and z, both positive. Why? Not yz, but W-H-Y . Why am I focusing on the variables y and z being positive? Because we started out defining them in terms of k and l, which were both positive, and alpha and l, which were both positive. Can we find solutions to this equation that have x and y negative? Sure. But they don't mean anything in terms of our original problem. So to map onto solutions of our original problem, we want to focus on the positive values of y and z. Cool? OK. So that's this one. The solutions lie on circles. So given a value of y, you can find a solution of z. But we want to also find a solution of this equation. And this equation is a little more entertaining to plot. Here's y. Here's z. So what does z tangent z do? Oh, shoot. I want to plot y vertical. Otherwise, it's going to a giant pain. Happily, this plot can be left identical. Let's plot y vertically. So the reason I want to plot y vertically is that this is z tangent z. So first off, what does tangent z look like? Yeah. This is awesome. Yeah, it looks like this. Yes, exactly. So tangent is sine over cosine. Sine is zero, and cosine is one. So it does this, as you go to a value where the argument, let's call the argument z. So if we just plot tangent-- OK. So when z is equal to pi over 2, then the denominator cosine vanishes, and that diverges. Oops. OK. So here's pi over 2. Here's pi. Whoops. Pi, pi over 2, and here's 3pi over 2, and so on. Now, we're only interested in the first quadrant. So I'm just ignore down here. OK. So this is pi over 2. This is pi, 3pi over 2. OK. But this is not what we're interested in. We're not interested in tangent of z. We're interested in z tangent z. And what does z tangent z look like? Well, it's basically the same. Right? z tangent z, it has an extra factor of zero here and remains extra small at the beginning. But it still curves off roughly like this. And z is just nice and linear, nice and regular throughout this. So it doesn't change the fact that we have a divergence at pi over 2. And it doesn't change the fact that it vanishes again at pi and becomes positive again. It just changes the shape of the curve. And in fact, the way it changes the shape of the curve is this becomes a little fatter around the bottom. It's just a little more round. And when we get out to large values of z, it's going to have a more pronounced effect because that slope is, at every example where it crosses z, that slope is getting larger because the coefficient of z is getting larger. OK. So it's just going to get more and more sharp. But anyway, with all that said, here's 0. Here's pi. Here's pi over 2. Here's pi. Here's 3pi over 2. The second plot we want to plot, y is z tangent z. We know how to plot this. Cool? And what we want to find are simultaneous solutions of this, values of y and z, for which this equation is solved and this equation is solved for the same value of y and z. This is a graphical solution. So let's combine them together. And the combined plots look like this. First we have pi over 2. So let's plot the tangents. And then we have these circles for various values of r. So for a particular value of r, for example, suppose this is the value of r. This is r0. So how many solutions do we have? One. One set of common points where at y and z solve both equations. So we immediately learn something really lovely. What happens to the radius of that circle as I make the well deeper? Yeah, as I make the well deeper, that means v0 gets larger and larger magnitude, the radius gets larger. So does the circle. So if I make the well deeper, I make this the circle larger. Will I still have a solution? Yeah, I'll still have a solution. But check this out. Now, I'll have a new solution. And you can even see the critical value of the depth and the width of the well. In order to have exactly a new bound state appearing, what must the value of r0 be? Well, it's got to be that value, such that r0 squared is pi. Yeah? And similarly, let me ask you the following question. As I make the well deeper and deeper and deeper, holding the width, and make it deeper and deeper and deeper, does the number of states increase or decrease? It increases. If you make it deeper and deeper, the radius of that circle is getting bigger and bigger. There are more points where this circle intersects this point. So here's another one. We've got one here, one here, one here, three solutions. And the number of solutions just goes. Every time we click over a new point by increasing the radius of the circle, we get a new solution. We get another bound state. But here's the thing that I really want to focus on. Let's make the well less and less deep. Let's make it shallower and shallower. At what depth do we lose that first bound state? We never do. Right? There is no circle so small that it doesn't intersect this curve. In a 1D, finite well potential, there is always at least one bound state. There are never zero bound states. This will turn out not to be true in three dimensions, which is kind of interesting. But it's true in one dimension that we always have at least one bound state. And in fact, you can decorate this. You can use this and fancy it up a bit to argue that in any potential in 1D, there's always at least one bound state unless the potential is constant, I mean any potential that varies and goes to zero infinity. Yeah? And so we still don't have any numbers. But we know an awful lot about the qualitative structure of the set of energy eigenvalues of the spectrum of the energy. Questions? Yeah? AUDIENCE: So what happens if r is bigger than pi or y is bigger than pi and you get two solutions? PROFESSOR: Great. So when we have two solutions, what does that mean? Well, you've got to bound states, two different energies. Right? It's a good question. Every solution here corresponds to some particular value of y and some particular value of z. But those values of y and z are just telling you what k and alpha are. And so that's determining the energy. So a different value of k is going to give you a different value of the energy. So we can just eyeball this in particular. Let's look at alpha. Alpha is just e. Alpha squared is just e. Yes? So here's a quick question. If alpha is just e, and alpha l is y-- So this is our y value. y is roughly alpha, the width, which means it's roughly the energy square root. So this value, the vertical value of each of these intersection points on a given circle corresponds to the square root of the energy times some coefficients. So which state has the largest value of the energy? Absolute value, which state is most deeply bound on this circle? Yeah, the first one. Right? Because it's got the largest value of alpha. So this is nice. We see that the first state always has a higher value of alpha than the second state, which always has a higher value of alpha than the third state. And every time we add a new state, we make the depth of these guys the binding energy of the already existing states. We make it just a little bit deeper. We make them a little more tightly bound. And only eventually then do we get a new bound state appearing. And what's the energy of that new bound state when it appears? Zero energy. It's appearing just at threshold. OK. So we'll explore that in more detail in the problem set. But for now, let me leave it at that. Questions? Other questions? Yeah. AUDIENCE: You said that this can be generalized to any nonconstant function that you'd like. So there's always going to be at least one bound state. What about, like with delta function? PROFESSOR: Excellent question. What about the delta function? We're going to come back to that in just a few minutes. It's a very good question. So the question is, look, if any potential that goes to zero infinity and wiggles inside, if any potential like that in 1D has a bound state, what about the delta function? We briefly talked about that. So we're going to come back to that in just a few minutes. But it's a pressing question. OK. Other questions? Yes. AUDIENCE: So the energy is zero, but that's not possible. PROFESSOR: Thank you. OK. Good. So let me talk about that in a little more detail. So I wasn't going to go into this, but-- So when new bound states appear, so let's consider a point where our r0 is, let's say, it's just the right value so that r0 is equal to pi. OK. And we see that we're just about to develop a new bound state. So let's think about what that bound state looks like. So this is the new bound state. And I'm going to put this in parentheses because it's got bound state. And we say at threshold. OK. At threshold, i.e. at the energy is roughly zero, and r0 is equal to pi. So this is really what we mean. This new state, when r0 is pi and we have a solution on that second branch. Cool? So what does this wave function look like? What does it look like when you have a wave function that just appeared? It's just barely bound. Well, first off, what does it mean to be bound? Let's just step back and remember for now. What does it mean to be a bound state? It means you're an energy eigenfunction and you're localized. Your wave function falls off at infinity. Now, if it falls off at infinity, do these guys fall off at infinity? These wave functions, sure, they fall of with an exponential damping. And in particular, let's look at the right hand side of the well. This new bound state is appearing just at zero energy. So out here, what is the wave function? It's e to the minus alpha x. But what's alpha? Zero, right? There it is, zero. So this is e to the minus alpha x where alpha is equal to 0. This is constant. So what does the wave function look like? Well, the wave function, again over the same domain-- Here's 0, l. And here's the value zero. We know that in this domain it's oscillatory, and in this domain, it's constant. And actually, since we know that it's the first excited state, we know that it does this. So if we make the well ever so slightly deeper, ever so slightly deeper, which means making the radius of the circle ever so slightly larger, we will get a nonzero value for the alpha of this solution. Right? It'll be just tiny. But it'll be nonzero. So we make the well just a little tiny bit deeper. We get something. OK, good. So what's going to happen to the wave function? Well, instead of going flat. This is going to curve. It's got a little more energy. There's a little more curvature. So it curves just a little tiny bit more. And then it matches onto a very gradually decaying exponential. OK. So what's happening as we take this bound state, the second bound state, and we make the well a little more shallow? We make the well a little more shallow. It's a little less curved inside. And the evanescent tails, the exponential tails become longer and longer and longer and broader until they go off to infinity, until they're infinitely wide. And is that normalizable anymore? No, that's not normalizable. So the state really isn't strictly localized at that point. It's not really a normalizable state. And just when the state ceases to be normalizable, it disappears. We make the well just a little deeper, and there's no state there at all. OK. So this tells you a very nice thing. It's a good bit of intuition that when states are appearing or disappearing, when states are at threshold as you vary the depth, those threshold bound states have exceedingly long evanescent tails, and they're just barely bound. OK? This turns out to have all sorts of useful consequences, but let me move on. Did that answer your question? Good. Yeah? AUDIENCE: So in this case, the radius, as you say, is proportional to the length. Right? But also we have intuition that as we increase the length of the well, the energy's going to keep increasing-- PROFESSOR: Fantastic. OK. So what's up with that? Right. So the question is this. We already have intuition that is if we take a finite well and we make it a little bit wider, the ground state energy should decrease. The energy of the ground state should get deeper and deeper, or the magnitude should increase, another way to say it. Right? That was our intuition. So let's check if that's true. What happens if I take the ground state some value of the radius, and then I make the well a little bit wider. Well, if I make the well a little bit wider, what happens to r0, the radius of the circle? Well, if I double the length of the well, the width of the well, then it will double r0, and it will double the radius of the circle. So if I make it wider, r0 gets bigger, and we go to a bigger circle. And what happened to the energy of this state? Yeah, it got deeper and deeper and deeper. And meanwhile, as we make it wider, as we make the well wider, the circle is getting bigger again. And we're going to get more and more states. So as we make the well wider, holding the depth fixed, we get more and more states. As we make the well deeper, holding the width fixed, we get more and more states. And so how do you trade off? If I make it twice as wide, how much-- So here's a good question. Suppose I take a well, and it has n states. Suppose I then make it twice as wide. What must I do to the energy so that it still has the same states with the same energies? AUDIENCE: Divide it by 4. PROFESSOR: Yep, exactly. I've got to divide it by 4. Because I've doubled the length, that means the radius has gone up by four. But I get exactly the same solutions if I just bring this out. Well, that's almost true. So if I put a factor of 1/4 here, that's almost true, except for the value of y is unchanged, but the alpha hasn't changed. Sorry. The alpha has changed because there's an l. So y is fixed. But alpha's changed because of the l. So the reason that it's useful to write things in these dimensionless forms is that you can see the play off of the various different dimensionful parameters in changing the answer. Other questions. OK. So a couple of comments. So the first is let's just check to make sure that this makes sense. We have already solved this problem. We solved this problem a while ago, but we solved it in a particular limit. We solved it in the limit that the potential one's arbitrarily deep. Right? So when the potential one's are arbitrarily deep, holding the width fixed, that was the infinite well. That was the first problem we solved. So let's make sure that we recover this in that limit. So what happens as we make the well arbitrarily deep, holding the length fixed or the width fixed. So if we make this arbitrarily deep, v0 is getting arbitrarily large. That means r0 is getting-- We've got a huge circle. So what do these solutions look like when we have a huge circle. Let me not do that here. Let me do that here. So if we make the potential nice and deep, Let's think about what that plot looks like. So again, that first plot looks identical with the tangents. So on and so forth. And what I want to do is I want to dot, dot, dot. OK. So this is way up there. So these guys are basically vertical lines at this point. So for very, very large values of y, and in particular, for very large values that are of order the gigantically deep radius r0, what does the circle look like? So what does the second equation look like, the second curve? Well, again it's circles. But now it's a gigantic circle. Yeah, exactly. If it's a gigantic circle, it's basically flat. It's not exactly flat, but it's almost flat. It's a circle. So what are the values? And here's the key question. What are the values of-- Did do that right? Yeah, OK. Good. So what are the values of the curve, where we get a solution? The values now of z, are just exactly on these vertical lines, on the separatrices. The value of z is at pi over 2. and 3pi over 2. And then another one at 5pi over 2, and so on and so forth. Right? So what we find is that the allowed values of z-- Sorry. kl, yes, good. --allowed values of z are equal to 2n plus 1 over pi. Whoops 2n plus 1 over 2 times pi. So let's just check that. So n is 0. That's 0 1/2 pi. n is 1. That's 3/2 pi. Good. So these are the values of z, which says that kl is equal to 2n plus 1 upon 2 pi or k is equal to 2n plus 1 over 2l, which is the width of this well because it's from minus l to l pi. So is this the correct answer for the infinite square well? Are these the allowed values of k inside the well for the infinite square well? Almost. Instead of 2n plus 1, it should just be n plus 1. We seem to be missing about half of the energy eigenvalues. AUDIENCE: That's only the even ones. PROFESSOR: Yeah, thank you. This is only the even ones. We started out saying, oh, look. Let's look only at the even ones. Where do you think the odd ones are going to be? Ah, well the odd ones, so this should be k even. So what about k odd? Well, we know the answer already-- Whoops. Odd, that's an odd spelling. --should be equal to 2n over 2l plus 2 over 2l. Whoops. 2 capital L, pi. OK. This is our guess just from matching on to the infinite square well. So what does that mean? Well, that means it should be this one and this one. So when you go through this exercise on your problem set and you find the solutions for the odd, and you repeat this analysis for the odd ground states. What should you expect? Well, you should expect to find this. And what do you think the curves are that you're going to use int he graphical solution to do your transcendental equation? Yeah, it's really tempting to say, look. It's just going to be something shifted, like this. OK. So these are going to be the odd question mark, question mark. OK. So you'll check whether that's correct intuition or not on your problem set. OK. Questions? Yeah? AUDIENCE: What about the lower end of these states, where it won't have gone off high enough? Is that [INAUDIBLE]? PROFESSOR: Excellent. So that's a really good question. How to say it? What we've done here to make this an infinite well-- So the question, let me just repeat the question. The question is, look, what about all the other states? OK, it's true that r0 as gigantic. But eventually, we'll go to a large enough z, where it's the circle coming down over here too. So what's up with that? What are those states? Where are they? What do they mean in terms if the infinite square well? Well, first off, what are the energies of those states? AUDIENCE: Very low. PROFESSOR: They're very low in magnitude, which means they're close to what in absolute value? AUDIENCE: Zero. PROFESSOR: Zero. They're close to zero. So they're at the top of the finite well. These are the states bound at the top of the finite well. These are the states bound at the bottom the finite well. But how many states are bound at the top of the finite well when we take the limit that the well goes infinitely deep? Yeah, none of them. Right? So when we make the well infinitely deep, what we're saying is, pay no attention to the top of the well. Look only at the bottom of the well. And if it's really deep, it's a pretty good approximation. So that's what we're doing here. We're saying, look. Pay no attention. There is no top of the well. There's just the bottom. And look at the energy eigenvalues. Does that make sense? So what we're saying is if you have a preposterously deep well, the energies of a preposterously deep well should be a good approximation to the low-lying energies of an infinitely deep well. Because it's way up there, what difference can it make? And that's what we're seeing work out. Did that answer your question? AUDIENCE: Wait. I might have this backward. But when it says the high energy instead of looking right at the top of the well-- PROFESSOR: Yeah. OK, good. So this is an important bit of intuition. So when we say this is the energy zero, and the potential has a minimum at minus v0, and we're measuring the energies relative to zero, then the states at the top of the potential well are the states with energy close to zero. And the states at the bottom of the potential well are those states with the energy of order v0. cool? And what are the energies of all these states? They're order of v0. Right? Because this is-- Are they exactly v0? No, because this isn't linear. It's actually a circle. So there's going to be a correction, and the correction is going to be quadratic. If you work out that correction, it's correct. The depth above the bottom of the potential is correct for the energy of the corresponding infinite well problem. I'll leave that to you as an exercise. Other questions? OK. So there's another limit of this system that's fun to think about. So this was the infinite well limit. What I want to do is I want to take advantage of the observation we made a second ago that as we make the well deeper, we get more states. As we make the well more narrow, we get fewer states. To trade that off, consider the following limit. I want to take a potential well, which has a ground state. What does the ground state look like? So the ground state wave function is going to be-- So here's zero. And here's exponentially growing. Here's exponentially decreasing. And if it's a ground state, how many nodes will it have inside? How many nodes will the ground state have? AUDIENCE: Zero. PROFESSOR: Zero. Good. OK. So the ground state will look something like this. We more conventionally draw it like this. But just for fun, I'm going to draw it in this fashion. In particular, it has some slope here. And it has some slope here. Oh, shoot. Did I? Yes, I did. Dammit. I just erased the one thing that I wanted to hold onto. OK. So there's my wave function. It has some particular slope here. It has some particular slope here. And this is the ground state wave function, with some energy. I don't know. I'll call it this. Now, what I want to do is we've already shown that as we make the well more and more shallow and more and more narrow, the energy of the ground state gets closer and closer to zero. But there remains always a bound state. There is always at least one bound state. We proved that. Proved, as a physicist would. So I want to do that. I want to take this seriously. But here's the limit I want to consider. Consider the limit that we make the potential v goes to infinity, v0, while making l go to zero. So what I want to do is I want to take this thing, and I want to make it deeper and deeper, but more and more narrow. If I do this repeatedly, eventually I will get a delta function. And I will get a delta function if I hold the area of this guy fixed. Yeah? So if I do so, holding the area under this plot fixed, I will get a delta function. Everyone cool with that? So let's think, though, quickly about what's going to happen to the ground state wave function. So as I make the potential, let's take this wave function, and let's look at this version of the potential. So as I make the potential deeper, what happens to the rate of the oscillation inside or to the curvature inside? It increases. Right? So the system is oscillating more, it changes more rapidly, because phi double dot or phi double prime is equal to v minus e phi. It oscillates more rapidly. So to make it deeper, the system tends to oscillate more rapidly. However, as we make it more narrow, the system doesn't have as far to oscillate. So it oscillates more rapidly, but it doesn't oscillate as far. So what's going to happen? Well, as we make it more and more narrow and deeper and deeper, we again have the wave function coming in. And now it oscillates very rapidly. Let's do it again. The wave function comes in, and it oscillates very rapidly. And the it evanescent tail out. And now as we have a delta function, exponential damping, it oscillates extremely rapidly over an arbitrarily short distance and gives us the kink that we knew at the very beginning we should expect when the potential is a delta function. Right? From our qualitative structure of the wave function at the very beginning we saw that when we have a delta function potential, we should see a kink in the wave function. Because again, if we have phi prime prime is delta function discontinuous, phi prime is the integral of this. This is a step, and phi is continuous. And so here we have a step function. We get a discontinuity in the second derivative. Here we have a delta function in the potential, and we get a discontinuity in the first derivative if we get a kink in the wave function. Yeah? AUDIENCE: Would we get a jump there? PROFESSOR: Sorry. Say again. For e1? AUDIENCE: Yeah. Would we get a jump? PROFESSOR: Very good question. So let me do this more seriously. Let's do this more carefully. So the question is, for the first excited state, do we get a jump? Do we get a discontinuity? What do we get for the first excited state? Right? So let's talk about that in detail. It's a very good question. Example v is equal to minus v0 delta of x. Now, here I want to just warn you of something. This is totally standard notation for these problems, but you should be careful about dimensions. What are the dimensions of v0, the parameter of v0? It's tempting to say energy. That's an energy. That's an energy. But wait. What are the dimensions of the delta function? AUDIENCE: Length. PROFESSOR: Whatever length right? Because we know that delta of alpha x is equal to 1 over norm alpha delta of x. So if I write delta of x, which is always a slightly ballsy thing to do because this should really be dimensionless, but if I write delta of x, then this has units of 1 over length, which means this must have units of length times energy. OK. Just a little warning, when you check your answers on a problem, you always want to make sure that they're dimensionally consistent. And so it will be important to make sure that you use the energy times the length for the dimensions of that beast. So my question here is, is there a bound state? So for this example, for this potential, the delta function potential bound state, which again is this guy, is there a bound state? So again, we just ran through the intuition where we made the potential deep and deeper and deeper, v0 divided by epsilon over width epsilon. So v0, in order for this to be an energy has to be an energy times a length because we're going to divide it by length. So this is going to give us a delta function potential in the limit. We have an intuition that we should get a bound state with a kink. But let's check that intuition. We want to actually solve this problem. So we'll do the same thing we did before. We now write the general solution in the places where the potential is constant, which is on the left and on the right. And then we want to impose appropriate boundary conditions at the interface and at infinity, where these are going to be normalizable, and this is whatever the right boundary conditions are. So we are going to have to derive the appropriate boundary conditions. So let's just do that quickly. So phi with definite e is equal to a-- So in this region in the left, it's either growing or decreasing exponential. So ae to the alpha x plus be to the minus alpha x. And this is x less than 0. And ce to the alpha x plus de to the minus alpha x for x greater than 0. So first off, let's hit normalizability. What must be true for normalizability? Yeah, they'd better be converging to zero here and converging to zero here, which means that c had better vanish and b had better vanish. OK. So those guys are gone from normalizability. And meanwhile, if this is symmetric, what is going to be true of the ground state? It's going to be symmetric. So a must be equal to d. Great. So a is now just some overall normalization constant, which we can fix from normalization. So it looks like this should be the solution. We have an exponential. We have an exponential. But there's one more boundary condition to fix. We have to satisfy some matching. We have to satisfy the boundary conditions at the delta function. So what are those? What are those matching conditions? So we can get that from the energy eigenvalue equation, which says that phi prime prime is equal to h bar squared upon 2m. Sorry. Get your dimensions right. So it's 2m over h bar squared v minus e. In this case, v is equal to minus v0 delta function. That's very strange. So minus 2m over h bar squared v0 delta of x minus e-- I pulled out the minus. --so plus e phi. So this must be true at every point. This of course, is zero everywhere, except for at the origin. So what we want to do is we want to turn this into a boundary condition. And we know what the boundary condition is. If v is a delta function, that means that phi prime prime is also a delta function or proportional to a delta function. That means that phi prime is a step function. And how did I get that? I got that by integrating phi prime prime. You integrate a delta function, you get a step function. Well, that's cool. How do we figure out what step function discontinuity gives us? Let's integrate. Let's integrate right across the delta function and figure out what the discontinuity is. So let's take this equation, integrate it from minus epsilon to epsilon, where epsilon is a very small number. And that's epsilon to epsilon. So what is this going to give us on the left? Well, integral of a total derivative is just the value of the thing at a value to the point. So this is going to be phi prime at epsilon minus phi prime and minus epsilon. What does that mean? The difference between the derivative just after the origin and just before the origin. This is the discontinuity for very small epsilon. This is the discontinuity of the derivative at the origin at the delta function. And we already expected it to have a step discontinuity. And there it is. And how big is it? Well on the right hand side we have 2m minus 2m upon h bar squared. And we're going to get two terms. We get a term from integrating the first term. But over this narrow window, around, let's say epsilon was here, over this narrow window we can treat the wave function as being more or less constant. But in any case, it's continuous. And this is a delta function. So we know what we get from the integral of the delta function. We just get the value v0 phi at the delta function. Phi of the zero at the delta functions, so phi 0. We get a second term, which is plus the energy integrated against phi. The energy's a constant. And phi is continuous. So this, whatever else you can say about it, is roughly the constant value of phi at the origin times the energy times the width, which is epsilon. So plus-order epsilon terms. Everyone cool with that? So now what I'm going to do is I'm going to take the limit as epsilon goes to zero. So I'm just going to take that the discontinuity just across zero. So this is going to give me, of this gives me the change in the slope at the origin. OK. The derivative just after the origin minus the derivative just before the origin is equal to-- These order epsilon terms go away. --minus 2m upon h bar squared v0 phi. So that's my continuity. That's the condition for continuity of the derivative and appropriate discontinuity of the first derivative at the origin. And so this, when we plug-in these values of this form for the wave function, when we take a derivative, all we're going to do is we're going to pick up an alpha. And so when we work all of this out-- I'm not going to go through the algebra. You're going to go through it on the problem set. --when we take this condition, when we impose this condition with this wave function, it gives us a very specific value for alpha. This is only solvable if alpha is equal to mv0 upon h bar squared. Good. So let's just check the units. So this is momentum times length, momentum times length. This is mass. This is an energy times a length. So this has overall units of, p squared over m, overall units of 1 upon length, which is what we wan. So that's good. So we get alpha is equal to mv0 upon h bar. And that gives us the form of the potential. And it also tells us that the energy, plugging this back in, is equal to minus h bar squared alpha squared upon 2m, which we could then plug-in the value of alpha and solve for v0. So what we found is that there is a single bound state of the delta function potential, which we could have gotten by just taking this limit. It's a fun way to rederive the same result. It's a nice check on your understanding. So we find that there's a single bound state of the delta function potential. Now, what about an odd bound state? We assumed at an important point that this was even. What if I assume that it was odd? One node, what if we had assumed that it was odd? What would be true of the wave function? Well for odd, this would be a, and this would be minus a. So the value of the wave function at the origin is what? Zero. So that tells us the value of the wave function is zero. What's the discontinuity? Zero. So it's as if there's no potential because it has a zero right at the delta function. Yeah? But that means that this wave function, an odd wave function, doesn't notice the delta function potential. So is there a bound state? No. So how many bound states are there? Always exactly one for the single isolated delta function. On your problem set, you're going to use the result of the single isolated delta function, and more broadly you're going to derive the results for two delta functions. So you might say, why two delta functions? And the answer is, the two delta function problem, which involves no math-- Right? It's a totally straightforward, simple problem. You can all do it right now on a piece of paper. The two delta function problem is going to turn out to be an awesome model for the binding of atoms. And we're going to use it as intuition on your problem set to explain how quantum mechanical effects can lead to an attractive force between two atoms. See you next time. [APPLAUSE]
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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: Very good. So today, we'll begin with a study of one dimensional potentials and the energy eigenstates or properties. And we'll discuss this for about half of the lecture. And then go into the variational principle. So let me go back then to our discussion of last time, where we were talking about energy eigenstates and the Schrodinger equation in one dimension. So there's a term that we use all the time. This term is called the bound state. And bound state seems something sort of very non-trivial. But mathematically, we can say quite clearly what it is. A bound state is something that is not spread all over space, basically. And the way we therefore use the terminology for bound state it that we only speak of bound states when there are energy eigenstates. So an energy eigenstate may or may not be a bound state, but any bound state is an energy eigenstate. So an energy eigenstate is a bound state if the wave function goes to zero when you go sufficiently far away. So it's a simple definition, but it helps us understand that the basic idea is at the state is just not spread all over the world. Now remember we're trying to find energy eigenstates, and that is to find wave functions, time independent wave functions, that solve the time independent Schrodinger equation. Which I have written there for convenience. This is the same equation we wrote last time. For a time independent potential, so that the full wave function can be written as a phase that contains the information of the energy times a function of x, sine of x. That still may be complex, but it doesn't have to be complex. As you can see, this is a real equation. Now in order to clean up some of the constants in this equation, it's-- yes? AUDIENCE: Why is it not V x minus e? PROFESSOR: Well you have to look back at the Schrodinger equation, and bring the things to the other side, and take care of the signs. It is correct as stated. Check it. Now we have, in order to do this and just to write things with a little more clarity, we scale the energy. So we define a calligraphic energy, which is 2m over h squared times energy E. And a calligraphic V with the same scaling factor. So that the Schrodinger equation then takes the form-- this same equation-- sine. And I'll use the double prime notation for two derivatives. Plus now E minus cal of x sine equals zero. So this is how our equation looks. It's relatively simple. And can be treated without carrying constants all over the place. Now we're going to discuss a few results that are extremely important that you keep in mind for the future. It helps you think clearly. So these results are little theorems that I'm going to state. And the proofs are given in the notes. I will skip some of the proofs, although I will tell you what the strategy is for the proofs and then show how the theorem implies, as a corollary, interesting things that could easily be misconstrued. So let me give you the first theorem. Theorem one. This is just the statement that if you are talking about bound states in one dimensional potentials, there are no degeneracies. So let's write it. There is no degeneracy for bound states of one dimensional potentials. So it's a fundamental result. We'll see some way to understand it a little later. And you have seen this in 804, I believe. So what is a strategy to a proof? So we want a rigorous proof. So what is a strategy? I will not go through the proof. The strategy is to assume there is a degeneracy. So strategy, assume a degeneracy. So sine 1 and sine 2 different with the same energy. And then you write this equation for sine 1 and sine 2, and they have the same energy. Then you take the two equations, and you multiply the first by something, the second by another thing. Subtract it. Do a bit of algebra. And you suddenly can prove that one solution is equal to a constant times the other solution. So you do the work, and that's what comes out. Once you've shown that one is proportional to the other, we've declared in quantum mechanics that two wave functions that differ by a constant have exactly the same physics. You can normalize it, and normalize them. And the constant can be just a phase, but the phase doesn't matter. So those two wave functions are really the same. So you get a contradiction. You thought they were different. But the difference is just too trivial, and therefore they are really the same wave function. Because we've said that two wave functions that differ by a constant should be identified. So that's this theorem. And it's a key theorem. Let's go to the next theorem that is also important. So theorem two. Energy eigenstates sine f of x can be chosen to be real. So here it is. We mentioned that this differential equation allows for real solutions. There's no contradiction, because there's no i in there. But now we say more. That even if you find a complex solution, you can work with real solutions. So what is the strategy here? Again strategy. To have you get a complex solution. Sine f of x. And then you go about to prove that actually this complex solution implies the existence of two real solutions. So this complex solution implies existence of two real solutions. And moreover, two real-- I should say it better here-- degenerate solutions. How could you prove that? Well, it probably would occur to you that you would take a solution and say, oh, this solves it. Then you would show that sine star also solves this equation. And then by linearity of the Schrodinger equation, you could form what we would call the real solution formed by taking sine plus sine star and adding them. Or the real solution that comes from the imaginary part. You see the real and imaginary parts of a complex number are real. So is the case here as well. That they can take sine and sine star that are different. And the sum is real by construction. You put the star, and recalling that star of star gives nothing, you see that this function is equal to its star. So it's real. So is this one. This function, which is the imaginary part of the solution that you had, is also real. So you get two real solutions. Moreover, you can show that sine and sine star have the same energy. Therefore sine r and sine m have the same energy. So I'm really skipping very little. But I don't want to clutter the blackboard with the whole derivation at this moment. So here is the strategy. A complex solution implies two real solutions. And therefore, since you have real solutions, you can always choose them to be real. No loss of generality. You had a complex solution. You say, I like just real solutions. And you chose your two real solutions. You give them. You've given the whole thing. But then there is a nice corollary for the case of one dimension bound states. So this is the part that is perhaps somewhat interesting. And the corollary is that if you're talking bound states of one dimension, any solution is equal to a real solution up to a constant. You may say, well what's the difference? Let's write it down and look at it. Corollary for bound states of one dimensional potentials. Any solution is, up to a phase, equal to a real solution. So how does that go? Maybe we can use more of this blackboard. Why would any solution be, up to a phase, equal to a real solution? OK. Suppose you have these two solutions here. Well this is real, so that's trivially true. And this is real. But what if I put a linear combination of them? Or take the original complex solution? Why would it be, up to a phase, equal to a real solution? The answer is the following. Because of theorem and, that there's no degeneracy, you've got two solutions that claim to be different and degenerate. But that can't be. So by theorem one, you must have that-- in the notations, I have here-- sine imaginary of x must be equal to a constant times sine real of x. Now if these function's anyway are real, the constant here must be real. Therefore, even if you form the original complex solution, sine, which is sine real plus i sine imaginary. You can check that. It's clear by definition that that's the way it should be. If you think this, this is equal to 1 plus i times c times sine real of x. Therefore this is equal to a constant times the phase of-- this is a complex number. So this is yet another constant that we could say, the norm of 1 plus i c times some phase times sine r of x. So you see that any solution, any linear combination, in fact, with any numbers of sine r and sine imaginary, will be a constant times a phase times this thing. So it is really just real anyway. So you can just have one solution. And any solution that you may write, that represents a bound state, is, up to a phase, equal to a real solution. OK. So that's our theorem two. Theorem three. It's also very famous. If the potential V of x is even-- that is, V of minus x is equal to V of x-- the eigenstate can be chosen to be even or odd under x2 minus x. OK. So here is another claim. So the potential is even. There's no theorem for all the potentials. No clear theorems, no simple theorems for all the-- but if the potential is even, the claim is that you can choose eigenstates to be even or odd. The word chosen is very important here. Otherwise, this would not be a precise statement. You cannot say the eigenstates are even or odd. You can choose them to be. So how does the proof go? Strategy begin with a wave function, sine of x, that is neither even nor odd. And then you do a little work with the Schrodinger equation. Take the Schrodinger equation, change all x's to minus x's and show that, in fact, not only is sine of x a solution, but sine of minus x is also a solution. With the same energy. So prove that sine of minus x is a solution with the same energy. And in this case, of course, we can already have shown that we can choose these wave functions to be real. So we can choose all of these wave functions to be real. And what do we do next? If we have these two solutions with the same energy, then you can build of sine s, which is 1/2 of sine of x plus sine of minus x. And of sine a. s for symmetric, and a for anti-symmetric. And of sine a, that is 1/2 of sine of x minus sine of minus x. And this tool would be this even, under the exchange of x for minus x. This one odd, under the exchange of x for minus x. And both would be solutions by superposition. And both would have the same energy. So that's the end of the theorem because then these things are even or odd and have the same energy. So the solutions can be chosen to be even or odd under x. So if you've proven this, you've got it already. But now we get the corollary. For bound states in one dimension, the solutions not anymore the word chosen. We can delete the word chosen. The solutions are either odd or even. So it's a lot stronger. It's not anymore, you can choose them to be, but a general one is neither odd nor even. No. You try to find the solution that is neither odd nor even, and you can't find it. So it's very strong. Yes? AUDIENCE: Is this for even potentials? PROFESSOR: Even potentials. Yes. For bound states in one dimension with even potentials. V of x. V of minus x equal V of x. Yes. So how do we show that? Well, again, you've got two solutions here that are degenerate that have the same energy. Sine of x and sine of minus x. So given that there's no degeneracy in one dimensional problems, this thing that they have the same energy, the only thing that can be happening is that sine of minus x is equal to a constant times sine of x. Where this constant is real. Why real? Because we said already by the previously theorem, wave functions can be chosen to be real. So you've got this thing already that this is true. Sine of minus x is equal to c times sine of x. Which, if you use this property again by saying, oh, but sine of x is equal to c times sine of minus x. Basically, what this says is that you can change the sign of the argument by putting an extra c. So you do it again, sine of x is equal to this. So this is c squared times sine of minus x. So c squared must be equal to 1. And therefore c is equal to either plus or minus 1. No other option. So the functions are either even or odd, but they can't be arbitrary. This point is sufficiently settled that part two general exam at MIT 10 years ago had a question like that. And the person that invented the problem claimed that there would be a solution that could be neither even nor odd. So even faculty members at MIT sometimes get this wrong. It's not as weak as this, that can be chosen. But it's really either or other in the case you have one dimension. OK. So these are our main theorems. And we're going to proceed now by clarifying a little more the nature of the spectrum. So are there questions? Yes? AUDIENCE: Can you give an example of state that's not bound? PROFESSOR: OK. The question is can I give an example of a state that is not bound? Yes. We can give such state. You have a potential like this. And you have an energy like that. And then the wave function could look like this. Then do something like that. And then look like that. It just doesn't vanish when you go to infinity. AUDIENCE: So that can't be normalized? PROFESSOR: Can't be normalized. That's right. If it's not bound, it can't be normalized. Other questions? Yes? AUDIENCE: So you can't really represent-- it's doesn't really represent single particles, more like a stream of particles? PROFESSOR: Yes. So it doesn't represent a single particle. Now trying to interpret it as a stream of particles is a little delicate. So what we usually do is we build superpositions of those states that can represent a localized thing. But it's true that, morally speaking, it seems to represent more than one particle. OK. So now we talk a little about the nature of the spectrum. So what do we want to say here? We want to go back to the Schrodinger equation here. And just move one term to the right hand side. And just see what can happen in terms of singularities and discontinuities. So first of all, we always begin with the idea that sine must be continuous. And the reason sine must be continuous is that we don't want singularities worse than delta functions in potentials. If a function is continuous, the derivative might be discontinuous, and the second derivative would have a delta function. So the second derivative would have a delta function. But if the function is not even continuous, the second derivative would have derivatives of delta functions. So somehow the potential would have to have that. And we don't want it. So to simplify our life, we say that sine must be continuous. Now if sine is continuous, we will consider several possibilities, possibilities for V, possibilities for V of x. So first possibility-- one, V is continuous. So psi is continuous, and V is continuous. If psi is continuous and V is continuous, this product is continuous. Psi double prime is continuous. And psi prime is continuous. Two, V has finite jumps. Well, if V has finite jumps, and psi is continuous, this product has finite jumps. So psi double prime has finite jumps. If psi prime has finite jumps, the worst is that psi prime still must be continuous. But it changes. Psi prime could look like that, could have a corner. But it cannot be worse than that. Because if V has finite jumps, if psi double prime has finite jumps, and if psi prime is not continuous, it would have delta functions. So for these two conditions, continuous or even finite jumps, psi prime is still continuous. Things change qualitatively if three, V has delta functions. If V has a delta function, then psi double prime has a delta function. And psi prime therefore jumps. Psi prime is not continuous. Psi prime jumps. This may be reminiscent to you whenever you had to solve the problem of a bound state of a delta function, you got a wave function that looked like this in which psi prime jumps. And it has to jump. Because psi double prime has a delta function. Another case in which psi prime jumps is for if V has a hard wall. That is, the potential suddenly at one point becomes infinite, and it prevents the particle from moving across. A hard wall is a place, as you remember in the infinite square well, in which the wave function vanishes, but the derivative doesn't vanish. So you could say that psi is 0 here outside. Psi prime, then, jumps in that it's non-0 here and it's 0 afterwards. So now, you could object to what I'm saying by explaining that, well, we shouldn't talk about the wave function beyond the hard wall. And in some sense, you're right. But suppose we do and we say the wave function is just 0 everywhere, we see that psi prime jumps. So really, psi prime jumps. So this is as bad as things can get. So we can summarize this in just one sentence. And the sentence reads, psi and psi prime are continuous unless V has delta functions or hard walls, in which case psi prime can have finite jumps. So basically, psi and psi prime, continuous. Exceptions-- delta functions and hard walls, and psi prime can change. So questions. There was a question before. Maybe it's still there. Yes. AUDIENCE: Do you absolutely need that function [INAUDIBLE]? Or do we just assume that [INAUDIBLE]? PROFESSOR: We assume it-- so the question is, do I need that in absolute generality to do quantum mechanics? I don't think so. I presume you could discuss some potentials that lead to psi that are not continuous and still make sense. But we will not discuss them. And actually, I don't think I've encountered any of those. Yes. AUDIENCE: Can you give an example of a physical system whose potential is well approximated by a delta function [INAUDIBLE]? PROFESSOR: Yes, there are systems like that. For example, this one that is somewhat well approximated by a delta function. For example, a nucleus sometimes is considered to be like a spherical cavity in which particles are bound by a deep potential and don't escape. So the nuclei are moving like that and don't escape. In that case, it would be a three dimensional delta function that vanishes at the origin. I presume there are many examples. Any potential that sort of begins like a finite square well that is sufficiently deep will start to look like a delta function after awhile. AUDIENCE: [INAUDIBLE] PROFESSOR: Well, it's again neither-- yeah, I guess so. But it depends how big is this psi. So yeah, probably you're right. This looks a little more like an analog of a hard wall. But if a hard wall is very narrow and very deep, it looks like a delta function. So it's idealizations for sure. But I'm sure we could get a better example. And I'll try to find one. Now, the next thing we want to do is give you intuition for this incredible result that there's no degeneratives in one dimensional potentials. That is not to say that the proof is not good enough. It is just to say that we can illustrate that without going into a mathematical proof that is more complicated. So how do we do that? We'll consider the following case, a simple case, an example of a potential of this form. V of x-- this is x. And here is V of x. And we will try to find a solution with some energy that is like that, an energy that is right there below the barrier. So this would be a bound state. Why? Because solutions here are exponentials that decay, exponentials that decay. And here, the wave function would be oscillating presumably. So the wave functions go to 0 and infinity. You could get a bound state. So let's see how we get a bound state. Now, the argument I'm going to follow is just an elaboration of something you can read in Shankar. And it's a nice and simple argument. So we want to understand why we would get here no degeneracies. Or even more-- in fact not just no degeneracies, but the spectrum is quantized. That is, you find one energy, and then another energy maybe, and another energy. So how do we see that? Well, you look at the way you can write solutions and count the parameters of the solutions and try to see how many conditions you have to satisfy. So here, the wave function would be a decay in exponential. A decay in exponential is of the form alpha e to the K, kappa, x. Because x here is negative. So this decays as x goes to minus infinity if kappa is positive. And that's how a solution looks. You need one coefficient here to determine this solutions. So I'll put a 1 here. Now, in here, the solution is oscillatory. So it's a sine plus cosine. So you need two coefficients. In here, the solution must again be decaying. And therefore, you just need one coefficient. Again, this time it would be a beta e to the minus Kx. The fact that this potential looks symmetric-- I'm not assuming it is. Yes. AUDIENCE: Won't one of the coefficients be unconstrained by normalization? Isn't one just the normalization factor? PROFESSOR: OK, how about normalization? Indeed, we have one, two, and two and one, so a total of four coefficients, four parameters. But indeed, suppose you wrote your whole solution. You could say, look, let me divide this solution by 3. That's an equivalent solution. I'm just checking if it solves the Schrodinger equation. That's all I have to check. I don't have to check normalization. Normalization, in fact, is sort of irrelevant here. You just need to know if a bound state exists. So indeed, even though you have these four parameters, given that you can multiply the solution by a constant, there's just three constants to fix. Four parameters via normalization or the multiplication by any constant-- just three constants to fix. But this potential is nice enough that psi and psi prime must be continuous. So you get two conditions here and two conditions here. So four conditions-- continuity of psi and psi prime, continuity of psi and psi prime, four conditions. So what did we get? We got in trouble. We've shown that this is unsolvable in general. Because there are more conditions than parameters. Now, this equation could to be a little peculiar. Maybe this is not completely general. But you seem to have more conditions than parameters. But here comes the catch. The solution really is determined. Like kappa-- do you know kappa? Well, you know kappa if you know the energy of the solution. Kappa is determined by the energy of the solution. So some parameters in the solution depend on the energy. So the way we have to think of this is that in fact three constants to fix, but four conditions. So we really need four constants to fix. And the fourth constant is the energy. So the energy is the fourth constant to fix. And with four conditions, these three constants that we had there and the energy, they can just be fixed. So the solution should fix the energy and should fix this coefficient. So the solution exists for some energy, or possibly some values of the energies, but not all values of the energy. So this shows, or at least very clearly illustrates, that you are going to find sets of energies for which you have solutions depending on how the equations look, and one solution each time. So you get what is called a discrete non-degenerate spectrum. Now, there are more cases to discuss, the case in which you have just the step, or the case in which you have three bound states. And I will not do them but state the results. Again, all that I don't do explicitly can be found in the notes. So you would look at them later. And so here is the second case, a potential like this and an energy like that, energy level. And what you get here is that in fact, doing the counting and analyzing the boundary conditions, you should do it by yourselves. But you will see the answers in the notes. You get here continuous spectrum, non-degenerate. So you will get a solution for every value of the energy-- that's to mean, continuous spectrum-- and one solution each time. Finally, this case-- if you have an energy like this, e, you get continuous spectrum and doubly degenerate, so two solutions. Now, after this, there's one more result that qualifies as a theorem. And it's hard to prove rigorously. I will not attempt to prove it here nor even in the notes. It's hard enough. So this theorem has to do with nodes. Theorem-- so if you have the discrete bound state spectrum of a one dimensional potential, and you list the energies E1 less than E2 less than E3 like that, E1 is the ground state energy. Remember, this spectrum is discrete. So this is less than E2, less than E3, and it goes on like that, the ground state energy. Then, you have associated wave functions, energy eigenstates psi 1 of x, psi 2 of x, psi 3 of x. Here is the theorem. The theorem tells you something about the vanishing of the wave function. It says that psi 1 has no nodes. Psi 2 has one node. Psi 3 has two nodes. And so it goes so that psi n has n minus one node. So psi n is greater than-- well, it's correct. Any n greater or equal to 1 has n minus 1 nodes. Now, there are several ways people show this. Mathematicians show it in a rather delicate analysis. Physicists have an argument as well for this, which is based on approximating any potential by infinite square wells to begin with. So suppose you have a potential like that. Well, think of it as first being a little potential like that, an infinite square well. And you start making the window of the square well bigger. The argument-- it's a neat argument. Maybe you can discuss it in recitation. I would suggest that. It's a good argument. But it's not rigorous. But still, one can do something like that. You make it grow. And what you know is that the infinite square well, the first wave function has no node. And as you change the screen to make the potential really what it's supposed to be and not just that of a square well, the wave function cannot gain a node. On the other hand, what you will show in the homework is something that is a partial result which says that the solution with n plus 1 has at least one more node than the solution with n. So it's part of what you can show. And it doesn't take too much effort. And you can prove it rigorously. So we will assign that eventually for a homework to do. In the homework that you have for the first homework, you also have a problem with delta functions. And I suggest that you read the notes that will be posted today. Because there's an example there with delta functions. If you study that example, you'll find the problem quite easy to solve. You may have solved already. Some of you are very eager to get going with the homework. So it's something you can study first, and then make your life a little easier. So what we're going to do now for the rest of the lecture is consider the variational problem, which is something you probably haven't seen before, the variational problem. This problem has to do with calculus of variations. Now, calculus of variations is something considered fairly advanced. And therefore, as you will see, we will avoid some of the major difficulties of calculus of variations in our discussion of the variational principle. But I wanted to mention a little story about this problem. So this calculus of variations is a more complicated version of maxima and minima in which in maxima and minima of functions you look at the function. And if you could plot it, you could say, here's a maximum, here's a minimum. If you want to figure out where they are, you know. You take a derivative, set it equal to 0, you find the maxima and minima. So the typical calculus problem is one in which you have a function, and you want the maxima and minima. The variational problem is a problem in which you want to maximize or minimize something. But what you don't know is not where the maximum or minimum occurs, but which kind of function will give you this maximum or minimum. So your unknown is not a point where there's a maximum or a minimum but a function where there is a maximum and a minimum. So it's slightly more complicated. So this is the calculus of variations. And people wonder when did it start. It actually seems to have first been discussed by Newton. And it's quite an interesting story. Newton was trying to understand apparently the following problem. If you would have a cross sectional area like this, he asked the question, how should you make a solid out of this by tapering it and ending with this, tapering it, in such a way that as it moves in a viscous fluid, the resistance is the minimum possible-- very complicated problem. And as you can imagine, this is a complicated problem because you're trying to find a shape-- not just a maximum or a minimum of a function but what shape maximizes or minimizes this. So apparently, he solved the problem and wrote it in Principia but didn't explain his solution. And people for years were trying to figure it out. And nobody could figure out how he did it. Then, the story goes that this mathematician Johann Bernoulli in 1696 came up with a challenge to all mathematicians. At that time, people would announce a problem and challenge to see who's smart, who can solve this problem. So Johann Bernoulli in around 1696 poses a problem of, you're given two points in the plane, in the vertical plane like this blackboard, point this, A, and point B in here. You must design the curve of shortest time for fall, so some curve here. If you put an object and let it fall, it will get the fastest to that point, so maybe something that looks like this. It's a complicated curve, or at least not all that simple. And he asked all the people to solve it, gave them one year to solve it. So who was around at that time? Well, one person that got the letter was Leibniz. He got it on the 9th of June of that year, 1696. And he answered it, sent an email back, by the 16th of June with a letter with a complete solution. It's a funny thing that actually apparently Newton was very busy and didn't receive this letter. Or something happened, and he got it one day, and he actually solved the problem in one night. It took him one full night to solve it. Now, you say, well, how brilliant. And true, but given that he had solved this problem, he was criticized as being really slow and-- how come you took 12 hours to solve this problem? So it's quite amazing. There's a lot of Bernoullis. And apparently, this question by Jacob Bernoulli, the main purpose of this question was to demonstrate to everybody that his older brother, Jacob Bernoulli, who had invented the Bernoulli numbers, was actually an incompetent person that could not solve this problem. So that was apparently what he wanted to do. It's a rather famous family. But they obviously didn't get along. But apparently, Jacob did manage to solve the problem. So Jacob Bernoulli, Leibniz, and Newton all solved the problem. Johann Bernoulli, the one that started this problem-- and I think it's maybe with a double N, I'm sorry-- his son is Daniel Bernoulli. And engineers know him, because that's the Bernoulli of the Bernoulli fluid dynamics stuff. So the problem is not all that easy. And calculus of variations determines this shape. So the calculus of variation applied to quantum mechanics asks, here, this function is determined by the principle that it minimizes time. So you have the Schrodinger Equation. And you could ask, you have all these Eigenfunctions. What do they minimize? Is there something they're minimize? And the answer is yes. And this is what you'll understand in the next few minutes. So what is the problem we want to solve? We want to solve the problem h psi equal e psi. So some Hamiltonian. Now my notation will be such that it applies to three dimensions as well. So I'll put just arrows on top of it, and you would have to write the proper Hamiltonian. It will not be necessary for you to know the Hamiltonian. So I'll put psi of x here, meaning that this is equally valid for more than one dimension. Now we want to find solutions of this equation. And you can say, what do they maximize or minimize? Well we won't get to it until 15 minutes. First let's try something simpler. How about, can we learn something about the ground state energy of the system? So let's try to think about the ground state energy. State energy. Now consider ground state energy and we'll consider an arbitrary-- arbitrary is the most important word here-- psi of x that is normalized. Is normalized. So integral the x of psi squared is equal to 1. And here comes the claim. The first claim that we can make. You see, this wave function doesn't solve the Schrodinger equation. That's what we mean by arbitrary. It's just any function of space that is normalizable. Doesn't solve the Schrodinger equation. Never the less, you are commanded to compute the following quantity. This quantity is also by definition what we call the expectation value of the Hamiltonian in the state psi. I love the fact, the remarkable fact that we're going to show now, is that this thing provides an upper bound for the ground state energy for all psi. So let me try to make sure we understand what's happening here. Here it says you don't know the ground state energy but you're going to learn something about it. Something that's interesting is if you know that it has an upper bound, so the ground state energy is definitely not higher than this one, so you learn something. Would be ideal if you had also lower bound so you knew it's in this range. But an upper bound is a nice thing to have. And the claim here is that each time you try an arbitrary function, you put anything here, you ever write, you've got an upper bound. So how is this used? You try arbitrary functions that you think look possibly like the wave function of a bound state. And you get numbers and you already know that the ground state energy smaller than some number. So it's a rather nice way of getting some information about the ground state energy. So this psi effects is called a trial wave function. Is a trial wave function. So is the statement of this theorem clear? Not the proof, but the statement. Do we have questions? Yes. AUDIENCE: Is there any statement about how using the wave function will give us how accurate an estimate [INAUDIBLE] PROFESSOR: No. We're going to become good and figure out some nice way of choosing wave functions, but no. Once you tried you got some information. And you may not know so easily whether you could go much lower. You can try a little, but there's no clear way to know. This is just some partial information. OK. So let me first prove this. We'll prove it and then explain a little more what it all means. So the proof. Now it's a proof under quotation marks. I will make a few assumptions. Basically, that I don't have a continuous spectrum. Now that assumption is done for me to write a simpler proof, not because the result doesn't hold. So the proof is good, but I will just consider for notational purposes no continuous spectrum. So we'll have a ground state energy which is e1 that is maybe less than or equal to e2 less than or equal e3 like that. So you even may consider the [? genorisies. ?] And we have h psi n is equal to en psi n. So what do we have? We have a trial wave function. So your trial wave function since it's an arbitrary function of x should be expandable by completeness as a serious or a superb position of the energy eigenstates. Let me clarify this point. This is a trial wave function. Doesn't solve the Schrodinger equation. So this Doesn't solve this energy eigenstate equation. So in fact, it doesn't solve it because this is a superb position of many in here. So that's consistent with this, and the fact that this wave function as given in here just can be represented using the energy eigenstates. But being a superb position, it's not an energy eigenstate which is true because a trial wave function is something that you invent out of your head. It's not a solution. If you had a solution, you wouldn't need this. So you don't have a solution. You invent the trial wave function, and you have this. A couple of things. The psi squared integral being one. You can do that integral, that condition, is the sum of their bn-2nd is equal to 1. This I use the orthonormality. You can elevate this. It's sort of the kinds of things we're doing last time. Please make sure you know how to do that. Then there's the other computation that we also have sketched last time, which is that the integral of psi star h psi which is what we want to compute, h hat psi is actually bn-2nd en. So that was something we're doing towards the end of last lecture. And this computation takes a few lines, but it was there. It's in the notes. And now comes the kind of thing we want to say. Look at this sum. It has b1, e1, b2, e2, b3, e3, but e2, e3, e4, all those are bigger, or at most, equal to e1. So if I did, here, the following bad joke of substituting en for e1, which is not the same, if I put here bn squared n equals 1 to infinity. I put here e1, well this is bigger than that because e2 is possibly bigger than e1, e3 is bigger than e1. But it may be equal. But at this moment, e1 can go out of the sum. So this is e1 times this sum which is 1 because bn is equal to 1. And e1 is the ground state energy by definition. So the ground state energy is less than this which is the expectation value of the Hamiltonian. Pretty simple, in fact, the proof is really a little too simple. Where do we go from now? Well let's make a more general statement of the variational principal. Again, sometimes it's not all that convenient to have normalized wave functions. So recall that if psi of x is not normalized, psi of x over the square root of integral of psi-2nd dx is. Therefore if you hand me an arbitrary psi of x that is really arbitrary. You don't even bother to normalize it. Then when I plug here in this formula it is supposed to be normalized. So I plug the second expression there. So therefore I get that egs less than or equal to the integral of psi star h psi the x over integral of psi star psi the x. This is actually nicer in one way because you don't have to work with normalized wave functions. And that result must be true still. Yes? AUDIENCE: [INAUDIBLE] PROFESSOR: Sure. It cannot be completely arbitrary, the function should we normalizable. Doesn't have to be normalized but normalizable. So here you got. And here let me introduce just a little name. f of psi. f of psi is what is called a functional. f of is a functional. What is a functional? A functional is a machine or an expression whose input is a function and whose output is a number. So here f of psi is a functional. And maybe I should use brackets. Many times people with brackets denote that, watch out, this is not a function, it's a functional. And here it is. No dash there. You give me a psi of x, which is a function, and then this is a number because you've done the integrals. So that is like the Brachistochrone problem, that a funny name for it. Here it is. There is a function now which is the time that it takes to go here. You give me a path, a function, and I can calculate the time it will take the mass to get here. So this was the issue of finding a critical point of a functional. So actually we start to see, it seems that the ground state energy is the minimum of this functional. And what this interesting as well is that when you get the minimum you will have gotten a ground state wave function. So the ground state wave function actually is the thing that minimizes this functional and gives you some value, the ground state energy. A little more in a couple of minutes. Let's do an example. How do we use this so now if you think about this carefully it's kind of dizzying because what is sell functional, really? It's, in some sense, a function in an infinite dimensional space because a function itself is specified by infinitely many numbers that you can change. So how many dimensions you have is how many ways you can move your hands, but they are linearly independent. But if you have a function, and you can change it, you can change it here, you can change it there, or you can change it there, and those are all orthogonal directions. You're finding a critical point. When you find the critical point, you should imagine that you're plotting a function that is not one dimensional function or two dimensional function, but it's infinitely dimensional function. Direction one, direction two, infinitely many directions. And suddenly in this infinitely many directions you find the critical point. It's really incredible that one can do these things. So you're at that critical point, and you can deform the energy eigenstate by making it a little fatter here or thinner here or up there. And those are the infinite directions in any direction that you will the energy goes up because you're at the global minimum it's pretty amazing. Something that you will prove in the homework is that actually it's even more. Every single eigenstate is a critical point of this functional. So you've got the lowest energy state and in that infinite dimensional space in every direction that you move you go up. The first excited state is another critical point. But it will not be an absolute minimum. It will be a [? saddle ?] an infinite dimensional [? saddle ?] which are infinitely many directions in which you go up. There's one direction in which you go down because you could flow towards the ground state. So the first excited state is the [? saddle ?] but these are all stationary points of this functional. So we'll conclude by doing this example. Sorry, what is the question? AUDIENCE: [INAUDIBLE] PROFESSOR: I didn't assume it's non-degenerate. So if you have two things that have the same ground state, the functional will have, in fact, the degeneracy there will be two solutions that have the same energy. And any linear combination of them will have the same energy. The proof that I did here doesn't assume non-degeneracy, it's even true with degenerate things. So the example is an example for illustration, not for solving something that you can't do otherwise. So it's a delta function potential. v of x is minus alpha delta of x with alpha positive. And the ground state energy is well known. It's minus m alpha squared over 2h-2nd. You've solved this problem many times in 804. So trial wave function. Well you know how it should look, but let's assume you don't. And you say, it's some sort of [INAUDIBLE]. So trial. It would be psi of x equals e to the minus x squared. While this would do, you're going to work hard and you're not going to reap all the benefits of this calculation. So what you should do at this moment is put a constant here. Minus beta squared x squared and I'll put the minus one half. This is our trial wave function. You see, by this, you're going to get an expression. You calculate this number, and you're going to get the function of beta. Beta is not going to disappear. And therefore, you're going to know that the ground state energy is less than this function of beta. And then you can adjust beta to get the best bound. So beta is put as a parameter to begin with, and we hope to use it. So note that integral of psi squared dx in this case is equal to square root of pi over beta. So we have to calculate this whole functional. So this integral of psi star-- well I don't have to bother with psi star because it's real. h psi over psi psi, and what do we get? Well the denominator is easy. So we beta over square root of pi, and let me write the whole thing here. dx the psi would have e to the minus one half beta squared x squared minus h squared over 2m d-2nd dx2nd minus alpha delta of x. And another wave function, e to the minus one half beta squared x-2nd. OK. So you have to evaluate that. And that's the part that is not so much fun. For any integral that you have in 805, we believe that you can use Mathematica or Maple or MATLAB or whatever and do it. The only reason not to use any of these things is if you think you could not do the integral. But once you realize, oh, this is an integral, I know how to do, don't waste time. Use any of those programs. Now, this part of the integral is kind of easy because the delta function just picks the value of 0. So this part of the integral gives you minus beta alpha over square root of pi. The other part of the integral, however, is a little more complicated. Believe it or not, it makes a big difference whether you take the two derivatives of this function or you integrate by parts. If you integrate by parts, you save a lot of work. So let me integrate by parts. This becomes, plus beta over square root of h, h-2nd over 2m, integral dx of ddx of e to the minus one half beta squared x-2nd squared. So you integrate by parts one of the ddx and then you have the other ddx, so it's the thing squared. And that's an easier integral to do. We don't want to bother with this, but this whole thing then becomes minus beta over square root of pi, alpha plus beta squared h squared over 4m. That's the whole answer And that's the valuation of this whole thing. So look what you get. You got a function of beta indeed. So how does that function of beta look? It's 0 for beta equals 0 is 0 for some other possible beta, it's going to look like this. So there's a point at which this function is going to have a minimum, and we should choose that point to get the best upper bound on the function. Our claim is that following from the variational theorem that we've proven is that the e ground state is less than or equal than the minimum value over beta of this beta squared h squared over 4m minus beta square root of pie alpha. So you minimize over beta, and yet still the ground state energy must be a little smaller than that. Well what do you get? You do this, this minimization gives beta equal 2m alpha over h squared square root of pi. It's a little messy but not terrible. And you substitute, and you get that e ground state is less than or equal than m alpha squared over pi h squared. And to better write it as 2 over pi times minus m alpha squared over 2h-2nd which is the true ground state energy. So let's just make sure we understand what has happened. Here is the energy. Here is zero. Energy as just a vertical plot here is zero. The true ground state energy of this problem is this one, let's call it egs, is negative. And we go 2 over pi times that. 2 over pi is about 0.64. So the bound says that here is 0.64 egs. So that's what the bound told you. The bound state energy must be lower than this quantity. We got close. Not impressively close, but the work functional was not all that bad either. Question? AUDIENCE: [INAUDIBLE] always going to be a constant times incorrect energy value or is it just the closest approximation? PROFESSOR: Is it going to be what? AUDIENCE: Is it always going to be a constant times the correct energy value or is it just-- PROFESSOR: Well it typically is like that because of dimensional units. You're looking for a constant because you're not looking for the function. So you will get a number times a correct value, yes, indeed. That's an illustration of the problem of wave functions. You know the variational principle tells you things about the ground state but allows you to find the first excited state as well if the potential is symmetric, will allow you to prove that any attractive potential has a bound state. And you will prove in the homework that stationary points of these things are the eigenfunctions. See you next Monday.
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https://ocw.mit.edu/courses/3-054-cellular-solids-structure-properties-and-applications-spring-2015/3.054-spring-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 view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. LORNA GIBSON: OK, so we should probably start. So last time we finished up talking about energy absorption in foamy cellular materials. And today I wanted to start a new topic. We're going to talk about sandwich panels. So sandwich panels have two stiff, strong skins that are separated by some sort of lightweight core. So the skins are typically, say, a metal like aluminum, or some sort of fiber composite. And the core is usually some sort of cellular material. Sometimes it's an engineering honeycomb. Sometimes it's a foam. Sometimes it's balsa wood. And the idea is that what you're doing with the core is you're using a light material to separate the faces, and if you think about an I-beam-- so if you remember when we talk about bending and we talk about I-beams, the whole idea is that in bending, you want to increase the moment of inertia. So you want to make as much material as far away from the middle of the beam as possible to increase the moment of inertia. So if you think about an I-beam, you put the flanges far apart with the web, and that increases the moment of inertia. And the sandwich panels and the sandwich beams essentially do the same thing, but they're using a lightweight core instead of a web. And so the idea is you use a lightweight core. It separates the faces. It increases the moment of inertia. But you don't add a whole lot of weight because you've got this lightweight core in the middle. So I brought some examples that I'll pass around and we can play with. So these are some examples up on the screen, and some of those I have down here. So for instance, the top-- turn my little gizmo on-- the top left here, this is a helicopter rotor blade, and that has a honeycomb core in it. This is an aircraft flooring panel that has a honeycomb core and has carbon fiber faces. So that's this thing here. I'll pass that around in a minute. This is a downhill ski. This has aluminum faces and a polyurethane foam core. And that's the ski here. And it's quite common in skis now to have these sandwich panels. This is a little piece of a small sailing boat. It had, I think, glass fiber faces and a balsa wood core. And I don't know if any of you sail, but MIT has new sailing boats. Do you sail? AUDIENCE: I do not much these days. LORNA GIBSON: OK, but those little tech dinghies that you see out in the river, those have sandwich panel holes to them. So those are little sandwich panels. This is an example from a building panel. This has a dry wall face and a plywood face and a foam core, and the idea with panels for buildings is that usually they use a foam core because the foam has some thermal insulation. So as well as sort of separating the faces and having a structural role, it has a role in thermally insulating the building. The foams are a little less efficient than using a honeycomb core. So for the same weight, you get a stiffer structure with a honeycomb core than a foam core. But if you want thermal insulation as well as a structural requirement, then the foam cores are good. And these are a couple examples of sandwiches in nature. This is the human skull. And your skull is a sandwich of two dense layers of the compact bone, and you can see there's a little thin layer of the trabecular bone in between. So your head is like a sandwich, your skull is like a sandwich. And I don't know if I'll get to it next time, but in the next couple of lectures, I'm going to talk a little bit about sandwich panels in nature, sandwich shells in nature. You see this all the time. And this is a bird wing, here. And so you can see there's got the dense bone on the top and the bottom, and it's got this kind of almost trust-like structure in the middle. And obviously birds want to reduce their weight because they want to fly, so reducing the weight's very important. And so this is one of the ways that birds reduce their weight, is by having a sandwich kind of structure. So I have a couple of things here. These are the two panels at the top there. This is the ski, and you can yank those around. I also have a few panels that people at MIT have made. And I have the pieces that they're made from. So you can see how effective the sandwich thing is. So this was made by a guy called Dirk Moore. He was a graduate student in ocean engineering. And it has aluminum faces and a little thin aluminum core. So you can see, if you try and bend that with your fingers, you really can't bend it any noticeable amount. And this panel here is roughly the same thickness of the face on that. And you can see how easy it is for me to bend that-- very easy. And this is the same kind of thing as the core. It's thicker than that core, but you can see how easy it is for me to bend this, too. So each of the pieces is not very stiff at all. But when you put them all together, it's very stiff. So that's really the beauty of this. You can have lightweight components, and by putting them together in the right way, they're quite stiff. So here's another example here. This is a panel that one of my students, Kevin Chang made. And this has actually already been broken a little bit, so it's not quite as stiff as it used to be. And you can kind of hear, it squeaks. But you can feel that and see how stiff that is. And this is the face panel here. And you can see, I can bend that quite easily with my hands. Doodle-doot. And then this is the core piece here. And again, this is very flexible. So it's really about putting all those pieces together. So you get this sandwich construction and you get that effect, OK? [? Oop-loo. ?] All right, so what we're going to do is first of all look at the stiffness of these panels, calculate their deflection. We're going to look at the minimum weight design of them. So we're going to look at how for, say, given materials in a given span, how do we minimize the weight of the beam for a given stiffness? And then we're going to look at the stresses in the sandwich beams. So there's going to be one set of stresses in the faces, and a different kind of stress distribution in the core. So we'll look at the stress distribution. And then we'll talk about failure modes, how these things can fail, and then how to figure out which failure mode is dominant, which one occurs at the lowest load. And then we'll look at optimizing the design, minimizing the design for a certain strength and stiffness. So we're not going to get all that way through everything today, but we'll kind of make a start on that. OK. So let me start. So the idea here is we have two stiff, strong skins, or faces, separated by a lightweight core. And the idea is that by separating the faces, you increase the moment of inertia with little increase in weight. So these are particularly good if you want to resist bending, or if you want to resist buckling. Because both of those involve the moment of inertia. And they work like an I-beam. So the faces of the sandwich are like the flanges of the I-beam, and the core is like the web. And the faces are typically made of either fiber reinforced composites or metals. So typically, something like aluminum, usually you're trying to reduce the weight if you use these things, so a lightweight metal like aluminum is sometimes used. And the cores are usually honeycombs, or foams, or balsa. And when they use balsa wood, what they do is-- I brought a piece of balsa here-- what they do is they would take a block like this and chop it into pieces around here. And then they would lay those pieces on a cloth mat. So typically the pieces are maybe two inches by two inches. They lay them on a cloth mat, and because they're not one monolithic piece, they can then shape that mat to curved shapes. So it doesn't have to be just a flat panel. They can curve it around a curved surface if they want. So we'll say the honeycombs are lighter than the foams for a given stiffness or strength. But the foams provide thermal insulation as well as a mechanical support. And the overall mechanical properties of the honeycomb depend on the properties of each of the two parts, of the faces and the core, and also the geometry of the whole thing-- how thick's the core, how thick's the face, how dense is the core? That kind of thing. And typically, the panel has to have some required stiffness or strength. And often what you want to do is minimize the weight for that required stiffness or strength. So often these panels are used in some sort of vehicle, like we talked about the sailboat, or like a helicopter, or like an airplane. They're also used in like refrigerated trucks-- they would have a foam core because they'd want the thermal insulation. So if you were going to use it in some sort of a vehicle, you want to reduce the mass of the vehicle and you want to have the lightest panel that you can. Yup? AUDIENCE: So if you saw the base material and you'd have the [INAUDIBLE] sandwich panel, that piece [INAUDIBLE] the sandwich panel with something [? solid ?] in the middle? LORNA GIBSON: Well-- AUDIENCE: So, as we're getting [INAUDIBLE] aluminum piece that's was as thick as a sandwich panel. LORNA GIBSON: Yeah. AUDIENCE: [INAUDIBLE]. LORNA GIBSON: Oh, well, if you have the solid aluminum piece that was as thick as the sandwich, it's going to be stiffer, but it's going to be a lot, lot heavier. So the stiffness per unit weight would not be as good. OK? So we're going to calculate the stiffness in just one minute. And then we're going to look at how we minimize the weight, OK? OK. So what I'm going to do is set this up as kind of a general thing. We're just going to look at sandwich beams rather than plates, just because it's simpler. But the plates, everything we say for the beams basically applies to the plates. The equation's just a little bit more complicated. So we're going to start with analyzing beams. And I'm just going to start with a beam, say, in three-point bending. So there's my faces there. Boop. And I've made it kind of more stumpy than it would be in real life, just because it makes it easier to draw it. And then if I look at it the other way on, it would look something like that. So say there's some load P here. Say the span of the beam is l. Say the load's in the middle, so each of the supports just sees a load of P/2. And then let me just define some geometrical parameters here. I'm going to say the width of the beam is b. And I'm going to say the face thicknesses are each t. So the thickness of each face is t. And the thickness of the core is c, OK? So that's just sort of definitions. And I'm going to say the face has a set of properties, the core has a set of properties, and then the solid from which the core is made has another set of properties. So the face properties that we're going to use are a density of the face. We'll call that row f. The modulus of the face, Ef, and some sort of strength of the face, let's imagine it's aluminum and it yields, that would be sigma y of the face. And then the core similarly is going to have a density, rho star c. It's going to have a modulus, E star c. And it's going to have some strength, I'm going to call sigma star c. And then the solid from which the core is made is going to have a density row s, a modulus Es, and some strength, sigma ys, OK? So the core is going to be some kind of cellular material, a honeycomb, or a foam, or balsa. And typically, the modulus of the core is going to be a lot less than the modulus of the face. So I'm just going to say here that the E star c is typically much greater than Ef. And we're going to use that later on. So we're going to derive some equations, for example, for an equivalent flexural rigidity for the section, an Ei equivalent. And that has several terms. But if we can say the core stiffness is much less than the face thickness, and also if we can say the core-- the stiffness is less and also the thickness of the core is much greater than the thickness of the face, a lot of the expressions we're going to use simplify. So we're going to make those assumptions. So let me just draw the shear diagram here. So V is shear, so that's the shear diagram. We have some load P/2 at the support. There's no other load applied until we get to here. Than the shear diagram goes down by P, so we're at minus P/2. Then there's no load here, so this just stays constant, and then we go back up to 0. And then let me just draw bending moment diagram. The bending moment diagram for this is just going to look like a triangle. Remember, if we integrate the shear diagram, we get the bending moment diagram. And that maximum moment there is going to Pl over 4. OK. So initially, I'm going to calculate the deflections. And I don't really need those diagrams for that, but then I'm going to calculate the stresses, and I'm going to need those diagrams for the stresses. So just kind of keep those in mind for now. So to calculate the deflections, sandwich panels are a little bit different from homogeneous beams. In a sandwich panel, the core is not very stiff compared to the faces. And we've got some shear stresses acting on the thing. And the shear stresses are largely carried by the core. So the core is actually going to shear, and there's going to be a significant deflection of the core and shear as well as the overall bending of the whole panel. So you have to count for that. So we're going to have a bending term and a shear term-- that's what those two terms are there. So we're going to say there's a bending deflection and a shear deflection. And that shearing deflection arises from the core being sheared and the fact that the core, say, Young's modulus or also the shear modulus, is quite a bit less than the face modulus. So if you think of the core as being much more compliant than the face, then the core is going to have some deflection from that shear stress. OK, so we're going to start out with this term here, the bending term. And if I just had a homogeneous beam in three-point bending, the central deflections-- so these are all the central deflections I'm calculating here-- with Pl cubed over it turns out to be 48 is the number, and divided by EI. And because we don't have a homogeneous beam here, I'm going to call that equivalent EI. And to make it a little bit more general, instead of putting 48, that number, I'm just going to put a constant B1. And that B1 constant is just going to depend on the loading geometry. So any time I have a concentrated load on a beam, the deflection's always Pl cubed over EI, and then the sum number in the denominator and that number just depends on the loading configuration. So for three-point bending, it's 48. For the flexion of a cantilever, B1 would be 3. So think of that as just a number that you can work out for the particular loading configuration. So here we'll say B1 is just a constant that depends on the loading configuration. And I'll say, for example, for three-point bending, B1 is 48. For a cantilever end deflection, then B1 would be 3. So it's just a number. So the next thing we have to figure out is what's the EI equivalent. So if this was just a homogeneous beam, and it was rectangular, E would just be E of the material and I would be the width B times the height H cubed divided by 12. So here we don't quite have that because we have two different materials. So here we have to use something called the parallel axis theorem, which I'm hoping you may have seen somewhere in calculus, maybe? But, yeah, somebody is nodding yes. OK, so what we do, what we want to do is get the equivalent EI-- I'm going to put it back up, don't panic-- of this thing here, right? So I want-- this is the neutral axis here, and I want the EI about that neutral axis there. So, OK, you happy? There. OK, so I've got a term for the core. OK, the core, that is the middle of the core, right? So for the core, it's just going to be E of the core times bc cubed over 12. Remember, for a rectangular section, it's bh cubed over 12 is the moment of inertia. And here our height for the core is just c, OK? And then if I took the moment of inertia for, say, one face about its own centroidal axis, I would get E of the face now times bt cubed over 12. So that's taking the moment of inertia of one face about the middle of the face. And I have two of those, right? Because I have two faces. And the parallel axis theorem tells you what the moment of inertia is going to be if you move it, not to the-- you don't use the centroid of the area, but you use some other parallel axis. And what that tells you to do is take the area that you're interested in-- so the area of the face is bt, and you multiply by the square of the distance between the two axes that you're interested in. Oop, yeah. Let me change my little brackets. Boop. So, oop-a-doop-a-doop. Maybe I'll stick this, make a little sketch over here again. OK, all right. So this term here, Ef bt cubed over 12, that would be the moment of inertia of this piece here, about the axis that goes through the middle of that, right? Its own centroidal axis. But what I want to do is I want to know what the moment of inertia of this piece is about this axis here. This is the neutral axis. So let's call this the centroidal axis. And the parallel axis theorem tells me what I do is I take the area of this little thing here, so that's the b times t, and I multiply by the square of the distance between those two axes. So the distance between those axes is just c plus t over 2, and I square it. And then I multiply that whole thing by 2 because I've got two faces. Are we good? AUDIENCE: [INAUDIBLE]. LORNA GIBSON: Yeah? AUDIENCE: The center [INAUDIBLE] and the [INAUDIBLE], are those Ed's or Ef's? LORNA GIBSON: These are Ef's because this is the face now, right? So this term here is for the core. So here the core is E star c. And these two Ef's are for the face up there, OK? Because you have to account for the modulus of the material of the bit that you're getting the moment of inertia for. Are we good? OK. So now I'm just going to simplify these guys a little bit. Doodle-doodle-doodle-do-doot. OK? So I've just multiplied the twos, and maybe I'll just write down here this is the parallel axis theorem. Doot-doot-doot. Yes, sorry? AUDIENCE: So for the term that comes from the parallel axis theorem, why do we only consider Ef and not [INAUDIBLE]. LORNA GIBSON: Because I'm taking-- what I'm looking at-- so the very first term, this guy, here-- AUDIENCE: Yeah, [INAUDIBLE]. LORNA GIBSON: Accounts for this, right? And these two terms both account for the face. AUDIENCE: Oh, OK, so the face acting-- LORNA GIBSON: Yeah, about this axis. So the parallel axis theorem says you take the moment of inertia of your area about its own centroidal axis, and then you add this term here. But it's really referring to that face, OK? Let me scoot that down and then scoot over here. And this is where we get to say the modulus of the face is much greater than the modulus of the core. And also, typically c, the core thickness, is much greater than t, the face thickness. So if that's true, then it turns out this term is small compared to that one. And also this term is small compared to this one. And also this term, instead of having c plus t squared, if c is big compared to t, then I can just call it c squared, OK? So you can see here, if Ec is small, then this is going to be small compared to these. If t is small, then this guy is going to be small. So even though it looks ugly, many times we can make this simpler approximation. OK, so we can just approximate it as Ef times btc squared over 2. So then this bending term here, we've got everything we need now to get that bit there. So the next bit we want to get is the shearing deflection. So what's the shearing deflection equal to? So say we just thought about the core, and all we're interested in here is what's the deflection of the core and shear? And so say that's P/2, that's P/2, that's l/2. We'll say that's-- oops. That's our shearing deflection there. We can say the shear stress in the core is going to equal the shear modulus times the shear strain, so we can say P over the area of the core is going to be proportional to the Young's modulus times delta s over l. And let's not worry about the constant just yet. So delta s is going to be proportional to-- well, let me [? make it ?] proportional at this point. Delta s is going to equal Pl divided by some other constant that I'm going to call B2, and divided by the shear modulus of the core, and essentially the area of the core. And here B2 is another constant. So again, B2 just depends on the loading configuration. Yeah, this is a little bit of an approximation here, but I'm just going to leave it at that. OK, so then we have these two terms and we just add them up to get the final thing. Start another board. OK. So that would give us an equation for the deflection. And one thing to note here is that this shear modulus of the core, if the core is a foam, then we have an equation for that. We also could use an equation if it's a honeycomb. But I'm just going to write for foam cores. Whoops. This is for-- that will be for open-cell foam cores. Oops, don't want to-- and get rid of that. We won't update just now, thank you. OK, so the next thing I want to think about is how we would minimize the weight for a given stiffness. So say if we're given a stiffness, we're given P over delta, so I could take out the two P's here. If I divide it through by P, delta over P would be the compliance, P over delta would be the stiffness. So imagine that you're given the face and core materials, and you're told how long the span has to be, you're told how wide the beam is going to be, and you're told the loading configuration. So you know if it's three-point bending, or four-point bending, or a cantilever-- whatever it is. And you might be asked to find the core thickness, the face thickness, and the core density that would minimize the weight. So I have a little schematic here. I don't know if you're going be able to read it. So I'm going to walk through it and then I'll write things on the board. Whoops, hit the wrong button. OK, so we start with the weight equation here. The weight's obviously the sum of the weight of the faces, the weight of the core, so those two terms there. So I'll write that down in a minute. And then we have the stiffness constraint here. So this equation here is just this equation that I have down here on the board, OK? Then what you do is you solve that stiffness constraint for the density of the core. So this equation here just solves-- we're solving this equation here in terms of the density, and we get the density by substituting in this equation here for the shear modulus of the core. So you substitute that there. It's kind of a messy thing, but you solve that in terms of the density. Then you put that version of the density here in terms of this weight equation up here. So then you've eliminated the density out of the weight equation, now you've just got it in terms of the other variables. And then you take the partial derivative of the weight with respect to the core thickness c, set that equal to 0, and you take the partial derivative of the weight with respect to the face thickness, t, and you set that equal to 0. And that then gives you two equations and two unknowns. You've got the core thickness and the face thickness are the two unknowns. And you've got the two equations, so then you solve those. So the value you get for the core thickness is then the optimum, so it's going to be some function of the stiffness, the material properties you started with in the beam geometry. And similarly, you get some equation for the optimum face thickness, t. And again, it's a function of the stiffness and the material properties in the beam geometry. Then you take those two values for c and t, those two optimum values, and plug it back into this equation here, and get the optimum value of the core density. And so what you end up are three equations for the optimal values of the core thickness, the face thickness, and the core density in terms of the required stiffness, the material properties, and then the loading geometry. So I'm going to write down some more notes, because I'll put this on the Stellar site. But it's hard to read just here. So let me write it down and I'll also write out the equations so that you have the equations for calculating those optimum values. So before I do that, though, one of the interesting things though is if you figure out the optimal values of the core thickness and the face thickness and the core density, and you substitute it back into the weight, and you calculate this is the weight of the face relative to the weight of the core, no matter what the geometry is, and what the loading configuration is, the weight of the face is always a quarter of the weight of the core. So the ratio of how much material is in the core and the face is constant, regardless of the core-- of the loading configuration. And this is the bending deflection relative to the total deflection. It's always 1/3. And the shearing deflection relative to the total deflection is always 2/3. So regardless of how you set things up, the ratio of what weight the face is relative to the core and the amount of shearing and bending deflections is always a constant at the optimum. OK, so let's say we're given the face and the core materials. So that means we're given their material property, too. And say we're given the beam length and width and the loading configuration. So that means we're given those constants, B1 and B2. If I told you it was three-point bending, you would know what B1 and B2 are. So then what you need to do is find the core thickness, c, the face thickness, t, and the core density, rho c, to minimize the weight of the beam. So there's two faces, so the weight of the face is 2 rho f g times btl. And then the weight of the core is rho c g times bcl. So I'm going to write down the steps and then I'll write down the solution. So you solve. So you put this equation for the shear modulus of the core into here, and then you rearrange this equation in terms of the density of the core here. So you have an equation for the core density in terms of that stiffness, and then you solve the partial derivatives of the weight equation with respect to the core thickness, c, and put that equal to 0. And then the partial of dw [? over ?] dt and set that to 0. And if you do that, you can then solve for the optimal values of the face and core thicknesses. Yes? AUDIENCE: [INAUDIBLE] for weight, what is g? LORNA GIBSON: Gravity. AUDIENCE: OK. LORNA GIBSON: Just density is mass, mass times gravity-- weight. That's all it is. And then you've got a version of this that's in terms of the core density. You can substitute those values of the optimum face and core thicknesses into that equation and get the optimum core density. And then in the final equations, you get, when you do all that, and I'm going to make them all dimensionless, so this is the core thickness normalized by the span of the beam is equal to this thing, here. So you can see each of these parameters here, the design parameters that we're calculating the optimum of. I've grouped the constants B1 and B2 together that describe the loading configuration so you'd be given those. C2 is this constant-- oop, which I just rubbed off-- that relates the shear modulus of the foam core. So you'd be given that. These are the material properties of the-- you know, say, it's a polyurethane foam core, this would be the density of the polyurethane. Say it's aluminum faces, that would be the density of the aluminum. so you'd be given that. You'd be given the stiffnesses of the two materials, the solid from which the core is made and the face material. And then this is the stiffness here that you're given, just divided by the width of the beam, B. So the stiffness, you'd be given the width B. So you're given all those things, then you could calculate what that optimum design would be. So the next slide here just shows some experiments. And these were done on sandwiches with aluminum faces and a rigid polyurethane foam core. And here we knew what the relationship was for the shear modulus. We measured that. And what we did here was we designed the beams to all have the same stiffness, and they all had the same span in the width, B, then we kept one parameter at the optimum value and we varied the other ones. So here, on this beam, this set of beams here, the density was at the optimum. And we varied the core thickness, and we varied the face thickness, and the solid line was our model or our sort of optimization. And the little X's were the experiments. So you can see there's pretty good agreement there. Then the second set here, we kept the face thickness at the optimum value and we varied the core thickness, we varied the core density. So the same thing, the solid line is the sort of theory and the X's are the experiments. And here we had the core thickness of the optimum value, and we varied the face thickness and the core density. So you can kind of see how you can see this here. And over here, just because I forgot to say it, this is the stiffness per unit weight, over here, OK? So these are the optimum designs here, all right? So there was pretty good agreement between these calculations and what we measured on some beams. Do I need to write anything down? Do you think you've got that? Yeah? AUDIENCE: I was just going to ask, for the optimum design column that you have there, do those numbers like fall out of these equations if you do the math? LORNA GIBSON: They do, yeah. I mean, it's-- yeah, exactly. So if you remember the equation we had for the weight, so the weight is equal to 2 rho f gbtl plus the density of the core, bcl, so if you plug these things into there, then-- so this is the way to the face, that's the way to the core, then it drops out to be a quarter. So it's kind of magical. I mean, you have this big, long, complicated gory thing, and then, poof, everything disappears except a factor of 1/4. And the same for the bending deflection. So we had those two terms, so there was the bending and the shear. If you just calculate each of those terms and take the ratio of 1 over the total, or the one over the other, everything drops out except that number. So that's why I pointed it out, because it seemed kind of amazing that everything would drop out except for that one thing. OK, so then the next thing-- so that's the stiffness in optimizing the stiffness. Are we happy-ish? Yeah? OK. So the next thing-- oh, well, let's see. I don't think I need to write any. I think if you have that graph, I don't really need to write much down. So the next thing then is the strength of the sandwich beams. So let me get rid of that. You guys OK? Yeah? AUDIENCE: Yeah. LORNA GIBSON: Yeah, but you're shaking your head like this is very, very helpful for me. AUDIENCE: [INAUDIBLE]. LORNA GIBSON: Oh OK, that's OK. AUDIENCE: [INAUDIBLE]. LORNA GIBSON: That's OK, you can do that. I don't mind. But as long as you don't have questions for me. OK, and so the first step in trying to figure out about this strength is we need to figure out the stresses in the beams. So we need to find out about the stresses. And we're going to have normal stresses and we're going to have shear stresses. So I'm going to do the normal stresses first and then we'll do the shear stresses. So you do this in a way that's just analogous to how you figure out the stresses in a homogeneous beam. So we'll say the stresses in the face-- normally it would be My over I. M is the moment, y is the distance from the neutral axis, I is the moment of inertia. So this time, instead of having a moment of inertia, we have this equivalent moment of inertia. And we multiply by E of the face. So you can think of this as being the strain essentially. And then you multiply by E of the face to get the stress. The maximum distance from the neutral axis, we can call c/2. So that's y. Then EI equivalent we had Ef btc squared over 2. And then I have a term of Ef here. c squared. So one of the c's goes, the 2's go, the Ef's go. Then you just get that the normal stress in the face is the moment at that section divided by the width, b, the face thickness, t, the core thickness, c. And I can do the same kind of thing for the stress in the core, except now I multiply by the core modulus. So if I go through the same kind of thing, it's the same factor of M over btc, but now I multiply times E of core over E of the face. And since E of the core is a lot smaller than E of the face, typically these normal stresses in the core are much smaller than the normal stresses in the face. So the faces carry almost all of the normal stresses. And if you look at an I-beam, the flanges of the I-beam carry almost the normal stresses. So I want to do one more thing here. I want to relate the moment to some concentrated load. So let's say we have a beam with a concentrated load, P. So for example, something in three-point bending, typically we're interested in the maximum stresses, so we want the maximum moment. So M max is going to be P times l over some number. And this B3 is another constant that depends on the loading configuration. So if it was three-point bending, B3 would be 4. If it was a cantilever, B3 would be 1. So if I put those things together, the normal stress in the face is Pl B3 divided by btc. OK, so that's the normal stresses. And then the next thing is the shear stresses, and the shear stresses are going to be carried largely by the core. And if you do all the exact calculations, they vary parabolically through the core. But if we make those same approximations that the face is stiff compared to the core, and that the face is thin compared to the core, then you can say that the shear stress is just constant through the core. So we'll say the shear stresses vary parabolically through the core. But if the face is much stiffer than the core and the core is much thicker than the face, then you can say that the shear stress in the core is just equal to the sheer force over the area of the core, bc. So here, V is the shear force of the cross-section you're interested in. And bc is just the area of the core. And we could say the maximum shear force is just going to be V over-- actually, let's make it P, P over yet another constant. And B4 also depends on the loading configuration. So if I was giving you a problem, I would give you all these B1, B2, B3, B4's and everything. So the maximum shear stress in the core is in just the applied load, P, divided by this B4 and divided by the area of the core. OK, so this next figure up here just shows those stress distributions. So here's a piece of the cross-section here. So there's the face thickness and the core thickness. You can think of that as a piece along the length, if you want. This is the normal stress distribution, here. So this is all really from saying plane sections remain plane. These are the stresses, the normal stresses in the core. And you can see they're a lot smaller in this schematic than the ones in the face. And then this is the parabolic stress in the core. And similarly, there'd be a different parabola in the face. And these are the approximations. Typically these approximations are made so the normal stress in the face is just taken as a constant. The normal stress in the core is often neglected. And here the shear stress in the core is just a constant here. So the two things you need to worry about are the normal stress for the face and the shear stress for the core. Are we good? We're good? Yeah, good-ish. OK, so if we want to talk about the strength of the beam, we now have to talk about different failure modes. And the next slide just shows some schematics of the failure modes. So there's different ways the beam can fail. Say it's in three-point bending just for the sake of convenience. One way it can fail is, say it had aluminum faces. This face here would be in tension, and the face could just yield. So you could just get yielding of the aluminum. That would be one way. It could be a composite face and you could have some sort of composite failure mode. You can get more complicated failure modes for composites, but there could be some sort of failure mode. This face up here is in compression, and if you compress that face, you can get something called face wrinkling. You get sort of a local buckling mode. So imagine you have the face, that you're pressing on it, but the core is kind of acting like an elastic foundation underneath it. And you can get this kind of local buckling, and that's called wrinkling. That's another mode of failure. You can also get the core failing in shear. So here's these two little cracks, denoting shear failure in the core. And there's a couple of other modes you can get, but we're going to not pay much attention to those. The whole thing can delaminate, and, as you might guess, if the whole thing delaminates, you're in deep doo-doo. Because, remember when I passed those samples around, how flexible the face was by itself and how flexible the core is by itself. If the whole thing delaminates, you lose that whole sandwich effect and the whole thing kind of falls apart. We're going to assume we have a perfect bond and that we don't have to worry about that. The other sort of failure mode you can get is called indentation. So imagine that you apply this load here over a very small area. The load can just transfer straight through the face and just kind of indent the core underneath it. We're going to assume that you distribute this load over a big enough area here, that you don't indent the core. So we're going to worry about these three failure modes here-- the face yielding, the face wrinkling, and the core failing and shear, OK? So let me just write that down. And then you also can have debonding or delamination, and we're going to assume perfect bond. And then you can have indentation, and we're going to assume the loads are applied over a large enough area that you don't get-- So you can have different modes of failure, and the question becomes which mode is going to be dominant? So whichever one occurs at the lowest load is going to be the dominant failure mode. So you'd like to know what that lowest failure mode is. So we want to write equations for each of these failure modes and then figure out which one occurs first. So we'll look at the face yielding here. And face yielding is going to occur just when the normal stress in the face is equal to the yield stress of the face. So this is fairly straightforward. So this was our equation for the stress in the face. And when that's equal to the face yield strength, then you'll get failure. And the face wrinkling occurs when the normal compressive stress in the face equals a local buckling stress. And people have worked that out by looking at what's called buckling on an elastic foundation. So the core acts as elastic support. You can think that as the face is trying to buckle into the core, the core is pushing back on the face. And so the core is acting like a spring that pushes back, and that's called an elastic foundation. So people have calculated this local buckling stress, and they found that's equal to 0.57 times the modulus of the face to the 1/3 power times the modulus of the core to the 2/3 power. And here, if we use our model for open cell foams, we can say the core modulus goes as the relative density squared times the solid modulus. And so you can plug that in there. So then the wrinkling occurs when the stress in the face, the Pl over the B3 btc is equal to this thing here. OK, so one more failure mode that's the core shear, and that's going to occur when the shear stress in the core is just equal to the sheer strength of the core. So the shear stress is P over B4 times bc, and the shear strength is some constant, I think it's C11, times the relative density of the core to the three halves power times the yield strength of the solid. And here, this constant is about equal to 0.15, something like that. So now we have a set of equations for the different failure modes, and we could solve each of them, not in terms of a stress, but in terms of a load P. The load P is what's applied to the beam, right? So we could solve each of these in terms of the load, P. And then we can see which one occurs at the lowest load, P. And that's going to be the dominant failure mode. So one way to do it would be to, for every time you wanted to do this, to work out all these three equations and figure out which one's the lowest load. But there's actually something called a failure mode map, which we're going to talk about. So let me just show you it and we'll start now. I don't know if we'll get finished this. But there's a way that you can manipulate these equations and plot the results as this failure mode map. And you'll end up plotting the core density on this plot, on this axis here, and the face thickness to span ratio here, and so this will kind of tell you, for different configurations of the beam, different designs, for these ones here, the face is going to wrinkle, for those ones there, the face is going to yield, and for these ones here, the core is going to shear. So I'm going to work through these equations, but I don't think we're going to finish it today. So this is just kind of where we're headed is to getting this map. So we'll say the dominant failure mode is the one that occurs at the lowest load. So the question we're going to answer is how does the failure mode depend on the beam design? And we're going to do this by looking at the transition from one failure mode to another. So at the transition from one mode to another, the two modes occur at the same load. So I'm going to take those equations I had for each of the failure modes, and instead of writing this in terms of, say, the stress in the face, I'm going to write it in terms of the load, P. So using that first one over there, the load for face yielding, I'm just rearranging that. It's B3 times bc times t/l times the yield strength of the face. And similarly for face wrinkling, I can take this equation down here and solve it for this P here, OK? And then I can take that equation at the top and solve that for P2 for the core shear, and that's equal to C11 times B4 times bc times sigma ys times-- oops, wrong thing-- times the relative density to the 2/3 power. OK? And then the next step is to equate these guys. So you get a transition from one mode to the other when two of these guys are equal to each other, right? So there's going to be a transition from face yielding to face wrinkling when these guys are equal. And I'm not going to start that because we're going to run out of time. But let me just say that I can pair these two up and say there's a transition between those two. And that transition is going to correspond to this line here, OK? So at this line here, that means you get face yielding and face wrinkling at the same load, OK? And then if I paired up-- let's see here. If I paired up face wrinkling and core shear, these two guys here, I'm going to get this equation here on that plot. And then if I paired up these two guys here, the face shielding and the core shear, I would get that line there, OK? So once I have those lines, that tells me, you know, anything with a lower density core and a smaller face thickness is going to fail by face wrinkling. Anything with a bigger density is going to fail by face yielding. And anything with a larger face thickness and a larger density is going to fail by core shearing. And so you can start to see that it-- I'll work out the equations next time, but you can start to see that it kind physically makes sense. Intuitively, this face wrinkling, it depends on the normal stress in the face, in compression. So obviously the thinner the face gets, the more likely that's going to be to happen. So it's going to happen at this end of the diagram. And it also depends on that elastic foundation, on how much spring support the foundation has, right? So the lower the core density, the more likely that is to happen. Then if you, say you have small t, so the face is going to fail before the core, as you increase the core density, you're making that elastic foundation stiffer and stiffer, and you're making it harder for the buckling to occur. It can't buckle into the elastic foundation, so then you're going to push it up to the yielding. And then as you make the face thickness bigger, as t gets bigger, then the face isn't going to fail and the core is going to fail. So you can kind of see just looking at the relative position of those things, they all kind of make physical sense. So I'm going to stop there for today and I'll finish the equations for that next time. And we'll also talk about how to optimize for strength next time. And we'll talk about a few other things on sandwich panels.
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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: I'm going to write this e to the ikz somewhat differently so that you appreciate more what it is. So e to the ikz, I'll write it as square root of 4 pi over k. You say, where does that k come from? We'll see in a second. Sum over l, square root of 2l plus 1, i to the l, yl0, 1 over 2i. Basically what I'm going to do is I'm expanding this jl function for large x. So I'm going to take this equation and I'm going to expand it so that the argument is large, which means r is large. I want to describe for you those waves that are really going on here. So since the argument is kr and you have a kr here, that's the origin of the k that I pulled out in here. And then you have e to the i kr minus l pi over 2 over r, minus e to the minus i kr minus l pi over 2 over r. And this is valid for r much bigger than a. It's an approximate thing, because we approximated the Bessel function. This is exact. This equation is exact for all r. But that equation, now, is not. That equation is showing you that you have here an outgoing wave because is e to the i kr times e to the minus i et, the energy terms time. So that's an outgoing wave. This is an ongoing wave. So this is ingoing and this is outgoing. So I'm going to say a couple of things about this that are going to play some role later. It's a series of comments. Because this is a particular expansion and due to the fact that this is a general expansion, you could say that this is an approximate solution of the Schrodinger equation far away. But in fact, each term, each value of l is an approximate solution. This is not an approximate solution because we added over many ls and somehow they helped to create an approximate solution. This is an approximate solution because each l is an approximate solution, because to get a solution, you could have said l is equal to 50, and that's all that I'm going to use, and that's an exact solution. And therefore when I look far away it will look like that with l equal 50, and that would be an approximate solution that can be extended to a full solution. So each l term here is actually independent. If you tell me a wave looks like that far away, I would say good, yes, that's possible and that comes from a solution. You're getting an approximate solution. And you could make it exact by turning the exponentials into Bessel functions. Even more, we will show-- we'll discuss it a little later-- I could get a solution that is approximate by considering either this wave or that wave. Suppose we just have the outgoing wave. The fact that the Schrodinger equation works for this one approximately doesn't require the ingoing wave. The Schrodinger equation doesn't mix, really, the ingoing waves and the outgoing waves. It keeps them separate. So this is an approximate solution and this is an approximate solution as well. Every term here, independently, every l and every ingoing and every outgoing wave is a good approximate solution of the Schrodinger equation. Now this is important to emphasize because this is called partial wave. So the term partial wave, waves, refers to the various l's that work independently. Now to motivate what we're going to do, I'm going to go back to the case of one dimension so that you need the map to orient yourself in the argument we're going to do now. So in order to understand partial waves and how they work, let's think about the one-dimensional case for a minute. I used to do that in 804. I don't know if that has been done by other instructors, but we've done in 804 that stuff. So let me tell you about it and how it works in that case. So this is an aside 1D case of scattering. So in scattering in one dimension, one usually puts a hard wall at x equals 0. There is some potential up to x equals a, and then there's nothing. It's a finite range potential. So even in the case of d equals 1, you look for a solution when there is no potential. This is the v of x. This is the x-axis. If v is equal to 0, you would have a solution. Phi of x would be a solution, energy eigenstate, and would be sine of kx. This is the analog of our e to the ikz, that we call it, phi of r. Here you have a phi of x, which is the sine of kx, and it's your energy eigenstate. That's the solution if v is equal to 0. If v is equal to 0, the wave must vanish at the hard wall, at x equals 0. So that's your solution. On the other hand-- or, well, we can also write it as 1 over 2i e to the ikx minus e to the minus ikx. And this wave is the incoming wave, and this is the outgoing wave. Very analogous to above, because this x is defined to be positive and plays the role of radius. In this problem, incoming means going down in x. Incoming in spherical dimensions means going down in r. So we're doing something very analogous. That's why the study of scattering in one dimension is a good preparation for the study of scattering in three dimensions. So you have an incoming wave and an outgoing wave, also because, again, there's an e to the minus iet over h bar. So here it is. What we try to do is to write a scattering solution corresponding to the same physics-- so the scattering solution, psi of x, is the full solution-- is going to have the same incoming wave because that's the physics that we're trying to understand. We put the same incoming wave and we try to see what happens. That's what do we do when we do scattering. So if we want to compare our solution to the case when v is equal to 0, we put the same incoming wave. But then solving the scattering problem means finding the outgoing wave, which now may be different. Because, sadly, when you have a potential, you have something in, that you have control what you send in, but you have no control what comes out. That's solving the scattering problem. So the thing that you have here must be an outgoing wave, e to the ikx, with the same energy because energy is conserved. We're talking about an energy eigenstate anyway, so it must have the same energy, and it must have the same probability flux as the incoming wave. So the magnitude of this wave and the magnitude of this wave must be the same. It almost seems that you'd have to write the same thing that I wrote here, but no. I can add one more little factor. It's conventionally written as 2i delta of k. A phase shift, delta of k, an extra phase, and that's the claim. That's the whole thing. You have to solve what's going on here. But outside-- and this is only valid for x greater than a-- outside, where the potential is zero, you must have the solution is of this form. Here, this is for 0 potential everywhere. So this was valid everywhere, but in particular is also valid for x greater than a, and here we compare these two things. So this is the general solution for scattering here. And now we can also add another definition. We can say that psi of x is going to be equal to phi of x plus the scattered wave. Our intuition is that this wave here has the same incoming component as the reference wave, but nevertheless, it has the different outgoing component and that's the scattered wave. It is exactly analogous to what we did here, in which we have the total wave being a reference incoming wave. Well, actually, incoming and outgoing wave. The e to the ikz has incoming and outgoing wave and needs a solution when the potential is equal to 0, just like this one, this phi. So psi is equal to 5 plus scattering wave. Here it is the scattering wave. It's purely outgoing, because the incoming part is already taken care of. So completely analogous thing. So let's solve for the scattered wave here. The scatter wave. So in this equation, we have 1 over 2i, e to the ikx plus 2i delta k, minus e to the minus ikx is equal to 1 over 21, e to the ikx minus e to the minus ikx plus psi s of x. Well, these two terms cancel. It is the fact that they are ingoing waves are the same. And then I can subtract this other two to find psi s is equal to 1 over 2i. I can factor the e to the ikx. That tells me it's an outgoing wave. And then I have e to the 2i delta k minus 1. That's a term there. I think I got everything there. It's written-- I'll write it here, there's a little bit of space-- psi s as e to the ikx, e to the i delta k, sine of delta k. I factor out 1 times i delta k, so this becomes e to the i delta k minus e to the minus i delta k, the form together with a 2i in front, a sign of delta k. So this is your outgoing scattered wave. That's the shape of the outgoing scattered wave in terms of the phase shift. OK, so now we have to redo this in the slightly more complicated case of three dimensions and get it right. So that is our task. That's what we have to figure out how to do.
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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: Good morning, good morning. So last time, we talked about the most remarkable biochemical purification procedure ever invented-- cloning. You remember the issue with biochemistry. You're going to grind up a cell, you're going to take the contents and run it over different kinds of separation columns or centrifuge it or things, in order to separate some things from other things. The problem with purifying a gene that way, away from all the rest of DNA in a cell, is that the gene has exactly the same biochemical properties as all the other genes. How in the world are you going to do it? The solution was devilishly simple and devilishly complex at the same time. All you have to do to purify something is dilute it. If I take any substance an I add enough water, it's very dilute. And in any little drop in any little test tube there will only be one DNA molecule. It's now purified. I could do that for every molecule there is. The problem is, it's not very useful unless I have a way to take that molecule and make extra copies of it. But if I could do that, dilution is purification. The trick we use initially for making more copies is we ask E. coli to do it for us. That, as usual, is the solution to most issues. Find something in life that already does it. And so, just to remember what we did, we took total DNA. Maybe it was yeast DNA, maybe it was human DNA, maybe it was zebra DNA. And we cut it up with some restriction enzyme. Our restriction enzyme we used was EcoR1 And it therefore cut at the EcoR1 sites, G, A, A, T, T, C. We could have used a different restriction enzyme. We then added it to a vector. The vector was cut open at an EcoR1 site. The vector had an origin of replication, ORI, and the vector had a resistance marker, some resistance gene that made a protein that could break down some antibiotic found in nature. We combine these two pieces. Vector now gets its insert. We call this vector, we call this insert. We attach them together using what? AUDIENCE: Ligase. PROFESSOR: Ligase, the enzyme that ligates DNA. We then take this, we transform it into a bacterium. The scale is obviously off, right? This DNA plasmid here is tiny compared to this bacteria. But if I draw it to scale, it won't be very helpful. So we then transform it in here. We do that in a test tube. We treat the bacteria a little roughly so it likes to slurp up the DNA. We then plate it on a plate. And those bacteria that have acquired our plasmid containing the antibiotic resistance gene are able to grow on this plate that has antibiotic on it. And we call this thing a library. So that's it. We can make a library. And we have in effect, then, what I said. We've deluded the individual molecules out, and each one went into its own bacteria, and each one got replicated by that bacteria. Those plasmids are replicated by the bacteria. In fact, we choose plasmids that are called multicopy plasmids, where there's not just one copy but the cell might make 50 copies of it. We grow up a whole colony of it and there you go. We talked about some of the issues with it. Where do we get the restriction enzymes, the ligases, the vectors, et cetera? It's in the catalog, right? We used to purify them ourselves, but they're all in the catalog. So any questions about this? We had some questions. I have a question about this. How come when we add ligase the vector itself just doesn't close up into a circle without an insert in it? It might. What would happen then? AUDIENCE: You'd just have your original vector. PROFESSOR: You'd just have your original vector. And what happens when I transform into the bacteria? AUDIENCE: It'll still survive. PROFESSOR: It'll still survive. So will the unimolecular closure of that circle be more common than the bimolecular interaction between a vector and an insert? AUDIENCE: Probably. PROFESSOR: Probably, because the two ends of the circle are pretty close to each other. So what do I do? Yeah? AUDIENCE: Because [INAUDIBLE] are the same. [INAUDIBLE] same restriction enzymes because if they're two different enzymes so they don't match. PROFESSOR: Ooh. That's cute trick. Two different restriction enzymes so that they couldn't reclose. But then my fragments better have those two different sites, too, and only be able to clone those fragments that have the two different ones. But that could work. So you want a trick for making sure it doesn't reclose. Any other tricks? But the problem is, I won't get all the fragments, only the ones that have, say, an Eco at a Bam site. But that's-- So I bring this up not because it's particularly important, but to tell you the kind of engineering that really does have to go on in molecular biology. What happens is when you have your DNA put in here, we have a sugar-phosphate backbone in both cases. And if we look up close, one of these sides has the phosphate, the other has a hydroxyl. Phosphate, hydroxyl, phosphate, hydroxyl, right? And ligase comes in and joins that. So this guy has a phosphate on this strand, and this guy has a phosphate on that strand, hydroxyl there, hydroxyl there. What would happen if I got rid of the phosphates on the vector? Could it reclose? AUDIENCE: No. PROFESSOR: No. So if there were an enzyme that removed phosphates, I could treat my vector first with the phosphate-removing enzyme. And now it couldn't possibly reclose on itself. And is there such an enzyme? And what's it called? Phosphatase, because it takes phosphates so easy. Phosphatase, it takes off the phosphates. And then it can't reclose anymore. And where do we get phosphatase? It's in the catalog. Exactly. So now what happens is that the vector only has an OH here. What happens to ligase? When I put an insert in here, ligase can make a covalent join on this strand. But it can't actually make a covalent join on the other strand. But does it matter? No, because I've closed up my circle on that strand, and I close up the circle there on the other strand, and I just throw it into E. coli. And you know what happens when it goes into E. coli? It's got a nick, obviously, on that strand. It hasn't closed up. But what does E. coli do when it sees that DNA? Must be damaged DNA. I'll fix it. So E. coli actually does the last little trick of closing that up for you with its own enzymes for repairing its own DNA. I bring this up not because it's crucial that you should worry about it, but because I want to know that there's a whole layer of these interesting engineering tricks that get developed. Every one of them exploits enzymes that we know. Every one of them deals with questions like, will my vector reclose on itself, how do I avoid that? And there's a vast cooking book of protocols in molecular biology. And we constantly are just cribbing from things life does to make our protocols more and more efficient. So I bring it up more because it's kind of a cool thing that all that goes on, and because it helps you remember that these phosphates are very important to joining things up. That's a digression. Now, let's go to the topic. How do we actually read the library? How do we read our library? How do we use the library, read from the library? Well let's say we're going to try to find the arginine gene. We talked about the gene for arginine in yeast. So I'd like to clone the gene for ARG1. We found mutants before that were unable to grow without supplemental arginine. They somehow had a defect in producing their own origin. It's a mutant. I want to find the gene, please. How do I do it? We've got to think about what's our insert DNA. What are our vectors? What insert DNA should we start with, zebra? No. Human? No. How about yeast? We're trying to clone a gene from yeast, right? So let's start with yeast. OK. So we're going to start with yeast DNA. We're going to cut up yeast DNA. We're going to attach it to our vector, we're going to transform it into E. coli, E. coli will grow up on our plates. And one of these guys, I happen to know it's that one there, contains the ARG1 gene. The problem is I happen to know it, but you don't. How are you going to find out where the ARG1 gene is? Any takers? Yeah? AUDIENCE: It could be like when you put the gene in make it flourescent. PROFESSOR: A fluorescent tag? So I should just attach the fluorescent tag to the ARG1 gene? AUDIENCE: Yes. PROFESSOR: How do I do that? All the DNA looks the same in the tube. How do I know where to attach the fluorescent tag? AUDIENCE: Maybe you could size it [INAUDIBLE]. PROFESSOR: There's a lot of pieces of DNA there. And my eyes are not that good. AUDIENCE: Separate it? PROFESSOR: Separate it. But will I know which one is ARG1? See, I don't actually know anything about ARG1. All I know is I made a mutant. The mutant is unable to grow without arginine. I haven't got a clue what that gene is. I don't know what it encodes, I don't know how big it is, I don't know nothing. All I know is that whatever it is, it's a gene which when mutated prevents you from growing without arginine. AUDIENCE: Could you plate all of your colonies onto a-- could you put [INAUDIBLE]? PROFESSOR: Minimal medium. What if I plate on minimal medium? Now what? What are you hoping for? AUDIENCE: The one that has the ARG1 gene will not grow. PROFESSOR: The one that has the ARG1 gene won't be able to grow-- oh wait, yeah. But something like that. Let's work it through. We've got my idea here. What are we going to do with it? AUDIENCE: [INAUDIBLE] PROFESSOR: I've got a working ARG1 Mutate ARG1 afterwards? AUDIENCE: [INAUDIBLE] PROFESSOR: OK. How will that work? I'm open for-- got an idea here? AUDIENCE: If you have your different colonies [INAUDIBLE], you could have a secondary plate them all over to one of middle medium. The ones that die are the ones that already [INAUDIBLE]-- PROFESSOR: So guys, I have a concern. I'm just transferring this into E. coli. E.coli grows just fine without arginine. I mean, I'm going to take this yeast DNA. I'm going to put it in E. coli. E. coli was kind enough to grow it for me. But frankly, E. coli doesn't need this ARG1 gene. E. coli can grow without arginine. I can plate this with and without arginine, E. coli grows just fine. But you're on to something. You're onto the idea that somehow, the only thing we know about ARG1 is that the functional, wild-type copy of that gene confers an ability to grow without arginine. And who does it confer it on? And what kind of yeast? Haploid mutant yeast. Ah. So suppose I put a working copy of ARG1, a good copy, a wild-type copy, into a mutant yeast. Now what would happen to that mutant yeast? What would happen? The mutant yeast before, could it grow without arginine? No. If I put in a working copy of the ARG1 gene, what will happen? It grows. Now let's design a scheme. Do I want to use E. coli at all? No. What do I want to use? I want to use a yeast. So let's get rid of E. coli and let's instead use yeast. And which yeast should we use, wild-type or mutant? Mutant yeast. What mutation? ARG1 mutant yeast. ARG1 minus yeast. Now, if I plate ARG1 minus yeast on minimal medium, what happens? It doesn't grow. It dies. What DNA should I be putting in? Yeast DNA. Mutant or wild type? Why wild type? Because it'll have a working copy of ARG1. So I want yeast, wild type. Now what happens? One of these guys, and only one of these guys here, this one, has a ARG1 gene. That's ARG1. When it goes in, that plasmid has the ARG1 plus gene, whereas other plasmids don't. That cell that inherits that gene there, that gets that gene, is not green. I just drew it green for you. But it has the ARG1 gene. And when I plate this on minimal medium, what's distinctive about it? It grows. That's how you can clone the ARG1 gene. You clone it by the only thing you know about it. Namely, it confers a function. This is called cloning by function. Or, what did we do when we crossed two mutants together to see if things were in different genes? It was a test of complementation. Really what we're asking is, is there a plasmid that complements the defect? In effect what's happening is in this cell, right here, we have a yeast cell. And the yeast cell has a defect in its ARG1. But the plasmid has a working ARG1. So for that one gene, this cell could be thought of maybe a little bit like a diploid, just at that one gene. And we've done a complementation, just a teeny little complementation for one gene. And we could call this cloning complementation. It's essentially cloning by function. Any questions? Yes? AUDIENCE: [INAUDIBLE] --they all have functioning [INAUDIBLE]? PROFESSOR: Yep. AUDIENCE: So why does that then have a function? PROFESSOR: Oh, oh. You see, the yeast genome has about 4,000 different genes. I chop it up with my EcoR1. Some plasmids get ARG1 but most of them get leucine 2 or [INAUDIBLE] or other things. And each yeast cell in my library only picks up a plasmid with one chunk of DNA, one gene. So it turns out that the yeast cells in my library, each one has one of thousands of alternative possibilities. And it's just the guy who gets ARG1 who grows. AUDIENCE: But you're saying that the yeast [INAUDIBLE] plasmid. All code for-- PROFESSOR: That's right. AUDIENCE: But I don't get why you would end up with [INAUDIBLE]. PROFESSOR: What do I--? AUDIENCE: Why do you end up with a strain that has ARG1 in it? PROFESSOR: Oh. ARG1 is the ARG1 working copy. In the yeast, I'm talking about this yeast here has the working copy of ARG1, ARG1 plus. So this guy has an ARG1 plus. It also has lots of other genes. Each of these plasmids gets one gene. Some of them get an ARG plus. Some of them happen to get a leucine gene or some other gene that's irrelevant. And the plasmids that contain the working copy of the gene, they, when they go into the cell, give the cell the ability to grow. So that's why. All right. So that's how we get a gene by 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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We already showed that the torque about a point can also be thought of as a decomposition. We take the vector from the point P to the center of mass and apply all the forces acting on the particle at the center of mass. And we can calculate the torque about the center of mass due to the action of some forces where we're having forces acting about the center of mass. Now if we choose the point P to equal the center of mass, then we know that the vector r center of mass to the center of mass is 0. So the torque about the center of mass is just equal to the forces about that point. We know that torque is always just Lcm/dt. Now, again, how could we justify that statement that because we're only calculating the torque about the center of mass, it's only the rotational angular momentum about the center of mass that's changing. We saw before that if we thought of Lp, again, as a translational and a rotational angular momentum-- I'm sorry, rotational angular momentum, omega-- and the point p was equal to the center of mass, then this first piece would be 0. And L about p is only Icm omega and dLp/dt is equal to Icm alpha for a fixed axis. And that's exactly equal to dLcm/dt. And so the point here is that when we're applying problems involving rotation and translation, we can just analyze the torque about the center of mass and only consider how the angular momentum about the center of mass is changing. And then for the center of mass motion-- so this gives us our rotational dynamics. And for our linear dynamics, we will still apply F equals the total mass times Acm. So that's our linear dynamics. And this is our overall decomposition of rotational motion. To analyze it, we study the rotational dynamics and the linear dynamics.
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https://ocw.mit.edu/courses/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2013/8.333-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. PROFESSOR: Let's start. Are there any questions? We would like to have a perspective for this really common observation that if you have a gas that is initially in one half of a box, and the other half is empty, and some kind of a partition is removed so that the gas can expand, and it can flow, and eventually we will reach another equilibrium state where the gas occupies more chambers. How do we describe this observation? We can certainly characterize it thermodynamically from the perspectives of atoms and molecules. We said that if I want to describe the configuration of the gas before it starts, and also throughout the expansion, I would basically have to look at all sets of coordinates and momenta that make up this particle. There would be some point in this [? six ?], and I mention our phase space, that would correspond to where this particle was originally. We can certainly follow the dynamics of this point, but is that useful? Normally, I could start with billions of different types of boxes, or the same box in a different instance of time, and I would have totally different initial conditions. The initial conditions presumably can be characterized to a density in this phase space. You can look at some volume and see how it changes, and how many points you have there, and define this phase space density row of all of the Q's and P's, and it works as a function of time. One way of looking at how it works as a function of time is to look at this box and where this box will be in some other instance of time. Essentially then, we are following a kind of evolution that goes along this streamline. Basically, the derivative that we are going to look at involves changes both explicitly in the time variable, and also increasingly to the changes of all of the coordinates and momenta, according to the Hamiltonian that governs the system. I have to do, essentially, a sum over all coordinates. I would have the change in coordinate i, Qi dot, dot, d row by dQi. Then I would have Pi, dot-- I guess these are all vectors-- d row by dPi. There are six end coordinates that implicitly depend on time. In principle, if I am following along the streamline, I have to look at all of these things. The characteristic of evolution, according to some Hamiltonian, was that this volume of phase space does not change. Secondly, we could characterize, once we wrote Qi dot, as dH by dP, and the i dot as the H by dQ. This combination of derivatives essentially could be captured, and be written as 0 by dt is the Poisson bracket of H and [? P. ?] One of the things, however, that we emphasize is that as far as evolution according to a Hamiltonian and this set of dynamics is concerned, the situation is completely reversible in time so that some intermediate process, if I were to reverse all of the momenta, then the gas would basically come back to the initial position. That's true. There is nothing to do about it. That kind of seems to go against the intuition that we have from thermodynamics. We said, well, in practical situations, I really don't care about all the six end pieces of information that are embedded currently in this full phase space density. If I'm really trying to physically describe this gas expanding, typically the things that I'm interested in are that at some intermediate time, whether the particles have reached this point or that point, and what is this streamline velocity that I'm seeing before the thing relaxes, presumably, eventually into zero velocity? There's a lot of things that I would need to characterize this relaxation process, but that is still much, much, much less than all of the information that is currently encoded in all of these six end coordinates and momenta. We said that for things that I'm really interested in, what I could, for example, look at, is a density that involves only one particle. What I can do is to then integrate over all of the positions and coordinates of particles that I'm not interested in. I'm sort of repeating this to introduce some notation so as to not to repeat all of these integration variables, so I will call dVi the phase place contribution of particle i. What I may be interested in is that this is something that, if I integrate over P1 and Q1, it is clearly normalized to unity because my row, by definition, was normalized to unity. Typically we may be interested in something else that I call F1, P1 Q1 P, which is simply n times this-- n times the integral product out i2 to n, dVi, the full row. Why we do that is because typically you are interested or used to calculating things in [? terms ?] of a number density, like how many particles are within some small volume here, defining the density so that when I integrate over the entire volume of f1, I would get the total number of particles, for example. That's the kind of normalization that people have used for f. More generally, we also introduced fs, which depended on coordinates representing s sets of points, or s particles, if you like, that was normalized to be-- We said, OK, what I'm really interested in, in order to calculate the properties of the gases it expands in terms of things that I'm able to measure, is f1. Let's write down the time evolution of f1. Actually, we said, let's write down the time evolution of fs, along with it. So there's the time evolution of fs. If I were to go along this stream, it would be the fs by dt, and then I would have contributions that would correspond to a the changes in coordinates of these particles. In order to progress along this direction, we said, let's define the total Hamiltonian. We will have a simple form, and certainly for the gas, it would be a good representation. I have the kinetic energies of all of the particles. I have the box that confines the particles, or some other one particle potential, if you like, but I will write in this much. Then you have the interactions between all pairs of particles. Let's write it as sum over i, less than j, V of Qi minus Qj. This depends on n set of particles, coordinates, and momenta. Then we said that for purposes of manipulations that you have to deal with, since there are s coordinates that are appearing here whose time derivatives I have to look at, I'm going to simply rewrite this as the contribution that comes from those s particles, the contribution that comes from the remaining n minus s particles, and some kind of [? term ?] that covers the two sets of particles. This, actually, I didn't quite need here until the next stage because what I write here could, presumably, be sufficiently general, like we have here some n running from 1 to s. Let me be consistent with my S's. Then I have Qn, dot, dFs by dQn, plus Pn, dot, dFs by dPn. If I just look at the coordinates that appear here, and say, following this as they move in time, there is the explicit time dependence on all of the implicit time dependence, this would be the total derivative moving along the streamline. Qn dot I know is simply the momentum. It is the H by dPn. The H by dPn I have from this formula over here. It is simply Pn divided by m. It's the velocity-- momentum divided by mass. This is the velocity of the particle. Pn dot, the rate of change of momentum is the force that is acting on the particle. What I need to do is to take the derivatives of various terms here. So I have minus dU by dPn. What is this? This is essentially the force that the particle feels from the external potential. If you are in the box in this room, It is zero until you hit the edge of the box. I will call this Fn to represent external potential that is acting on the system. What else is there? I have the force that will come from the interaction with all other [? guys. ?] I will write here a sum over m, dV of Qm minus Qn, by dQn-- dU by dQm. I'm sorry. What is this? This Is basically the sum of the forces that is exerted by the n particle on the m particle. Define it in this fashion. If this was the entire story, what I would have had here is a group of s particles that are dominated by their own dynamics. If there is no other particle involved, they basically have to satisfy the Liouville equation that I have written, now appropriate to s particles. Of course, we know that that's not the entire story because there are all these other terms involving the interactions with particles that I have not included. That's the whole essence of the story. Let's say I want to think about one or two particles. There is the interaction between the two particles, and they would be evolving according to some trajectories. But there are all of these other particles in the gas in this room that will collide with them. So those conditions are not something that we had in the Liouville equation, with everything considered. Here, I have to include the effect of all of those other particles. We saw that the way that it appears is that I have to imagine that there's another particle whose coordinates and momenta are captured through some volume for the s plus 1 particle. This s plus 1 particle can interact with any of the particles that are in the set that I have on the other side. There is an index that runs from 1 to s. What I would have here is the force that will come from this s plus 1 particle, acting on particle n the same way that this force was deriving the change of the momentum, this force will derive the change of the momentum of-- I guess I put an m here-- The thing that I have to put here is now a density that also keeps track of the probability to find the s plus 1 particle in the location in phase place that I need to integrate with both. I have to integrate over all positions. One particle is moving along a straight line by itself, let's say. Then there are all of the other particles in the system. I have to ask, what is the possibility that there is a second particle with some particular momentum and coordinate that I will be interacting with. This is the general set up of these D-B-G-K-Y hierarchy of equations. At this stage, we really have just rewritten what we had for the Liouville equation. We said, I'm really, really interested only one particle [? thing, ?] row one and F1. Let's focus on that. Let's write those equations in more detail In the first equation, I have that the explicit time dependence, plus the time dependence of the position coordinate, plus the time dependence of the momentum coordinate, which is driven by the external force, acting on this one particle density, which is dependent on p1, q1 at time t. On the right hand side of the equation. I need to worry about a second particle with momenta P2 at position Q2 that will, therefore, be able to exert a force. Once I know the position, I can calculate the force that particle exerts. What was my notation? The order was 2 and 1, dotted by d by dP1. I need now f2, p1, q2 at time t. We say, well, this is unfortunate. I have to worry about dependence on F2, but maybe I can get away with things by estimating order of magnitudes of the various terms. What is the left hand side set of operations? The left hand side set of operations describes essentially one particle moving by itself. If that particle has to cross a distance of this order of L, and I tell you that the typical velocity of these particles is off the order of V, then that time scale is going to be of the order of L over V. The operations here will give me a V over L, which is what we call the inverse of Tau u. This is a reasonably long macroscopic time. OK, that's fine. How big is the right hand side? We said that the right hand side has something to do with collisions. I have a particle in my system. Let's say that particle has some characteristic dimension that we call d. This particle is moving with velocity V. Alternatively, you can think of this particle as being stationary, and all the other particles are coming at it with some velocity V. If I say that the density of these particles is n, then the typical time for which, as I shoot these particles, they will hit this target is related to V squared and V, the volume of particles. Over time t, I have to consider this times V tau x. V tau xn V squared should be of the order of one. This gave us a formula for tau x. The inverse of tau x that controls what's happening on this side is n V squared V. Is the term on the right hand side more important, or the term on the left hand side? The term on the right hand side has to do with the two body term. There's a particle that is moving, and then there's another particle with a slightly different velocity that it is behind it. In the absence of collisions, these particles would just go along a straight line. They would bounce off the walls, but the magnitude of their energy, and hence, velocity, would not change from these elastic collisions. But if the particles can catch up and interact, which is governed by V2, V on the other side, then what happens is that the particles, when they interact, would collide and go different ways. Quickly, their velocities, and momenta, and everything would get mixed up. How rapidly that happens depends on this collision distance, which is much less than the size of the system, and, therefore, the term that you have on the right hand side in magnitude is much larger than what is happening on the left hand side. There is no way in order to describe the relaxation of the gas that I can neglect collisions between gas particles. If I neglect collisions between gas particles, there is no reason why the kinetic energies of individual particles should change. They would stay the same forever. I have to keep this. Let's go and look at the second equation in the hierarchy. What do you have? You have d by dT, P1 over m d by d Q1, P2 over m, P d by d Q2. Then we have F1 d by d Q1, plus F2, d by d Q2 coming from the external potential. Then we have the force that the involves the collision between particles one and two. When I write down the Hamiltonian for two particles, there is going to be already for two particles and interactions between them. That's where the F1 2 comes from. F1 2 changes d by the momentum of particle one. I should write, it's 2 1 that changes momentum of particle two. But as 2 1 is simply minus F1 2, I can put the two of them together in this fashion. This acting on F2 is then equal to something like integral over V3, F3 1, d by dP1, plus F3 2, d by dP2. [INAUDIBLE] on F3 P1 and Q3 [INAUDIBLE]. Are we going to do this forever? Well, we said, let's take another look at the magnitude of the various terms. This term on the right hand side still involves a collision that involves a third particle. I have to find that third particle, so I need to have, essentially, a third particle within some characteristic volume, so I have something that is of that order. Whereas on the left hand side now, I have a term that from all perspectives, looks like the kinds of terms that I had before except that it involves the collision between two particles. What it describes is the duration that collision. We said this is of the order of 1 over tau c, which replaces the n over there with some characteristic dimension. Suddenly, this term is very big. We should be able to use that. There was a question. AUDIENCE: On the left hand side of both of your equations, for F1 and F2, shouldn't all the derivatives that are multiplied by your forces be derivatives of the effects of momentum? [INAUDIBLE] the coordinates? [INAUDIBLE] reasons? PROFESSOR: Let's go back here. I have a function that depends on P, Q, and t. Then there's the explicit time derivative, d by dt. Then there is the Q dot here, which will go by d by dQ. Then there's the P dot term that will go by d by dP. All of things have to be there. I should have derivatives in respect to momenta, and derivatives with respect to coordinate. Dimensions are, of course, important. Somewhat, what I write for this and for this should make up for that. As I have written it now, it's obvious, of course. This has dimensions of Q over T. The Q's cancel. I would have one over T. D over Dps cancel. I have 1 over P. Here, dimensionality is correct. I have to just make sure I haven't made a mistake. Q dot is a velocity. Velocity is momentum divided by mass. So that should dimensionally work out. P dot is a force. Everything here is force. In a reasonable coordinate-- AUDIENCE: [INAUDIBLE] PROFESSOR: What did I do here? I made mistakes? AUDIENCE: [INAUDIBLE] PROFESSOR: Why didn't you say that in a way that-- If I don't understand the question, please correct me before I spend another five minutes. Hopefully, this is now free of these deficiencies. This there is very big. Now, compared to the right hand side in fact, we said that the right hand side is smaller by a factor that measures how many particles are within an interaction volume. And for a typical gas, this would be a number that's of the order of 10 to the minus 4. Using 10 to the minus 4 being this small, we are going to set the right hand side to zero. Now, I don't have to write the equation for F2. I'll answer a question here that may arise, which is ultimately, we will do sufficient manipulations so that we end up with a particular equation, known as the Boltzmann Equation, that we will show does not obey the time reversibility that we wrote over here. Clearly, that is built in to the various approximations I make. The first question is, the approximation that I've made here, did I destroy this time reversibility? The answer is no. You can look at this set of equations, and do the manipulations necessary to see what happens if P goes to minus P. You will find that you will be able to reverse your trajectory without any problem. Yes? AUDIENCE: Given that it is only an interaction from our left side that's very big, that's the reason why we can ignore the stuff on the right. Why is it that we are then keeping all of the other terms that were even smaller before? PROFESSOR: I will ignore them. Sure. AUDIENCE: [LAUGHTER] PROFESSOR: There was the question of time reversibility. This term here has to do with three particles coming together, and how that would modify what we have for just two-body collisions. In principle, there is some probability to have three particles coming together and some combined interactions. You can imagine some fictitious model, which in addition to these two-body interactions, you cook up some body interaction so that it precisely cancels what would have happened when three particles come together. We can write a computer program in which we have two body conditions. But if three bodies come close enough to each other, they essentially become ghosts and pass through each other. That computer program would be fully reversible. That's why sort of dropping this there is not causing any problems at this point. What is it that you have included so far? What we have is a situation where the change in F1 is governed by a process in which I have a particle that I describe on the left hand side with momentum one, and it collides with some particle that I'm integrating over, but in some particular instance of integration, has momentum P2. Presumably they come close enough to each other so that afterwards, the momenta have changed over so that I have some P1 prime, and I have some P2 prime. We want to make sure that we characterize these correctly. There was a question about while this term is big, these kinds of terms are small. Why should I basically bother to keep them? It is reasonable. What we are following here are particles in my picture that were ejected by the first box, and they collide into each other, or they were colliding in the first box. As long as you are away from the [? vols ?] of the container, you really don't care about these terms. They don't really moved very rapidly. This is the process of collision of two particles, and it's also the same process that is described over here. Somehow, I should be able to simplify the collision process that is going on here with the knowledge that the evolution of two particles is now completely deterministic. This equation by itself says, take two particles as if they are the only thing in the universe, and they would follow some completely deterministic trajectory, that if you put lots of them together, is captured through this density. Let's see whether we can massage this equation to look like this equation. Well, the force term, we have, except that here we have dP by P1 here. We have d by dP 1 minus d by dP2. So let's do this. Minus d by dP2, acting on F2. Did I do something wrong? The answer is no, because I added the complete derivative over something that I'm integrating over. This is perfectly legitimate mathematics. This part now looks like this. I have to find what is the most important term that matches this. Again, let's think about this procedure. What I have to make sure of is what is the extent of the collision, and how important is the collision? If I have one particle moving here, and another particle off there, they will pass each other. Nothing interesting could happen. The important thing is how close they come together. It Is kind of important that I keep track of the relative coordinate, Q, which is Q2 minus Q1, as opposed to the center of mass coordinate, which is just Q1 plus Q2 over 2. That kind of also indicates maybe it's a good thing for me to look at this entire process in the center of mass frame. So this is the lab frame. If I were to look at this same picture in the center of mass frame, what would I have? In the center of mass frame, I would have the initial particle coming with P1 prime, P1 minus P center of mass. The other particle that you are interacting with comes with P2 minus P center of mass. I actually drew these vectors that are hopefully equal and opposite, because you know that in the center of mass, one of them, in fact, would be P1 minus P2 over 2. The other would be P2 minus P1 over 2. They would, indeed, in the center of mass be equal and opposite momenta. Along the direction of these objects, I can look at how close they come together. I can look at some coordinate that I will call A, which measures the separation between them at some instant of time. Then there's another pair of coordinates that I could put into a vector that tells me how head to head they are. If I think about they're being on the center of mass, two things that are approaching each other, they can either approach head on-- that would correspond to be equal to 0-- or they could be slightly off a head-on collision. There is a so-called impact parameter B, which is a measure of this addition fact. Why is that going to be relevant to us? Again, we said that there are parts of this expression that all of the order of this term, they're kind of not that important. If I think about the collision, and what the collision does, I will have forces that are significant when I am within this range of interactions, D. I really have to look at what happens when the two things come close to each other. It Is only when this relative parameter A has approached D that these particles will start to deviate from their straight line trajectory, and presumably go, to say in this case, P2 prime minus P center of mass. This one occurs [? and ?] will go, and eventually P1 prime minus P center of mass. These deviations will occur over a distance that is of the order of this collision and D. The important changes that occur in various densities, in various potentials, et cetera, are all taking place when this relative coordinate is small. Things become big when the relative coordinate is small. They are big as a function of the relative coordinate. In order to get big things, what I need to do is to replace these d by dQ's with the corresponding derivatives with respect to the center of mass. One of them would come be the minus sign. The other would come be the plus sign. It doesn't matter which is which. It depends on the definition, whether I make Q2 minus Q1, or Q1 minus Q2. We see that the big terms are the force that changes the momenta and the variations that you have over these relative coordinates. What I can do now is to replace this by equating the two big terms that I have over here. The two big terms are P2 minus P1 over m, dotted by d by dQ of F2. There is some other approximation that I did. As was told to me before, this is the biggest term, and there is the part of this that is big and compensates for that. But there are all these other bunches of terms. There's also this d by dt. What I have done over here is to look at this slightly coarser perspective on time. Increasing all the equations that I have over there tells me everything about particles approaching each other and going away. I can follow through the mechanics precisely everything that is happening, even in the vicinity of this collision. If I have two squishy balls, and I run my hand through them properly, I can see how the things get squished then released. There's a lot of information, but again, a lot of information that I don't really care to know as far as the properties of this gas expansion process is concerned. What you have done is to forget about the detailed variations in time and space that are taking place here. We're going to shortly make that even more explicit by noting the following. This integration over here is an integration over phase space of the second particle. I had written before d cubed, P2, d cubed, Q2, but I can change coordinates and look at the relative coordinate, Q, over here. What I'm asking is, I have one particle moving through the gas. What is the chance that the second particle comes with momentum P2, and the appropriate relative distance Q, and I integrate over both the P and the relative distance Q? This is the quantity that I have to integrate. Let's do one more calculation, and then we will try to give a physical perspective. In this picture of the center of mass, what did I do? I do replaced the coordinate, Q, with a part that was the impact parameter, which had two components, and a part that was the relative distance. What was this relative distance? The relative distance was measured along this line that was giving me the closest approach. What is the direction of this line? The direction of this line is P1 minus P2. This is P1 minus P2 over 2. It doesn't matter. The direction is P1 minus P2. What I'm doing here is I am taking the derivative precisely along this line of constant approach. I'm taking a derivative, and I'm integrating along that. If I were to rewrite the whole thing, what do I have? I have d by dt, plus P1 over m, d by dQ1, plus F1, d by dP1-- don't make a mistake-- acting on F1, P1, Q1, t. What do I have to write on the right hand side? I have an integral over the momentum of this particle with which I'm going to make a collision. I have an integral over the impact parameter that tells me the distance of closest approach. I have to do the magnitude of P2 minus P1 over n, which is really the magnitude of the relative velocity of the two particles. I can write it as P2 minus P1, or P1 minus P2. These are, of course, vectors. and I look at the modulus. I have the integral of the derivative. Very simply, I will write the answer as F2 that is evaluated at some large distance, plus infinity minus F2 evaluated at minus infinity. I have infinity. In principle, I have to integrate over F2 from minus infinity to plus infinity. But once I am beyond the range of where the interaction changes, then the two particles just move away forever. They will never see each other. Really, what I should write here is F2 of-- after the collision, I have P1 prime, P2 prime, at some Q plus, minus F2, P1, P2, at some position minus. What I need to do is to do the integration when I'm far away from the collision, or wait until I am far after the collision. Really, I have to just integrate slightly below, after, and before the collision occurs. In principle, if I just go a few d's in one direction or the other direction, this should be enough. Let's see physically what this describes. There is a connection between this and this thing that I had over here, in fact. This equation on the left hand side, if it was zero, it would describe one particle that is just moving by itself until it hits the wall, at which point it basically reverses its trajectory, and otherwise goes forward. But what you have on the right hand side says that suddenly there could be another particle with which I interact. Then I change my direction. I need to know the probability, given that I'm moving with velocity P1, that there is a second particle with P2 that comes close enough. There is this additional factor. From what does this additional factor come? It's the same factor that we have over here. It is, if you have a target of size d squared, and we have a set of bullets with a density of n, the number of collisions that I get depends both on density and how fast these things go. The time between collisions, if you like, is proportional to n, and it is also related to V. That's what this is. I need some kind of a time between the collisions that I make. I have already specified that I'm only interested in the set of particles that have momentum P2 for this particular [? point in ?] integration, and that they have this kind of area or cross section. So I replace this V squared and V with the relative coordinates. This is the corresponding thing to V squared, and this is really a two particle density. This is a subtraction. The addition is because it is true that I'm going with velocity P1, and practically, any collisions that are significant will move me off kilter. So there has to be a subtraction for the channel that was described by P1 because of this collision. This then, is the addition, because it says that it could be that there is no particle going in the horizontal direction. I was actually coming along the vertical direction. Because of the collision, I suddenly was shifted to move along this direction. The addition comes from having particles that would correspond to momenta that somehow, if I were in some sense to reverse this, and then put a minus sign, a reverse collision would create something that was along the direction of P1. Here I also made several approximations. I said, what is chief among them is that basically I ignored the details of the process that is taking place at scale the order of d, so I have thrown away some amount of detail and information. It is, again, legitimate to say, is this the stage at which you made an approximation so that the time reversibility was lost? The answer is still no. If you are careful enough with making precise definitions of what these Q's are before and after the collision, and follow what happens if you were to reverse everything, you'll find that the equations is fully reversible. Even at this stage, I have not made any transition. I have made approximations, but I haven't made something to be time irreversible. That comes at the next stage where we make the so-called assumption of molecular chaos. The assumption is that what's the chance that I have a particle here and a particle there? You would say, it's a chance that I have one here and one there. You say that if two of any P1, P2, Q1, Q2, t is the same thing as the product of F1, P1, Q1, t, F1, P2, Q2, t. Of course, this assumption is generally varied. If I were to look at the probability that I have two particles as a function of, let's say, the relative separation, I certainly expect that if they are far away, the density should be the product of the one particle densities. But you would say that if the two particles come to distances that are closer than their separation d, then the probability and the range of interaction d-- and let's say the interaction is highly repulsive like hardcore-- then the probability should go to 0. Clearly, you can make this assumption, but up to some degree. Part of the reason we went through this process was to indeed make sure that we are integrating things at the locations where the particles are far away from each other. I said that the range of that integration over A would be someplace where they are far apart after the collision, and far apart before the collision. You have an assumption like that, which is, in principle, something that I can insert into that. Having to make a distinction between the arguments that are appearing in this equation is kind of not so pleasant. What you are going to do is to make another assumption. Make sure that everything is evaluated at the same point. What we will eventually now have is the equation that d by dt, plus P1 over n, d by dQ1, plus F1, dot, d by dP1, acting on F1, on the left hand side, is, on the right hand side, equal to all collisions in the particle of momentum P2, approaching at all possible cross sections, calculating the flux of the incoming particle that corresponds to that channel, which is proportional to V2 minus V1. Then here, we subtract the collision of the two particles. We write that as F1 of P1 at this location, Q1, t, F1 of t2 at the same location Q1, t. Then add F1 prime, P1 prime, Q1 t, F1 prime, P2 prime, Q2, t. In order to make the equation eventually manageable, what you did is to evaluate all off the coordinates that we have on the right hand side at the same location, which is the same Q1 that you specify on the left hand side. That immediately means that what you have done is you have changed the resolution with which you are looking at space. You have kind of washed out the difference between here and here. Your resolution has to put this whole area that is of the order of d squared or d cubed in three dimensions into one pixel. You have changed the resolution that you have. You are not looking at things at this [? fine ?] [? state. ?] You are losing additional information here through this change of the resolution in space. You have also lost some information in making the assumption that the two [? point ?] densities are completely within always as the product one particle densities. Both of those things correspond to taking something that is very precise and deterministic, and making it kind of vague and a little undefined. It's not surprising then, that if you have in some sense changed the precision of your computer-- let's say, that is running the particles forward-- at some point, you've changed the resolution. Then you can't really run backward. In fact, to sort of precisely be able to run the equations forward and backward, you would need to keep resolution at all levels. Here, we have sort of removed some amount of resolution. We have a very good guess that the equation that you have over here no longer respects time reversal inversions that you had originally posed. Our next task is to prove that you need this equation. It goes in one particular direction in time, and cannot be drawn backward, as opposed to all of the predecessors that I had written up to this point. Are there any questions? AUDIENCE: [INAUDIBLE] PROFESSOR: Yes, Q prime and Q1, not Q1 prime. There is no dash. AUDIENCE: Oh, I see. It is Q1. PROFESSOR: Yes, it is. Look at this equation. On the left hand side, what are the arguments? The arguments are P1 and Q1. What is it that I have on the other side? I still have P1 and Q1. I have introduced P1 and b, which is simply an impact parameter. What I will do is I will evaluate all of these things, always at the same location, Q1. Then I have P1 and P2. That's part of my story of the change in resolution. When I write here Q1, and you say Q1 prime, but what is Q1 prime? Is it Q1 plus b? Is it Q1 minus b? Something like this I'm going to ignore. It's also legitimate, and you should ask, what is P1 prime and Q2 prime? What are they? What I have to do, is I have to run on the computer or otherwise, the equations for what happens if I have P1 and P2 come together at an impact parameter that is set by me. I then integrate the equations, and I find that deterministically, that collision will lead to some P1 prime and P2 prime. P1 prime and P2 prime are some complicated functions of P1, P2, and b. Given that you know two particles are approaching each other at distance d with momenta P1 P2, in principle, you can integrate Newton's equations, and figure out with what momenta they end up. This equation, in fact, hides a very, very complicated function here, which describes P1 prime and P2 prime as a function of P1 and P2. If you really needed all of the details of that function, you would surely be in trouble. Fortunately, we don't. As we shall see shortly, you can kind of get a lot of mileage without knowing that. Yes, what is your question? AUDIENCE: There was an assumption that all the interactions between different molecules are central potentials [INAUDIBLE]. Does the force of the direction between two particles lie along the [INAUDIBLE]? PROFESSOR: For the things that I have written, yes it does. I should have been more precise. I should have put absolute value here. AUDIENCE: You have particles moving along one line towards each other, and b is some arbitrary vector. You have two directions, so you define a plane. Opposite direction particles stay at the same plane. Have you reduced-- PROFESSOR: Particles stay in the same plane? AUDIENCE: If the two particles were moving towards each other, and also you have in the integral your input parameter, which one is [INAUDIBLE]. There's two directions. All particles align, and all b's align. They form a plane. [? Opposite ?] direction particles [? stand ?] in the-- PROFESSOR: Yes, they stand in the same plane. AUDIENCE: My question is, what is [INAUDIBLE] use the integral on the right from a two-dimensional integral [? in v ?] into employing central symmetry? PROFESSOR: Yes, you could. You could, in principle, write this as b db, if you like, if that's what you want. AUDIENCE: [INAUDIBLE] PROFESSOR: Yes, you could do that if you have simple enough potential. Let's show that this equation leads to irreversibility. That you are going to do here. This, by the way, is called the Boltzmann equation. There's an associated Boltzmann H-Theorem, which restates the following-- If F of P1, Q1, and t satisfies the above Boltzmann equation, then there is a quantity H that always decreases in time, where H is the integral over P and Q of F1, log of F1. The composition of irreversibility, as we saw in thermal dynamics, was that there was a quantity entropy that was always increasing. If you have calculated for this system, entropy before for the half box, and entropy afterwards for the space both boxes occupy, the second one would certainly be larger. This H is a quantity like that, except that when it is defined this way, it always decreases as a function of time. But it certainly is very much related to entropy. You may have asked, why did Boltzmann come across such a function, which is F log F, except that actually right now, you should know why you write this. When we were dealing with probabilities, we introduced the entropy of the probability distribution, which was related to something like sum over iPi, log of Pi, with a minus sign. Up to this factor of normalization N, this F1 really is a one-particle probability. After this normalization N, you have a one-particle probability, the probability that you have occupation of one-particle free space. This occupation of one-particle phase space is changing as a function of time. What this statement says is that if the one-particle density evolves in time according to this equation, the corresponding minus entropy decreases as a function of time. Let's see if that's the case. To prove that, let's do this. We have the formula for H, so let's calculate the H by dt. I have an integral over the phase space of particle one, the particle that I just called one. I could have labeled it anything. After integration, H is only a function of time. I have to take the time derivative. The time derivative can act on F1. Then I will get the F1 by dt, times log F1. Or I will have F1 times the derivative of log F1. The derivative of log F1 would be dF1 by dt, and then 1 over F1. Then I multiply by F1. This term is simply 1. AUDIENCE: Don't you want to write the full derivative, F1 with respect [INAUDIBLE]? PROFESSOR: I thought we did that with this before. If you have something that I am summing over lots of [? points, ?] and these [? points ?] can be positioned, then I have S at location one, S at location two, S at location three, discretized versions of x. If I take the time derivative, I take the time derivative of this, plus this, plus this, which are partial derivatives. If I actually take the time derivative here, I get the integral d cubed P1, d cubed Q1, the time derivative. This would be that partial dF1 by dt is the time derivative of n, which is 0. The number of particles does not change. Indeed, I realize that 1 integrated against dF1 by dt is the same thing that's here. This term gives you 0. All I need to worry about is integrating log F against the Fydt. I have an integral over P1 and Q1 of log F against the Fydt. We have said that F1 satisfies the Boltzmann equation. So the F1 by dt, if I were to rearrange it, I have the F1 by dt. I take this part to the other side of the equation. This part is also the Poisson bracket of a one-particle H with F1. If I take it to the other side, it will be the Poisson bracket of H with F1. Then there is this whole thing that involves the collision of two particles. So I define whatever is on the right hand side to be some collision operator that acts on two [? powers ?] of F1. This is plus a collision operator, F1, F1. What I do is I replace this dF1 by dt with the Poisson bracket of H, or H1, if you like, with F1. The collision operator I will shortly write explicitly. But for the time being, let me just write it as C of F1. There is a first term in this sum-- let's call it number one-- which I claim to be 0. Typically, when you get these integrations with Poisson brackets, you would get 0. Let's explicitly show that. I have an integral over P1 and Q1 of log of F1, and this Poisson bracket of H1 and F1, which is essentially these terms. Alternatively, I could write it as dH1 by dQ1, dF1 by dt1, minus the H1 by dt1, dF1, by dQ1. I've explicitly written this form for the one-particle in terms of the Hamiltonian. The advantage of that is that now I can start doing integrations by parts. I'm taking derivatives with respect to P, but I have integrations with respect to P here. I could take the F1 out. I will have a minus. I have an integral, P1, Q1. I took F1 out. Then this d by dP1 acts on everything that came before it. It can act on the H1. I would get d2 H1 with respect to dP1, dQ1. Or it could act on the log of F1, in which case I will get set dH1 by dQ1. Then I would have d by dP acting on log of F, which would give me dF1 by dP1, then the derivative of the log, which is 1 over F1. This is only the first term. I also have this term, with which I will do the same thing. AUDIENCE: [INAUDIBLE] The second derivative [INAUDIBLE] should be multiplied by log of F. PROFESSOR: Yes, it should be. It is Log F1. Thank you. For the next term, I have F1. I have d2 H1, and the other order of derivatives, dQ1, dP1. Now I'll make sure I write down the log of F1. Then I have dH1 with respect to dQ1. Then I have a dot product with the derivative of log F, which is the derivative of F1 with respect to Q1 and 1 over F1. Here are the terms that are proportional to the second derivative. The order of the derivatives does not matter. One often is positive. One often is negative, so they cancel out. Then I have these additional terms. For the additional terms, you'll note that the F1 and the 1 over F1 cancels. These are just a product of two first derivatives. I will apply the five parts process one more time to get rid of the derivative that is acting on F1. The answer becomes plus d cubed P1, d cubed Q1. Then I have F1, d2 H1, dP1, dQ1, minus d2 H1, dQ1, dP1. These two cancel each other out, and the answer is 0. So that first term vanishes. Now for the second term, number two, what I have is the first term vanished. So I have the H by dt. It is the integral over P1 and Q1. I have log of F1. F1 is a function of P1, and Q1, and t. I will focus, and make sure I write the argument of momentum, for reasons that will become shortly apparent. I have to multiply with the collision term. The collision term involves integrations over a second particle, over an impact parameter, a relative velocity, once I have defined what P2 and P1 are. I have a subtraction of F evaluated at P1, F evaluated at P2, plus addition, F evaluated at P1 prime, F evaluated at P2 prime. Eventually, this whole thing is only a function of time. There are a whole bunch of arguments appearing here, but all of those arguments are being integrated over. In particular, I have arguments that are indexed by P1 and P2. These are dummy variables of integration. If I have a function of x and y that I'm integrating over x and y, I can call x "z." I can call y "t." I would integrate over z and t, and I would have the same answer. I would have exactly the same answer if I were to call all of the dummy integration variable that is indexed 1, "2." Any dummy variable that is indexed 2, if I rename it and call it 1, the integral would not change. If I do that, what do I have? I have integral over Q-- actually, let's get of the integration number on Q. It really doesn't matter. I have the integrals over P1 and P1. I have to integrate over both sets of momenta. I have to integrate over the cross section, which is relative between 1 and 2. I have V2 minus V1, rather than V1 minus V2, rather than V2 minus V1. The absolute value doesn't matter. If I were to replace these indices with an absolute value, [? or do a ?] V2 minus V1 goes to minus V1 minus V2. The absolute value does not change. Here, what do I have? I have minus F of P1. It becomes F of P2, F of P1, plus F of P2 prime, f of P1 prime. They are a product. It doesn't really matter in which order I write them. The only thing that really matters is that the argument was previously called F1 of P1 for the log, and now it will be called F1 of P2. Just its name changed. If I take this, and the first way of writing things, which are really two ways of writing the same integral, and just average them, I will get 1/2 an integral d cubed Q, d cubed P1, d cubed P2, d2 b, and V2 minus V1. I will have F1 of P1, F1 of P2, plus F1 of P1 prime, F1 of P2 prime. Then in one term, I had log of F1 of P1, and I averaged it with the other way of writing things, which was log of F-- let's put the two logs together, multiplied by F1. So the sum of the two logs I wrote, that's a log of the product. I just rewrote that equation. If you like, I symmetrized It with respect to index 1 and 2. So the log of 1, that previously had one argument through this symmetrization, became one half of the sum of it. The next thing one has to think about, what I want to do, is to replace primed and unprimed coordinates. What I would eventually write down is d cubed P1 prime, d cubed P2 prime, d2 b, V2 prime minus V1 prime, minus F1 of P1 prime, F1 of P2 prime, plus F1 of P1, F1 of P2. Then log of F1 of P1 prime, F1 of P2 prime. I've symmetrized originally the indices 1 and 2 that were not quite symmetric, and I end up with an expression that has variables P1, P2, and functions P1 prime and P2 prime, which are not quite symmetric again, because I have F's evaluated for P's, but not for P primes. What does this mean? This mathematical expression that I have written down here actually is not correct, because what this amounts to, is to change variables of integration. In the expression that I have up here, P1 and P2 are variables of integration. P1 prime and P2 prime are some complicated functions of P1 and P2. P1 prime is some complicated function that I don't know. P1, P2, and V, for which I need to solve in principle, is Newton's equation. This is similarly for P2 prime. What I have done is I have changed from my original variables to these functions. When I write things over here, now P1 prime and P2 prime are the integration variables. P1 and P2 are supposed to be regarded as functions of P1 prime and P2 prime. You say, well, what does that mean? You can't simply take an integral dx, let's say F of some function of x, and replace this function. You can't call it a new variable, and do integral dx prime. You have to multiply with the Jacobian of the transformation that takes you from the P variables to the new variables. My claim is that this Jacobian of the integration is, in fact, the unit. The reason is as follows. These equations that have to be integrated to give me the correlation are time reversible. If I give you two momenta, and I know what the outcomes are, I can write the equations backward, and I will have the opposite momenta go back to minus the original momenta. Up to a factor of minus, you can see that this equation has this character, that P1, P2 go to P1 prime, P2 prime, then minus P1 prime, minus P2 prime, go to P1, and P2. If you sort of follow that, and say that you do the transformation twice, you have to get back up to where a sign actually disappears to where you want. You have to multiply by two Jacobians, and you get the same unit. You can convince yourself that this Jacobian has to be unit. Next time, I guess we'll take it from there. I will explain this stuff a little bit more, and show that this implies what we had said about the Boltzmann equation.
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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 far so good. So here is the kind of very entertaining thing that happens when you try to do some physics with this. And this was done by two physicists, Elitzur and Vaidman in Tel Aviv, they invented or fantasized about some sort of bombs-- things that explode. So they're called Elitzur-Vaidman bombs. And you could invent different things, but here is what an Elitzur-Vaidman bomb is-- some sort of bomb, and the way it works is with a photon detector. So there's a little tube in the bomb, and there's a photon detector. And you have your bomb, and you want to detonate it-- you send the photon in, you send the photon in through the tube. And the photon, it's detected by the detector. And the bomb explodes. On the other hand, if the bomb is defective, the photon goes in, and the detector doesn't work. The photon goes out. Just goes through. So that's an Elitzur-Vaidman bomb. And here is the puzzle for you-- suppose you have those bombs, and unfortunately, those bombs, after time, they decay. And sometimes the detectors go wrong, and they don't work anymore. So you have 10 bombs, and you know, maybe five have gone wrong. And now, you have maybe a very important mission and you need the bomb that really works. So what do you do? Let's assume you cannot break apart the detector-- it's just too complicated. So you have the bombs, and you want to test them. If you send in a photon and nothing happens, the bomb is not working. But if you send in a photon, and the bomb is working-- explodes. So you cannot use it anymore. So the question that Elitzur and Vaidman pose, is there a way to certify that the bomb is working without exploding it? Can you do that? The answer looks absolutely impossible. And certainly, in classical physics, it's completely impossible. You either do the measurement to see if the detector works, and if it works, your lab goes off. It's totally destroyed. And if it doesn't work, well, OK, it's not a good bum anyways. So there's no way out. But there is a way out, and it is to insert this bomb in the mass [? center ?] interferometer. So here we go. We put the mass [? center ?] interferometer, and we insert the bomb here with the detector along this place. D 0 and D 1 are still here. And now, you put this, and you send in a photon. So let's see what happens if you send in a photon. Suppose the bomb is defective-- bomb is defective. So what are the possible outcomes? Outcome and probability. Photon goes to D 0-- 0. Photon to D 1, bomb explodes. Well, we said the bomb is defective. So if the bomb is defective, we said it's like a detector that doesn't work, and lets the photons go through. So if the bomb is defective, it's as if there no bomb here, and you have the situation where all is open. So there will be a probability of 1 to get the photon to D 0-- a probability of 0 to get the photon to D 1. And the bomb, of course, doesn't explode-- probability of 0. On the other hand, suppose the bomb is good-- bomb is good. And then, what are the outcomes? And what are the probabilities? Well, you know, more or less, what's happening already. The bomb is good means there is a detector that never fails to detect the photon. And if a photon comes in, it will capture it-- it will block it. The bomb will explode. So you have your mass [? center ?] interferometer, and you've really done the equivalent of this-- if the bomb is really working. You've put a block of concrete-- it's going to absorb the photon. So if the bomb is really working, the outcome are the following-- well, I'm sorry to say, your lab will explode half of the times, because the photon on the block happens, and bomb explodes with probability 1/2. On the other hand, in this situation, it is possible that the photon-- photon-- at D 0, and bomb doesn't explode-- not explode. And there is a probability 1/4. And there is a probability, 1/4, that the photon is at D 1, and the bomb does not explode. But here is the catch now-- yes, half of the bombs exploded, we're sorry about that. But if the bomb doesn't work, there is no way a photon can reach D 1, because if a bomb doesn't work, all photons go to D 0. So the fact that some photons, a quarter percent of the time, 1/4-- 25% of the time-- reach D 1 implies that photon is at D 1, and bomb did not explode. But the bomb is good. So look what has happened-- it's really strange. The photon went-- the bomb was here, it was ready to explode. The photons kept the bomb, and ended at D 1, and you still know that the bomb works now-- even though the photon never went through the detector. It never touched here, it never went inside and get detected. Somehow it went through the other way, but you know that the bomb is working, with a quarter percent efficiency. We will do exercises, and it's possible to raise the efficiency to 50%. And if you put the bomb inside a cavity, a resonant cavity with photons going around, you can reach 99% efficiency. So the probability of blowing up MIT goes down to 1%. [LAUGHTER] I don't know if we can live with that-- I don't think so. But anyway, this is a true fact-- experiments without bombs have been done, and it shows that in quantum mechanics, you can do very surprising measurements.
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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 we have this integral. And with-- let me go here, actually. With the counter gamma equal to C1, this counter over here, and the constant c equal to 1. So C1 and the constant c equal to 1. This psi that we have defined, psi of u, is in fact the airy function of u. A i of u. I is not-- I think I tend to make that mistake. I doesn't go like a subscript. It's A i like the first two letters of the name. So that's the function A i of u. And now you could ask, well, what is the other solution? Now, in fact, this diagram suggests to you that there's other solutions because you could take other counters and make other solutions. In fact, yes, there are other ways. For example, if you did a counter-- do I have a color? Other color? Yes. If you did a counter like this, yellow and yellow, this is not the same solution. It is a solution because of the general argument and because the endpoints are in these regions where things vanish at infinity. So the yellow thing is another solution of the differential equation. So the other airy function is defined, actually, with this other counter. It's defined by taking the yellow counter like this. This is going to be called C2. A counter like that. It just comes parallel to this one and then goes down. And, actually, in order to have a nicely defined function, one chooses for the function B i of u the following. Minus i times the integral over the counter C1 of the same integrant. So I will not copy it. Always the same integrant. Plus 2 i terms the integral over the counter C2 of the same thing. So the B i function is a little unusual in that it has kind of a little bit of the A i function because you also integrate over C1. But you integrate as well over C2. That guarantees that-- actually, this second airy function behaves similar to A i for negative u, and while A i goes to 0 for positive u, this one will diverge. There are expressions for this function. I'll give you an integral. 2 integrals that are famous are A i of u equals 1 over pi. Integral from 0 to infinity. d k cosine. k cubed over 3 plus k u. And for B i of u. 1 over pi-- this is a little longer. Integral from 0 to infinity as well. d k. And you have an exponential of minus k cubed over 3 plus k u, plus the sine of k cubed over 3, plus k u. And that's it. It's kind of funny. One is the cosine and one is the sine. And it has this extra different factors. So these are your two functions. And of the relevance to our w k b problem is that they're necessary to connect the solutions, as we will see. But we need a little more about these functions. We need to know there are asymptotic behaviors. Now there are functions, like the exponential function, that has a Taylor series, e to the z. 1 plus z plus z squared over 2. And it's valid, whatever these angles-- the argument of z is. That's always the same asymptotic expansion, or the same Taylor series. For this functions, like the airy function, for some arguments of u, there's one form to the asymptotic expansion. And for some other arguments, there's another form. That's sometimes called the Stokes phenomenon. And for example, the expansion for positive u is going to be a decaying exponential here. But for negative u, it will be oscillatory. So it's not like a simple function, like the exponential function has a nice, simple expansion everywhere. It just varies. So one needs to calculate this asymptotic expansions, and I'll make a small comment about it of how you go about it. The nice thing about these formulas is that they allow you to understand things intuitively and derive things yourself. Here you see the two airy functions. They make sense. The other thing that is possible to do with this counter representations is to find the asymptotic expansions of these functions. And they don't require major mathematical work. It's kind of nice. So let's think a little about one example. If you have the airy function A i of u that is of the form integral 1 over-- well, it begins v k over 2 pi. The integral over counter C1. Maybe I should have done them curly. Curly C1 would have been clearer. e to the i. k cubed over 3 plus k u. That is your integral. And this is the phase of integration. Phase. Now, in order to find the asymptotic expansion for this thing, we'll use a stationary phase condition. So the integral is dominated by those places where the phase is stationary. So the 5 prime of k is k squared plus u. And we want this to be equal to 0. So take, for example, u positive. Suppose you want to find the behavior of the airy function on the right of the axis. Well, this says k squared is equal to u. So k squared is equal to minus u. And that's-- the right hand side is negative because u is positive. And therefore, k-- the points where this is solved are k equal plus minus i square root of u. So where are those points in the k plane? They're here and there. And those are the places where you get stationary phase. So what can you do? You're supposed to do the integral over this red line. C1. Well, as we argued, this can be lifted and we can make the integral pass through here. You can now do this integral over here. It's the same integral that you had before. In this line, we can say that k is equal to i square root of u plus some extra k tilde. We say here is i square root of u for u positive. There is this other place, but that-- we cannot bring the counter down here because in this region, the end points still contribute. So we cannot shift the counter down, but we can shift it up. So we have to do this integral. So what happens? You can evaluate that phase 5k under these conditions. It is a stationary phase point-- this one-- so the answer is going to have a part independent of k tilde, a linear part, with respect to k tilde that will vanish because at this point the phase is stationary. And then a quadratic part with respect to k tilde. So the phase, 5k, which is k cubed over 3 plus k u. When you substitute this k here, it's going to give you an answer. And the answer is going to be 2 over 3 i u to the 3/2, plus i square root of u k squared tilde, plus k tilde cubed over 3. That's what you get from the phase. You say, well, that's pretty nice because now our integral psi has become the integral d k tilde over 2 pi. So you pass from k to k tilde variable. e to the minus 2/3. u to the 3/2. That is because you have i times 5k, so you must multiply by i here. Then minus square root of u k squared. Then plus k cubed. OK. Tildes of course. AUDIENCE: [INAUDIBLE] PROFESSOR: Um. Yes. i k. This should be like that. Yes. OK. So now what happens? If u is large enough, this quantity over here is going to mean that this integral is highly suppressed over k tilde. It's a Gaussian with a very narrow peak. And therefore, we can ignore this term. This term is a constant. So what do we get from here? We get 1 over 2 pi from the beginning. This exponential. e to the minus 2 over 3. u to the 3/2. 3/2. And then from this Gaussian, we get square root of pi over square root of square root of u. So it's u to the 1/4. So I think I got it all correct. And therefore, the function A i of u goes like-- or it's proportional to this decaying exponential. Very fast decaying exponential. 1 over 2. 1 over square root of pi. 1 over u to the 1/4. e to the minus 2/3. u to the 3/2. And this is when u is greater than 0 and, in fact, u much greater than 1, positive and large. So this shows you the power of this method. This is a very powerful result. And this is what we're going to need, pretty much, now in order to do our asymptotic solutions and the matching of w k b. There is an extra term, or an extra case, you can consider. What happens to this integral when u is negative? When u is negative, there's two real roots, and the stationary points occur on the real line. The integral is, therefore, a little more straightforward. You don't have to even move it. And we have another expansion, therefore, which is A i of u is actually equal to 1 over the square root of pi. 1 over u, length of u to the 1/4. Cosine of 2/3. Length of u to the 3/2 minus pi over 4. This is for u less than 0. So the airy function becomes an oscillatory function on the left. So how does this airy function really look like? Well, we have the true airy function behaves like this. It's a decaying exponential for u positive. And then a decaying oscillatory function for u negative. For u negative, it decays eventually. The frequency becomes faster and faster. And that's your airy function A i of u. Here is u. Your asymptotic expansions, the ones that we found up there for u greater than 1, match this very nicely. But eventually they blow up, so they're a little wrong. And they actually match here quite OK. But here they go and also blow up because of this factor. So they both blow up, but they're both quite wrong in this area, which you would expect them to be wrong. These are the regions where this two asymptotic expansions make no sense. They were calculated on the hypothesis that u is much bigger than 1, or much less than 1. Minus 1. And therefore, you get everything under control except this area. So now let's do the real work of w k b. That was our goal from the beginning, so that's what we'll try to do now.
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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: OK, so welcome back to computational systems biology, I'm David Gifford. I'm delighted to be with you here here today. And today we're going to be talking about a topic that is central to modern high throughput biology, which is understanding how to do short read alignment, sometimes called read mapping. Now it's very important to me that you understand what I'm about to say today, and so I'm hopeful that you'll be uninhibited to raise your hand and ask questions about the fine points in today's lecture if you have any, because I'd be totally delighted to answer any questions and we have enough time today that we can spend time looking at one aspect of this problem and understand it thoroughly. An associated topic is the question library complexity. How many people have heard of sequencing libraries before? Let's see a show of hands. OK, how many people have heard of read alignment before, read mapping? OK, great, fantastic. Let's start with what we're going to be talking about today. We're going to first begin talking about what a sequencing library is and what we mean by library complexity. We'll then turn to what has been called a full text minute size index, sometimes called a burrows Wheeler transform index, a BWT index, an FM index, but this is at the center of most modern computational biology algorithms for processing high throughput sequencing data. And then we'll turn how to use that type of index for read alignment. So let's start now with what a sequencing library Is. Let's just say that you have a DNA sample, we'll be talking about various ways of producing said samples throughout the term. But we're going to assume that we have a bunch of different DNA molecules. And I'll illustrate the different molecules here in different colors. And we have three different types of molecules here. Some molecules are duplicated, because as you know, typically, we're preparing DNA from an experiment where there are many cells and we can get copies of DNA from those cells or the DNA could be amplified using PCR or some other technique. So we have this collection of molecules, and to make a library, we're going to process it. And one of the things that we'll do when we process the library is we'll put sequencing adapters on. These are short DNA sequences that we put on to the end of the molecules to enable them to have defined sequences at the ends which permits sequencing. Now, if somebody hands you a tube of DNA like this, there are a couple questions you could ask. You could check the DNA concentration to find out how DNA is there, you could run a gel to look at the size of the fragments that you're sequencing. We'll be returning to that later, but these are typically called the insert sizes of the library that you're sequencing, the total length of the DNA excluding the adapters. But we could also ask questions about how complex this library is, because it's possible to run experiments where you produce libraries that are not very complex, where they don't have very many different types of molecules. Now that typically is a failure of the experiment. So an important part of quality control is characterizing the library complexity where we want to figure out here complexity is equal to 3. There are three different types of molecules. And we sample these molecules. And when we sample them, we get a bunch of DNA sequence reads. And typically the number of reads that we get is larger than the complexity of the library. Here we have a total of 12 different reads. And when we sequence a library, we're sampling from it. And so the probability that we get any one particular molecule is going to be roughly speaking equal to 1 over c, which is the complexity. And thus, we could use the binomial distribution to figure out the likelihood that we had exactly four of these type one molecules. However, as n the number of sequencing reads grows to be very large, typical numbers are a hundred million different reads, the binomial becomes cumbersome to work with. And so we typically are going to characterize this kind of selection process with a different kind of distribution. So one idea is to use a Poisson, where we say that the rate of sequencing is going to be n over c. And we can see that here shown on the slide above is the same process where we have the ligation of the adapters. We have a library and we have reads coming from the library. We have a characterized library complexity here, there are four different types of molecules. And the modeling approach is that assuming that we have c different unique molecules, the probability that we'll get any one of them when we're doing the sequencing is 1 over c. And if we do end sequencing reads, we can find out the probability that we'll get a certain number of each type of molecule. Let's just stick with the first one to start, OK? Now part of the challenge in analyzing sequencing data is that you don't see what you don't sequence. So things that actually occur 0 times in your sequencing data still may be present in the library. And what we would like to do is from the observed sequencing data, estimate the library complexity. So we have all of the sequencing data, we just don't know how many different molecules there are over here. So one way to do with this is to say that let us suppose that we make a histogram of the number of times we see distinct molecules and we're going to say that we can observe molecules that are sequenced or appear l times up through r times. So we actually can create a version of the distribution that characterizes just a part of what we're seeing. So if we do this, we can build a Poisson model and we can estimate lambda from what we can observe. We don't get to observe things we don't see. So for sure, we know we can't observe the things that are sequenced 0 times. But for the things that are sequenced at least one time, we can build an estimate of lambda. And from that estimate of lambda, we can build an estimate of C. So one way to look at this is that if we look at the total number of unique molecules that we sequence, which is equal to m, then the probability that we observe between l and r occurrences of a given individual sequence times c is going to be equal to the total number of unique molecules that we observe. Another way to look at this is the very bottom equation where we note that if we look at the total complexity of the library and we multiply it by 1 minus the probability that we don't observe certain molecules, that will give an estimate of the total number of unique molecules that we do see. And thus we can manipulate that to come up with an estimate of the complexity. Are there any questions about the details of this so far? OK, so this is a very simple model for estimating the complexity of a library based upon looking at the distribution of reads that we actually observe for quality control purposes. And let us suppose that we apply this to thousands genomes data, which is public data on human. And suppose we want to test whether this model works or not, so what we're going to do is we're going to estimate the library complexity from 10 percent of the sequencing reads, so we'll pick 10 percent of the reads of an individual at random, we'll estimate the complexity of the library, and then we'll also take all of the region the individual and estimate the complexity. And if our estimator is pretty good, we should get about the same number, from 10 percent of the reads and from all of the reads. Will people go along with that? Think that seems reasonable? OK, so we do that. And this is what we get. And it's hard to see the diagonal line here, but there's a big oops here. And the big oops is that if we estimate the library complexity from just 10 percent of the reads, it's grossly underestimating the number of unique molecules we actually have. In fact, it's off by typically a factor of two or more. So for some reason, even though we're examining millions of reads in this subsample, we're not getting a good estimate of the complexity of the library. Does anybody have any idea what could be going wrong here? Why is it that this very simple model that is attempting to estimate how many different molecules we have here based upon what we observe is broken? Any ideas at all? And please say your name first. AUDIENCE: I'm Chris. PROFESSOR: Hi Chris. AUDIENCE: Is it because repeated sequences, so there could be a short sequence at the end of one molecule that's the beginning of another one, middle of another one, so [INAUDIBLE]. PROFESSOR: Chris, you're on the right track, OK? Because what we have assumed at the outset was that all of these molecules occur with equal probability. Right? What would happen if in fact there are four copies of this purple one and only two copies of the other molecules? Then the probability of sampling this one is going to be twice as high as the probability of sampling one of these. If there's non uniformity in the original population, that's going to mess up our model big time. And that could happen from repeated sequences or other kinds of duplicated things, or it could be that there's unequal amplification. It might be that PCR really loves a particular molecule, right, and amplifies that one a lot, and doesn't amplify another one that's difficult to amplify. So somewhere in our experimental protocol pipeline, it could be that there's non uniformity and thus we're getting a skewness to our distribution here in our library. So the other thing that's true is in a Poisson, lambda, which is equal to the mean, is also equal to the variance. And so our Poisson's only one knob we could turn to fit the distribution. So coming back to this, we talked about the idea that the library complexity still may be four but then there may be different numbers of molecules of each type. And here's an idea for you, right? The idea is this. Imagine that the top distributions are the number of each type of molecule that are present. And it might be that our original assumption was that it was like the very top, that typically there are two copies of each molecule in the original sequencing library, and that's a fairly tight distribution. But it could be, in fact, that the number of molecules of each type is very dispersed. And so if we look at each one of those plots at the top, the first four, those are going to be our guesses about the distribution of the number of copies of a molecule in the original library. And we don't know what that is, right? That's something we can't directly observe, but imagine that we took that distribution and used it for lambda in our Poisson distribution down below for sampling. So we have one distribution over the number of each type of molecule we have and we have the Poisson for sampling from that, and we put those two together. And when we do that, we have the Poisson distribution at the top, the gamma distribution is what we'll use for representing the number of different species over here and their relative copy number. And when we actually put those together as shown, we wind up with what's called the negative binomial distribution, which is a more flexible distribution, it has two parameters. And that negative binomial distribution can be used, once again, to estimate our library complexity. And when we do so, we have lamba be the same, but k is a new parameter. It measures sort of the variance or dispersion of this original sequencing library. And then when we fit this negative binomial distribution to that 1,000 genomes data, it's going to be hopefully better. Let's start with a smaller example. If we have a library that's artificial with a known million unique molecules and we subsample, it gives you 100,000 reads, you can see that with different dispersions here in the left, k with different values from 0.1 to 20, the Poisson begins to grossly underestimate the complexity of the library as the dispersion gets larger, whereas the negative binomial, otherwise known as the GP or gamma Poisson, does a much better job. And furthermore, when we look at this, in the context of the thousand genomes data, you can see when we fit this how much better we are doing. Almost all those points are almost exactly on the line, which means you can take a small sampling run and figure out from that sampling run how complex your library is. And that allows us to tell something very important, which is what is the marginal value of extra sequencing. So for example, if somebody comes to you and they say, well, I ran my experiment and all I could afford was 50 million reads. Do you think I should sequence more? Is there more information in my experimental DNA preparation? It's easy to tell now, right? Because you can actually analyze the distribution of the reads that they got and you can go back and you could estimate the marginal value of additional sequencing. And the way you do that is you go back to the distribution that you fit this negative binomial and ask if you have r more reads, how many more unique molecules are you going to get? And the answer is that you can see that if you imagine that this is artificial data, but if you imagine that you had a complexity of 10 to the 6 molecules, the number sequencing regions is on the x-axis, the number of observed distinct molecules is on the y-axis, and as you increase the sequencing depth, you get more and more back to the library. However, the important thing to note is that the more skewed the library is, the less benefit you get, right? So if you look at the various values of k, as k gets larger, the sort of the skewness of the library increases, and you can see that you get fewer unique molecules as you increase the sequencing depth. Now I mention this to you because it's important to think in a principled way about analyzing sequencing data. If somebody drops 200 million reads on your desk and says, can you help me with these, it's good to start with some fundamental questions, like just how complex is the original library and you think that these data are really good or not, OK? Furthermore, this is a introduction to the idea that certain kinds of very simplistic models, like Poisson models of sequencing data can be wrong because they're not adequately taking into account the over dispersion of the original sequencing count data. OK, so that's all there is about library complexity. Let's move on now to questions of how to deal with these reads once we have them. So the fundamental challenge is this. I hand you a genome like human. 3 times 10 to the ninth bases. This will be in fast a format, let's say. I had you reads. And this will be-- we'll have, say, 200 base pairs times 2 times 10 to the eighth different reads. And this will be in fast q format. The q means that there are-- it's like fast a except that our quality score's associated with each particular base position. And the PHRED score which is typically used for these sorts of qualities, is minus p minus 10 times log base 10 of the probability of an error. Right, so a PHRED score of 10 means that there's a 1 in 10 chance that the bases is an error, a PHRED score of 20 means it's one in a 100, and so forth. And then the goal is today if I give you these data on a hard drive, your job would be to produce a SAM file, a Sequence Alignment and Mapping file, which tells us where all these reads map in the genome. And more pictorially, the idea is that there are many different reasons why we want to do this mapping. So one might be to do genotyping. You and I differ in our genomes by about one base in a thousand. So if I sequence your genome and I map it back or align it to the human brain reference genome, I'm going to find differences between your genome and the human reference genome. And you can see how this is done at the very top where we have the aligned reads and there's a G, let's say, in the sample DNA, and there's a C in the reference. But in order to figure out where the differences are, we have to take those short reads and align them to the genome. Another kind of experimental protocol uses DNA fragments that are representative of some kind of biological process. So here the DNA being produced are mapped back to the genome to look for areas of enrichment or what are sometimes called peaks. And there we want to actually do exactly the same process, but the post processing once the alignment is complete is different. So both of these share the goal of taking hundreds of millions of short reads and aligning them to a very large genome. And you heard about Smith Waterman from Professor Berg, and as you can tell, that really isn't going to work, because its time complexity is not going to be admissible for hundreds of millions of reads. So we need to come up with a different way of approaching this problem. So finding this alignment is really a performance bottleneck for many computational biology problems today. And we have to talk a little bit about what we mean by a good alignment, because we're going to assume, of course, fewer mismatches are better. And we're going to try and align to high quality bases as opposed to low quality bases and note that all we have in our input data are quality scores for the reads. So we begin with an assumption that the genome is the truth and when we are aligning, we are going to be more permissive of mismatches in read locations that have higher likelihood of being wrong. So is everybody OK with the set up so far? You understand what the problem is? Yes, all the way in the back row, my back row consultants, you're good on that? See, the back row is always the people I call on for consulting advice, right? So yeah. You're all good back there? Good, I like that, good, that's good, I like that, OK. All right. So now I'm going to talk to you about what are the most amazing transforms I have seen. It's called the Burrows Wheeler Transform. And it is a transform that we will do to the original genome that allows us to do this look up very, very quickly. And it's worth understanding. So here's the basic idea behind the Burrows Wheeler Transform. We take the original string that we want to use as our target that we're going to look things up in, OK, so this is going to be the dictionary looking things up in and it's going to be the genome sequence. And you can see the sequence on the left hand side, ACA, ACG, and the dollar sign represents the end of string terminator. OK. Now here's what we're going to do. We take all possible rotations of this string, OK? And we're going to sort them. And the result of sorting all the rotations is shown in the next block of characters. And you can see that the end of string character has the shortest sorting order, followed by A, C, and G and that all the strings are ordered lexically by all of their letters. So once again, we take the original input string, we do all of the possible rotations of it, and then we sort them and wind up with this Burrows Wheeler Matrix as it's called in this slide, OK? And we take the last column of that matrix and that is the Burrows Wheeler Transform. Now yo might say, what on earth is going on here? Why would you want to take a string or even an entire genome? We actually do this on entire genomes, OK? Consider all the rotations of it, sort them, and then take the last column of that matrix. What could that be doing, OK? Here's a bit of intuition for you. The intuition is that that Burrows Wheeler Matrix is representing all of the suffixes of t. OK, so all the red things are suffixes of t in the matrix. And when we are going to be matching a read, we're going to be matched it from its end going towards the beginning of it, so we'll be matching suffixes of it. And I'm going to show you a very neat way of using this transform to do matching very efficiently. But before I do that, I want you to observe that it's not complicated. All we do is we take all the possible rotations and we sort them and we come up with this transform. Yes. AUDIENCE: What are you sorting them based on? PROFESSOR: OK, what was your name again? AUDIENCE: I'm Samona. PROFESSOR: Samona. What are we sorting them based upon? We're just sorting them alphabetically. AUDIENCE: OK. PROFESSOR: So you can see that if dollar sign is the lowest alphabetical character, that if you consider each one a word, that they're sorted alphabetically, OK? So we have seven characters in each row and we sort them alphabetically. Or a lexically. Good question. Any other questions like that? This is a great time to ask questions, because what's going to happen is that in about the next three minutes if you lose your attention span of about 10 seconds, you're going to look up and you'll say, what just happened? Yes. AUDIENCE: Could you explain the suffixes of t? PROFESSOR: The suffixes of t? Sure. Let's talk about the suffixes of tr. They're all of the things that end t. So a suffix of t would be G, or CG, or ACG, or AACG, or CAACG, or the entire string t. Those are all of the endings of t. And if you look over on the right, you can see all of those suffixes in red. So one way to think about this is that it's sorting all of the suffixes of t in that matrix. Because the rotations are exposing the suffixes, right, is what's going on. Does that make sense to you? Now keep me honest here in a minute, OK, you'll help me out? Yes. Your name first? AUDIENCE: [INAUDIBLE]. [INAUDIBLE] PROFESSOR: [INAUDIBLE]. AUDIENCE: What is dollar sign? PROFESSOR: Dollar sign is the end of string character which has the lowest lexical sorting order. So it's marking the end of t. That's how we know that we're at the end of t. Good question. Yes. AUDIENCE: Can you sort them non-alphabetically, just different ways to sort them [INAUDIBLE] algorithm? PROFESSOR: The question is, can you sort them non alphabetically. You can sort them any way as long as it's consistent, OK. But let's stick with alphabetical lexical order today. It's really simple and it's all you need. Yes. in red is the suffixes in the last colored group on the right? PROFESSOR: No, no. AUDIENCE: What's in red? PROFESSOR: What's in red are all the suffixes of T on the very far left. OK? AUDIENCE: On the right, last column group? PROFESSOR: The right last column group. That last column in red, that is the Burrows-Wheeler Transform, read from top to bottom. OK? And I know you're looking at that and saying, how could that possibly be useful? We've taken our genome. We've shifted it all around. We've sorted it, we take this last thing. It looks like junk to me, right? But you're going to find out that all of the information in the genome is contained in that last string in a very handy way. Hard to believe but true. Hard to believe but true. Yes. Prepare to be amazed, all right? These are all great questions. Any other questions of this sort? . OK. So, I'm going to make a very important observation here that is going to be crucial for your understanding. So I have reproduced the matrix down on this blackboard. What? That's usually there under that board, you know that. You guys haven't checked this classroom before, have you? No. It's always there. It's such a handy transform. So this is the same matrix as the matrix you see on the right. I'm going to make a very, very important assertion right now. OK? The very important assertion is that if you consider that this is the first a in the last column that is the same textual occurrence in the string as the first a in the first column. And this is the second a in the last column, that's the same as the second a in the first column. And you're going to say, what does he mean by that? OK, do the following thought experiment. Look at the matrix. OK? And in your mind, shift it left and put all the characters on the right hand side. OK? When you do that, what will happen is that these things will be used to sort the occurrences a on the right hand side. Once again, if you shift this whole thing left and these pop over to the right, then the occurrence of these a's will be sorted by these rows from here over. But these are alphabetical. And therefore they're going to certain alphabetical order. And therefore these a's will sort in the same order here as they are over there. So that means that when we do this rotation, that this textual occurrence of a will have the same rank in the first column and in the last column. And you can see I've annotated the various bases here with their ranks. This is the first g, the first c, the first end of line, end of string character. First a, second a, third a, second c. And correspondingly I have the same annotations over here and thus the third a here is the same lexical occurrence as the third a on the left in the string, same text occurrence. Now I'm going to let that sink in for a second, and then when somebody asks a question, I'm going to explain it again because it's a little bit counterintuitive. But the very important thing is if we think about textual recurrences of characters in that string t and we put them in this framework, that the rank allows us to identify identical textual recurrences of a character. Would somebody like to ask a question? Yes. Say your name and the question, please. AUDIENCE: Dan. PROFESSOR: Dan. AUDIENCE: So in your original string though those won't correspond to the same order in the transformed string. So like the a's in the original string in their order, they don't correspond numerically to the transformed string. PROFESSOR: That's correct. Is that OK? The comment was that the order in BWT, the transform is not the same as the order in the original string. And all I'm saying is that in this particular matrix form, that the order on the last column is the same as the order in the first column for a particular character. And furthermore, that these are matching textual occurrences, right? Now if I look at a2 here, we know that c comes after it, then a, then a, and c and g, right? Right. OK, so did that answer your question that they're not exactly the same? AUDIENCE: Yes. I don't understand how they're useful yet. PROFESSOR: You don't understand how it's useful yet. OK. Well, maybe we better get to the useful part and then you can-- OK. So let us suppose that we want to, from this, reconstruct the original string. Does anybody have any ideas about how to do that? OK. Let me ask a different question. If we look at this g1, right? And then this is the same textual occurrence, right? And we know that this g1 comes right before the end of character, in end of string terminator, right? So if we look at the first row, we always know what the last character was in the original string. The last character is g1, right? Fair enough? OK Where does g1 would occur over here? Right over here, right? What's the character before g1? c2. where is c2 over here? What's the character before c2? a3. What's the character before a3? a1. Uh, oh. Let me just cheat a little bit here. a1 a3 c2 g1 $. So we're at a1, right? What's the character before a1? c1, right? What's the character before c1? a2. And what's the character before a2? That's the end of string. Is that the original string that we had? Magic. OK? Yes. AUDIENCE: Wouldn't it be simpler to look at-- to just remove the dollar sign [INAUDIBLE] or do you mean reconstruct from only the transformed? PROFESSOR: We're only using this. This is all we have. Because I actually didn't use any of these characters. I was only doing the matching so we would go to the right row. Right? I didn't use any of this. And so, but do people understand what's going on here? If anybody has any questions, now is a great time to raise your hand and say-- here we go. We have a customer. Say your name and the question, please. AUDIENCE: My name is Eric. PROFESSOR: Thanks, Eric. AUDIENCE: Can you illustrate how you would do this without using any of the elements to the left of the box? PROFESSOR: Absolutely, Eric. I'm so glad you asked that question. That's the next thing we're going to talk about. OK, but before I get to there, I want to make sure, are you comfortable doing it with all the stuff on the left hand side? You're happy about that? OK. if anyone was unhappy about that, now would be the time to say, I'm unhappy, help me. How about the details? Everybody's happy? Yes. AUDIENCE: So, you have your original string in the first place, though, so why do you want to create another string of the same length? Like, how does this help you match your read? PROFESSOR: How does this help you match your read? How does this help you match your read, was the question. What is was your name? AUDIENCE: Dan. PROFESSOR: That's right, Dan. That's a great question. I'm so glad you asked it. First we'll get to Eric's question and then we'll get to yours. Because I know if I don't give you a good answer that you're going to be very mad, right? OK? Let's talk about the question of how to do this without the other things. So we're going to create something called the last to first function that maps a character in the last row, column, I should say, to the first column. And there is the function right there. It's called LF. You give it a row number. The rows are zero origined. And you give it a character and it tells you what the corresponding place is in the first column. And it has two components. The first is Occ, which tells you how many characters are smaller than that character lexically. So tells you where, for example, the a's start, the c's start, or the g's start. So in this case, for example Occ of c is 4. That is the c's start at 0, 1,2, 3, the fourth row. OK? And then count tells you the rank of c minus 1. So it's going to essentially count how many times c occurs before the c at the row you're pointing at. In this case, the answer is 1 and you add 1 and that gets you to 5, which is this row. So this c2 maps here to c2 as we already discussed. So this LF function is a way to map from the last row to the first row. And we need to have two components. So we need to know Occ, which is very trivial to compute. There are only five elements, one for each base and one for the end of line terminator, which is actually zero. So it will only have integers and count, which is going to tell us the rank in the BWT transform and we'll talk about how to do that presently. OK. So did that answer your question, how to do this without the rest of the matrix? Eric? AUDIENCE: Could you show us step by step on the blackboard how you would reconstruct it? PROFESSOR: How do we reconstruct it? AUDIENCE: Yeah. PROFESSOR: You mean something like this? Is this what you're suggesting? AUDIENCE: Somehow I get a feeling that the first column doesn't help us in understanding how the algorithm work only using the last column. PROFESSOR: OK. Your comment, Eric, is that you feel like the first column doesn't help us understand how the algorithm works, only using the last column, right? OK. AUDIENCE: [INAUDIBLE] going back to the first column of data. PROFESSOR: OK. Well let's compute the LF function of the character and the row for each one of these things, OK? And that might help you, all right? Because that's the central part of being able to reverse this transform. So this is, to be more clear, I'll make it more explicit. This is LF of I and BWT of i, where i goes from 0 to 6. So what is that value for this one right here? Anybody know? Well it would be Occ of g, which is 6, right? Plus count of of 6 n g, which is going to be 0. Or I can look right over here and see that in fact it's 6, right? Because this occurrence of g1 is right here. So this LF value is, it's 6 4 0 a1 is in 1, a2 is in 2, a3 is in 3, c2 is in 5. So this is the LF function, 6 4 0 1 2 3 5. And I don't need any of this to compute it. Because it simply is equal to, going back one slide, it's equal to Occ of c plus count. So it' going to be equal to where that particular character starts on the left hand side and its rank minus 1. And so these are the values for LF. This is what I need to be able to take this string and recompute the original string. If I can compute this, I don't need any of that. And to compute this, I need two things. I need Occ and I need count. All right? Now, I can tell you're not quite completely satisfied yet. So maybe you can ask me another question and it would be very helpful to me. AUDIENCE: How did you get G1's last and first functions score being 6? PROFESSOR: OK. Let's take that apart. We want to know what LF of 6 and where was that G1? G1 is 1 and 0, right? Sorry. LF of 1 and g is equal to, right? Is that g and 1 or 0? Oop, sorry it's in 0. So this is what you like me to compute, right? OK what's Occ of g? It's how many characters are less than g in the original string? I'll give you a clue. It's 1, 2, 3, 4, 5, 6. AUDIENCE: [INAUDIBLE]. PROFESSOR: No, it's how many characters are less than g in the original string. How many things are going to distort underneath it? Where do the g's begin in the sorted version? The g's begin in row 6. OK? So OCC of g is 6. Is that-- are you getting hung up on that point? AUDIENCE: Yes. How do you know that without ever referencing back to the first 5 columns? PROFESSOR: Because when we build the index we remember. AUDIENCE: Oh, OK. PROFESSOR: So we have to, I haven't told you this, but I need to compute, I need to remember ways to compute Occ and count really quickly. Occ is represented by four values only. They're where the a's start, the c's start, the t's start, and the g's start. That's all I need to know, four integers. OK? Are you happy with that? Occ, the a's start at 1. The c's start at 4 and the g's start at 6. OK? That's all I need to remember. But I precompute that. OK? Remember it. Are you happy with that no? AUDIENCE: Yes. PROFESSOR: OK. And then this business over here of count of zero and g. Right? Which is how many g's, what's the rank of this g in the right hand side? 1. Minus 1 is 0. So that's 0. That's how we computed it. OK? These are great questions because I think they're foundational. Yeah. AUDIENCE: Occ and count are both precomputed as you're building on this last column [INAUDIBLE]. PROFESSOR: They are precomputed and well, I have not told you, see, you're sort of ahead of things a little bit in that I'd hoped to suspend disbelief and that I could actually build these very efficiently. But yes, they're built at the same time as the index is built. OK? But yes. OK and if I have Occ and count and I have the string, then I can get this. But I can still need to get to Dan's question because he has been very patient over there. He wants to know how I can use this to find sequence region of genome. And he's just being so good over there. I really appreciate that, Dan. Thanks so much for that. Are you happier? OK you're good. All right. So and this is what we just did. The walk left algorithm actually inverts the BWT by using this function to walk left and using that Python code up there, you can actually reconstruct the original string you started with. So it's just very simple. And we went through it on the board. Are there any questions about it at all? AUDIENCE: Yes. Can you actually do this any column? Like, why are you using the last column instead of like, why can't you just change it like the [INAUDIBLE], the equation and make it work for-- PROFESSOR: Because a very important thing is that this is actually a very important property right? Which is all the suffixes are sorted here. And if we didn't do that though, I couldn't answer Dan's question. And he'd be very upset. So please be respectful of his interest here. Now, the thing is, the beauty of this is, is that I have all these suffixes sorted and what you're about to see is the most amazing thing, which is that we're going to snap our fingers and, bang, we can map 200 million reads in no time at all. You like that? You're laughing. Oh, oh, that's not a good That's not a good sign. Let's press ahead fearlessly, OK? And talk about how we're going to use this to map read. So we're going to figure out how to use this index and this transform to rapidly aligned reads to a reference genome. And we're not talking about one read or 10 reads or 1 million reads. We're talking about hundreds of millions of reads. So it has be very efficient indeed, OK? So here's the essential idea. There's the core algorithm on the slide. Which is that what we do is we take the original query that we have the read that we're trying to match and we're going to process it backwards from the end of the read forwards. And we begin by considering all possible suffixes from row zero to, in this case, it would be row seven. Which is the length of the entire transform. And we iterate and in each iteration we consider suffixes that match the query. So in the first step, right here, let me see if I can get my point working, there we are. So in the first step here, we matching this c. OK? And we compute the LF of the top, which is this row and of the bottom, which is down here pointing off the end, and that takes us to the first d here and to this point. Here are the two c's that could be possible matches to our query, which ends in a c. We then say, oh, the next character we have to match is an a. So we look here at the a we need to match, and starting from this row, which is row four, and this row, which is row six, we compute the LF of each one of these to figure out what rows in a precedes these c's. And the way we compute the LF is that we use the character a to be able to figure out which rows have the a preceding the c. You can see, when we compute those LF functions, what we wind up with are these rows where we have a followed by c. So we're beginning to match the end of our read, as we go from right to left. We then compute the same thing once again, considering the first a and ask what rows are going to allow us to put this a in front of the ac to form our entire read. And we compute the LF once again of these things. And you can see that here it takes us to this specific row aac. So that row represents a suffix that is matching our query exactly. So we iterate this loop to be able to match a read against the index. And we're using the LF function to do it. And it's a really beautiful algorithm. And remember, we only have the transform. We don't have the rest of this matrix. So before I press ahead and talk about other details, I think it's important to observe a couple of things that are a little bit counterintuitive about this. One counterintuitive aspect of it is, that when I'm over here for example, and for example when I'm computing the LF here, I'm computing the LF of row two with respect to a. But there's a dollar sign there. Right? So I'm using this to the LF function, to tell me where a suffix would be that actually follows my constraint of having to have an a be the prefix of ac, where I am right now. This code is actually not fake code. It's the actual code that's in a matcher, for matching a read against the index. Now let me just stop right here for a second and see if there any other questions. Dan is getting now his answer to his question, right? About how you actually use this for matching reads. You do this once for every read. And it is linear time. Right? It's the length of the read itself is all the time it takes to match in a huge genome. So once we've built the index of the genome, in fact, most of the time when you're doing this sort of mapping, you don't build the index. You download the index off of a website. And so you don't have to pay for the time to build this index. You just download the index and you take your reads and the time to match all of your sequencing reads against a certain build of whatever genome you're using is simply linear in the number of bases you have. Questions? Yes. And say your name and the question, please. AUDIENCE: How did [INAUDIBLE] come up with intuition [INAUDIBLE]? It seems like they just pulled it out of a hat. PROFESSOR: You know, I asked Mike that the other day. I saw him in a meeting and he sort of surprised at how this has taken off. And he told me some other interesting facts about this, which you probably could deduce. Which is that if you only want to match reads that are four long, you only have to sort this matrix by the first four characters. But there are other little tricks you can play here. Any other questions? Yes. AUDIENCE: I'm Deborah. What is the FM index? PROFESSOR: What is the FM index. Well, the guys who thought this up have the last initials of F and M, but that's not what it stands for, contrary to popular opinion. It stands for full text minute size. That's what they claim. So if you hear people talking about full text minute size indices, or FM indices, the Fm index is actually the part that was being asked over here, the Occ part and the LF part, how you actually compute those quickly. That was what FNM contributed to this but, generically when we're talking about this style of indexing, it's called FM indexing or you might hear, I'm using a BWT. Some people will say that. But that's what FM stands for. Does that answer your question? Excellent. OK. All right. Any-- these are all great questions. Yes. AUDIENCE: [INAUDIBLE]. PROFESSOR: Oh, you don't know that a and c are there, except that remember, if you look at the way that this is working, is that you're not actually reconstructing strings, you're only trying to find them. Right? And so at the end, top and bottom are going to point to the row that contains the suffix where your original read was. And now your next question is going to be, where is that in the genome? This doesn't do me any good. I mean, the number 1 doesn't help me out here, doesn't mean anything. Not good, right? So where is it in the genome is the next question. So we'll get to that in a second. What happens if you give me a read that doesn't match anywhere in this index? Well if you give me a read that doesn't match anywhere in this index, what happens is the top and bottom become the same. So on top and bottom become the same, it's a failed look up. All right? And that's because the suffix doesn't exist in the index. And once top and bottom become the same, they remain the same throughout that loop. Yes. AUDIENCE: I'm Sally. My main question is that this doesn't provide any leeway for errors. You kind of have to be able to present all of your rates. PROFESSOR: Sally, you're absolutely correct. I'm so glad you asked that question. Your observation is it does not provide any leeway for mismatches. And so unlike all the other algorithms we study, which had these very nice matrices and ability to assign weights to mismatches, this is only doing exact matching. And so what you need help understanding is, how we can deal with mismatches in the presence of this. And I'll get to that in less than 10 minutes. And it won't be quite as elegant as what you saw from Professor Berg but it's what everybody does. So that's my only excuse for it OK? Yes. AUDIENCE: What is the bottom set of arrows doing? What's its significance? PROFESSOR: The significance of top and bottom, that's a great question. What the significance of top and bottom? Top and bottom bracket in that original matrix, the suffixes that are matching the original query. OK? And so between top and bottom minus 1 are all of the rows that have a suffix that match our original query. And if top equals bottom, there are no matching suffixes. But assuming there are matching suffixes, those are rows that contain a matching suffix. And as we progress along, top and bottom change as the suffixes change as we expand the suffix to contain more and more bases. OK? OK. Any other questions? OK. So now back to the question over here, which is that OK, I know that we've matched, and I know that we have this hit. The question is, where is it in the Genome because the fact that it matched row one of my BWT matrix does me absolutely no good at all. Right? Anybody have any ideas about how we could figure out where it is the genome? Given what I've told you so far, which is that you have the BWT, you have Occ, and you have count, and you can compute the LF function. Any ideas? Doesn't matter how slow it is. OK well how could I figure out what came before aac in the genome? Yes. AUDIENCE: So, for, like at the beginning, we rebuilt this whole string, starting at the end. You could rebuild the whole string starting at, rebuild the whole genome starting at the 1 and see-- PROFESSOR: You could rebuild the entire genome that prepends or occurs before aac, right? AUDIENCE: Exactly. PROFESSOR: Exactly. So that's what we can do. We could actually do our walk left algorithm. We can walk left from there, find out that we go two steps until we hit dollar sign, and therefore, the offset is two, where it occurs in the genome. So we can give a match position by walking left. Does everybody see that, that we could walk left to figure where it is? It's not fast, but it works. Yes. AUDIENCE: I'm Ted. PROFESSOR: Hi, Ted. AUDIENCE: So now our function first has to take the read and it has to align it and the same, where the position is built into the end of the function, the speed of the function is now dependent on position as well. Is that right? Because the longer it takes, PROFESSOR: I was being a little bit glib find if it matches or not this linear time. Now you're saying, hey, wait a minute. I want to know where it is in the genome. That's a big bonus, right? And so you'd like to know where that is? Yes. But I still can do that in linear time. And we'll show you how to do that in a second. This is not linear time. This actually needs to walk back, order the length of genome for every single query. That's not good. All right? Well, What we could do is, we could store what's called a suffix array with each row and say, where in the genome that position is. Where that row starts. And then maybe a simple look-up. That when you actually have a hit in row one, ah, OK, start your position two of the genome. But then the problem with that is that it actually takes a lot of space. And we want to have compact indices. So the trick is what we do is, instead of storing that entire suffix array, we store every so many rows, like every 25 rows. And all we do is, we walk left until we hit a row that actually has the value, and then we add how many times we walked left, plus the value and we know where we are in the genome. So we can sample the suffix array, and by sampling the suffix array, we cut down our storage hugely and it's still pretty efficient. Because what we can do is, we just walk left until we hit a sample suffix array location and then add the two numbers together. All right? So that's how it's done. OK? So that's how we actually do the alignment and figure out where things are. The one thing I haven't told you about is how to compute count efficiently. Now remember what count does. Count is a function-- but this is putting it all together where we're matching this query, we do the steps. We get the match. Then we do walk left once and then we look at the suffix to figure out where we are, right? The business about count is that what we need to do is to figure out the rank of a particular base in a position in the transform. And one way to do that is to go backwards to the whole transform, counting how many g;s occur before this one, and that's very expensive, to compute the rank of this particular g. Remember the rank is simply the number of g's that occur before this one in the BWT. Very simple metric. So instead of doing that, what we can do is, we can build a data structure that every once in awhile, counts how many a's, c's, g's, and t's have occurred before now in the BWT. And so we're going to sample this with these checkpoints, and then when you want to compute count at any point, you can go to the nearest checkpoint, wherever that is, and make an adjustment by counting the number of characters between you and that checkpoint. Very straightforward. All Right So this, coming back to question, it's Time, right? --asked you need to build this checkpointing mechanism at the same time you build the index, as well as the sampling of the suffix array. So a full index consists of the transform itself, which is the genome transformed into its BWT. And they literally take the entire genome and do this. Typically they'll put dollar signs between the chromosomes. So they'll transform the whole thing. It takes a sampling of the suffix array we just saw and it takes the checkpointing of the LF function to make a constant time. And that's what is inside of an FM index. OK? Now it's small, which is one of the nice things, compared to things like suffix tree, suffix arrays, or even other kinds of hash structures for looking for seeds, it really is not even twice the size of the genome. So it's a very compact index that is very, very efficient. And so it's a wonderful data structure for doing what we're doing, except we have not dealt with mismatches yet, right? And so once again, I want to put a plug in for BWA which is really a marvelous aligner. And we'll talk about tomorrow in recitation if you want to know all of the details of what it actually takes to make this work in practice. Now, it finds exactness matches quickly, but it doesn't really have any allowances for mismatches. And the way that bow tie and other aligners deal with this, and they're all pretty consistent, is in the following way, which is that they do backtracking. Here's the idea. You try and match something or match a read and you get to a particular point in the read, and you can't go any further. Top is equal to bottom. So you know that there's no suffix in the genome that matches your query. So what do you do? Well, what you can do is you can try all of the different bases at that position besides the one you tried to see whether it matches or not. I can see the horror coming over people. Oh, no, not backtracking, not that. But sometimes it actually works. And just to give you order of magnitude idea about how this works in practice, when reads don't match, they limit backtracking to about 125 times in these aligners. so they try pretty hard to actually match things. And yes, it is true that even with this backtracking, it's still a great approach. And sometimes the first thing you try doesn't work, and you have to backtrack, trying multiple bases at that location until you get one that matches. And then you can proceed. OK And you eventually wind up at the alignment you see in the lower right hand corner, where you're substituting a g for an a, an a for a g, excuse me, to make it go forward. Do people understand the essential idea of this idea of backtracking? Does anybody have any comments or questions about it? Like ew, or ideas? Yes. AUDIENCE: What about gaps? PROFESSOR: What about gaps? BWA, I believe, processes gaps. But gaps are much, much less likely than missed bases. The other thing is that if you're doing a sequencing library, and you have a read that actually has a gap in it, it's probably the case you have another read that doesn't. For the same sequence. So it is less important to process gaps than it is to process differences. The reason is that differences mean that it might be a difference of an allele. In other words, it might be that your base is different than the reference genome. Indels are also possible. And there are different strategies of dealing with those. That would be a great question for Hang tomorrow about gaps. Because he can tell you in practice what they do. And we'll get into a little bit of that at the end of today's lecture. Yes. Question? AUDIENCE: I'm Levy. PROFESSOR: Hi, Levy. AUDIENCE: How do you make sure when you're backtracking that you end up with the best possible match? Do you just go down the first-- PROFESSOR: The questions is how do you guarantee you wind up with the best possible match? The short answer is that you don't. There's a longer answer, which we're about to get to, about how we try to approximate that. And what judgment you would use to get to what we would think is a practically good match. OK? But in terms of theoretically optimal, the answer is, it doesn't attempt to do that. That's a good question. Yes. AUDIENCE: In practice, does this backtracking method at the same time as you're computing the matches or-- PROFESSOR: Yes. So what's happening is, you remember that loop where we're going around, where we were moving the top and bottom pointers. If you get to a point where they come together, then you would at that point, begin backtracking and try different bases. And if you look, I posted the BWA paper on the Stellar website. And if you look in one of the figures, the algorithm is there, And you'll actually see, if you can deconvolute what's going on, that inside the loop, it's actually doing exactly that. Yes. Question. AUDIENCE: In practice, is the number of errors is small, would it make sense just to use [INAUDIBLE]? PROFESSOR: We're going to get to that. I think the question was, if the number of errors is small, would it be good to actually use a different algorithm, drop into a different algorithm? So there are algorithms on FM index assisted Smith Waterman, for example. Where you get to the neighborhood by a fast technique and then you do a full search, using a more in depth principles methodology, right? And so there's some papers I have in the slides here that I referenced that do exactly that. These are all great questions. OK. Yes. AUDIENCE: If you're-- If you only decide to backtrack a certain number of times, like 100 times, then wouldn't like the alignment be biased towards the end of the short read? PROFESSOR: I am so glad you asked this question. The question is, and what was your name again? AUDIENCE: Kevin. PROFESSOR: Kevin. Kevin asks, gee, if you're matching from the right to the left, and you're doing backtracking, isn't this going to be biased towards the right into the read, in some sense, right? Because if the right into the read doesn't match, then you're going to give up, right? In fact, what we know is the left end of the read is the better end of the read. Because sequences are done five prime to three prime and thus typically, the highest quality scores or the best quality scores are in the left hand side of the read. So do you have any idea about how you would cope with that? AUDIENCE: You could just reverse one of them. But you'd reverse-- PROFESSOR: Exactly what they do. They execute the entire genome and they reverse it and then they index that. And so, when they create what's called a mirror index, they just reverse the entire genome, and now you can match left to right, as opposed to right to left. Pretty cool, huh? Yeah. So backtracking, just note that there are different alignments that can occur across different backtracking paths. And this is not optimal in any sense. And to your question about how you actually go about picking a backtracking strategy, assuming we're matching from right to left again for a moment, what you can do is, if you hit a mismatch, you backtrack to the lowest quality based position, according to PHRED scores. We talked about PHRED scores earlier, which are shown here on the slide. And you backtrack there and you try a different base and you move forward from there. So you're assuming that the read, which is the query, which is associated quality scores, is most suspect where the quality score is the lowest. So you backtrack to the right to the leftmost lowest quality score. Now it's a very simple approach. Right? And we talked a little bit about the idea that you don't necessarily want to match from the right side and thus, typically the parameters to algorithms like this include, how many mismatches are allowed in the first L bases on the left end, the sum of the mismatch qualities you're going to tolerate, and so forth. And you'll find that these align yourself with a lot of switches that you can set. And you can consult with your colleagues about how to set switches, because it depends upon the particular type of data you're aligning, the length of the reads and so forth. But suffice it to say, when you're doing this, typically we create these mirror indices that actually reverse the entire genome and then index it. So we can either match either right to left or left to right. And so for example, if you have a mirror index, and you only tolerate up to two errors, then you know that either, you're going to get the first half right in one index or the mirror index. And so you can use both indices in parallel, the forward and the reverse index of the genome, and then get pretty far into the read before you have to start backtracking. There are all these sorts of techniques, shall we say, to actually overcome some the limitations of backtracking. Any questions about backtracking at all? Yes. AUDIENCE: Is it trivial knowing the BWT originally to find the mirror BWT? Like for example, PROFESSOR: No, it's not trivial. AUDIENCE: So it's not like a simple matrix transforming [INAUDIBLE]. PROFESSOR: No. Not to my knowledge. I think you start with the original genome, you reverse it and then you compute the BWT with that. Right? That's pretty easy to do. And Hang was explaining to me today how you compute his new ways of compute BWT, which don't actually involve sorting the entire thing. There are insertion ways of computing the BWT that are very [INAUDIBLE], and you could ask him this question tomorrow if you care to come. All right just to give you an idea on how complex things are, to build an index like this, takes, for the entire human genome, we're talking five hours of compute time to compute an index to give you an order of magnitude time for how to compute the BWT. the LF checkpoints, and the suffix array sampling. Something like that. So it's really not too bad to compute the index of the entire genome. And to do searches, you know, we're talking about, like on a four processor machine, we're talking about maybe upwards of 100 million rads per hour to map. So if you have 200 million reads and you want to map them to a genome, or align them as it's sometimes called, it's going to take you a couple hours to do it. So this is sort of the order of magnitude of the time required to do these sorts of functions. And there are a couple fine points I wanted to end with today. The first is we haven't talked at all about paired, erred, and read alignment. In paired read alignment, you get, for each molecule, you get two reads. One starting at the five prime end on one side, , and one starting from the five prime end on the other side. So typical read links might be 100 base pairs on the left and 100 base pairs on the right. What is called the insert size is the total size of the molecule from five prime end to five prime end to read. And the stuff in the middle is not observed. We actually don't know what it is. And we also don't know how long it is. Now when these libraries are prepared, size selection is done, so we get a rough idea of what it should be. We can actually compute by looking at where things align on the genome, what it actually is. But we don't know absolutely. If we were able to strictly control the length of the unobserved part, which is almost impossible to do, then we would get molecular rulers. And we would know exactly down to the base, whether or not there were indels between the left read and the right read when we did the alignment. We actually don't have that today. The sequencing instrument actually identifies the read pairs in its output. That's the only way to do this. So when you get an output file, like a fast Q file, from a sequencing instrument, it will tell you, for a given molecule, here's the left read and here's the right read. Although left and right are really sort of misnomers because there really is no left and right, right? This is one end and then this is the other end. Typical ways of processing these paired reads, first you align left and right reads. And they could really only be oriented with respect to a genome sequence where you say that one has a lower coordinate than the other one when you're actually doing the alignment. And if one read fails to align uniquely, then what you can do is, you know what neighborhood you're in because you know, roughly speaking, what the insert size is, so you can do Smith Waterman to actually try and locate the other read in that neighborhood. Or you can tolerate multiply mapped reads. One thing that I did not mention to you explicitly, is that when you match the entire query, and top and bottom are more than one away from each other, that means you've got many places in the genome that things map. And thus you may report all of those locations or I might report the first one. So that's one bit of insight into how to do a map paired reads. And these are becoming very important because as sequencing costs go down, people are doing more and more paired and sequencing because they give you much more information about the original library you created and for certain protocols can allow you to localize events in the genome far more accurately. Final piece of advice on considerations for read alignment. We talked about the idea that some reads will map or align uniquely to the genome and some will multimap. You know that the genome is roughly 50% repeat sequence. And thus it's likely that if you have a particular read molecule, there's a reasonable chance that it will map to multiple locations. Is there a question here? No. OK. You have to figure out what your desired mismatch tolerance is when you're doing alignment and set the parameters to your aligner carefully, after reading the documentation thoroughly, because as you could tell, there's no beautiful matrix formulation like there is with a well established basis in the literature, rather it's more ad hoc. And you need to figure out what the desired processing is for paired reads. So what we've talked about today is we started off talking about library complexity and the idea that when we get a bunch of reads from a sequencer, we can use that collection of reads to estimate the complexity of our original library and whether or not something went wrong in the biological processing that we were doing. Assuming it's a good set of reads, we need to figure out where they align to the genome. So we talked about this idea of creating a full text minute size index, which involves a Burrows-Wheeler transform. And we saw how we can compute that and throw away almost everything else except for the BWT itself, the suffix array checkpoints, and the FM index checkpoints to be able to reconstruct this at a relatively modest increase in size over the genome itself and do this very, very rapid matching, albeit with more problematic matching of mismatches. And then we turned to the question of how to deal with those mismatches with backtracking and some fine points on paired end alignment. So that is the end of today's lecture. On Tuesday of next week, we'll talk about how to actually construct a reference genome, which is a really neat thing to be able to do, take a whole bunch of reads, put the puzzle back together again. I would encourage you to make sure you understand how this indexing strategy works. Look at the slides. Feel free to ask any of us. Thanks so much for your attention. Welcome back. Have a great weekend. We'll see you 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: OK, let's get started. Everyone can go ahead and take 10 more seconds on today's clicker question. This is about hybridization, which we will go over more to start today, but just to see where we are as a starting point. All right, great. So most of you got this. The question is what was the hybridization of an atom if we know that it has two hybrid orbitals. This is really, really important, so I'm just going to write this out for that 20% of you that didn't see this right away. So we know that when we have hybrid orbitals that come from s and p orbitals, these are our atomic orbitals we have to choose from to hybridize. So if we know, for example, that we need two hybrid orbitals, that means we needed to have started with two atomic orbitals. So that means what we would form are two s p hybrid orbitals and we'd be left with our other two p orbitals left over. So any time you figure out that you're going to need two hybrid orbitals, and we'll go over when you would find this out in just a minute, but any time that happens you know it has to come from an s and a p, so that's why you'll end up with s p hybrid orbitals there. And exam 2 is happening in just one week from today, so next Wednesday. The semester moves really fast. So I can't believe we're already up to exam 2 talk, but we are. So, you should have all received two handouts when you walked in today, your notes and then an exam 2 info sheet. Did everybody get that? If you didn't get it, just raise your hand and a TA will come to you with that. I see there's a few around, if one of the TA's can jump up and fill those in. All right, so basically, as with exam 1, all the information you need to know about what's going to be on exam 2 is on this handout. Specifically exam 2 is going to cover lectures 10 through 17. Today is lecture 16, so that means it's going to cover today's lecture and then a bit into Friday's lecture as well, so I'll be really clear on Friday where exam 2 material ends, where you can kind of switch off the exam 2 part of your studying. And it will also be on problem-sets 4 and 5. So it's going to be on two problem-sets instead of three problem-sets, but as you know, there are a lot of different concepts covered in these two problem-sets. So it's still a lot to be thinking about and you need to make sure that you really understand all of these different concepts. So this is explicitly listed in this sheet here, so make sure you look it over. But I just want to say is well, the concepts that we've covered since exam 1 have been talking about covalent bonds, going along with that are all the Lewis structures that we've done. Then we talked about ionic bonds, so you need to make sure you can solve those ionic bond problems. And then we started talking about different types of bonding in terms of thinking about MO theory, and also vesper theory and hybridization. And then today, and a little bit on Friday, you'll be responsible for some of this new material, which is going to be a little bit of thermochemistry. Not too much on exam 2, most of that will be saved for exam 3 material, but an intro to thermochemistry, that will also be on the exam, just as far as these two lectures. So, basically this is all written out here, and I also want to mention in terms of your MO diagrams and MO theory, I sent out an email about this, but I also wrote it out here so you have it written down in terms of what you are responsible for in terms of ordering the energy levels of molecular orbitals. So that's all written down explicitly right here, this is what you need to know for the exam. Also it's helpful for the problem-set as well. And also, in terms of MO diagrams, I think that you will already gone over this pretty explicitly either last week in recitation or yesterday, in terms of what you need to do to get full credit when you're actually drawing an MO diagram. Just in case you don't have that information or you didn't write it down, I wrote this down for you here as well in terms of things like making sure to draw your energy arrows, or remembering to label all of your atomic orbitals. We don't want anyone getting points off on the exam for doing something silly where they do understand what's going on, but they're just not writing it in the correct form. So all this information is on your sheet here. And also on this sheet mentions that you will be getting optional extra problems on Friday. This is just like for exam 1. The way that you should approach these extra problems, you don't need to turn them in, you're not required to do them, but if you want to do well on the exam we really, really encourage that you do these problems. And in terms of having it be worth your time when you're solving these problems, what you want to do is make sure you're at the point where you can put your notes away, where you can sit alone without your friends, and really approach these extra problems as if they're exam problems. This way if there's some conceptual leap that you haven't quite made yet, but you don't realize it because you've been using your notes or using the conversations with your friends, you want to make that leap and run into those problems alone in your room when you have two hours to figure it out. You don't want to do this leap on an exam. That's not the ideal situation. So especially with Lewis structures where sometimes you can run into confusions that you need to work through and make sure it all makes sense in terms of your head how to do this, make sure you take the time to do that before the exam and don't have to end up struggling with that on the exam in a time situation. All right, so the extra problem solutions will be posted on Sunday. And in terms of office hours for next week, my office hours are going to be on Monday from 2 to 4, I'm moving them up, and your TAs will move up their office hours as well, so check with them in recitation or on the website. So that's pretty much it for exam 2. Are there any questions about what you're responsible for or anything regarding exam 2? OK, and in terms of the room it's in Walker again, so at least this part should be familiar. You'll be used to going to your row, it'll be the same row, it's all written here again, but you can just do exactly what you did for exam 1. All right, so one last class announcement, and this is that if you've been looking at our first sheet of the whole year, sort of the course overview sheet, you know that our class in terms of grading is out of 750 points, and 50 of those points are for attendance and something called occasional in-class quizzes. So what we're going to do is start filling in some of those points because we only have 36 classes worth of attendance. So we have a couple of points to make up in terms of quizzes, and these are going to be clicker quizzes, and let me explain how this is going to work. Basically, in at least for my part of the class, what we're going to do, because what I really care about is that people are answering the clicker questions, it's really valuable for me. I don't necessarily care if you get them right or wrong during class, because this is the point at which you're still learning them. So these quiz points, which could be up to 2 points per class, will be in addition to attendance points, and you'll get full credit if you answer the quiz question. So you need to be answering the questions because the quizzes will not be announced. So this is another reward for people that are always participating. These clicker questions are really valuable to us in terms of the feedback, it really gives me an idea of where the class is as a whole. So really, it's important that you are answering these questions and this is a way to get your quiz points up, so in addition to attendance, you'll get 2 quiz points. We'll have our first unannounced quiz today, so you won't know what problem it is beforehand, so I suggest you make sure you're answering all the clicker questions, which most of you do anyway. And I will just say I recognize that some of you are just scurrying across campus to get here on time. So I'll never make that first question a quiz question, because I know how hard it is to get from one place to another on this campus, and I know most of you do a great job of sneaking in right at the last minute when you have that long run. So we won't make the first question a quiz question, but be on the look out for a quiz question today. All right. So let's get to today's topic. We're going to finish talking about valence bond theory and hybridization. So what we've already done is really cover all the theory that we're going to cover behind how hybridization works, and behind how valence bond theory works. But what we haven't covered is how do we actually solve problems doing this in the really quick way. So we're going to do that. This is going to be a more practical lesson in hybridization. And once we do that, we'll move on to now talking about energies and enthalpies of chemical reactions. So this is a real shift. We spent a while talking about single atoms, and then we've spent most of the material, since exam 1, talking about bonding. So now we're taking it one step further and we're going to actually get to talk about some chemical reactions. So we'll do that once we finish up with our hybridization unit here. All right, so how do we go about determining hybridization in complex molecules? It's actually incredibly simple, and all that you need to do is think about for the given atom that you're looking at, what you want to do is add up the number of atoms that are bonded to your central atom, or the atom you're considering, and add to that the number of lone pairs. And what you end up with is the number of hybrid orbitals that you need. All right, so the theory behind hybridization and picturing everything, that can be pretty complicated, and I do want you to understand that. But if, for example, on an exam situation, you don't necessarily have to think through everything, you can just use this very quick way to figure out the number of hybrid orbitals that you need. And what we just said is that any time we need two hybrid orbitals, we know that the hybridization of that atom is going to be s p, because it's made up of one s and one p orbital. So let's say we need three hybrid orbitals. What is our hybridization of our atom in this case? Yeah, it's s p 2. We want to take three atomic orbitals to make three hybrid orbitals, so it's going to be s p 2. So what if we need four hybrid orbitals, what's the hybridization going to be in this case? STUDENT: S p 3. PROFESSOR: S p 3. All right, great. So, really that's it. Hopefully hybridization just got a lot simpler for any of you that have been struggling a little bit with it. But let's go ahead and do an example to make sure that we can all do this. And let's take a look -- first let me mention an exception here. The one exception to thinking about how many hybrid orbitals that you have are in the case where you have single bonded terminal atoms. So in the case of a terminal atom that is a single bond, you're not going to hybridize it. So in that case, just don't even change anything it's just going to be one of the p orbitals, or unless you're talking about hydrogen in which case will be an s orbital that overlaps. All right, so we can illustrate this with an example of formal chloride. So this is a good example because we're just dealing with our carbon as our central atom here, and we have three terminal atoms, two of which are single bonded, so we're not going to hybridize those, and one which is double bonded, the oxygen, so we will hybridize that one. So in terms of thinking about the carbon, what is a hybridization around the carbon atom? STUDENT: [INAUDIBLE] PROFESSOR: All right, I'm hearing some mixed answers. So let's think. So the carbon atom is bonded to three different atoms, and it has no lone pairs. So what's the hybridization? STUDENT: S p 2. PROFESSOR: Good s p 2. So, if we're talking about the c h bond, we're going to say that it's carbon, it's a sigma bond because it's a single bond, and then it's carbon 2 s p 2 bonded to a hydrogen 1 s orbital. All right, so let's take a look at the carbon chlorine bond here. Again, carbon is still 2 s p 2, it's the same carbon atom, and it's going to be a sigma bond because it's a single bond, but what about the chlorine, what atomic orbital is going to be bonding in the chlorine? STUDENT: [INAUDIBLE] PROFESSOR: All right. I'm hearing a little bit of mixes, they're all p's, which is a good start. It turns out that it's going to be the chlorine 3 p z. So the reason that it's p is because we're not hybridizing it, and the reason that I specifically say that it's the p z instead of the p x or the p y, is remember that the z-axis, that's our bonding axis, that's our internuclear axis. So any time we have a p orbital that's involved in a sigma bond, it has to be the z orbital because that's the only one that has the right orientation to overlap along that z-axis. So we're going to say it's carbon s p 2, and then chlorine 3 p z. All right, so let's look at the last bond here, which is a carbon oxygen double bond. We know that any time we have a double bond it's made up of one sigma bond plus one pi bond. So, when we look at the carbon again, that's 2 s p 2. But what about the oxygen, what's the hybridization of this oxygen atom? S p 2. So we're going to have carbon 2 s p 2, and then oxygen 2 s p 2. It's s p 2 because the oxygen is bonded to one atom plus two lone pairs, so we're going to have a total of three hybrid orbitals. All right, so this is not our only bond, we have a double bond, so we also need to talk about the pi bond. So if we talk about the pi bond, we can say that's carbon 2 p y, then the oxygen 2 p y. We could alternatively say, and be correct, that it was the carbon 2 p x and the oxygen 2 p x. Either one of those is fine. All right, so that's an example of how we can assign very quickly and very easily what the hybridization is of a given atom. And then also fully describe the symmetry and the atomic or hybrid orbitals that make up bonds. So this is a lot of your problems on the p-set, so if you haven't finished that section, hopefully you'll be able to get through that section pretty quickly when you go back to it. And let's take a look at a little bit of a more complex molecule here, which is ascorbic acid or vitamin C. And this is a good example because it's starting to look a little more complicated, but not too much more complicated. But the reality is if you know how to do assigning the hybridization this way, even if I gave you a 1,000 atom protein, you should still be able to get it completely correct. Vitamin C is a really important molecule to actually think about in terms of thinking about its shape and its hybridization. Vitamin C is an antoxidant. So we might remember that from when we were talking about free radicals. Free radicals can cause oxidative damage. These antioxidants, some of our vitamins, including vitamin C are antioxidants. So that's something that a lot of people in the general population know if they're into their health. But another really important thing about ascorbic acid is that it's an enzyme cofactor. And a cofactor just means that it's some type of molecule or atom, in this case it's a molecule, that's required by an enzyme in order for the enzyme to carry out its chemistry. And the chemistry that the enzymes that vitamin C are a cofactor for are responsible for putting oxygen, o h groups, onto collagen molecules in order to form what is the collagen triple helix. And collagen is the main constituent of bones, of joints, of connective tissue, of many important structural parts of our bodies. So you can imagine that vitamin C is very important because we need to keep our collagen intact. Collogen is actually one of the most prevalent proteins in our entire body, and it makes up a large part of our cells. So you can imagine if we don't have collagen that is in this triple helix, we're going to run into some problems. Does anyone, or I'm sure many of you do know, but can someone tell me what the name is of the disease if you have a deficiency in vitamin C? STUDENT: Scurvy. PROFESSOR: Yeah, scurvy. So scurvy, most often associated with ships because of the 1500's and 1600's and 1700's when boating technology was far above biochemistry technology, so people could make ships and go on these long journeys, but people did not yet understand they needed to keep certain vitamins in their bodies, such as vitamin C. Maybe also, they didn't know about -- well they didn't even have the structure, so they couldn't have realized that vitamin C is polar and won't stay in their bodies for very long, it will get excreted out through their urine. So they were not thinking about how quickly they could run into trouble in the long sea voyages. And I don't know how many of you had a unit on explorers in elementary school -- I did in fifth grade, and one that always sticks in my head is Magellan's trip around the world, who led the first fleet of ships around the globe. What they did not mention to us in fifth grade was that 90% of his men died on that journey. And I understand we were young kids, I didn't want to know that at the time, but it's interesting and it's really important, and the reason for many of these deaths was scurvy. And it's totally preventable. And, in fact, the cure for scurvy, which is as simple as taking vitamin C in any form, was known for -- since the fifth century people started figuring this out, but the problem was there was so much false information as well that it was not a general practice on ships to be treating men that were suffering with vitamin C or using it as any kind of preventative measure. The other somewhat interesting thing I want to say about vitamin C is that it was part of the first ever published clinical study. And this was a controlled clinical study that was done in the 1700's, the first one that's ever been reported, and not surprisingly it was done on a ship and it was done by James Lind who was a Scottish naval surgeon. And basically, he took 12 of his men, so this was not a very large clinical study like they do today, but he took 12 men that were suffering from scurvy, and he gave them a diet mostly of just carbohydrate mush, because when you have scurvy, one of the first signs is bleeding gums and teeth falling out, they couldn't really eat too much. But he supplemented their diet. He put them in pairs of two and supplemented with 1 of 6 things. So two of them got a pint of cider a day to see if that helped. Two of them got half a pint of seawater -- maybe not the best group to be in. Even worse group to be in is two of them got diluted sulfuric acid. Tuned out that didn't help so well. Other things where some got spices, not so bad. And then two of them actually got citrus fruit. So no big surprise ending to this clinical study, the people with citrus fruit were completely cured and back to duty within a week, and I'm not sure exactly what happened to everyone else, but let's hope they got switched over as well. But it's just neat to think about even back then, people were doing these controlled scientific studies, this was one of the first cases. Unfortunately, it was another 40 years between this study being published and the British Navy requiring supplements of citric acid. So, really, we've come a long way in terms of disseminating scientific knowledge to the community, and that's a really important part of being scientists -- not just making these discoveries, but also passing along, for example, to these ships full of people that need to know it. So, in terms of scurvy, we don't see too much of it today, but who should be concerned about suffering from scurvy? Do we have to worry about our cats or dogs at home, are they getting enough of their vegetables and vitamin C? It turns out we don't have to worry. Primates are who should be concerned. So we need to be concerned, other primates need to be concerned. It turns out that most other mammals actually biosynthesize vitamin C, so we don't have to worry about most of our pets at home. Unless you have a guinea pig and they don't, in fact, biosynthesize vitamin C, so I think they supplement most guinea pig pellets with vitamin C, but maybe keep an eye on your guinea pig. We primates and guinea pigs are the only types of mammals that don't actually biosynthesize their vitamin C. So again, vitamin C is a cofactor, that means it needs to bind to an enzyme, it means that the shape of vitamin C and the hybridization are going to be really important. So let's take a look and talk about some of these carbon atoms here. And you can look at your structure of vitamin C in your class notes while you're doing this, and first I want you to tell me what the hybridization is of that carbon a in the vitamin C molecule. And let's take 10 more seconds on that. OK, 78%. We'll do another question like this in a minute. So if we can switch back to the notes we'll take a look at that. I think we can get better, I think we can get to the 90's with these hybridization, we just need to follow the rules. So if we're thinking about vitamin C and we're talking about carbon a, how many things is carbon a bonded to? STUDENT: [INAUDIBLE] PROFESSOR: four things. We have two hydrogen atoms, an oxygen, and then another carbon. So that means we need, if it's bonded to four things, we need four different hybrid orbitals, so it needs to be s p 3 hybridized. All right, let's try this again but just shouting out, what is the hybridization of carbon b? Good. OK, I'm going to say that was about 90% of you I heard, which is excellent. S p 3. What about carbon c here? Yup, s p 3 again. Bonded to four things it's s p 3. What about carbon d? S p 2, great. Carbon e? STUDENT: S p 2. PROFESSOR: All right, and this last one, the last 10% of you can join in. Carbon f? Great, s p 2. All right, so we can also think about shape once we know hybridization and how many atoms a carbon atom is bonded to. What would we say that the shape is of these s p 3 orbitals -- or the shape of the carbon atom. Yeah, it's tetrahedral. So this is going to be tetrahedral. And what about carbon d, e, and f? What is the geometry of carbon d, e, and f? Good, trigonal planar. All right. So let's actually take this one step further from just talking about the hybridization of the individual atom. Let's talk about a couple of these bonds. I won't go through all of them. I think you can get the idea with just a few, but you can go back and try all of them later. So let's start with thinking about carbon b hydrogen bond. Is this going to be a sigma or a pi bond? Sigma. We know it's sigma, because any time we have a single bond it's a sigma bond. So carbon b, we just said, is carbon, it's going to be 2 s p 3. And then hydrogen, what's the atomic orbital we're talking about here? 1 s. All right, great. So let's take a look at another one, let's look again at carbon b, but now let's talk about it bonding to its oxygen there. So again, are we going to be talking about a sigma or a pi bond? A sigma bond, because again, it's a single bond here. So what we'll say again is carbon 2 s p 3, and let's take a look at this oxygen that it's bound to. Is this oxygen -- what is the hybridization of this oxygen right here? What is it? OK, good, it's s p 3. It's bound to two atoms, plus it has two lone pairs, that's a total of four things. So it needs to have three different hybrid orbitals, so we'll say it's oxygen 2 s p 3. So let's skip to a different carbon now, let's talk about carbon d, specifically the carbon d oxygen bond. So let's do a clicker question for this one. Let's get into the 90's, if you don't mind, this would be very good to do. It'll make me feel better about you finishing your problem-sets tonight. So if you talk about the symmetry and the hybrid or atomic orbitals that contribute, we're talking about the c d oxygen bond. So go ahead and look at your notes and see what that bond is. We already identified the hybridization of carbon d, so we really just need to think about the oxygen here. All right, let's take 10 more seconds. OK, I'll take that, 84%. That's not bad. So again, we're looking at a sigma bond here. It looks like the one people got it confused with was talking about oxygen s p 2 versus s p 3. Remember, you just need to look at what it is actually bound to, so for carbon d, the oxygen's bound to two atoms plus two lone pairs, so it's going to be s p 3 hybridized. So we'll call this a sigma bond where we have carbon 2 s p, now it's s p 2, and oxygen 2 s p 3. All right, let's take a look at just one more here. So, let's look at the carbon d bond with carbon e. So if we can switch back to the class notes just to see that, though you guys can see it on your notes as well. So in terms of that, we're talking about a double bond now, so we know that we have to have one sigma bond and one pi bond to completely describe our double bond. So for the sigma bond, again, it's going to be carbon 2 s p 2. And now we're talking about the other carbon. What is the hybridization of the other carbon? Yup, s p 2. So carbon 2 s p 2. All right, but we're not done there. We also need to talk about the pi bond. So for the pi bond, we'll talk about carbon 2 p y, and then the second carbon 2 p y. All right, so I filled in a few more in your notes, so you can maybe cover those up and make sure that you get all of those correct is a good self test before you finish off any in your problem-set. Does anyone have any questions about figuring out hybridization? Yes? STUDENT: [INAUDIBLE] PROFESSOR: I'm sorry, what is that? In this bond here? Oh, OK, that's a good question. So the question was why is there not hybridization in these p orbitals here. So in order to form a double bond, we need to have a pi bond forms, and a pi bond forms from two unhybridized p orbitals. And if you can kind of picture those p orbitals, if our molecules are this way and they're coming up here, we need to have electron density above the bond and below the bond. So if they're hybrid, then they're going to be spread out into that tetrahedral geometry. We want to have them parallel to each other and they need to be the p bonds, they need to be the p orbitals. Does that makes sense? So any time we have a double bond, the pi part of the bond is going to be p orbitals. Yes? STUDENT: [INAUDIBLE] PROFESSOR: No, absolutely not. It would be absolutely correct to put 2 p x and 2 p x here, you just can't put z, because that's the one that's going to be involved in the sigma bond. And the other important thing is if you do put x for one, you have to make sure you put x in the other, because they do need to be able to interact with each other. You can't have one x and one y, but sure, you can have both x or both y. OK, so let's shift gears a little bit and start talking about bonding in terms of chemical reactions. So we're going to start today with talking about bond energies, and also something called bond enthalpies. And the first point is just to bring us back to something we're very familiar with. When we talked about covalent bonding, a concept that we have been discussing is bond dissociation energy. That's just the energy that's required to put into a molecule in order to break that bond. This is something that we saw when we were talking just at the very beginning of our unit on covalent bonds. That energy difference is the energy difference between when we have the c h 4, if we're talking about methane, versus one of those hydrogen bonds breaking, one of those c h bonds breaking. So we've talked in the past about bond energy, but what I want to introduce to you today is a very related concept, which is called bond enthalpy. So that's delta h here is what we call bond enthalpy. So, this is talking about instead of the change of energy accompanied with a bond breaking, we're talking about a change in heat accompanied with a bond breaking. So whether when that bond breaks it requires heat, or whether it gives off heat when you break that bond or talking about a reaction. That's what we're talking about when we're talking about enthalpy. It turns out that enthalpy is, in fact, very related to energy, and we can relate it with this equation here that bond enthalpy is equal to bond energy plus the change in pressure times volume. So we could, in fact, go back and forth between bond energies and bond enthalpies, but the reality is if we're talking about gases, which we are in many cases, then what we find is the difference between bond enthalpy and bond energy is only 1% to 2%. So it's not very significant. And, in fact, if we're talking about solids or liquids, now we can say that the difference between bond energy and bond enthalpy is going to be negligible. So, for the rest of this class today, we're going to stop talking about bond energy, which I did just for a moment, to orient you back to a discussion that was familiar, we're going to switch our discussion to bond enthalpies. One reason that we like to talk about enthalpies is unlike energy, which can be a little bit more tricky, bond enthalpies are easy to measure. It's easy to measure how much heat a reaction gives off or how much heat a reaction takes in. So we can also talk about something, which is called the standard bond enthalpy, so any time you see this symbol here, which looks like a knot, so it looks like a delta h knot, that refers to the standard bond enthalpy. Any time we we're talking about a standard, whether it's enthalpy or, as we'll see in the next lecture, standard entropy or a standard free energy, we're just saying that the molecules that we're talking about are in their standard states, which means they're in their pure form. If they're a gas it means they're at one bar. Usually this is referring to room temperature, so 298 kelvin. It's often that when you look up tables of different bond enthalpies, you'll see them in their standard state and it will be at room temperature. So if we talk about standard bond enthalpy, let's talk about what this is for these c h bonds, which we just saw in methane. So if we're going from c h 4 to breaking one of these c h bonds, what we see is that the bond enthalpy is 438 kilojoules per mole. The fact that it's positive tells us that we need to put in that much heat into the system in order to break that bond. So we now know the enthalpy of a c h bond, but, of course, methane is not the only kind of c h bond that you can envision. We could talk about the c h bond, for example, in ethane or trifluoromethane or trichloro or tribromomethane, and what you'll find is that the amount of enthalpy that that reaction is associated with depends on the exact type of c h bond that you have. So, for example, we can see that we have slightly different, not vastly different, but slightly different changes in enthalpy depending on which kind of c h bond that were breaking. And what I want to point out also is that all of these delta h's, all of these standard bond enthalpies are positive. Any time we have a positive delta h, we call this an endothermic reaction. And again, endothermic just means that the reaction takes in energy. It's very similar to what when we're talking about energy whether something's taken in or released, whether it's positive or negative. If you have positive delta h, it's endothermic, it takes in heat in order for the reaction to go. So we can think about instead of talking about all of these individual c h bonds, instead we can talk about what the average value would be for all c h bonds. So we can say that the average of any c h bond is 412 kilojoules per mole. So you can imagine if you're looking these things up in terms of a reference table or maybe in the appendix of your textbook, it would be an impossibly huge number of pages to write every single different kinds of c h bond. So that's why instead they talk about the average bond enthalpies. So for c h, again, that's 412, and all of the enthalpies we just saw fall within about 8% of that. And in your book you can look up any type of bond that's listed here. So we saw c h is 412. You can look up a c c bond, you can look up a c c double bond, a c c triple bond, and so on. And you'll note that what these are are mean bond enthalpies, so they're the average of all different types of c h bonds or c c bonds that you can imagine. So, you might be asking what's so important about being able to look up and think about these different bond enthalpies. And the reason that it's important is because if you're looking at a reaction, no matter how complicated that reaction is, you can actually figure out what the enthalpy of the entire reaction is by adding up all the individual mean bond enthalpies of the products, and all the individual mean bond enthalpies of their reactants and thinking about the difference between those two. So we'll do that in just a second, and the reaction that we'll do it with is the oxidation of glucose here. So, c 6 h 12 o 6, one mole of glucose plus six moles of oxygen, gives us six moles of carbon dioxide and six moles of water. So what we can find out, and hopefully what we will match up when we look at using the different bond enthalpies, is that the enthalpy of this entire reaction is negative 2816 kilojoules per mole. So in this case we're saying that delta h is negative. Does anyone know what it's called when delta h is negative? Everyone knows, great. So it's exothermic. This is an exothermic reaction, the reaction releases heat. So I just want to mention before we go on, if you look at almost any freshman chemistry textbook, what you'll find is this oxidation of glucose reaction is used a lot in talking about thermochemistry. And one reason it's talked about is because it's very convenient to talk about something where we start with one mole of glucose and end up with 12 moles of products. That's going to be helpful what we're doing a practice problem to see exactly how you deal with it when there's different numbers of moles. But the other reason you always see this reaction is because it's an incredibly important reaction. The oxidation of glucose is going on all the time in our body, this is our main source of energy for all animals. So let's think a little bit about why this reaction is so important and so prevalent. It turns out that if we're talking about plants, plants do the reverse reaction. So plants take carbon dioxide and water and they turn it into glucose or energy for us, energy stored in the bonds of glucose plus oxygen. So if we do the reverse reaction, this is actually going to require energy. Where do plants get this energy? Yeah, this is just photosynthesis here. This is the photosynthesis where plants are turning carbon dioxide and water into sugar and into oxygen. So what happens when we eat the plants or when we eat animals that have eaten the plants is that we perform the reverse reaction now, which is what I just showed you, the oxidation of glucose. And even though it's not the products of the reaction that are particularly valuable to us, we just breathe out the c o 2, and actually we'd rather have less of that than more in our environment, but what's important here is instead the energy or the enthalpy that's given often in this reaction. So as I said, this reaction has a negative enthalpy of 2816 kilojoules per mole. That's a lot of enthalpy and a lot of energy. So we actually end up using that energy to fuel most of what is going on in our bodies. So instead of storing it as sugar, once we oxidize the sugar, now we just store it as ATP, and as you know, ATP is the currency of energy in the cells. So this is why you see this reaction again and again and again in just about any chemistry textbook that you open up as sort of a general reaction. The reason it's used is it's just so important and so prevalent in terms of thinking about our bodies and how we're staying alive and using energy. All right, so let's go ahead and use this as an example of what we just said, which is using the bond enthalpies of the products and the reactants to figure out the enthalpy of the entire reaction. So the way that we do this is we add up all of the individual mean bond enthalpies of the reactants, and we subtract from that all of the individual bond enthalpies of the products. So we can think about what this will tell us. If you think about the fact if the bonds are stronger in the products than they were in the reactants, you can go ahead and click in and tell me if you think we'll have a negative or a positive delta h here. All right, let's take 10 more seconds on this. OK, great, so negative is correct and some people got totally mixed up. Didn't just get one thing -- so let's just focus on the right answer to start with. The correct answer is that it's negative, it's an exothermic reaction. And actually, lets switch to our class notes to explain why. And also, this was the quiz question for today, so whether you got the answer correct or incorrect, you get full quiz points if you did, in fact, answer. But let's still focus on the right answer here and see why it's correct. So if we're talking about the bonds being stronger in the products, that basically means that we ended up releasing a lot of heat when we made those bonds and the products, and we didn't have to use up too much of that heat to break all the bonds and the reactants. So that's why we say that delta h is going to be negative. You could also just do it not conceptually, just plugging it into the equation, but it's better to kind of understand exactly why that is. It's because it takes more energy -- you gain more energy forming the products than you take breaking up the reactants. So if we have the opposite case here, if the bonds are stronger in the reactants, now what we're going to find is that the delta h of the reaction is positive and what you're dealing with is an endothermic reaction. So let's go ahead and do this with our example of the oxidation of glucose. So, I've just written out the glucose molecule so you can see all of its individual bonds here since we're going to be using that information. So let's start by talking about the bonds that are broken in terms of thinking about all of the reactants here. So if we're talking about the sugar molecule itself, what we have is we have seven c h bonds. And if we add up all the o h bonds, we have five of those. How many c o single bonds do we have? What do people think? Yeah. five c o single bonds. What about c c single bonds? five of those as well. C o double bonds? Just one c o double bond. And then we have our oxygen molecule to worry about and we have six o o double bonds. So if we add up all of the bonds are broken, and we subtract from that the bonds that are formed, those strengths, the c double bond o, we have 12 of those. And how many o h bonds do we have? Right, 12 as well. So if we add up all the bonds broken, what we end up with is 12,452 kilojoules per mole, and it's talking about per mole of glucose that's oxidized. And in terms of bonds formed, what we see is 15,192 kilojoules per mole. So all we need to do to figure out the change in enthalpy, and when it has this subscript r here, I meant to mention, that means the enthalpy for the reaction. We just subtract these bonds here from the ones that we ended up forming. So basically, what we're saying -- excuse me, these are the ones that we broke. It's 12,452 minus 15,192, which we formed. So we end up with an enthalpy of reaction of negative 2,740 kilojoules per mole of glucose that's oxidized. All right, so if you remember the number that I told you before, it doesn't exactly match up with what we had said is the exact number. The exact number is 2,816. It is within 3% though, that's pretty good. Because remember, we're not using exact bond enthalpies here, what we were using is average bond enthalpies. So it makes sense that we're going to be a little bit off. But if all you have in front of you is the information on bond enthalpies, mean bond enthalpies, this is a great way to figure out the enthalpy of an overall reaction. We can think of a different way, however, to think about the enthalpy of an overall reaction, and this is talking about the heat of formation. And the heat of formation or delta h formation here is equal to the heat of -- or the enthalpy of reaction, if we're talking about forming one mole of compound from its pure elements, which are in their standard states. So basically, there's a table that you can look up which tells you the enthalpy of formation for any compound that you're interested in, and this is actually an appendix to of your textbooks, and you will need to use this to solve your problems on p-set. But let me say that, for example, if we're talking about water here, that's formed from hydrogen and oxygen, if we're talking about the elements in their pure forms at one bar, standard states, and room temperature. And if we look this up in our textbook, we would find that the delta h of formation for water is negative 286 kilojoules per mole. Similarly we can look up the same thing, for example, for carbon dioxide, and what we'll find here is that our delta h of formation that we look up is negative 393 . 5 kilojoules per mole. We can also look up or think about what our delta h of formation would be for oxygen. Does anyone have a guess here? Yup, it's actually going to be zero. Oxygen already is in its most stable state, so any time we're talking about an element in their most stable state, it's going to be zero. That's going to be the delta h of formation, there is none, it's zero. We can also look up what it was for sugar, for glucose, c 6 h 12 o 6. So that coming from its most stable forms, it was the most pure form of the elements, is going to be negative 1260 kilojoules per mole. All of this information has been tabulated and is in tables, you can refer to them, and they're also in the back of your textbook, so you can refer to those in terms of your problem-set. So let's go ahead and solve a problem actually using the delta h's of formation. Let's see if we can do any better to coming close to the reality for the oxidation of glucose. So if we're going to calculate the delta h for the reaction for the oxidation of glucose, or actually for any reaction at all, what we want to do is take the delta h of formation of the products and subtract from that the delta h of formation of the reactants. So let's go ahead and do that. And in terms of talking about the oxidation of glucose, if we talk about delta h of this reaction, what we need to take is 6 times delta h of formation of c o 2, since we're forming, that's in our products, we're forming six moles of c o 2. Plus 6 times delta h formation of h 2 o, since we're forming six moles of h 2 o. And we subtract from that the heat of formation of all of our reactants. So we only have one mole, so we just say delta h formation of c 6 h 12 o 6, and then in addition to that we have six moles delta h formation of oxygen, of o 2. All right, so this is our equation here and at this point what we would do is we would look up what all of the delta h formation values are for c o 2, h 2 o, sugar, and oxygen. So what we would find is that we end up having 6 times negative 393 . 5 -- that's what we had just looked up and told you for c o 2. Plus 6 times negative 285 . 8 for the water. Minus 1 times 1260, so it's negative 1260 for our sugar here. And then what we're going to end up with is having minus, and what is it for oxygen again? So minus 0. So what we end up with in terms of the delta h for the entire reaction here, is we end up with negative 2816, and it's going to be kilojoules per mole of glucose. All right, so let's see how this matches up, and hopefully you can actually remember that this matches up actually perfectly here. So what we are going to see is that, in fact, what we calculated versus what is experimental is dead-on the same. So there's actually one more way to figure out the enthalpies of a reaction, I'm going to go over it just very briefly. And that's based on the fact that enthalpy is a state function, and by state function what I mean is that it doesn't matter how you got from point a to point b, all that matters is the difference between the two. So another example of a state function is altitude, for example, on a mountain. So if you're talking about altitude, you go from point a to point b on the mountain, and it doesn't matter how you got there -- you could have climbed all the way up the top of the mountain, then went back down and eventually landed on point b. Or you could have gone straight from point a to point b. The change in altitude between point a and point b do not depend on the path, they're independent of path. So this difference right here does not matter on how you got there, it does not make a difference how you got there, altitude is a state function. So similarly, we can say that enthalpy is a state function as well, if we're talking about part reaction a here, which has our reactants going to our products in 2 b here. It doesn't matter how we got there. We can actually calculate the change in enthalpy at all different points here. All that we actually have to worry about at the end of the day is the difference between a and b. So that's what we mean by state function. Actually, we're not going to have a chance to get to fully explaining the consequence of this, which is Hess's law, which allows us to add and subtract different reactions. So what I'm going to say is the last two problems on your problem-set you won't have to do because we're not going to get a chance to cover in class on Friday. So I'll send out an email about this as well, so that'll be pushed back. So, it will be on the exam, so you want to do them at some point, but you don't need to actually turn those in. So those are the last two problems, I believe.
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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: I'm Sanjoy Mahajan. This course is Teaching College-Level Science and Engineering. The course is intended for graduate students some towards the end of their PhD who are intending to go onto academic teaching careers. But all graduate students are welcome, and even undergraduates who have a strong interest in teaching. The goal of the course is to give some preparation and background into the principles of teaching, so graduate students and in fact as undergraduates too, all get the same for their specific research area. For example, in physics, they'll learn the principles of conservation of momentum and energy. But in teaching, everyone is basically just thrown to the wolves and figured, well, if you know your subject, you can teach or given very little preparation. So the idea is to actually show that teaching and thinking about teaching is a very serious intellectual activity. And there's general ideas and principles that can be used to make teaching much more successful. And therefore, students learn much more and enjoy what they're learning. Many departments encourage their students who they know are going on to teaching careers. They say, well, you might actually find this course interesting. It's paired also with another new program at MIT, called the TA certificate program, which is a non-credit option where students do about six or seven seminars of an hour and a half each and do work for each seminar. Here, this is the sort of four-credit cousin of that course. And it's for students who want to do everything in one semester and want something on their transcript as well. So at the end of each lecture, I give out a short sheet that has three questions on it. One is what was most confusing today or what do you wonder about most? What example or teaching technique worked well or not well? And any other comments? And so at the end of each session, the students have one or two minutes to fill that out. And at the end of the next session, I answer all the questions. I was just amazed at the thoughtfulness of the questions and how right on they were and how many useful suggestions I got for when I teach the class the next time. Many students have told me they think everyone who teaches should take it. And some told me also that they wish some of their professors had taken those classes, their professors from when they were undergraduates. But on the other hand, if it were required, then it wouldn't be such a lively atmosphere as it is now, where we have students who really are fascinated by teaching and are therefore much more motivated to contribute. But the theme that I got from many, many students was, wow, this has really helped them thinking about their teaching and said oh, finally, this may be the only preparation I get in teaching and I'm really glad I had that.
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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. So welcome back. I had a little bit of a break. Before I get into-- I go over my quick recap, you hopefully all have seen the announcements that we're going to delay the due date of the next problem set until Tuesday. That's in part because some of the material that appears on it is what we're going to talk about in today's lecture. So I want to make sure you've at least seen a lecture on all the topics before you try to do some of the problems on it. In truth, you could probably-- based on things I talked about in the previous lecture, you could probably deduce a lot of what you need to do, but still, it's good to go over it a little bit more methodically. And I'm trying to think. Is there anything else I want to say? Yeah. So then the next problem set will be posted a week from-- is that right? Yeah, I think we post a week from today. So let me just get back to what you were talking about last time. So I began introducing some geometric concepts and some quantities that we use to describe matter. We've done a lot of things so far that's appropriate for describing kind of the kinematics of particles, but that's rather restrictive. We want to talk about a broader class of things than that. And I began by introducing something which is, again, fairly simple, but it's a useful tool for beginning to think about how we are going to mathematically categorize certain important types of matter. Pardon me while I put some of my notes away. Also, pardon me. I'm recovering from a-- I caught a terrible cold over the long weekend, and I'm coughing incessantly. So I will be occasionally sucking on a cough drop. So I introduced the quantity called the number four-vector. And that is given by-- imagine you have sort of a-- it's best to think about this in the context of something like dust. So you have some kind of non-interacting little agglomeration of tiny particles. And in the rest frame of an element of this dust-- so imagine you go into-- well, I'm going to define the rest frame of that element by saying imagine you've got a cubic nanometer of it or something like that. And you go into the frame where everything in that cubic nanometer is on average at rest. We'll call n sub 0 the rest density. So I forgot to write that down. So that's the rest density. The rest-- pardon me-- number density. So I go into that rest frame. That tells me how many little dust particles there are per cubic volume, per element of volume. Multiply that by the four-vector that-- excuse me-- the four-velocity describing that element. And we're going to call that capital N. So that's a vector that describes-- its components describe the number density in some other frame of reference, in any frame of reference. And the flux, talking about how those dust particles flow from one element to another. So we spent a little bit of time talking about how to define volume elements in a covariant fashion. I'm not going to go through that, but part of the punchline of doing that is that we're going to define a sort of what I call a covariant formulation of conservation of number. It's going to be the spacetime divergence of that vector is equal to 0. That holds-- this equation holds in all frames of reference. If I choose a particular frame of reference, in other words, I define a particular time, I define a particular set of spatial coordinates, I can then break this up and say that this is equivalent to saying that the time derivative of the t component of that plus the spatial divergence, the spatial confluence is that they sum to 0. I can also take this thing and integrate over a-- oops-- it would help to have one element defined here. I integrate this over a four-dimensional volume. And by my conservation law, I must get 0 when I do that. Again, having chosen a particular frame, defined what time means, defined what space means, I can then split this out into a statement that the time derivative of the volume integral of the time component of this thing is balanced by the flux of the number across the boundaries of a particular three-volume. I remind you this notation, this sort of delta V3 means the boundaries of the V3 that's used under this integral. So we talked a little bit about a few other four-vectors that are used. In particular I introduced some stuff that we use in electricity and magnetism. But I want to switch gears. In particular, I want to introduce and discuss in some detail one of the most important tensors we are going to use all term. So I motivated this by saying imagine I have a cloud of dust. And I won't recreate one, because I am getting over a cold. And I don't want to breathe chalk dust. So I imagine I make a little cloud of these things here. So far we've characterized this thing by just counting the number of dust particles that are in there. But that's not all there is to it, right? That dust has other properties. And so the next thing which I would like to do is let's consider the energy and the momentum of every particle of dust in that cloud. So let's just imagine, again, we're going to start with something simple, and then we'll kind of walk up from there to a more generic situation. Let's imagine that my cloud consists of particles that are all identical, and each dust particle has the same rest mass. So suppose this guy has a rest mass m. So one of the things that I might want to do is in addition to characterizing this dust by saying that the number density-- [COUGHING] excuse me-- in addition to saying that the number density in a particular frame is n or n0, let's go into the rest frame of this thing. I might be interested in knowing about the rest energy density. So I'm going to denote the rest energy density by rho sub 0. Each particle has a rest energy of m. Remember, c is equal to 1. So mc squared, if you want to hold to the formula that we've all known and loved since we were babies. So that's the rest energy. And then how many-- if I want to get the rest energy density, I count up the number in each volume, or the number per unit volume. So that's m times n0. So that's the rest energy density of this thing. Great. Let's now ask, OK, so that's what it looks like in the rest frame of this dust element. Let's imagine that I now bop into a frame that is moving with some speed relative to this rest frame. In this frame, the energy density-- I'm not going to say rest energy density, because I'm no longer in the rest frame. But the energy density in this frame, which I will denote by rho without the subscript 0 is going to be the energy of each particle. So that's gamma times m. And the volume is going to be length-- it's going to be a little bit larger. Sorry. The volume is smaller because of a length contraction. And so this guy actually gets boosted up to gamma and 0. So in this frame, the energy density is larger than the rest energy-- excuse me-- the rest energy density with a gamma squared factor. Straightforward algebra. But I want you to stop and think about what that's telling us. If this rho were a component of a four-vector, is there any way I could get a gamma squared out of this thing by bopping between reference frames? No, right? When I do that, it's linear in the Lorentz transformation matrix. There's no way with a linear transformation that I would pick up two powers of gamma. This is not how a-- you guys are probably all used to from Newtonian physics of thinking of energy density as a scalar. This is not a Lorentz scalar. A scalar is the same in all frames of reference. So this is a different quantity. So it's neither a four-vector component nor a scalar. It's got two of those things, so it's not going to surprise you that what's actually going on here is we are picking out-- what we've done here is we've picked out a particular component of a tensor. Let's think a little bit more methodically about what tensor that must be. So when I originally wrote down my rho 0 here, I sort of just argued on physical grounds that it is the energy of every particle times the number of particles per unit volume of that thing. Well, the energy that I use in this thing doing it in both the rest frame and in the other frame, those are the time-like components of a particular four-vector. So we assembled rho by combining energy, which is the time-like component of the four-momentum with number density, which is the time-like component of the number vector that I just recapped a few minutes ago. So those are two time-like components of four-vectors here. So that rho is something like-- well, let's put it this way. I can write this as pt nt, which tells me that it belongs to a tensor. So let's define this as some tt of a tensor that I've not carefully introduced yet. There is some underlying tensor that we have built by looking at the-- there are various names for this. We'll call it the tensor product of two different four-vectors. So if I write this in the kind of abstract tensor notation that I've used occasionally-- in some of your textbooks this would be written with a bold faced capital T, I will use a double over line for this. We might say that this t is the tensor product of the number vector with the four-momentum. Now, the number vector is itself just n0 times the four-velocity. And my momentum, since all the particles, I'm assuming, are the same is just the rest mass times the four-momentum. This thing actually looks kind of like the four-velocity times four-velocity with a prefactor And that prefactor is nothing more than the rest density. So another way to say this is that if I make my cloud of dust, and I go in and I look at every little element in it, I take the rest energy density of every element in that dust. I construct the tensor that comes from making the tensor product of the four-velocity of that dust with itself. That is what this geometric object is equal to. If you want to write this in index notation, which is how we will write it 99.8% of the time, we would say that component alpha beta of this quantity t is that rest density times u alpha u beta. So this is-- in terms of the physics we're going to do this term, this is perhaps the most important quantity that we're going to talk about. So it's worth understanding what this thing is really telling us. So if I wanted to get these components, the alpha beta, out of this thing from this sort of abstract notation of the tensor, I'll remind you that we can do this by taking the tensor and plugging into its slots the basis 1 forms, which tell me something about-- basis 1 forms are really useful for sort of measuring fluxes across in a particular direction. And in fact, this quantity has the geometric interpretation-- think of this as the flux of momentum component alpha in the beta direction. So if we look at this component by component, it's worth doing that. So let's do the tt or 00 component. So this is the one we've already done. This is just rho 0 ut ut. It's just rho in that thing. According to the words I've written down here, this is the flux of pt in the t direction. So pt tells me about energy. The flux of energy in the energy direction refers to-- so here's my water flowing in the time-like direction, just sitting there apparently doing nothing, but no, it's moving through time. This is energy flowing through time. It's energy density, all sort of locked up in stable equilibrium here. But if somebody comes in someday with a nice cup of anti-water and combines it, we get to have all that energy released. And we can enjoy that for a few femtoseconds before we evaporate. Let's look at the other components. So T0i, you can take the definition and plug it in in terms of the components. It's obviously ut times ui, but still use the definition by words. So this is the flux of p sub t in the xi direction. This is talking about energy moving in a particular direction. This is nothing more than energy flux, energy sort of flux that we are used to think about, not flowing through time, but flowing through space. Now, T0i, this is the flux of momentum component i in the t direction. So this tells me about momentum density. So if I had this stuff flowing, there's some momentum associated with that flow. You can count up the amount of mass and every little volume. Divide by that volume. Take some ratios. That is the density of momentum. And finally, this last one, flux of p i in the xj direction, there's really no great wisdom for that. This is nothing more than momentum flux. If I have a bucket of water and things are sort of sloshing around, there's some momentum moving, and the whole assembly is moving in some direction, we can get a flow of momentum going in kind of a non-normal direction. As long as it goes in a normal direction, it might be moving along its flow. I'm going to talk about that in a few moments when I talk about a few different kinds of stress energy tensors. The key thing which I want you to be aware of is that under the hood of this thing, we're going to talk about different kinds of stress energy tensors for a little later in this class. But this basic interpretation of the way to think about the different components, it holds for all of them. So this is good intuition to have. One thing which is worth noting is if you look at-- if you actually-- let's take the form of the stress energy tensor that corresponds to the dust. So I'm going to write out all four of my components. I've already given you T00. T0i, I can write that as gamma squared rho 0 vi T0i equals gamma squared rho 0 vi. This guy is gamma squared rho 0 vi vj. There's two elements of symmetry here, which I want to emphasize. Notice energy density and momentum flux. Sorry. Energy flux and momentum density, those words are important. Energy flux and momentum density, they're exactly the same. That is actually true in all physics. It's clouded by the fact, though, that in most units that we measure things, we don't typically use speed of light equal to 1. So when you look at these kind of quantities in the physics that you are more used to, you would find your energy, your energy flux, your momentum density will have different units, because you'll have factors of c that come into there that convert from 1 momentum, and energy, and time flow, and space flow. Also, you don't usually include rest mass and rest energy in the energy flows and momentum densities that you've probably been familiar with in the past. So it's when we do relativity, and we set the speed of light equal to 1 that this symmetry between energy flux and momentum density becomes apparent. Question. AUDIENCE: The second one, do you mean rho 1 will be there? SCOTT HUGHES: Sorry. This one? AUDIENCE: Yeah, on the right-hand side. SCOTT HUGHES: On the right-hand side. This is-- AUDIENCE: Rho i. SCOTT HUGHES: No, this is definitely what I mean to write down here. So this is-- what I've done, I've taken advantage of the fact that I can write-- let's see. Do I have it down somewhere? Yeah. Yeah. So what I've done over here is I have written the fact-- so u sub t, ut is just the gamma factor. And I didn't explicitly write out what t-- the i components, but you get a similar factor that way. So no, that is correct that way. AUDIENCE: So that's why you mean the first and the second are the same. SCOTT HUGHES: Yeah. Yeah. And I just want to emphasize-- I mean, the key thing which I want to emphasize is when you do it in the physics that you guys have generally all seen up till now, it won't look that way. And there's two causes of that. One is you're not used to necessarily working in units where the speed of light is 1. And two, rest energy is built into these definitions. That shifts things a little bit. The other thing which I want to emphasize is notice this is symmetric under exchange of indices. Now, it's obvious for dust. It comes right back to the fact that these guys-- that those indices enter like so. It's an exterior product of the four-velocity with itself. I'm not going to go through in detail, but there is a physical motivation that I will sketch in a few moments that argues that it must be symmetric like this in all physical cases. I have a detailed sketch of this. I keep using the word sketch, because my mind wants to say proof, but it ain't a proof. It's sort of a motivation. And so I will post it for everyone to go through, but I will sort of illuminate it in a couple of moments. Basically, if this is not true, if it's not the case this is symmetric, a physical absurdity can be set up. I'll describe that in a few moments. So most of the world is not dust. So this is a decent example for helping to understand things. But we are going to want to do more complicated and interesting examples. Pardon me. Just one second. Let me just check here with this page. Yeah. So for us, the way that we are going to do this, there is a fairly general recipe that one can imagine applying to this. I'm going to save it for a little later in the class. It sort of borrows some techniques from field theory. Basically, if you can write down a Lagrangian density that describes the fifth system that you're under study, there's a particular variation you can do, that stress energy tensor emerges from. But for now, the key thing which I want to say is that we basically are going to deduce what the stress energy tensor looks like by essentially going into a particular frame-- we'll call it the rest frame. It's usually the rest frame of some element of the material being studied-- and thinking carefully about this physical definition of what the different components mean. This is not the most rigorous way to do it, but it's a good way to get started and develop some intuition. So let me give one example that for many of us in astrophysics, this is probably the one stress energy tensor that we write down and use over, and over, and over again in our career. And it's rare we do anything more than this in many, many cases. This is called a perfect fluid. So what is perfect about a perfect fluid? It begs the question here. What is perfect referring to here? So a perfect fluid is a fluid in which there is no energy flow in what I will call the "rest" frame. I put rest in quotes, because you have to sort of define it from the context of what your fluid is here. Basically, what it means is I can find a frame in which each fluid element there is no energy flow there. If a frame exists where that happens, there's a candidate to be a perfect fluid. And I also require there to be no lateral stresses. Lateral stresses refer to this Tij when i and j are not equal to one another. So this sort of refers to-- imagine that I have some-- let's say going into the board is the y direction. So if there's like some y stress that is somehow being transported in the x direction, that would be a lateral stress. Physically, this is actually-- that kind of a stress is hugely important when one is studying fluids. And one characterizes it by a quantity known as the viscosity. Viscosity tells me about how stress gets transported-- how momentum, rather, gets transported in a non-normal direction, against the direction in which the fluid is moving. So my perfect fluid has no viscosity. And I'll just conclude. I want to make-- there's a very important point here. A fluid that has no viscosity is a fluid that doesn't get anything wet. So this refers to when you pour water on yourself, the reason your hand gets wet is that there's some viscosity that actually prevents it-- causes there to be sort of a shearing force, which causes the water to stick to your skin. So a perfect fluid, this has been described as the physics of dry water. So it is-- anyone who's in applied math will know. They'll sort of roll their eyes and say, OK, fine, we'll do this sort of infant version of fluid first. And then a lot of the action and a lot of the fun comes from putting in viscosity and doing the real fluids. For the purposes of our class, what this boils down to is this tells me that the physics of this quantity is totally dominated by the fluid's energy density and its pressure. The pressure is an isotropic spatial stress. So in this particular frame, where I have no energy being transported, then the stress energy tensor can be represented as energy density as usual up in the-- whoops, over there now-- up in the tt component. There is no energy flux. By symmetry, if there's no energy flux, there is no momentum of density. And my spatial stresses are totally isotropic, and none of them are lateral. So it just looks like this, the diagonal of rho p, p, p, p. As I said, this is actually something that we're going to use over, and over, and over again. This is actually a-- so you can actually consider my dust stress energy tensor to be a perfect fluid with no pressure. So this actually subsumes this other one. I will-- just illustrative purposes, I'll show an example of a case that cannot be thought of as a perfect fluid. But we will tend to use this a lot in our class. And as I will demonstrate in just a moment, we're going to find that this ends up playing an important role in generating gravitational fields. What's interesting about this is you guys are probably used to the idea that mass generates gravity, and then throw in a c square that tells you energy to generating gravity. But it's also-- we're going to see pressure generates gravity. And it's connected to the mathematical structure of this guy here. So notice I've written an equal dot here. So this is just the way this is represented in this particular frame. I would like to write this in a more covariant form, something that does not rely on me going into a particular frame of reference. So the trick which I'm going to use for this is when I think about that form there-- so the rho piece of it, clearly, what I'm doing there is I'm picking out-- that is-- can be thought of-- as the energy density multiplied by the tensor product of the four-velocity that describes a particular element of this fluid. So I can write, again, using this sort of abstract notation. This piece of it looks like this. And so if I go into the frame, if I go into the rest frame of the fluid, that's just my u is 1 and 0. And the spatial components, that builds my upper left-hand corner of this tensor. How do I get the rest of this? Well, to get the rest of it, these are all sort of picked out of components of the tensor that are orthogonal to u. And I put an Easter egg in p set 1. You guys developed a tensor that allows me to build-- it's a geometric object that allows me to describe things that are orthogonal to a given four-vector. So if I take the projection tensor that you guys built one piece at 1, which looks like the metric plus the extern-- the tensor product of a four-vector with itself, that gives me the p's that go into that component. Or if I write this in index notation, I can do it in two ways. I will also emphasize that there are few moments in this class where I sort of urge you to take the sort of long-term memory synapses and switch them on. This is one of those moments. In a couple of lectures, we're going to introduce the principle of equivalence, which is the physical principle by which we're going to argue how we go from formulas that work in special relativity to formulas that work in general relativity. And by invoking the principle of equivalence, we're going to see that when we want to describe perfect fluids in a general spacetime, not just in special relativity and general relativity, it's exactly this formula. I just need to modify what the metric means, but that will allow me to carry that over. Let me see. So my notes are a little bit disorganized here. There's a piece that I-- every year I want to clean this bit up, and every year at this time of year, I have 70 gajillion administrative tasks. And I end up getting behind schedule. So I will clean this up on the fly. So there's a couple of points which I want to make about this. Let me do this point first. So by virtue of taking a graduate physics class, I'm confident you guys know about Newtonian gravity. One can write the field equation for Newtonian gravity as essentially a differential equation for the potential that governs Newtonian gravitational interaction. So let's call phi sub g the Newtonian gravitational interaction. Whoops. And I can write a field equation that's governing it as essentially its Poisson's equation. So the Laplace operator acting on that potential is up to a constant equal to the-- we usually learn it in terms of mass density. We're working in units where the speed of light is equal to 1. So it could just as well be the energy density. So you've all kind of seen that. Now, if we think about how we're going to carry this forward and make gravity a relativistic interaction, this equation should right away make us suspicious, because we spent several minutes earlier today talking about the fact that this is not a scalar. This is a component of a tensor. A physical theory which tries to pick out just one component of a geometric tensor is not a healthy theory. It would be like if you had sort of learned in E&M that there was a preferred direction of the electric field. If there's a particular set of Maxwell's equations for Ex, and a different set of Maxwell's equations for Ey. That would just be nonsense. Nature doesn't pick out spatial components of any one being particularly having some weight over the others, and the same thing holds in relativity for spacetime. So when we make this into a relativistic theory, we're going to say if this component of a tensor plays a large role in gravity to make this into a geometric object that makes a large role in gravity, I'm going to have to-- so let's call this my Newtonian equation. My Einsteinian equation-- I'm going to put an equal sign in quotes here, because there's a lot of details to fill in. But what's going to go on the right-hand side of this has to be something that involves the stress energy tensor. Newton picks out one component stress energy tensor. Relativity doesn't let me pick out particular components. So whatever I get when I do Einstein's gravity, the whole stress energy tensor is going to be important in setting the source of my gravity. That then sort of says, well, then what the heck do you do with this left-hand side? That is, in fact, going to be starting probably on Tuesday the subject of the next couple of lectures, basically going all the way up to spring break. The week before spring break is when we complete the story of what goes on the right-hand side-- excuse me-- the left-hand side of this equation. But what I will tell you is that it is indeed going to involve two derivatives of a potential-like object. And the potential-like object is actually going to turn out to be the metric of spacetime. So that's kind of where we're going with this. Let's do a little bit more physics with the stress energy tensor. So I have somewhat more detailed notes, which I will post online, which I am not going to go through in great detail here. But I'm going to kind of sketch this. So one is that I would like to prove-- again, that word is a little bit of an overstatement, but at least motivate the symmetry of this tensor. I'm just going to focus on the spatial bits of this. That'll be enough. Like I said, you can kind of see that T0i and ti0 are the same thing by thinking about the physical meaning of four-momentum and what a flux of four-momentum is. This one, there's kind of a cute calculation you can do. So imagine you have some little cube of stuff. It could be immersed in some field or fluid, something that is described by a field of stress energy. And so to start out with, let's look at how the flux-- remember what t alpha beta, or really Tij tells me about is the flux of momentum in a particular direction. It's telling me about the amount of momentum going into this box on one side and coming out on the other. So I'm going to look at the momentum flux into and out of this box. And so what I'm going to do is let's look at the momentum going into the sides that are-- so let's call this the top and the bottom here. What I'm going to do then is number the four sides, the other four sides of the box, not the top and the bottom-- so let's call this side, which is sort of facing away from us here, that side one. The one that is facing us that sort of points out in the x direction, I'll call that two. This side here, I will call three. And the one that is on the back I will call four. Apologies for the little bit of a busy picture here. But just what you want to do is sort of imagine there's a cube in front of you, and you go around and label the four sides. What we'd like to do is calculate what is the force that is flowing through each of these four sides, 1, 2, 3, and 4. Just ignore the top and the bottom for just a second. I'll describe why that is in just a moment. So if I look at the force on face one, well, face one, it is-- pardon me-- I mislabeled my sides. I want to be synced up with my notes, and so I realize it's annoying. And I'm very sorry about that. But if I mess this up, I will get out of sync with what I have written down. So the one that is facing us is side one. Two is on the right-hand side. Three is the back. Four is on the left-hand side. My apologies for that. But I think it's important we get that right so I don't get out of sync with myself. So the total force on face one is what I get by basically adding up all the flux of momentum flowing through face one. So force on face one, that is-- the i-th component of that is what I get when I integrate that over face one, which is normal to the x-axis. Each side of the cube is little l. So it's Tix l squared. Force on face two, now this is the one that is normal for the y-axis. And so this guy looks like Tiy l squared. And you continue this F3-- if you look at, and you if you assume that this thing is small, it's approximately the same as the force on F1. It'll become equal in the limit of the q becoming infinitesimally small. And this is approximately the same as the force on face two. So that tells us that there's no unbalanced force on this thing, which is great. Basically, it means that there's no unbalanced force causing this little element to accelerate away. One of the reasons why I am focusing on these four sides, though, is I would also like to consider torques that are acting on this thing. And here's where I'm going to skip over a couple of details and just leave a few notes for you guys to look at, because we're a little short on time with other things. This calculation is straightforward, but it gets-- I already screwed up a detail here. I don't want to sort of risk messing up a few other things. What I want to do is consider the torques about an axis that sort of runs through the middle of this thing that goes from the top and the bottom of this. So go through and just add up all the torques associated with these little forces about an axis in the center of-- that goes through the center of this cube. So I'm going to-- like I said, I'm going to leave out the details, but basically, you go through, and you do the usual r cross f to get the forces, the torques associated with each of these different faces. And what you'll find when you do this is that there is a net torque that looks like l cubed times Txy minus Tyx. Again, though, this scales with the size of the cube. So you look at this and kind of go, well, who cares? It's a little bit there. Maybe it spins up a little bit. But in the limit, the thing going to 0, there's no net effect. Well, let's be a little bit careful about this. What is the moment of inertia of this cube? I don't know exactly, but I know that it's going to be something like the mass of this cube. We'll call that l cubed times its mass density, or its energy density. And it's going to involve two powers of the only length scale characterizing this thing. And there will be some prefactor alpha. There will be some number that's related to the geometry of this. An integral or two will easily work out what the alpha is and make this more precise. But the key bit which I want to emphasize is that this is something that scales as the size of the fifth power. So yeah, the torque vanishes as l goes to 0, but it does so with the cubed power. It doesn't vanish as quickly as the moment of inertia vanishes. And I'll remind you, the angular acceleration of the cube, theta double dot, is the torque divided by the moment of inertia. So this is something that is proportional to Txy minus Tyx divided by l squared. So yeah, the torque does vanish as l goes to 0, but the moment of inertia vanishes more rapidly as l goes to 0. That sort of suggests that if we're in a screw universe, little microscopic vortices are just going to start randomly oscillating in any cup of water that you pour in front of you. Now, I don't have a proof that nature abhors that, but it seems pretty screwy. And so the case that most textbooks make at this point is to say, physics indicates we must have Txy strictly equal to Tyx in all cases, independent of what the size of this thing is. That had better just be a bloody 0 in the numerator in order to prevent this physical absurdity from being set up. Repeat the exercise by looking at torques around the other axes. And that drives you to the statement that this thing must, in general, be spatially symmetric. And again, physics of the way that energy and momentum behave in relativity makes our sort of timespace component symmetric as well. I emphasize this is not a proof. This is really just a physical motivation. This is the kind of thing that I'm a bloody astrophysicist. I like this kind of stuff. There is a different way of developing the stress energy tensor, as I said, that comes from a variational principle sort of based on sort of quantum field theoretic type of methods. And the symmetry in that case is manifest. It really does sort of come up. But this is a good way of just motivating the fact that you're not going to have any non-diagonal-- or excuse-- non-symmetric stress energy tensors. If you did, you would have really bizarre matter. Now, the stress energy tensor has this physical interpretation that it tells me about the flow of energy and the flow of momentum in spacetime. As such, it is the tool by which we are going to put conservation of energy and conservation of momentum into our theory. And the way we're going to do this is with a remarkably simple equation. The spacetime divergence of t alpha beta-- pardon me-- must equal 0. This is a covariant formulation of both conservation of energy and conservation of momentum. And if you want to say, well, which one? Is it energy or is it momentum? You can't say that in general. I can only say that after I have picked a particular reference frame, because it's only once I have defined time and I've defined space that I've actually defined energy and momentum. Prior to choosing a particular reference frame, all I have is four-momentum. One observer's energy is another observer's superposition of energy and momentum. Once I have picked a particular frame-- so once I have picked a particular frame, then if I evaluate d alpha T alpha either 0 or t, this is what conservation of energy looks like in that frame. So in that frame, here is conservation of energy. I can fit it. Hang on a second. No, no, no. I had it right the first time. Yeah. Sorry. I was just trying to make sure I got my indices lined up properly. So apologies you can't quite read the bottom one so well. The top one is conservation of energy in that particular frame. The second one is conservation of momentum in that particular frame. And again, the key thing which I want to emphasize is the covariant statement basically puts both of them together into a single equation. I can only sensibly state conservation of energy and conservation of momentum according to some particular observer. Now, I can repeat the game that I had done before with the number vector and turned these conservation laws into integral equations as well. Let me do it for energy. So if I take that integral equation and integrate over a three-volume, or if you prefer, I can take the original covariant formulation and integrate that over a four-volume. With a little bit of manipulation akin to the way I manipulated the conserv-- the integrals associated with the number vector in the previous lecture, you can write down a law that looks like this. And so this-- again, what I've done here is I've chosen a particular set of timeline coordinates. And the language that we often use in relativity is we basically say I'm going to take a single slice of time. And I would say that the rate of change of energy-- so integrate energy over a volume. It's the total energy in that V3. The rate of change of that thing is balanced by the amount of energy flowing into or out of through the boundaries of that volume. Add an extra index here. So make this-- make one of these be a j. Make this guy be a j. And you've got a similar statement for the conservation of momentum. This is a particular trick. I have a set of typed up notes that just sort of clean this up a little bit. They're basically exactly this point, but I'll put them online after today's lecture. You're going to want to use this on one of the p set problems this week. So you guys are going to do a couple exercises that involve integrating and actually finding essentially moments of the left-hand side of this thing. You're going to do a few exercise where you take advantage of this formulation of conservation of energy and momentum to derive a few identities involved in the stress energy tensor, one of which is another Easter egg that we're going to use quite a bit in a future lecture. So let me just wrap up our discussion of the stress energy tensor by just doing two more examples. And then I'm going to sort of begin switching gears a little bit. So far I have only described-- essentially, I've really only described perfect fluids. Dust is a perfect fluid with no pressure. There are a lot of other kinds of materials that we want to work with in the universe. One of which-- it's unlikely many of you are going to use this, but it's actually the stress energy tensor on which I have built-- well, I guess Alex is going to use it a little bit-- I have built a big chunk of my career. Suppose you have-- we kind of talked a little bit about how the four-velocity and the four-momentum are really only good for talking about like the kinematics of particles. Well, actually, particles aren't a bad thing to focus on in some of your studies. And much of my research is actually based on the idea of thinking about a binary system as one member being a black hole, and its companion being a particle-like object that is a particular limit of it, a particle-like object that orbits it. So a really simple stress energy tensor-- and I'm just throwing it out here, because I think it nicely illustrates the principle. A point particle with rest mass, we'll call it m0. And I'm going to say it's moving on a particular world line through spacetime. So the way we define a world line is we just say it's some four-vector that describes the place of this thing from some chosen origin. And it's generally most convenient to parameterize it by the proper time of whatever object or creature is moving on that world line. And so it's a lot like dust, only there's no volume. It's sort of like one particle of dust. And so the stress energy tensor we use for this-- let me back up for just a second here. When you guys learned about electricity and magnetism, one of the first things you learn about are point charges. And then a little bit later you learn about charge distributions. And you have charge densities. And then, usually, at some point in a class-- it's often like junior level E&M-- we say, how do you describe the density of a point charge? That's where you learn about the Dirac delta function. Well, if I have a point particle, I'm going to need to describe this thing's energy density as essentially a Dirac delta function. And so what we do is imagining that these things-- so this is the four-velocity of that body. They might be functions of time as this guy is moving along here. What we do is we introduce a Dirac delta function as this thing moves along through spacetime. What this does-- you can check the dimensions. This gives me exactly what we need in order to have something that's dimensionally correct, and has all the symmetries and all the properties that describe a particle moving at a particular four-velocity through spacetime. Now, you might want to just-- it's the thing which I kind of want to pause on for a second. You go, what the hell do you do with this, right? That's kind of inconvenient. Well, the trick we use to sort of clean up that Dirac delta function, it's very much like what you do when you encounter multidimensional delta functions in basic physics. You just build it out of a bunch of one-dimensional delta functions. Likewise, you'll have a term with a y component and the z component. Then you use the rule that if I integrate a function of x against a delta function whose argument is itself a function of x, I evaluate that function at the 0's of g. So let's say x0 is where g has a 0. Normalizing to the first derivative of g evaluated at that 0. When you put all of that together, that basically means you can do this somewhat abstract integral formally. I mean, exactly. You can just do it analytically. And what you do is you choose one of the delta functions to apply it to. Traditionally, people apply it to the timeline component. And what is the derivative of the world line z component with respect to proper time? It's the 0 component of the four-velocity. So you just divide by the 0 component of the four-velocity, and then you're left with a three-dimensional delta function for the spatial trajectory of this thing through all space. So this is an example of one that just, again, kind of the intuition is one of the things I want to emphasize. Notice we had-- this has the symmetries that we wanted to have. It's not hard to show that this. You can think of this as essentially being kind like a gamma factor. This all ends up giving me just what we need for this thing to have the right transformation properties. And it does, in fact, play a role in some-- well, I'll say in some research that's near and dear to my heart. Let me do another example. Suppose you want to know the stress energy tensor associated with a given electric and magnetic field. Well, first, let me just quote for you the exact answer, which is most compactly written if we use that Faraday tensor F, which describes electromagnetic fields in a frame-independent fashion, the way that I introduced it in the last lecture. So in units where basically everything but pi is set equal to 1, it ends up turning into this. So that's a bit of a mouthful. Let's go in and look at particular components of it, though. So let's say I go into a particular frame. I fill in my Faraday tensor with the form of electromagnetic field that I introduced last time. And I'll just go through, and I'll evaluate all the different components of this thing. So what you find when you fill this in is your T00 component, 1 over 8 pi e squared plus b squared. That's good. Hopefully, y'all remember from basic E&M, the energy density of an E-field is e squared over 8 pi and the right system of units. Energy density of a b field is b squared over 8 pi if you work in God's units. Let's do the timespace component. So again, hack through that mess there. This is going to be something that is a vector. In fact, it's the Poynting vector. Could it be anything else? If you use the recipe that I sort of suggested as the easiest way to approach this, this is kind of what you would have guessed for something like that. The bit that's actually kind of hard is then trying to get the spatial stresses of this thing. And here I don't have any great intuition for this one to convey to you. It's derived in textbooks like Griffiths and Jackson. So I'll just quote for you the result. So you get one term that looks like E squared plus b squared on the diagonals. Then there's a correction, which looks like this. I want to just quickly call out one example so you can see what the significance of this example is. So suppose you have something like a pair of capacitors. And there's just a uniform electric field between them. You want to evaluate the stress energy tensor between those pairs of capacitors. So my spatial electric field, let's say it just points in the Ex direction. And it's constant. So when you actually evaluate this guy, there's no energy flow. There's no magnetic field. So there's no Poynting vector. You, of course, have E squared over 8 pi for your energy density. Very different from a perfect fluid. And this kind of makes sense. That's sort of telling you that there is a stress that if you have a pair of plain parallel capacitors, you tend to attract the plates to each other. But there is a pressure associated with that electric field that actually goes in the other directions. This is your x direction. This is y and z. And as a consequence of this-- so there's some stuff which we're not going to do too much with in this class, but I may give you some pointers on this. Electric and magnetic fields, they generate pressures, at least in certain directions. They kind of generate like an anisotropic pressure. And when we start coupling this to gravity, you can get electric fields and magnetic fields that contribute non-negligibly to the gravity of their object. That is it for the stress energy tensor. As I said, we are going to use this guy over, and over, and over again. And the reason for this does come back to that little motivation that I gave probably about 45 minutes ago where we sort of looked at the Newtonian field equation, and then said, picking out a particular scalar as the source of gravity makes no sense in a relativistic covariant theory. It's got to be the whole tensor. And indeed, this is the one-- well, not the E&M one, but the general notion of a stress energy tensor is the one that we are going to use for that. So we have about 10 minutes left. And so I would like to start the process of switching gears. Before I do that, are there any questions? I will clean the board. So the reason we are switching gears is we now have probably the most important physical tools that we need in order to start thinking about making a relativistic theory of gravity. But we need a few more mathematical tools. In particular, I'm going to argue in a couple of lectures that flat spacetime is not sufficient for us to build a theory of gravity. We're going to need to-- first of all, you're going to have to understand what that means. And we're not quite ready to go there. So for now, it's just fancy words. But I'm going to have to introduce some kind of a notion of curvature into things. What does that even mean really? We need to have the tools to do that. And so as a prelude to going in that direction-- I will call this my prelude to curvature-- what we're going to start doing is flat spacetime in curvilinear coordinates. And the importance of doing this, why this is going to be useful to us is that it will introduce a-- it'll keep the physics simple. It's still going to be special relativity, but it's now going to special relativity using a mathematical structure in which the basis vectors are no longer constant. So that's going to allow us to begin making a couple of the mathematical tools that are necessary to build gravity into this theory. So we'll start by replacing this with just simple plain polar coordinates mapped in the usual way. So if we want to go back and forth, well, at least one way transformation, I build x and y from r and phi in that usual way. There's an inverse mapping as well, which involves trig functions. So I'm not going to write it down. One point which I really want to emphasize here is we are going to continue to use a coordinate basis. And to remind you what that means, so a coordinate basis means that the differential displacement vector from an event a to a nearby event b is simply related by differentials of my coordinate contracted onto my basis vectors. But when I'm working in a curvilinear coordinate system like this, that means one of them has a somewhat different form from what you are used to. When you guys talk about the differential of a displacement in just about every physics class up to now, if you have a differential angle, you'd want to throw an r. So this has the dimensions of length. We ain't going to do that. And so what this means, since this is an angle, every component of this has the dimensions of length. That means that my basis vector is going to be something that has the dimensions of length associated with it. This in turn means that this is not going to be a normal basis. That's unfortunately a somewhat loaded word. What I mean by that is that it has not been normalized. But given that you guys have spent all of your career thinking about the dot product of a unit vector with itself-- of a basis vector with itself being equal to 1, the other meaning of normal might be good for you, too. The key thing which I want to emphasize here is e phi dot e phi does not equal 1. One other little bit of notation which I would like to introduce-- so we are going to want to talk about transformations between different representations. We've done this-- so far, we have generally focused on moving between different reference frames. I want to generalize this notion. And I'm going to tweak my notation a little bit to indicate the difference. So I'm going to call capital L alpha mu bar what I get when I just look at the variation of the alpha coordinate system-- sorry-- the unbarred coordinate system with the barred one. So this is just-- there's no deep things here. I just want you to be familiar with the notation which I'm going to use. I will always reserve lambda for the Lorentz Transformation. So in the interest of time, I'll just write down one of these. So this sort of means like, for instance, suppose that I'm transforming. Let's let barred indicate my polar coordinates. Unbarred be Cartesian. These things are surprisingly sticky sometimes. So if I want to transform in one direction between my barred and my unbarred ones, this matrix will go through it. And it looks like this. So this is the thing which I will call l alpha mu bar. In my notes, I also give the inverse transformation. And I will actually write-- you know what? I have minute. I will write that one down as well. If you want to know how to go in the other direction, it's just the matrix that inverts this. The reason I decided to take the extra 30 seconds or so to write this down is I want to call out the fact that in this kind of transformation, because of the fact that this is a coordinate basis that has this somewhat unusual property, different elements of the matrix have different dimensions. It's a feature, not a bug. So this is going to show up. I'm going to do a couple more calculations with this a little bit tomorrow-- sorry-- not tomorrow, on Tuesday. This is going to show up in the way the metric looks. The metric is going to have a very different character in this coordinate system. And we're going to see the way in which-- it basically boils down to it's going to pick up a non-trivial functional form. It's not going to just be a constant. And that fundamentally reflects the fact that the basis vectors are not constant anymore. We'll end it there.
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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 we are going to change perspective again and think in terms of lattice models. So for the first part of this course, I was trying to change your perspective from thinking in terms of microscopic degrees of freedom to a statistical field. Now we are going to go back and try to build pictures around things that look more microscopic. Typically, in many solid state configurations, we are dealing with transitions that take place on a lattice. For example, imagine a square lattice, which is easy to draw, but there could be all kinds of cubic and more complex lattices. And then, at each side of this lattice, you may have one microscopic degrees of freedom that is ultimately participating in the ordering and the phase transition that you have in mind. Could, for example, be a spin, or it could be one atom in a binary mixture. And what you would like to do is to construct a partition function, again, by summing over all degrees of freedom. And we need some kind of Hamiltonian. And what we are going to assume governs that Hamiltonian is the analog of this locality assumption that we have in statistical field theories. That is, we are going to assume that one of our degrees of freedom essentially talks to a small neighborhood around it. And the simplest neighborhood would be to basically talk to the nearest neighbor. So if I, for example, assign index i to each side of the lattice, let's say I have some variable at each side that could be something. Let's call it SI. Then my partition function would be obtained by summing over all configurations. And the weight I'm going to assume in terms of this lattice picture to be a sum over interactions that exist between pairs of sides. So that's already an assumption that it's a pairwise thing. And I'm going to use this symbol ij with an angular bracket along it, around it, to indicate the sum over nearest neighbors. And there is some function of the variables that live on these neighbors. So basically, in the picture that I have drawn, interactions exist only between the places that you see lines. So that this pin does not interact with this pin, this pin, or this pin, but it interacts with these four pins, to which it is near neighbor. Now clearly, the form of the interaction has to be dictated by your degrees of freedom. And the idea of this representation, as opposed to the previous statistical field theory that we had, is that in several important instances, you may want to know not only what these universal properties are, but also, let's say, the explicit temperature or phase diagram of the system as a function of external parameters as well as temperature. And if you have some idea of how your microscopic degrees of freedom interact with each other, you should be able to solve this kind of partition function, get the singularities, et cetera. So let's look at some simple versions of this construction and gradually discuss the kinds of things that we could do with it. So some simple models. The simplest one is the Ising model where the variable that you have at each side has two possibilities. So it's a binary variable in the context of a binary alloy. It could be, let's say, atom A or atom B that is occupying a particular site. There are also cases where there will be some-- if this is a surface and you're absorbing particles on top of it, there could be a particle sitting here or not sitting here. So that would be also another example of a binary variable. So you could indicate that by empty or occupied, zero or one. But let's indicate it by plus or minus one as the two possible values that you can have. Now, if I look at the analog of the interaction that I have between two sites that are neighboring each other, what can I write down? Well, the most general thing that I can write down is, first of all, a constant I can put, such as shift of energy. There could be a linear term in-- let's put all of these with minus signs. Sigma i and sigma j. I assume that it is symmetric with respect to the two sides. For reasons to become apparent shortly, I will divide by z, which is the coordination number. How many bonds per side? And then the next term that I can put is something like j sigma i sigma j. And actually, I can't put anything else. Because if you s of this as a power series in sigma i and sigma j, and sigma has only two values, it will terminate here. Because any higher power of sigma is either one or sigma itself. So another way of writing this is that the partition function of the Ising model is obtained by summing over all 2 to the n configurations that I have. If I have a lattice of n sides, each side can have two possibilities of a kind of Hamiltonian, which is basically some constant plus h sum over i sigma i kind of field that prefers one side or the other side. So it is an analog of a magnetic field in this binary system. And basically, I convert it from a description that is overall balanced to a description overall sides. And that's why I had put the coordination number there. It's kind of a matter of convention. And then a term that prefers neighboring sites to be aligned as long as k positive. So that's one example of a model. Another model that we will also look at is what I started to draw at the beginning. That is, at each site, you have a vector. And again, going in terms of the pictures that we had before, let's imagine that we have a vector that has n components. So SI is something that is in RN. And I will assume that the magnitude of this vector is one. So essentially, imagine that you have a unit vector, and each site that can rotate. So if n equals to one, you have essentially one component vector. Its square has to be one, so the two values that it can have are plus one and minus one. So the n equals to one case is the Ising model again. So this ON is a generalization of the Ising model to multiple components. n equals to two corresponds to a unit vector that can take any angle in two dimensions. And that is usually given the name xy model. n equals to three is something that maybe you want to use to describe magnetic ions in this lattice. And classically, the direction of this ion could be anywhere. The spin of the ion can be anywhere on the surface of a sphere. So that's three components, and this model is sometimes called the Heisenberg model. Yes. AUDIENCE: In the Ising model, what's the correspondence between g hat h hat j hat and g, h, and k? PROFESSOR: Minus beta. Yeah. So maybe I should have written. In order to take this-- if I think of this as energy pair bond, then in order to get the Boltzmann weight, I have to put a minus beta. So I would have said that this g, for example, is minus beta g hat. k is minus beta j hat. And h is minus beta h hat. So I should have emphasized that. What I meant by b-- actually, I wrote b-- so what I should have done here to be consistent, let's write this as minus beta. Thank you. That was important. If I described these b's as being energies, then minus beta times that will be what will go in the Boltzmann factor. OK So whereas these were examples that we had more or less seen their continuum version in the Landau-Ginzburg model, there are other symmetries that get broken. And things for which we didn't discuss what the corresponding statistical field theory is. A commonly used case pertains to something that's called a Potts model. Where at each side you have a variable, let's call it SI, that takes q values, one, two, three, all the way to q. And I can write what would appear in the exponent minus beta h to be a sum over nearest neighbors. And I can give some kind of an interaction parameter here, but a delta si sj. So basically, what it says is that on each side of a lattice, you put a number. Could be one, two, three, up to q. And if two neighbors are the same, they like it, and they gain some kind of a weight. That is, if k is positive, encourages that to happen. If the two neighbors are different, then it doesn't matter. You don't gain energy, but you don't really care as to which one it is. The underlying symmetry that this has is permutation. Basically, if you were to permute all of these indices consistently across the lattice in any particular way, the energy would not change. So permutation symmetry is what underlies this. And again, if I look at the case of two, then at each side, let's say I have one or two. And one one and two two are things that gain energy. One two or two one don't. Clearly it is the same as the Ising model. So q equals to two is another way of writing the Ising model. Q equals to three is something that we haven't seen. So at each site, there are three possibilities. Actually, when I started at MIT as a graduate student in 1979, the project that I had to do related to the q equals to three Potts model. Where did it come from? Well, at that time, people were looking at surface of graphite, which as you know, has this hexagonal structure. And then you can absorb molecules on top of that, such as, for example, krypton. And krypton would want to come and sit in the center of one of these hexagons. But its size was such that once it sat there, you couldn't occupy any of these sides. So the next one would have to go, let's say, over here. Now, it is possible to subdivide this set of hexagons into three sub lattices. One, two, three, one, two, three, et cetera. Actually, I drew this one incorrectly. It would be sitting here. And what happens is that basically the agile particles would order by occupying one of three equivalent sublattices. So the way that that order got destroyed was then described by the q equals to three Potts universality class. You can think of something like q equals to four that would have a symmetry of a tetragon. And so some structure that is like a tetragon getting, let's say, distorted in some particular direction would then have four equivalent directions, et cetera. So there's a whole set of other types of universality classes and symmetry breakings that we did not discuss before. And I just want to emphasize that what we discussed before does not cover all possible symmetry breakings. It was just supposed to show you an important class and the technology to deal with that. But again, in this particular system, let's say you really wanted to know at what temperature the phase transition occurs, as well as what potential phase diagrams and critical behavior is. And then you would say, well, even if I could construct a statistical field theory and analyze it in two dimensions, and we've seen how hard it is to go below some other critical dimension, it doesn't tell me things about phase diagrams, et cetera. So maybe trying to understand and deal with this lattice model itself would tell us more information. Although about quantities that are not necessarily inverse. Depending on your microscopic model, you may try to introduce more complicated systems, such as inspired by quantum mechanics, you can think of something that I'll call a spin S model in which your SI takes values from minus s, minus s plus 1, all the way to plus s. There's 2s plus 1 possibilities. And you can think of this as components of, say, a quantum spin of s along the zed axis. Write some kind of Hamiltonian for this. But as long as you deal with things classically, it turns out that this kind of system will not really have different universality from the Ising model. So let's say we have this lattice model. Then what can we do? So in the next set of lectures, I will describe some tools for dealing with these models. One set of approaches, the one that we will start today, has to do with the position space renormalization group. That is the approach that we were following for renormalization previously. Dealt with going to Fourier space. We had this sphere, hyper sphere. And then we were basically eliminating modes at the edge of this sphere in Fourier space. We started actually by describing the process in real space. So we will see that in some cases, it is possible to do a renormalization group directly on these lattice models. Second thing is, it turns out that as combinatorial problems, some, but a very small subset of these models, are susceptible to exact solutions. Turns out that practically all models in one dimension, as we will start today, one can solve exactly. But there's one prominent case, which is the two dimensionalizing model that one can also solve exactly. And it's a very interesting solution that we will also examine in, I don't know, a couple of weeks. Finally, there are approximate schemes that people have evolved for studying these problems, where you have series expansions starting from limits, where you know what is happening. And one simple example would be to go to very high temperatures. And at high temperatures, essentially every degree of freedom does what it wants. So it's essentially a zero dimensional problem that you can solve. And then you can start treating interactions perturbatively. So this is kind of similar to the expansions that we had developed in 8 333, the virial expansions, et cetera, about the ideal gas limit. But now done on a system that is a lattice, and going to sufficiently high order that you can say something about the phase transition. There is another extreme. In these systems, typically the zero temperature state is trivial. It is perfectly ordered. Let's say all the spins are aligned. And then you can start expanding in excitations around that state and see whether eventually, by including more and more excitations, you can see the phase transition out of the ordered state. And something that is actually probably the most common use of these models, but I won't cover in class, is to put them on the computer and do some kind of a Monte Carlo simulation, which is essentially a numerical way of trying to generate configurations that are governed by this weight. And by changing the temperature as it appears in that weight, whether or not one can, in the simulation, see the phase transition happen. So that's the change in perspective that I want you to have. So the first thing that we're going to do is to do the number one here, to do the position space renormalization group of one dimensional Ising model. And the procedure that I describe for you is sufficiently general that in fact you can apply to any other one dimensional model, as long as you only have these nearest neighbor interactions. So here you have a lattice that is one dimensional. So you have a set of sites one, two. At some point, you have i, i minus one, i plus one. Let's say we call the last one n. So there are n sites. There are going to be 2 to the n possible configurations. And your task is given that at each site, there's a variable that is binary. You want to calculate a partition function, which is a sum over all these 2 to the n configurations. Of a weight that is this e to the sum over i B, the interaction that couples SI and SI plus 1. Maybe I should have called this e hat. And B is the thing that has minus beta absorbed in it. So notice that basically, the way that I have written it, one is interacting with two. i minus one is interacting with i. i is interacting with i plus one. So I wrote the nearest neighbor interaction in this particular fashion. We may or may not worry about the last spin, whether I want to finish the series here, or sometimes I will use periodic boundary condition and bring it back and couple it to the first one, where I have a ring. So that's another possibility. Doesn't really matter all that much at this stage. So this runs for i going from one to n. There are n degrees of freedom. Now, renormalization group is a procedure by which I get rid of some degrees of freedom. So previously, I have emphasized that what we did was some kind of an averaging. So we said that let's say I could do some averaging of three sites and call some kind of a representative of those three. Let's say that we want to do a RG by a factor of b equals to two. So then maybe you say that I will pick sigma i prime and u sigma i to be sigma i plus sigma i plus 1 over 2, doing some kind of an average. The problem with this choice is that if the two spins are both pluses, I will get plus. If they're both minuses, I will get minus. If there is one plus and one minus, I will get zero. Why that is not nice is that that changes the structure of the theory. So I started with binary variables. I do this rescaling. If I choose this scheme, I will have three variables per site. But I can insist upon keeping two variables per site, as long as I do everything consistently and precisely. So maybe I can say that when this occurs, where the two sites are different, and the average would be zero, I choose as tiebreaker the left one. So then I will have plus or minus. Now you can convince yourself that if I do this, and I choose always the left one as tiebreaker, the story is the same as just keeping the left one. So essentially, this kind of averaging with a tiebreaker is equivalent to getting rid of every other spin. And so essentially what I can do is to say that I call a sigma i prime. Now, in the new listing that I have, this thing is no longer, let's say, the tenth one. It becomes the fifth one because I removed half of things. So sigma i prime is really sigma 2i minus 1. So basically, all the odd ones I will call to be my new spins. All the even ones I want to get rid of, I'll call them SI. So this is just a renaming of the variables. I did some handwaving to justify. Effectivity, all I did was I broke this sum into two sets of sums, but I call sigma i prime and SI. And each one of them the index i, rather than running from 1 to n in this new set, the index i runs from 1 to n over 2. So what I have said, agian, is very trivial. I've said that the original sum, I bring over as a sum over the odd spin, whose names I have changed, and a sum over even spins, whose names I have called SI. And I have an interaction, which I can write as sum over i, essentially running from 1 to n over 2. I start with the interaction that involves sigma i prime with si because now each sigma i prime is acting with an s on one side. And then there's another interaction, which is SI, and the next. So essentially, I rename things, and I regrouped bonds. And the sum that was n terms now n over 2 pairs of terms. Nothing significant. But the point is that over here, I can rewrite this as a sum over sigma i prime. And this is a product over terms where within each term, I can sum over the spin that is sitting between two spins that I'm keeping. So I'm getting rid of this spin that sits between spin sigma i prime and sigma i plus 1 prime. Now, once I perform this sum over SI here, then what I will get is some function that depends on sigma i prime and sigma i plus 1 prime. And I can choose to write that function as e to the b prime sigma i prime sigma i plus 1 prime. And hence, the partition function after removing every other spin is the same as the partition function that I have for the remaining spins weighted with this b prime. So you can see that I took the original partition function and recast it in precisely the same form after removing half of the degrees of freedom. Now, the original b for the Ising model is going to be parameterized by g, h, and k. So the b prime I did parameterize. So this, let's say, emphasizes parameterized by g, h, k. I can similarly parameterize this by g prime, h prime, k prime. And how do I know that? Because when I was writing this, I emphasized that this is the most general form that I can write down. There is nothing else other than this form that I can write down for this. So what I have essentially is that this e to the b prime, which is e to the g prime plus h prime sigma i prime plus sigma i plus 1 prime plus k prime sigma i prime sigma i plus 1 prime involves these three parameters, is obtained by summing over SI. Let me just call it s being minus plus 1 of e to the g plus h sigma 1 sigma i prime plus SI plus k sigma i prime SI plus g plus kh sigma SI plus sigma i plus 1 prime plus k SI sigma i plus 1. So it's an implicit equation that relates g prime, h prime, k prime, to g, h, and k. And in particular, just to make the writing of this thing explicit more clearly, I will give names. I will call e to the k to the x, e to the h to by, e to the g to bz. And here, similarly, I will write x prime e to the k prime, y prime e to the h prime, and z prime is e to the g prime. So now I just have to make a table. I have sigma i prime sigma i plus 1 prime. And here also I can have values of s. And the simplest possibility here is I have plus plus. Actually, let's put this number further out. So if both the sigma primes are plus, what do I have on the left hand side? I have e to the g prime, which is z prime. e to the 2h prime, which is y prime squared. e to the k prime, which is x prime. What do I have on the left hand side? On the left hand side, I have two possible things that I can put. I can put s to be either plus or s to be minus, and I have to sum over those two possibilities. You can see that in all cases, I have e to the 2g. That's a trivial thing. So I will always have a z squared. Irrespective of s, I have two factors of e to the h sigma prime. OK, you know what happened? I should've used-- since I was using b, this factor of h over 2. So I really should have put here an h prime over 2, and I should have put here an h over 2 and h over 2 because the field that was residing on the sites, I am dividing half of it to the right and half of it to the left bond. And since I'm writing things in terms of the bonds, that's how I should go. So what I have here actually is now one factor of i prime. That's better. Now, what I have on the right is similarly one factor of y from the h's that go with sigma 1 prime, sigma 2 prime. And then if s is plus, then you can see that I will get two factors of e to the k because both bonds will be satisfied. Both bonds will be plus plus. So I will get x squared, and the contribution to the h of the intermediate bond s is going to be 1e to the h. So I will get x squared y. Whereas if I put it the intermediate sign for s to be minus 1, then I have two pluses at the end. The one in the middle is minus. So I would have two dissatisfied factors of e to the k becoming e to the minus k. So it's x to the minus 2. And the field also will give a factor of e to the minus h or y inverse. So there are four possibilities here. The next one is minus minus. z prime will stay the way it is. The field has switched sign. So this becomes y prime inverse. But since both of them are minus-- minus, minus-- the k factor is satisfied and happy. Gives me x prime because it's e to the plus k. On the right hand side, I will always get the z squared. The first factor becomes the inverse. Now, s plus 1 is a plus site, plus spin, that is sandwiched between two minus spins. So there are two unhappy bonds. This gives me x to the minus 2. Why the spin is pointing in the direction of the field. So there's the y here. And here I will get x squared and y inverse because now I have three negative signs. So all the k's are happy. There's the next one, which is plus and minus. Plus and minus, we can see that the contribution to the field vanishes. Because sigma i prime and plus sigma i plus 1 prime are zero. I will still get z, the contribution, because plus and minus, the bond between them gives me e to the minus k prime. I have here z squared. There is no overall contribution of y for the same reason that there was nothing here. But from s, I will get a factor of y plus y inverse. And there's no contribution for x because I have a plus and a minus. And if the spin is either plus or minus in the middle, there will be one satisfied and one dissatisfied bond. Again, by symmetry, the other configuration is exactly the same. So while I had four configuration, and hence four things to match, the last two are the same. And that's consistent with my having three variables, x prime, y prime, and z prime, to solve. So there are three equations and three variables. Now, to solve these equations, we can do several things. Let's, for example, multiply all four equations together. What do we get on the left hand side? There are two x's, two inverse x's, y, and inverse y, but four factors of z. So I get z prime to the fourth factor. On the other side, I will get z to the eight, and then the product of those four factors. I can divide one, equation one, by equation two. What do I get? The x's cancel. The x's cancel. So I will get y prime squared. On the other side, divide these two. I will get y squared. x squared y plus x squared y inverse, x minus 2y plus x2 minus-- yeah, x2 y inverse. And finally, to get the equation for x prime, what I can do is I can multiply 1 and 2, divide by 3 and 4. And if I do that, on the left hand side I will get x prime to the fourth. And on the right hand side, I will get x squared y plus x to the minus 2 y inverse x minus 2y x squared y inverse divided by y plus y inverse squared. So I can take the log of these equations, if you like, to get the recurrence relations for the parameters. For example, taking the log of that equation, you can see that I will get g prime. From here, I would get 2g plus some function. There's some delta g that depends on k and h. If I do the same thing here, you can see that I will get an h prime that is h plus some function from the log of this that depends on k and h. And finally, I will get k prime some function of k and h. This parameter g is not that important. It's basically an overall constant. The way that we started it, we certainly don't expect it to modify phase diagrams, et cetera. And you can see that it never affects the equations that govern h and k, the two parameters that give the relative weight of different configurations. Whether you are in a ferromagnet or a disordered state is governed by these two parameters. And indeed, we can ignore this. Essentially, what it amounts to is as follows, that every time I remove some of the spins, I gradually build a contribution to my free energy. Because clearly, once I have integrated out over all of the energies, what I will have will be the partition function. Its log would be the free energy. And actually, we encountered exactly the same thing when we were doing momentum space RG. There was always, as a result of the integration, some contribution that I call delta f that I never looked at because, OK, it's part of the free energy but does not govern the relative weights of the different configurations. So these are really the two things that we need to focus on. Now, if we did things correctly and these are correct equations, had I started in the subspace where h equals to zero, which had up-down symmetries, I could have changed all of my sigmas to minus sigma, and for h equals to zero, the energy would not have changed. So a check that we did things correctly is that if h equals to zero, then h prime has to be zero. So let's see. If h equals to zero or y is equal to one, well if y is equal to one, you can see that those two factors in the numerator and denominator are exactly the same. And so y prime stays to one. So this check has been performed. So if I am on h equals to zero on the subspace that has symmetry, then I have only one parameter, this k or x. And so on this space, the recursion relation that I have is that x prime to the fourth is x squared plus x to the minus 2 squared. I think I made a mistake here. No, it's correct. y plus y inverse. OK, but y is one. So this is divided by 2 squared, which is 4. But let me double check so that I don't go-- yeah. So this is correct. So I can write it in terms of x being e to the k. So what I have here is e to the 4k prime. I have here e to the 2k plus e to the minus 2k divided by 2, which is hyperbolic cosine of 2k, squared. So what I will get is k prime is 1/2 log hyperbolic cosine of 2k. Now, k is a parameter that we are changing. As we make k stronger, presumably things are more coupled to each other and should go more towards order. As k goes towards zero, basically the spins are independent. They can do whatever they like. So it kind of makes sense that there should be a fixed point at zero corresponding to zero correlation length. And let's check that. So if k is going to zero-- it's very small-- then k prime is 1/2 log hyperbolic cosine of a small number, is roughly 1 plus that small number squared over 2. There's a series like that. Log of 1 plus a small number is a small number. So this becomes 4k squared over 2 over 2. It's k squared. So it says that if I, let's say, start with a k that is 1/10, my k prime would be 1/100 and then 1/10,000. So basically, this is a fixed point that attracts everything to it. Essentially, what it says is you may have some weak amount of coupling. As you go and look at the spins that are further and further apart, the effective coupling between them is going to zero. And the spins that are further apart don't care about each other. They do whatever they like. So that all makes physical sense. Well, we are really interested in this other end. What happens at k going to infinity? We can kind of look at this equation. Or actually, look at this equation, also doesn't matter. But here, maybe it's clearer. So I have e to the 4k prime. And this is going to be dominated by this. This is e to the 2k. e to the minus 2k is very small. So it's going to be approximately e to the 4k divided by 4. So what I have out here is that k is very large. k prime is roughly k, but then smaller by a factor that is 1/2 of log 2. So we try to take two spins that are next to each other, couple them with a very strong k. Let's say a million, 10 million, whatever. And then I look at spins that are too further apart. They're still very strongly coupled, but slightly less. It's a million minus log 2. So it very, very gradually starts to move away from here. But then as it kind of goes further, it kind of speeds up. What it says is that it is true. At infinity, you have a fixed point. But that fixed point is unstable. And even if you have very strong but finite coupling, because things are finite, as you go further and further out, the spins become less and less correlated. This, again, is another indication of the statement that a one dimensional system will not order. So one thing that you can do to sort of make this look slightly better, since we have the interval from zero to infinity, is to change variables. I can ask what does tang of k prime do? So tang k prime is e to the 2k prime minus 1 divided by e to the 2k prime plus 1. You write e to the 2k's in terms of this variable. At the end of the day, a little bit of algebra, you can convince yourself that the recursion relation for tang k has a very simple form. The new tang k is simply the square of the old tang k. If I plot things as a function of tang k that runs between zero and one, that fixed point that was at infinity gets mapped into one, and the flow is always towards here. There is no other fixed point in between. Previously, it wasn't clear from the way that I had written whether potentially there's other fixed points along the k-axis. If you write it as t prime tang is t squared, where t stands for tang, clearly the only fixed points are at zero and one. But now this also allows us to ask what happens if you also introduce the field direction, h, in this space. Now, one thing to check is that if you start with k equals to zero, that is independent spins, you look at the equation here. k equals to 0 corresponds to x equals to one. This factor drops out. You see y prime is equal to y. So if you have no coupling, there is no reason why the magnetic field should change. So essentially, this entire line corresponds to fixed points. Every point here is a fixed point. And we can show that, essentially, if you start along field zero, you go here. If you start along different values of the field, you basically go to someplace along this line. And all of these flows originate, if you go and plot them, from back here at this fixed point that we identified before at h equals to zero. So in order to find out what is happening along that direction, all we need to do is to go and look at x going to infinity. With x going to infinity, you can see that y prime squared, the equation that we have for y, y prime squared is y squared. And then there's this fraction. Clearly, the terms that are proportional to x squared will overwhelm those proportional to x to the minus 2. So I will get a y and a y inverse from the numerator and denominator. And so this goes like y to the fourth. Which means that in the vicinity of this point, what you have is that h prime is 2h. So this, again, is irrelevant direction. And here, you are flowing in this direction. And the combination of these two really justifies the kind of flows that I had shown you before. So essentially, in the one dimensional model, you can start with any microscopic set of k and h's that you want. As you rescale, you essentially go to independent spins with an effective magnetic field. So let's say we start with very close to this fixed point. So we had a very large value of k. I expect if I have a large value of k, neighboring spins are really close to each other. I have to go very far before, if I have a patch of pluses, I can go over to a patch of minuses. So there's a large correlation length in the vicinity of this point. And that correlation length is a function of our parameters k and h. Now, the point is that the recursion relation that I have for k does not look like the type of recursions that I had before. This one is fine because I can think of this as my usual h prime is b to the yh h, where here I'm clearly dealing with a b equals to two, and so I can read off my yh to be one. But the recursion relation that I have for k is not of that form. But who said that I choose k as the variable that I put over here? I have been doing all of these manipulations going from k to x, et cetera. One thing that behaves nicely, you can see, is a variable if I call e to the minus k prime is square root of 2 e to the minus k. At k equals to infinity, this is something that goes to zero on both ends. But if its k is large but not finite, it says that on the rescaling, it changes by a factor of root 2. So if I say, well, what is c as a function of the magnetic field, rather than writing k, I will write it as e to the minus k. It's just another way of writing things. Well, I know that under one step of RG, all my length scales have shrunk by a factor of two. So this is twice the c that I should write down with 2h and root 2 e to the minus k. So I just related the correlation length before and after one step of RG, starting from very close to this point. And for these two factors, 2 and root 2, I use the results that I have over there. And this I can keep doing. I can keep iterating this. After l steps of RG, this becomes 2 to the l, c of 2 to the lh, and 2 to the l over 2 e to the minus k. I iterate this sufficiently so that I have moved away from this fixed point, where everything is very strongly correlated. And so that means I move, let's say, until this number is something that is of the order of one. Could be 2. Could be 1/2. I really don't care. But the number of iterations, the number of rescalings by a factor of two that I have to do in order to achieve this, is when 2 to the l is of the order of e to the k. e to the 2k, sorry. And if I do that, I find that c should be e to the 2k some function of h e to the 2k. So let's say I am at h equals to zero. And I make the strength of the interaction between neighbors stronger and stronger. If I asked you, how does the correlation length, how does the size of the patches that are all plus and minus change as k becomes stronger and stronger? Well, RG tells you that it goes in this fashion. In the problem set that you have, you will solve the problem exactly by a different method and get exactly this form. What else? Well, we said that one of the characteristics of a system is that the free energy has a singular part as a function of the variables that we have that is related to the correlation length to the power d. In this case, we have d equals to one. So the statement is that the singular part of the free energy, as you approach infinite coupling strength, behaves as e to the minus 2k some other function of h, e to the 2k. Once you have a function such as this, you can take two derivatives with respect to the field to get the susceptibility. So the susceptibility would go like two derivatives of the free energy with respect to the field. You can see that each derivative brings forth a factor of e to the 2k. So two derivatives will bring a factor of e to the 2k, if I evaluate it at h equals to zero. So the statement is that if I'm at zero field, and I put on a small infinitesimal magnetic field, it tends to overturn all of the spins in the direction of the field. And the further you are close to k to infinity, there are larger responses that you would see. The susceptibility of the response diverges as k goes to infinity. So in some sense, this model does not have phase transition, but it demonstrates some signatures of the phase transition, such as diverging correlation length and diverging susceptibility if you go to very, very strong nearest neighbor coupling. There is one other relationship that we had. That is, the susceptibility is an integral in d dimension over the correlation length, which we could say in one dimension-- well, actually, let's write it in d dimension. e to the minus x over c divided by x. We introduce an exponent, eta, to describe the decay of correlations. So this would generally behave like c to the 2 minus eta. Now, in this case, we see that both the correlation length and susceptibility diverge with the same behavior, e to the 2k. They're proportional to each other. So that immediately tells me that for the one dimensional system that we are looking at, eta is equal to one. And if I substitute back here, so eta is one, the dimension is one, and the two cancels. Essentially, it says that the correlation length in one dimension have a pure exponential decay. They don't have this sub leading power load that you would have in higher dimensions. So when you do things exactly, you will also be able to verify that. So all of the predictions of this position space RG method that we can carry out in this one dimensional example very easily, you can also calculate and obtain through the method that is called transfer matrix, and is the subject of the problem set that was posted yesterday. Also, you can see that this approach will work for any system in one dimension. All I really need to ensure is that the b that I write down is sufficiently general to include all possible interactions that I can write between two spins. Because if I have some subset of those interactions, and then I do this procedure that I have over here and then take the log, there's no reason why that would not generate all things that are consistent with symmetry. So you really have to put all possible terms, and then you will get all possible terms here, and there would be a recursion relation that would relate. You can do this very easily, for example, for the Potts model. For the continuous spin systems, it becomes more difficult. Now let's say briefly as to why we can solve things exactly, let's say, for the one dimensionalizing model by this procedure. But this procedure we cannot do in higher dimensions. So let's, for example, think that we have a square lattice. Just generalize what we had to two dimensions. And let's say that we want to eliminate this spin and-- well, let's see, what's the best way? Yeah, we want to eliminate a checkerboard of spins. So we want to eliminate half of the spins. Let's say the white squares on a checkerboard. And if I were to eliminate these spins, like I did over here, I should be able to generate interactions among spins that are left over. So you say, fine. Let's pick these four spins. Sigma 1, sigma 2, sigma 3, sigma 4, that are connected to this spin, s, that is sitting in the middle, that I have to eliminate. So let's also stay in the space where h equals to zero, just for simplicity. So the result of eliminating that is I have to do a sum over s. I have e to the-- let's say the original interaction is k, and s is coupled by these bonds to sigma 1, sigma 2, sigma 3, sigma 4. Now, summing over the two possible values of s is very simple. It just gives me e to the k times the sum plus e to the minus k, that sum. So it's the same thing as 2 hyperbolic cosine of k, sigma 1 plus sigma 2 plus sigma 3 plus sigma 4. We say good. I had something like that, and I took a log so that I got the k prime. So maybe I'll do something like a recasting of this, and maybe a recasting of this will give me some constant, will give me some kind of a k prime, and then I will have sigma 1, sigma 2, sigma 2, sigma 3, sigma 3, sigma 4, sigma 4, sigma 1. But you immediately see that that cannot be the entire story. Because there is really no ordering among sigma 1, sigma 2, sigma 3, sigma 4. So clearly, because of the symmetries of this, you will generate also sigma 1 sigma 3 plus sigma 2 sigma 4. That is, eliminating this spin will generate for you interactions among these, but also interactions that go like this. And in fact, if you do it carefully, you'll find that you will also generate an interaction that is product of all four spins. You will generate something that involves all of the force. So basically, because of the way of geometry, et cetera, that you have in higher dimensions, there is no trick that is analogous to what we could do in one dimension where you would eliminate some subset of spins and not generate longer and longer range interactions, interactions that you did not have. You could say, OK, I will start including all of these interactions and then have a higher, larger parameter space. But then you do something else, you'll find that you need to include further and further neighboring interactions. So unless you do some kind of termination or approximation, which we will do next time, then there is no way to do this exactly in higher dimensions. AUDIENCE: Question. PROFESSOR: Yes. AUDIENCE: I mean, are you putting any sort of weight on the fact that, for example, sigma 1 and sigma 3 are farther apart than sigma 1 and sigma 2, or are we using taxi cab distances on this lattice? PROFESSOR: Well, we are thinking of an original model that we would like to solve, in which I have specified that things are coupled only to nearest neighbors. So the ones that correspond to sigma 1, sigma 3, are next nearest neighbors. They're certainly farther apart on the lattice. You could say, well, there's some justification, if these are ions and they have spins, to have some weaker interaction that goes across here. There has to be some notion of space. I don't want to couple everybody to everybody else equivalently. But if I include this, then I have further more complicated Hamiltonian. And when I do RG, I will generate an even more complicated Hamiltonian. Yes. AUDIENCE: [INAUDIBLE]. PROFESSOR: Question is, suppose I have a square lattice. Let's go here. And the suggestion is, why don't I eliminate all of the spins over here, maybe all of the spins over here? So the problem is that I will generate interactions that not only go like this, but interactions that go like this. So the idea of what happens is that imagine that there's these spins that you're eliminating. As long as there's paths that connect the spins that you're eliminating to any other spin, you will generate that kind of couple. Again, the reason that the one dimensional model works is also related to its exact solvability by this transfer matrix method. So I will briefly mention that in the last five minutes. So for one dimensional models, the partition function is a sum over whatever degree of freedom you have. Could be Ising, Potts, xy, doesn't matter. But the point is that the interaction is a sum of bonds that connect one site to the next site. I can write this as a product of e to the b of SI and SI plus 1. Now, I can regard this entity-- let's say I have the Potts model q values. This is q possible values. This is q possible values. So there are q squared possible values of the interaction. And there is q squared possible values of this Boltzmann weight. I can regard that as a matrix, but I can write in this fashion. And so what you have over there, you can see is effectively you have s1, ts2, s2, ts3, and so forth. And if I use periodic boundary condition like the one that I indicated there so that the last one is connected to the first one, and then I do a sum over all of these s's, this is just the product of two matrices. So this is going to become, when I sum over s2, s1 t squared s3 and so forth. You can see that the end result is trace of the matrix t raised to the power of n. Now, the trace you can calculate in any representation of the matrix. If you manage to find the representation where t is diagonal, then the trace would be sum over alpha lambda alpha to the n, where these lambdas are the eigenvalues of this matrix. And if n is very large, the thermodynamic limit that we are interested, it will be dominated by the largest eigenvalue. Now, if I write this for something like Potts model or any of the spin models that I had indicated over there, you can see that all elements of this matrix being these Boltzmann weights are plus, positive. Now, there's a theorem called Frobenius's theorem, which states that if you have a matrix, all of its eigenvalues are positive, then the largest eigenvalues is non-degenerate. So what that means is that if this matrix is characterized by a set of parameters, like our k's, et cetera, and I change that parameter, k, well the eigenvalue is obtained by looking at a single matrix. It doesn't know anything about that. The only way that the eigenvalue can become singular is if two eigenvalues cross each other. And since Frobenius's theorem does not allow that, you conclude that this largest eigenvalue has to be a perfectly nice analytical function of all of the parameters that go into constructing this Hamiltonian. And that's a mathematical way of saying that there is no phase transition for one dimensional model because you cannot have a crossing of eigenvalues, and there is no singularity that can take place. Now, an interesting then question or caveat to that comes from the very question that was asked over here. What if I have, let's say, a two dimensional model, and I regard it essentially as a complicated one dimensional model in which I have a complicated multi-variable thing over one side, and then I can go through this exact same procedure over here also? And then I would have to diagonalize this huge matrix. So if this is l, it would be a 2 to the l by 2 to the l matrix. And you may naively think that, again, according to Frobenius's theorem, there should be no phase transition. Now, this is exactly what Lars Onsager did in order to solve the two dimensionalizing model. He constructed this matrix and was clever enough to diagonalize it and show that in the limit of l going to infinity, then the Frobenius's theorem can and will be violated. And so that's something that we will also discuss in some of our future lectures. Yes. AUDIENCE: But it won't be violated in the one dimensional case, even if n goes to infinity? PROFESSOR: Yeah, because n only appears over here. Lambda max is a perfectly analytic 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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Let's apply the work energy theorem principle to the motion of a block sliding down an inclined plane. And here is an inclined plane at an angle, theta. And lets choose a coordinate system. We'll choose x equals 0 up here, or i-hat here. And here is our coordinate function. And suppose that the object starts at xi, that it ends at x final. If we want to calculate the work energy then what we're going to learn is that this theorem is two different sides. Calculating the change in kinetic energy is simply a description. This is 1/2 M v final squared minus 1/2 v initial squared. That's a property, parameters of the system, at the initial state, the speed, or the velocity's speed, which is speed squared. And the same property v final in the final state. Now over here we have two types of forces acting on this object. So here we need a free-body force diagram first. And this side is where the physics [INAUDIBLE] lie. And so we want to draw our force diagram. So if we have our block, we have the friction force, we have the normal force, we have mg. Recall, if the plane is inclined theta that's also the angle theta. If we chose i-hat, j-hat unit vectors, I just want to repeat that on my free-body diagram. Now we can think of this integral as just one-dimensional motion in the x direction. And so we have two different forces that we have to calculate. The friction force is in the minus x direction so we're integrating minus the friction force with respect to displacement from x initial to x final. And what about the gravitational force? Well the gravitational force has a component, mg sine theta in the x direction. So when we integrate that x component we have now plus because it's in the same direction. Remember, we're displacing a little bit dx down the inclined plane so we're going from x initial x final of mg sine theta, which is a constant, dx. And so when you're applying the work energy theorem you need to integrate your forces and actually calculate the work. Now again, if you looked in the j-hat direction, and we applied Newton's second law, n minus mg cosine theta 0, and our rule for friction is its mu k times n or mu k mg cosine theta, then what we have is in both instances we have a constant force. So it's just force times displacement. So we have minus mu k mg cosine theta times the displacement, which is x initial x final. Over here we have mg sine theta times x final minus x initial. And now we've calculated separately both sides of our work kinetic energy principle. As in all our physical laws the equal sign means the work is equal to the change in kinetic energy. I'll emphasize that by now placing the equal sign because of our physical law. And so equaling 1/2 M v final squared minus v initial mean squared. And now I have a relationship between the parameters of the initial state, which I'm calling x initial and v initial, and the parameters that describe the final state, x final, v final. And depending on which set of these parameters are given I can conceivably solve for the other ones. One thing I do want to point out when we do this example is we've described work as a dot product from A to B. Take the friction force. Well in this instance, if we wrote this out explicitly it would be minus fk i-hat dot dx i-hat from the initial to the final. i-had dot i-hat is 1. And so you see, we recover from x initial to x final of fk dx. And that was the first piece. The second piece, the gravitational force, dotted into ds from x initial to x final. Well, if you wrote down the gravitational force, mg, as a i-hat component and a negative mg cosine theta j-hat component, then when we take the dot product again where our ds-- here, we'll write ds as dx i-hat-- then when you dot product mg dot ds, mg dot ds, we have i-hat dot i-hat, which is one. But j-hat dot i-hat, they're perpendicular so that's 0, so the only piece that survives in the scalar product mg dot ds is these two pieces. And so we get the integral from x initial to x final of mg sine theta times dx. And that's precisely our second piece here. So here's the simple application of the work energy theorem.
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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 want to look at this pulley system. We want to find out what this force here is, for example, with which this block is being pulled. Now we have two massless pulleys here and two moving parts. And one key component of this problem is to derive the acceleration constraint. How are we going to do that? Well, we have to look at this string here. First of all, it's a fixed string length. And that will help us. Ultimately, this is just a distance that connects all of these objects. And if we eventually differentiate that twice, we get to an acceleration. So the first thing we're going to do is to figure out what the string length is to then derive the acceleration condition. Let's begin by first identifying the fixed length in this problem. So we have a little distance here. We call it sb. And we have a distance here that's fixed, sa. And we furthermore know that this block here is fixed to the ground. And we can choose a coordinate system origin. Let's say we do that here. We know that we have the distance d here to this little block. The moving parts, block 1 and block 2-- so we have to assign position functions. And actually this one here is x1. It goes up to here because we want to measure the string length, ultimately, and we know this portion here. So we can subtract it. And then of course we need also x1, so that goes here-- x2. So now we can tally up the length. Let's start with this portion here. That is d minus sa minus x2. d minus sa minus x2. And then we have a half circle here, pi r. And we know actually that there is another half circle that's going to come there, so we can just write 2 pi r immediately. And then we have this part here. That one is x1 minus sb, and then minus sa minus x2. And then finally we have this portion, and that is x1 minus sb minus x2. OK, so the next step is that we need to simplify this a little bit. So we have one x1 and another one here, 2x1 plus x2, x2, x2-- actually 3-- minus 3x2. And then we have all sorts of consonants. We have d, we have sa, sb, the 2 pi r. But, as you will see, if we differentiate this out here, that will actually all fall away. So we're going to make our life easy and just add a constant here. And so we want to now do the second derivative here of our string length with respect to time, because these all position functions. So we can differentiate those. What's important of course here is this string length is not changing with time. So actually we know that that derivative will be 0. And we can just write this up here because x differentiated twice is a. So 2a1 minus 3a2. And we immediately see from that that a1 equals 3/2 a2. So this is our constraint condition that we will need later. For now, we need to continue with setting up free body diagrams of all four objects. Let's start that with object 1. What's acting on object 1? Well, we have F here. Why don't we just write it as magnitude. We have F, and then we have here a tension that goes to the pulley B. That one is different from the tension in the string. So we're going to call this TB. And then we have object 2. Oh, and of course, i hat goes in this direction because it follows the motion of the object. We have here kind of the reverse. Now we have a tension of this string here that's attached to pulley A, but it's different from this string tension-- that's a specific one-- TA. And we also have a T from the string here. So we'll add a T. And then if we look at pulley A, we have two string tensions, T and T. And here we again have a TA. And pulley B, TB, and two T over here. So that means we can write down our equations of motions using Newton's Second Law, F equals ma. And since this is just going in the i hat direction, we can just write down the four equations following the four free body diagrams here. So we have F minus TB equals m1 a1. We have T plus TA equals m2 a2. Then we have 2T minus TA equals 0. That is 0 because we're dealing with two massless pulleys there. That means m is 0 and so our acceleration term is 0. So the m is 0 here. And finally, we have TB minus 2T. No, not 2 pi-- 2T. And that one is also 0 because it is a massless pulley. So here we have our four equations of motion that fully govern this pulley system. And we can use it then, for example, to find this pulling force here. We know what TB is from the equation down here-- so 2T. We know what TA is. It's also 2T. And we can then solve this for F. The only sticky part is that we have this a1 and a2 in here. But for that, we derived this constraint condition here. So with that one, we can fully solve this. Otherwise, we would have one too many unknowns. Alternatively, if one were to be interested in one of these accelerations, then this equation system can also be solved for a1 or a2.
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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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CATHERINE DRENNAN: One of the big challenges of a large lecture is having the graduate student TAs be really an integral part of this. And I think if the grad students TAs are not that excited about this teaching assignment, it's not so good. You want to have this sense of enthusiasm and energy. So one of the problems I had when I was first starting with this is that this TA assignment was not the most popular. So when the grad students are coming in in their first year, there are organic chemistry grad students and they want to teach organic chemistry. And there are physical chemistry students who want to teach thermo and kinetics. Biochem students want to teach biochemistry, and inorganic chemistry students want to teach inorganic chemistry. So who is there for general chemistry? And so often, it would be the booby prize assignment to get stuck in that class, and there might be a feeling like, oh, I got stuck because I wasn't the organic chemistry professor's first choice. So I wanted to change all that because to me, teaching freshmen in freshman chemistry is the most fun teaching ever. So I knew there were grad students out there that could engage and agree with me that this is going to be a lot of fun. I admit, it's more work than all of those other courses, but it also can be more fun. So I thought, we just need to get the right people in and I need to educate them about what this experience can be like. So I decided to once the students were accepted to come into MIT, I would send them an information packet about how they could be part of this exciting teaching opportunity, and I made it glossy and pretty and sent it to them and asked them to apply to be part of the class. And so we got some wonderful applications. I got a great group of people to be part of this class. So instead of being like, oh no, I have to do this, they're like, yes, freshman chemistry. And we brought them in early, and I got to know the group. We had extra TA training in the beginning. And t-shirts, and by the time we started, everyone knew each other. And it was an opportunity, also, for the TAs to make friends because they're first year graduate students and so they get to know other first year grad students and they build a sense of team as well. So for many years now when we've been really doing this-- having this extra TA training and applications to be part of the class-- we've started the semester with a group of people who are just ready to go and really excited about being part of that. And I think that has had such a wonderful impact on the course as well. They see this whole teaching team that really want to be there and are excited about this opportunity and care about each and every one of those students. And the comments that I've gotten later, both from the students and also sometimes from parents, that their child in their freshman year really felt a connection as part of this class and had this TA just checking in on them, not just about this class, but about how they were doing overall. And it was a support system that went beyond the teaching of chemistry, which made me really very happy.
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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 are going to recap some of the last results we had on scattering and then push them to the end and complete our discussion of this integral equation and how we approximate scattering with it. It's a pretty powerful method of thinking about scattering and gives you a direct solution, which is in a sense more intuitive perhaps than when you're dealing with partial waves. So let's see where we were last time. So we set up our solution of the Schrodinger equation in a funny way. We had an incoming wave plus a term that still had an integral over all of space of our Green's function integrated against a scaled version of the potential. Remember that v was defined to be equal to the potential up to some constants, h bar squared over m. And we had this integral equation where one could show that phi satisfies the correct Schrodinger equation. Now this function G plus was our Green function that we also discussed. And it was given by the following formula. It just depends on the magnitude of this vector argument of the Green's function, so 1 over 4 pi e to the ik, magnitude of that divided by magnitude of that vector. So in most applications, we look at the wave function. This should be phi of r. We look at the wave function far away. And if the potential is localized, this integral, since it involves the potential, can only be non-zero, have non-zero contributions to the integral on the region where the potential exists. So r prime, whenever you're doing this integral, must be small compared to r. This we can draw here. So assume we have a potential. And you have an observation point, r. We'll call the unit vector in this direction n. This is the vector r. And an integration point is a point, r prime here. And there is this distance, this vector, in between r minus r prime. And finally, on this graph, we could also have a direction that we could call the incident direction, perhaps the z-direction. And that's the direction which your wave is incident. This is the wave that were represented here. So we did a little approximation of this Green's function for the case where r prime where you're looking is much smaller than r. In that case, the distance, r minus r prime, is approximately equal to r minus-- so that's the distance, the vectors. When you don't put an arrow, we mean the distance. So this is r minus n dot r prime. To 0-th approximation, this is the most important term. To next approximation comes this term because that term is suppressed with respect to the original term by a factor of r prime over r. So if you factored, for example, the r here, you would say 1 minus n r prime over r. So you see that this ratio enters here, which says that the second term is smaller by this factor. We could even include more terms here. But those would not be relevant unless k is extremely high energy. So when k is very large, it would make a difference. But it has to be extremely large. And in general, this approximation isn't enough for any k for sufficiently far away. So this is really good enough. And the approximation is done on the Green's function at two levels. You need to be mildly accurate with respect to this quantity. But you need to be more accurate with respect to this phase because the phase can change rather fast depending on the wavelength of the object. So if you make an error in the total distance of 1%, that's not big deal. But if you make an error here, such that k dot times this distance is comparable to 1, the phase could be totally wrong. So the phase is always a more delicate thing. And therefore, with this approximation, your G plus of r minus r prime becomes minus 1 over 4 pi e to the ikr over r. So we replace this term by just r. And in this exponent, the absolute value is replaced by two terms. The first one is e to the ikr, is here. The second one is e to the minus ikn dot r prime. We also have here the incident vector. You can write e to the ikz as e to the ik incident dotted with r if k incident is a vector, k, times the unit vector in the z-direction. So that's a matter of just notation. Might as well use a vector for k incident because that is really the vector momentum that is coming in. And it's sometimes useful to have the flexibility, perhaps some moment. You don't necessarily want the wave to come in the z-direction. There's another vector that is kind of interesting. You're observing the wave in the direction of the vector r, the direction of the unit vector n. So it's interesting to call k an incident. I think people write it with just i, so ki for incident, k. And then, well, we're looking at the direction r. So we call the scattering momentum or the scattered momentum, kn. You're looking at that distance. So might as well call it that way. So with these things, we can rewrite the top equation making use of all the things that we've learned. And it's going to be an important equation, psi of r is equal to e to the iki dot r. So we have to replace the e to the ikz by this. Then we have the Green's function that has to be integrated. And you have to see, we've decided already to approximate the Green's function. So it's integral over r prime. So the only term here that has an r prime is this exponential. So all of this can go out. So the way I want to make it go out, we will have the minus 1 over 4 pi here and the integral d cubed r prime. And I'm going to have the rest of the Green's function placed in the standard position we've used for scattering. So we're still redoing the top integral, the G plus, which includes some of G plus. We've taken care of all of this part. And now we have that other factor. So it's e to the minus ikn. I could call it the scattering one. But I will still not do that. Then we have the u of r prime and the psi of r prime. So this is our simplified expression when we make use of the fact that we're looking far enough and the Green's function has simplified. So let's keep that for a moment and try to think how we could solve this integral equation. Now there's no very simple methods of solving them exactly. You have to do interesting things here. It's fairly nontrivial. But finding approximate solutions of integral equations is something that we can do. And it's relatively simple. So we will come back to this because in a sense, this is nice. But you must feel that somehow the story is not complete. We want to know the wave function far away. This has the form of this function that when we wrote psi is equal to e to the ikz plus f of theta phi e to the ikr over r. If we know this f of theta phi, we know everything about the scattering. But it almost looks like this is f of theta and phi. But we don't know psi. So have we made any progress? Well, the Born approximation is the way in which we see that we could turn this into something quite simple. So let's do Born approximation. So our first step is to rewrite once more for us this equation, psi of r is equal to e to the ikir plus integral d q bar prime G plus of r minus r prime u of r prime psi of r prime. OK. We wrote it. Good. Now let's rewrite it with instead of r, putting an r prime here. You'd say, why? Well, bear with me a second. Rewrite with r replaced by r prime. If we have an r prime, replace it with r double prime and so on. So let's rewrite this. So we have psi of r prime is equal to e to the iki r prime plus integral of d cubed r double prime G plus of r prime minus r double prime u of r double prime psi of r double prime. So I shifted. Whenever I had an r, put an r prime. Whenever I had an r prime, put an r double prime. The reason we do this is that technically, I can now substitute this quantity here into here and see what I get. And I get something quite interesting. I get the beginning of an approximation because I now get psi of r is equal to e to the ikir plus d cubed r prime G plus of r minus r prime-- I'm still copying the first equation-- u of r prime. And here is where the first change happens. Instead of psi of r prime, I'm going to put this whole other thing. So I'll write two terms. The first is just having here, e to the iki r prime. The second, I'll write this another line. So integral d cubed r prime G plus of r minus r prime u of r prime. And then the second term would be another integral, d cubed r double prime G plus of r prime minus r double prime u of r double prime psi of r double prime. So look at it and see what has happened. You have postponed the fact that you had an integral equation to a next term. If I would drop here this term, this would not be an integral equation anymore. Psi is given to this function that we know. Green's function we know. Potential we know. Function we know. It's a solution. But the equation is still not really solved. We have this term here. And now you could go on with this procedure and replace this psi of r double prime. You could replace it by an e to the iki r double prime plus another integral d r triple prime of G, like that. And then here you would have an e to the ik r double prime and then more terms and more terms. So I can keep doing this forever until I have this integral, G plus u incident wave. And here I have G plus u G plus u incident wave. And then I will have G plus u, G plus u, G plus u, incident wave. And it would go on and on and on. So this is called the Born approximation if you stop at some stage and you ignore the last term. So for example, we could ignore this term, in which we still have an unknown psi, and say, OK, we take it this way. And this would be the first Born approximation. You just go up to here. The second Born approximation would be to include this term with psi replaced by the incident wave. The third Born approximation would be the next term that would come here. So let me write the whole Born approximation schematically, whole Born approximation schematically. I could write three terms explicitly or four terms if you wish. But I think the pattern is more or less clear now. What do you have? Psi of r equal e to the ikir plus-- and now it becomes schematic-- integral-- so I put just integral-- G plus u e to the ikr. Don't put labels or anything. But that's the first integral. And if you had to put back the labels, you would know how to put them. You'd put an integral over some variable. Cannot be r because that's where you're looking at. So you do an r prime, the G of r minus r prime, the u at r prime, and this function at r prime. That's how this integral would make sense. The next term would be integral G u of another integral of G u e to the ikr. That's the next term. And the next term would be just G u G u G u e to the ikr. And it goes on forever. That's the Born series.
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https://ocw.mit.edu/courses/20-020-introduction-to-biological-engineering-design-spring-2009/20.020-spring-2009.zip
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>> Sally: Hey Dude, look who's here. It's my friend Izzy the iGEMer. She just stopped by for a visit. >> Dude: Whoa! I've never met a real iGEMer before! >> Izzy: Hey Dude, great to meet you. >> Sally: I need to split some cells in the tissue culture room, but I thought that you and Izzy could chat. Maybe she could tell you a bit about her project? >> Dude: Awesome! >> Dude: So what was your iGEM project? >> Izzy: Well we tried to engineer a bacterial arsenic detector. >> Dude: A what? >> Izzy: A bacterial cell that could tell us when arsenic is present in water. >> Izzy: See arsenic contamination in drinking water is a big problem for people living in Bangladesh and West Bengal. And since there's no easy way to know if arsenic is in the water, people there have been drinking contaminated water and getting sick . So we wanted to engineer a biological system that people can use to test their water for arsenic. >> Dude: I had no idea this was such a problem! What a great project to work on. How does your system work? >> Izzy: Well our goal or the system specification as engineers call it was to have people mix a little of their drinking water, with a culture of bacteria and know that there's arsenic contamination if the liquid changes color. >> Dude: Awesome!. But how did you get a bacteria to do that? I'm still trying to figure how to make my bacterial balloon work! >> Izzy: Well like all engineers, we used abstraction to help us. Let's take a look. >> Izzy: We knew we needed to build a system that could take as input arsenic and produce as output a change in the color. >> Izzy: So first, we compiled our system into devices. We needed one device that could detect arsenic and produce a signal. We called this our arsenic sensor and gave it a part number, BBA J33201. And we needed a second device that could produce a color when it got the signal from our arsenic sensor that there's arsenic around. We called this color reporter BBA J33202. >> Dude: I don't get it. Why are you using two devices? Wouldn't it be way easier to make just one: arsenic in and color out? >> Izzy: Well you could do it that way. But then the only thing that the device could be used for is to report on arsenic contamination with color. By splitting the system into two devices, a sensor and a reporter, we hope future iGEM teams can use one or the other for new projects. >> Dude: How? >> Sally: Well a team could connect the arsenic sensor device to a different reporter device that say, produces light instead of color. Or a team could build a new system that detects say, mercury contamination, and then produces a color as a response. >> Izzy: Yep, we designed our system so that people can more easily reuse and build on our work! >> Dude: Oh I get it! Maybe I could connect your arsenic sensor device to my balloon-making device! >> Izzy: Exactly! >> Dude: But shouldn't I know a little more about how the arsenic sensor works before I just hook it up to by balloon making device? Sally got kinda upset last time I didn't' stop the balloons from growing and growing... >> Izzy: That's why we entered both devices into the Registry of Standard Biological Parts. We also documented our project online at the iGEM wiki to make it easier for others to build upon our work. >> Sally: The goal is that eventually, for each device in the Registry, there will be a datasheet available that summarizes all the relevant information about that device so that you know how to use it in a higher level system like the arsenic detector. >> Dude: You mean just like this old datasheet I found for this a 7404 hex inverter chip? >> Sally: Yep. But I'm afraid there's still a lot of work to do before all the parts and devices in the Registry have datasheets. >> Dude: Great! Lots of work to do. Let's get started!
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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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NARRATOR: 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 OpenCourseWare at ocw.mit.edu. JOANNE STUBBE: So what I want to do is finish up purines today and talk about some interesting aspects of purine metabolism. I hope I'm going to be able to get through. I've given you handouts for pyrimidines and deoxynucleotide biosynthesis as well. The pyrimidines are pretty straightforward, much simpler than the purines. And so I think if I have time today, depending on when I get finished, I might talk a little bit about deoxynucleotide metabolism, since both Drennan's lab and my lab, both work in that area. So it would be good for you guys to know what's going on in the department and it's central to nucleotide metabolism. We started out-- we were drawing this. This is my notes that I tried to reproduce for you to look at. And I'm not going to read. So this was a big overview slide where we're going. And so central to everything. This wasn't in the original packet, but I will put this up. I'll try to get a better version of that. But PRPP is central. And we are talking about de novo purine bio-synthesis, but again, not only is de novo important, so is salvage. It depends on the cell type. You know, if you have cancer cells that are rapidly growing or B cells and T cells, de novo becomes really important. In other types of cells almost everything is salvage. And so I have that PRPP, at least in the purine case-- and I'll show you an example of that in a few minutes-- goes directly-- can make your nucleotide directly. That's a salvage pathway. And we'll see that the de novo pathway, which is what I was describing at the end-- and you've already seen this in recitation from last week-- is 10 steps to get to IMP. But then you need to get to GMP and AMP. And I showed you how all of this branches off with the cofactor folate between purines and pyrimidines. And in the end, we need both purines and pyrimidines. We need it in the nucleotide levels. So two hydroxyls, the two prime three prime cis hydroxyls, which in the diphosphate stage, that's also unusual. Most of the time you don't see high levels of diphosphates inside the cell, either they're monophosphates or triphosphates. So part of the complexity I think of nucleotide metabolism is figuring out where the kinases are and the phosotases are. And you'll notice that I've avoided that. And that's because every organism is different and every cell type is different, and the regulation is a little bit different. But I think it's important to realize that to make deoxynucleotides, which are required for DNA replication and repair, is done at the diphosphate level. So you make deoxynucleotides, but they still have to be converted to deoxy NTPs for DNA. And over here you need to again, make NTPs for RNA. So that's sort of the big picture. We have a purine pathway de novo. We're not going to be able to talk about pyrimidines, but the salvage pathway with pyrimidines is extremely important. It's a major target of cancer therapeutics now. I think only in the last few years has it been realized that in many cancers you have both pathways going on. And it turns out now with isotopic labeling and mass spec, metabolomics is coming into its own. So you can tell actually by feeding the cells this is all done in tissue culture. But you can tell by feeding the cells whether the deoxynucleotides came from de novo or whether they came from salvage. And so we're getting a really different view of nucleotide metabolism. And as I said in the very beginning, I think the next decade we're going to understand a lot more about how all these things interact and the kinases and phosatases that put the nucleotides into the correct phosphorylation state. So that's key to everything and it's complicated. So what I want to do now is briefly talk about the purine pathway. So we can look at the biology. I'm not going to write this down, because most of you already know this. So we'll just go through it again. Purine nucleotides are central to everything. So knowing where they come from and how you control them is really pretty important. And we don't understand that much. So I mean, NTPs and dNTPs are central to our genetic material. So we need to get them and we need to control them. If these levels become imbalanced, you have mutater phenotypes in DNA replication. And so fidelity of deoxynucleotide DNA replication is really important and regulated by ribonucelotide reductases. Building blocks for cofactors. We've seen flavins. We've seen NAD. We've seen CoA. None of this is an accident. Adenine can self-assemble from cyanide formate in the prebiotic world. And so that's why they're central to everything. So they're in a lot of the cofactors we've already talked about is they're not necessarily the business end, but they've got the phosphates and the adenines stuck on to end, which presumably helps in some way for binding. We're using GDP and ATP everywhere in the course of this semester. You've seen it in your macromolecular machines that you've talked about, especially in the first part of the course with the translational and protein folding and protein degradation all require energy ATP. We will see in today's pathway and today's lecture on purine biosynthesis de novo, it turns out 5 out of the 10 enzymes use ATP. So we'll see. And what you will hopefully now will know is what ATP does. I'm going to show you two examples. But you see the same thing really over and over and over again. So this should be sort of-- you might not know whether it uses ATP to get at the gamma position-- chemistry at the gamma position, or at the alpha position but the chemistry is the same over and over again. And so that part hopefully is part of your repertoire now, about thinking about the role of the ATP on primary metabolic pathways. And we've also seen in the last-- in the reactive oxygen species section, we are signaling by many mechanisms, signaling by phosphorylation is all over the place. And a lot of people are trying to understand. And I think one of the futures is how do you integrate signaling and primary metabolic pathways? And we're almost there. I decided-- I wrote a lecture on this and decided-- it's really still very phenomenological. But all of these key regulators and signaling linked to purines and pyrimidines in some way. I think the linkages aren't totally clear, in my opinion. So why else do we care about purines? When I was your age, purine and nucleotide metabolism was front and center. Why? Because people were successful at making drugs based on these molecules. The central role it plays in replication and repair has made them successful targets at many different levels. Here, this has both purines and pyramidines, but I'll just pick out a few. This guy acyclovir is what we use as an anti-herpes medicine. In fact, I think I've taken it. Here, mercaptopurine cures childhood leukemia. Clofarabine is something that's been studied my lab. It's a drug that-- it's not particularly effective, but it's used clinically against certain hematological cancers. And so these are all anti-metabolites not focused on signaling, which is what everybody is focused on. In reality, I think the success-- if there is success against cancer-- is going to be mixing the two. I think you need combinations of metabolic inhibitors. They're toxic. So is everything. But somehow figuring out how to use multiple approaches to avoid the resistance problem, which is a really important problem. And to combine the two once we understand the interconnections better, I think is where it will be to get successful, more successful. Therapeutics, but ultimately what we would like to do is catch it in the bud, rather than waiting to try to treat something where it's completely out of control. So I'll just show you one of my favorite ladies, Gertrude Elion. She worked at Burroughs Wellcome for many, many years. She went to Hunter College, as did many-- in New York City-- as did many outstanding women scientists. And she was involved at Burroughs Wellcome in discovery of mercaptopurines, acyclovir treating herpes, and AZT. She made several contributions. Never had a PhD. So anyhow-- So what I also wanted to show you, we're going to talk about de novo pathways. I just want to show you this is a typical-- in the case of the purines-- salvage pathway. So what does that mean? You get the bases, the nucleic acid bases from your diet. Or you're breaking down your DNA and your RNA. You have nucleic acid bases. Or you have nucleosides. So can you take those and make them into the right components to do RNA biosynthesis and DNA replication, make ATP, et cetera. And so here's an example of hypoxanthine reacting with our central phosphoribosyl pyrophosphate, which I had in the original slide that I talked about last time to make, in this case, the nucleotide. And why is this interesting? It's interesting because it turns out that many parasites like in malaria don't have any purines. So where do they get their purines from to replicate the DNA? They have to use salvage. So the salvage pathways have-- for treatment of those things-- have become front and center. Can you make specific inhibitors of phosphoribosyl pyrophosphate reaction with the bases? And we're pretty good at that actually. Vern Schramm's lab has done some beautiful work. And there's a lot of things in clinical trial targeting salvage pathways. So again, there's something different about the metabolism of us and whatever is invading us. That's not true in cancer. So cancer is a much tougher problem, because you get normal cells as well. It's a question of what the therapeutic index is. So that's all I want to say in the introduction to the biology. And then I want to talk about one cofactor. And then I'm going to talk about the pathway itself. So there's one cofactor, which I sort of told you I was going there in the first place. Let me break this down over here. So the one cofactor that I wanted to talk about is folate. Let me also show you. You don't have to sit and look at this. But I'm going to show you it's all written out. So you don't have to bob up and down. It's all written out on the handout. So this is folate. And let me just point out a few things. This is going to be the business end of the molecule. So I want you to know where the business end of the molecule is. I don't expect you to remember the structures. But what does this sort of look like? Anybody? This is the kind of chemistry-- I mean, I think there's a bunch of heterocyclic chemistry that you find in biology that most of you haven't been exposed to. And it's not intuitive what the most reactive positions are. This cofactor is much simpler than flavins, which we very briefly talked about before. So this has a polyglutamate on the end. So this is folate. And what you really need to know is that this is 5, 6, 7, 8. And this is 10. So the active part of this cofactor is here. So everything's going to happen at either N5-- if you have a copy of this, you can just circle N5, N10 and N5. That's where all the chemistry is going to happen. And it turns out the way this cofactor works-- so this is 1, 2, 3, 4. And so this is 4a, and this is 10a, 8a. It sort of looks like flavins. And it sort of looks like pterins. And pterins actually can undergo redox chemistry under certain sets of conditions. These molecules are only involved in one carbon transfer. So the major focus is one carbon transfers. And it can do it in the methyl state, in the aldehyde state, or it can do it in the acid state. So all three oxidation states from one carbon transfers. And so then how does it do it? And the chemistry actually is fairly simple compared to the chemistry that we've looked at before. And we looked at a little bit at hemes. We looked at a little bit at flavins. This is much simpler. And so what we're after in the end-- and I'll show you how we get there-- so here's N5 methyl. And we'll see this is tetrahydrofolate. So this ring is completely reduced. And so this is tetrahydrofolate. And this can undergo oxidation and reduction. And that becomes very important in the pyrimidine pathway to form thymidine, which is a major target of fluorouracil, which is a drug that's still used clinically. Anyhow, this is the reduced state here. So this is where the tetrahydro is. So both of these can be oxidized. And that would be folate. So you can make dihydrofolate, folate, and tetrahydrofolate. And the oxidations occur here and here. And we're not going to look at that, because we're not going to have time to look at pyrimidine metabolism. But the dihydrofolate plays an important role. It's the target of methotrexate. If you have rheumatoid arthritis, you take methotrexate is one of the drugs that people take nowadays. So what's unusual about this-- and this is key to the purine pathway, it's also key to the pyrimidine pathway-- that's why folate have been central. People made folates for decades. Even when I was your age, people were making folates for treatment therapies in cancer. And it's been successful. In fact, and if you've gone to Princeton's chemistry department, the whole department was funded on an anti-folate that Ted Taylor made 25 or 30 years ago. And they've tried it again under different conditions, and it's now being used clinically. So how does this work? So we have this oxidation state. We have this oxidation state. And then we'll see that this can ring open. And so this would be the aldehyde state. And this can hydrolyze. And that would be the acid state. So I'm going to show you in a second, I'm going to walk you through where those different states came from. So methyl state, aldehyde state, acid state. So there's the model, because I like to have the windows open, you probably can't see the model very well up there now. But you can pull it up on your computer if you want. I'm going to write out the model. So we start out over here with tetrahydrofolate. So this is tetrahydrofolate. And we have nothing here, which you notice was we could have something at N5, something at N10. We'll see the methyl group is always at N5. It could be at either, chemically, but it's always at N5. We'll see the aldehyde group is always at N10. It could be either chemically, but it's not. So these are going to be the key stages. And here we have no carbons. So somehow we have to get the carbons into the molecules. So we start out with this molecule, tetrahydrofolate. So what happens is you can start out here, and use formate. So formate is going to be the source of the one carbon in this case. So the names in this pathway are again, horrible, just like the purine pathway. And on the next slide I've written out the names. So it turns out that one enzyme can do three of these activities. So this is one of the enzymes. And so this is activity one. And it attaches a formate, so they call it a formate ligase. The names again, in my opinion, are horrible. But what it allows you to do is-- so what I'm going to draw out now is not the whole structure. I'm just going to focus on the business end of the molecule over here, and skip this ring over here. But that ring is there, and is key to making all of this work. So I'm just going to do this like that. And so what can happen is that you can formulate and form. And so this is now N10 formal tetrahydrofolate. So this is N10, and we'll call this R. So that's the first step. That's the enzyme. The same enzyme catalyzes the next step. And what you can picture happening here, if you watch me, is this nitrogen is juxtaposed to this imid. So can attack to form a tetrahedral intermediate and then lose a molecule of water. So that's called a cyclohydrolase. So this guy is attacking. And then you have loss of water. And this is a cyclohydrolase. So this is the same enzyme. So this is two. This is one. But they're both on the same polypeptide. So there are three of these on one polypeptide. You've seen that before in recitation last week. And so now what you've formed is-- and again, this is a cyclohydrolase-- now what you're formed is this structure. So we've lost a molecule of water. So you can draw NR. And so if you hydrolyze this, you can get back to the aldehyde stage. So if water adds here, this is an iminium system. Water can add, it can collapse, it can ring open, it can ring close. So the chemistry here-- we're going to see some really similar chemistry actually, because we can use N10 formal tetrahydrofolate in two steps in the purine pathway. So this chemistry I'm drawing right now is related to the pathway in general. And so this is called 5, 10 methylidine-- the names again, are horrible-- tetrahydrofolate. And then the third enzyme in this pathway is a dehydrogenase, so DH. And so what you can imagine you could do here is we have an iminium system. And NAD pH is the reductive. So you can reduce this down to methylene tetrahydrofolate. So this can be converted to an NADP. So this is the dehydrogenase. We've seen that used over and over again. This is the same enzyme. So this is also MTHFD. And I've given you the nomenclature on the next slide. So if you want to look at-- So this is tetrahydrofolate whatever. So it has of formal ligase, it has a cyclohydrolase, and it has a D hydrogenase all on one enzyme. And so what do you generate then? You generate-- so this is methylene tetrahydrofolate. And this is the key player in pyrimidine biosynthesis, which we are going to talk about. And it's an enzyme called thymidylate synthase, which makes thymidine, which is a major target for drugs in the treatment of cancer. So now you can even take this a step further and reduce this further. We're still now here. If you ring open this, you're at the aldehyde stage. You can reduce the aldehyde stage down to the methyl group. And that's then getting us into the methyl state, the aldehyde state, and the acid state. So I think when you sit down and look at this, it looks complicated at first. It's really not that complicated. So this can just ring open. And conceivably, it could ring open in either direction. It depends on the enzyme that's catalyzing it. But we always get N5 methyl tetrahydrofolate. That's what's used inside the cell. People don't find N10 methyl tetrahydrofolate, but chemically, that could happen. So what happens? This is now a new enzyme. And again, it's a dehydrogenase. So NADPH is going to NADP. So this is a new enzyme. I'm not going to write out the name. But this then reduces this to N5 methyl tetrahydrofolate. So what we've done then is, in the pathway I've drawn out here is, where do we get the one carbon from? Here, we got it from the formate. And we can change the oxidation states to get all three of these oxidation states, depending on what we need to do with it. You have to have the right enzymes and the right complexes to be able to make this all work. Now many of you might not recall this, but in the Benkovic paper you read for recitation last week, one of the controls with this tri-functional protein. And it does not exist in the purinosome. Benkovic's been interested never in these enzymes, and channeling of reactive intermediates in these systems. This does not exist in the purinosome. So then the question is how do you get back? And so there are three methylating agents inside the cell in a biology. Does anybody know what the other two are? So this is unusual, N5 methyl. So this is N5, this is again, N10. STUDENT: [INAUDIBLE]. JOANNE STUBBE: So S-adenosyl methionine is probably the most prevalent. What's another one? STUDENT: Methylcobalamin. JOANNE STUBBE: Yeah. So methylcobalamin. So S-adenosyl methionine is the universal methylating agent inside the cell. And then you also have-- I'm not going to draw the structure out. We're not going to talk about it, but methylcobalamin. And there's a single enzyme that uses all three of these methyl groups. And if I had another five lectures, I would talk about this enzyme. This was studied extensively by Rowena Matthews' lab, who was one of Cathy's mentors. And then Cathy was involved in getting the first structures many years ago with the little pieces. So it's one of these enzymes. It's huge. And it's got to juggle these three methyl groups to do the chemistry. It's really sort of fascinating. And so what it does is it takes homocysteine-- so this is homocysteine-- and converts it to methionine. I'm not going to draw the structure. So you methylate it. So you're going to methylate that cysteine. And then you're back to tetrahydrofolate. So there's another important reaction that I just want to put in here is that there's another way to go from tetrahydrofolate to this one, which is methylene tetrahydrofolate. And so in addition to being able to put on the one carbon with formate, does anybody have any idea what another major way-- it's probably the major way of doing one carbon transfers from metabolic labeling experiments? It comes from an amino acid. What amino acid could you use? So somehow we want to get from here to here. This is also a major target of therapeutics. Anybody got any ideas? We need to get one carbon out of an amino acid. What did you say? STUDENT: Thymine? JOANNE STUBBE: Thymine? That's not an amino acid. STUDENT: Thiamine. JOANNE STUBBE: Oh, thiamine. STUDENT: Methionine. JOANNE STUBBE: Oh, methionine. No. See, I guess I'm deaf. OK, I didn't hear you. No, that's, not it. So I'm not going to spend a lot of time, but serine-- so this is ours-- I'll draw this out, because I think this is really important. This can be converted into formaldehyde. Does anybody know what the cofactor would be that would do that? And then what you end up with is glycine. So this is the major way-- serine is a major one carbon donor. So seramine is going to generate the formaldehyde equivalent, which then can get picked up here and make methylene tetrahydrofolate. Anybody have any idea of how you would convert serine into glycine? You do learn about this cofactor. What is the cofactor that works on all amino acids, if you want to do something to it? There's only one. STUDENT: PLP. JOANNE STUBBE: PLP, yeah. So this isn't unusual-- PLP is sort of an amazing cofactor. It can do alpha decarboxylations, racemizations. It can do aldol reactions. And then it activates the beta positions so you can do beta eliminations replacements. It can do probably 10 or 15 different reactions. This one is unusual in that what you're doing is you're doing an aldol reaction. So you're cleaving that bond, and a reverse aldol reaction in this case. And then the other thing is if you want to link this into pyrimidines, you have dihydrofolate. So this is dihydrofolate. And that's a major player in pyrimidine metabolism to make thymidine. I'm not going to have time to talk about this. But folate is a central player in both purine and pyrimidine metabolism. And people have spent a lot of time thinking about it. And I think the chemistry of interconversions, once you sit and walk through this yourself, start over here and see if you can draw out the mechanisms. It's the same mechanisms we've seen over and over again, in addition to a carbonyl and loss of water. So that was the introductory part. And really what I want to do now is-- we can put that up here for those who still want to stare at it-- what I want to do now is talk about the pathway. And what I want to do is write out the pathway, and then use a PowerPoint to talk about a few features of the pathway that I think are the most interesting, and that you can make generalizations to other pathways, like, what is the role of glutamine? That's universally conserved. What is the role of ATP? And we're going to see the roles you see in the purine pathway are used in many metabolic pathways. So those are the ones I decided to focus on. So what I want to do is go step by step and just make a few comments. And then I'm going to use a PowerPoint over here so you can see what I have written down. I'm going to write down a few things. So that's the nomenclature. There's the pathway. We will start there. So I told you that the first step in this pathway is we start with phosphoribosyl pyrophosphate. That's central to a lot of things. It's chemically very unstable. It falls apart. It's hard to isolate. And the first step in this pathway-- we talked about this briefly in recitation-- is to make phosphoribosylamine. So the interesting thing about this pathway-- so is you start out-- and again, the nomenclature, I've written out. On the exam, you probably will have something about purines there. I will give you the pathway, and I will give you all the names and the enzymes. So you don't need to memorize that. I'm probably the only one that knows the names, because I've worked on it. Very confusing. So what's unique, again, and we've already mentioned this, is you start out with ribosyl phosphate. And what you're going to do-- and this is what we're going to walk through-- is that the first thing you do is you build up the imidazole moeity of your purine. So using sort of basic metabolites in ATP-- there are five steps out of the 10 that use ATP-- you make this amino imidazole ribonucelotide. And then what you do again, step by step, is convert this into the pyrimidine moiety. So you make your purine. So that's a step, one step at a time. And this was unraveled using metabolic labeling experiments. So the first enzyme I'll spend a little bit of time on, because I think it's a paradigm for many enzymes in metabolism in general, where do you get ammonia from most of the time? The major source of ammonia is glutamine. So that's something that you see in this pathway. So glutamine-- you all know glutamine has this part in this side chain-- is going to glutamate. And so you form glutamic acid. And the ammonia from the amid is going to interact with phosphoribosyl pyrophosphate, which is always bound to magnesium to form phosphoribosylamine. And so I'm now going to start being sloppier. Instead of writing phosphate here, I'm going to have a phosphorus with a circle around it. That means we always have the five prime phosphate. And furthermore, what I'm going to do is replace all of this with an R group, ribosylphosphate is present at every single step in the pathway. And in fact, one of the reasons I thought this pathway was interesting, every enzyme in the pathway has to have a binding site for ribosylphosphate. Well, have any of you ever thought about how metabolic pathways evolved? Where does it come from? You have these really complicated pathways. Where do you start? How do you think about that? Well, this might be a fantastic place to look at that. Why? Because you might have a ribosyl binding site for everything. So maybe it starts with something that binds ribosylphosphate. Anyhow, this is an unusual pathway in that you have something. You have a really good handle on to hang on to. And as we already talked about-- so this, we're going to call R-- what's unusual, there are a couple things I want to say about this. But we already talked about this a little in terms of channeling and this question of why you would ever want to have clustering enzymes. And that's because the half-life of this is about 10 seconds at 37 degrees. So it took a lot of effort to see this thing. I mean, you couldn't see it by normal methods. People inferred its presence because Buchanan actually was able to see the next intermediate in the pathway and inferred the existence of this. And many of these intermediates in the pathway, which is why Benkovic focused on this, are chemically unstable. Let's see if I have one of these. I don't have it in this pathway. Anyhow, I'll show you another one, which has a half-life of five seconds or something like that, that took forever for people to identify it, because when you try to work it up as a chemist, it falls apart. And I would say this is something any of you get into metabolomics, people are looking for metabolites now. There's one metabolite that people have found quite frequently, and it seems to be involved in regulation of glycolysis. It's this one. See, where am I? Aminoimidazole ribo-- this one. And that's because it's stable. And the lot of the other ones are not very stable. So I wouldn't be surprised if you ended up finding a lot more metabolites that are playing a central role in regulating enzymes in primary metabolism, because where does the serine come from? Does anybody know where the serine comes from that plays a key role in making this folate analog? Anybody have any idea? So serine is three phosphoglyceric acid in the glycolysis pathway. It's actually very straightforward to write a mechanism of how you get there. Intimately links the glycolysis pathway to purine metabolism. And we'll also see here of course, this is folate, but we also need glycine. That's the next step in small molecule in this pathway. It needs glycine. So everything is integrated. Once you see-- you sort of see the big picture and have central pictures of primary metabolism, everything becomes much more integrated. So how does this happen? So what I want to do is I want to talk a little bit about this enzyme. So here, let me just talk about this this. So if we call this Pur F, just so we have a name, Pur F is called an amidotransferase. And what it's going to do is it's going to take glutamine-- and it turns out these enzymes have a domain. They always have multiple domains. And the domain that uses the glutamine can be the same. There are actually two different convergent evolutions of glutamine binding domains that do the same chemistry. So what you do is-- we've seen this again many, many times-- so you form a covalent intermediate, which then hydrolyzes to glutamate regenerating ESH. And what happened during this reaction, you generate ammonia. So the goal of these amidotransferases in general, in many, many metabolic pathways, is to generate ammonia. And so to me, what's striking about this is the way nature evolved these metabolic enzymes that generate ammonia. And so what you see in a cartoon view-- so we are always going to have all of these enzymes. They may be a single polypeptide. They may be two polypeptides, but they all have a glutaminase domain. So the glutaminase is just generating the ammonia. But what do we have? We start out with phosphoribosyl pyrophosphate. So once we generate the ammonia, what can happen? You can now by-- it turns out by dissociated mechanism-- displace a pyrophosphate to form phosphoribosylamine. So all of these kinds of reactions involve dissociative rather than associative transition states. That's not important. But what it what's amazing about this is that PRPP, in this case, binds to one domain, and the glutamine binds to this second domain. So ammonia, what would happen to ammonia if it went out into solution? STUDENT: Protonated. JOANNE STUBBE: Yeah. Gets protonated really rapidly, becomes unreactive. I don't know why nature designed this. But what you see with all these enzymes is she makes a tunnel across the domain interface that's about 25 to 40 angstroms long. So the ammonia this released never gets out into solutions. This is another example of channeling a reactive intermediate, which we talked about as potentially a reason for channeling in the purine pathway. So there's a tunnel. And the tunnel can be 25-- we have a number of structures in the ammonia channels. And I have no idea-- I mean, this surprised the heck out of me. I thought the way nature would hold on to this is by hanging on to not the covalent intermediate, but the preceding tetrahedral intermediate. And then when the white substrate was there, release it and then bind it, sitting right next to it. But nature, in all designs, has done this thing where you have this channel. And here is an example of Pur F. This is the glutaminase domain up here. And here is where the phosphoribosyl pyrophosphate binds down here. You can't see the channel, but this is work of Jan Smith a number of years ago, was the first one that showed the channel in this pathway. So that's common. And we're going to look at another glutamine requiring enzyme in this pathway. It's the fourth enzyme in the pathway. Also is a channel. Again, it's distinct. It all does this glutaminase covalent intermediate, but the structure of the glutaminase domain is distinct. So what's the next enzyme in the pathway? So the next enzyme in the pathway again, is a paradigm for many, many unsungs and primary metabolic pathways. And if you look at the structure, let's just go back to the pathway. If you look at this pathway, what you now want to do-- so we keep the ribose 5-phosphate all the way through the whole thing. That's the scaffold. Now what are you going to add? You're going to add glycine. So here is your phosphoribosylamine. And you're going to add glycine. How do you inactivate an amino acid? You've seen activation of amino acids now many times. What are the two ways you can activate amino acids? STUDENT: [INAUDIBLE]. JOANNE STUBBE: So either adenylate or phosphorylate. So that's a paradigm that you see over and over again in nature. This enzyme uses ATP. This is one of the five enzymes. And it forms inorganic phosphate. So you're phosphorylating, not adenylating. And so I'll show you what the mechanism is up there. You've already seen this mechanism, but the idea is you phosphorylate this. You're going to form the phosphoanhydride. And then the phosphoanhydride can react with the amino group. And kinetically-- this is something that's one of my students working on a long time ago-- there was evidence that this intermediate, which is chemically unstable, could channel between the two proteins. So you don't generate this own solution where it can fall apart and it can anomerize. It gets transferred directly. In fact, in the early days when we invented the first biochemistry labs at MIT, they used this system. I really pushed them to the limit, because they were dealing with the substrate. They had a very short half-life Anyhow, they learned a lot from the exercise. So what you're going to have then is ribos-- I'm just going to call it R. And so here's our glycine. Whoops. Guess I'd better get the structure right. So this is from lysine. So what we will see is that this is another-- we're not getting very far-- but this is a member of the ATP grasp superfamily of enzymes. They all do the same chemistry. So let me just move forward a little bit. I'm not going to draw this out. You guys have seen this chemistry many times. So what's happening in this chemistry is you have a carboxylate. ATP phosphorylates it, and then you attack by a nucleophile, in this case, the nucleophile is the amino group of phosphoribosylamine. So what I just want you to see here if you look at this, there are four enzymes that are involved in purine metabolism that all have the same structure. They all have ATP grasp structures. They all go through phosoanhydride intermediates. And you can, from bioinformatics, pick these structures out. So this is again, an example. Once people defined-- there's almost a no sequence homology between these proteins-- but by knowing this chemistry, you can actually pick out that these are going to be family members. And if you know if they are organized in bacteria and operons, you can even guess at the substrate. And then you can test this model that they go through phosoanhydride intermediates. And I'm over, but the next step in this pathway-- the next step in this pathway-- we're going to come back. And what are we going to use? We're going to use N10 formal tetrahydrofolate. That's why I went through this. We're going to put a formal group here. And again, the chemistry is just the same. Go home and think about the chemistry of how you generate all the different oxidation states of the carbon. And then I think you can see the chemistry in this pathway actually is pretty simple, once a few basic reactions. So the ATP grasp family is interesting. The amidotransferase and the channel is interesting as being general in metabolism.
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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: All right. So in Monday's lecture, we talked about how cells replicate, OK? And today, I want to talk about how now an entire organ would essentially replicate. In this case, it's not going to divide, but it's going to regenerate or renew itself, OK? And so this involves adult stem cells and also apoptosis , which you've heard a little bit about earlier in the course. And to explain this to you, I'm going to have basically a model organ that we'll use. And we'll use it for a couple lectures. And the model organ is going to be the lining of the intestine, OK? So the lining of the intestine is an epithelial tissue. And I'll tell you a little bit about epithelia in just a minute. But you have the intestinal epithelium. And this is the-- these are the cells that are the lining of the intestine, OK? And one of the reasons that I've chosen this system is because the lining of your intestine has pretty remarkable regenerative capabilities, OK? So your small intestine renews about every four to five days, OK? So the vast majority of your cells in the intestine were basically derived in the last four to five days, OK? So humans aren't as cool as some organisms, like newts, in that you can't cut off your arm and have it grow back. But at least we have the intestine, which undergoes a pretty dramatic regeneration, OK? And this is not-- the intestine is unique in how rapid this is. But you have other tissues that also exhibit continual renewal, like your skin and your blood cells. And even the cells of your that line your insides of your lungs, they do exhibit renewal capabilities, OK? And so I'm going to use the intestine as a model system. It doesn't mean it doesn't happen in other tissues. But it just happens that we can-- we really understand the intestine system maybe a little bit more than many other systems. So I'm going to use it as an example. So in thinking about the lining of your intestine, it's important, and it has important functions. One important function of this lining is it has to absorb nutrients from food going through your intestine, OK? So it exhibits a nutrient absorption function. Now, the lining of your intimate intestine, much like your skin, is also a barrier between the inside of your body and the outside world, right? Because basically, the inside of your intestine is contiguous with the outside world, right? If you open your mouth, you can get down to the inside of your intestine, OK? So it serves also an important barrier function. And one point I want to make about this system right now is that the intestinal epithelium, like many of your other organs, are composed of multiple cell types, OK? So it also has multiple cell types. OK. So let's now consider the lining of the intestine. And the way the intestine morphologically looks is that there are a series of invaginations. So this is a cross-section view through the intestine. So you have to imagine that this is a cross-section view, but that this is a plane. And it's a plane with a lot of invaginations in the plane, but also protrusions out of the plane. And so you have to remember-- you have to think of this as a surface. And then it's wrapped up into a tube, OK? So this is just a very simple cross-section image of the intestine, OK? And the lining of the intestine is a sheet of cells. So this lining is composed of many, many cells. They're columnar in morphology. So what I've drawn here is just a small section of the intestine. This would be the lumen, out here. This is where the food is, or the food going through your intestine would be. And then below here, this is interstitial fluid inside of your body basically, interstitial, OK? So the food's up here. The rest of your body is down here. And you can see that there are there there's a structure to it. It's not just a flat surface that's wrapped up, but there are invaginations. And the invaginations are known as intestinal crypts, right? Much like you would-- if you bury something, like a body, it would go below ground. So the crypt is below the surface of the lumen. And this these projections, out here, are known as villi. So they're the villi of your gut, OK? Now, if we consider-- and this lining is made up of multiple cell types, which I've outlined up here and which are in your handout. So you don't have to write these down. But this is just making the point that there are various differentiated cell types. There are enterocytes, which are the absorptive cells. These are the cells that are taking in nutrients and transporting them into your body. There are enteroendocrine cells that play an important signaling function in the gut to regulate the biology of the gut system. There are a goblet cells, which secrete mucus into the intestine system, which protects these epithelial cells from digestive enzymes that are present in the lumen. And there is this last cell type, the paneth cell, which functions as an important role in regulating stem cells, as I'll outline in just a minute. So these are all the differentiated cell types. But there's a type of cell that's an undifferentiated cell type. And that's the intestinal stem cell, which will be the hero of today's lecture. All right. Now, if we consider just a small group of cells, what these cells look like is this, OK? So these are what are known as epithelial cells. And epithelial cells have certain properties. The first is you see that this end of the epithelial cell looks different from this end, OK? And this is called an apical-basal polarity. So much like neurons have a polarity where the on one side of the neuron there are dendrites and on the other side of the neuron there's an axon, these cells have another type of polarity, which is called apical-basal polarity. So the side facing the lumen is apical. So the lumen would be up here. And the basal side would be down here, OK? So this is basically oriented along this axis of the tissue, OK? where apical is on this side, basal is on this side, OK? And these projections from the individual cell, these are called microvilli. And essentially, these plasma membrane corrugations increase the surface area through which nutrients can be absorbed into these cells, OK? Now, one other defining feature of these cells is they have proteins that protrude from the plasma membrane. And these are adhesion proteins that couple the cells together. And actually, the cells are stuck together much tighter than I'm drawing here, such that the cells form a barrier so that things can pass unregulated from the lumen into the body, OK? But these proteins, which I'm drawing sticking out of the membranes of the cells, are adhesion proteins. And these adhesion proteins simply link the cells together such that they form a sheet of cells or tissue. So they link cells together, OK? So these are two of the key properties of epithelia. They have an apical-basal polarity. And they also exhibit cell-to-cell adhesion, such as the cells reach out and connect to each other. And they basically are glued together, or Velcroed together such that they don't easily come apart. OK. Now, in considering this system, what I'm going to tell you is that there is renewal of this lining. And the renewal starts at the base of the crypts, OK? So there's going to be renewal. And this cell renewal is at the base of the crypts, right? There are many of these crypts, right? You have you have like the surface, but there are many, many invaginations that are present in your gut. And the renewal is happening at the base of these crypts, OK? And that's because at the base of these crypts are where a type of cell, known as intestinal stem cells, lie, OK? So it's at the base of these crypts where there are what are known as intestinal stem cells. And I'm going to abbreviate these ISCs for this, so I don't have to write out intestinal stem cell whenever I tell you about them, OK? Now, that's where renewal occurs. And if you just had more and more cells getting put in the system without any removal of cells, then the organ would get bigger and bigger, right? And so our intestine more or less staying the same size at this point in our life. And so in addition to renewal, there's also cell death. And what happens when cell death occurs is that cells are shed from the intestinal lining into the lumen. So cells are shed into the lumen. And then they just go with the rest of the crap that's in your intestinal lumen. And it's eventually removed from the body, OK? So some cells are leaving the tissue and going into the lumen after they've existed for a few days in the intestine, OK? So to have an organ then the constant size renewal has to essentially be more or less equal to death, OK? And so you can think of this as a type of homeostatic state for this tissue where, if renewal equals death, you have what is known as tissue homeostasis, where the number of cells in the system as a whole is basically remaining the same, even though there's constantly new cells going into this system. But then the cells are also-- the older cells are being removed from the system. So what's really key in this process are these intestinal stem cells. So I'm first going to tell you about stem cells. And these are what I would define as adult stem cells, OK? So they're stem cells that are associated with a particular organ, OK? And I want to differentiate this right now between these types of stem cells and another type of stem cell that we're going to talk about later in the course, which is an embryonic stem cell. So adult stem cells-- these stem cells, like intestinal stem cells, are associated with a specific organ. And they only give rise to cell types that are present in that organ, OK? So they're not-- they can't give rise to any type of cell type. Your adult stem cells are really specific-- organ-specific, we'll say. In contrast, embryonic stem cells are-- these stem cells have more possibilities. They can give rise to pretty much any cell type in an entire organism, OK? So you can think of these adult stem cells as being more restricted in their fates than the embryonic stem cells. And so what adult stem cells are called is they're called multipotent, because they can give rise to multiple different cell types, but not all of the cell types that are present in an organism, right? An intestinal stem cell will not be giving rise to a blood cell or other cell types in other organs, right? It's restricted to just giving rise to cells that are present in that organ. In contrast, embryonic stem cells are what are known as totipotent, or sometimes pluripotent, where this type of cell really is capable of making any type of cell that is present in an adult organism, OK? So that's less restricted than the adult stem cells. And we're going to come back to these embryonic stem cells towards the end of the course. But for understanding cancer and also tissue homeostasis in the intestine, we're really needing to focus on the adult stem cells. OK. So these adult stem cells have two key properties. The first is what I just said. They're multipotent. And what multipotent means is that they can give rise to multiple different cell types. So this stem cell gives rise to multiple cell types. And this is associated usually with a single organ system. OK. So in the case of the intestinal stem cell, the intestinal stem cell is multipotent. And it can give rise to many different terminally differentiated cells. And you can see here I've just written in example cell types that the intestinal cell could generate. It can generate any of the four different types of cells that I introduced to you at the beginning of the lecture. The other key aspect of this system is that, in addition to generating all of these different cell types, the stem cell can also renew itself, OK? So the other key property is this self-renewal. So the intestinal stem cell also gives rise to another intestinal stem cell. So it basically duplicates itself such that you still have a stem cell in the organ. And then one of the daughter cells is self-renewed and remains a stem cell. The other daughter cell can go on to divide further and give rise to these differentiated cell types, OK? All right. So one question you might be asking yourself is, how is it that-- what regulates whether or not a cell will go on to differentiate or whether it will stay a stem cell? And the answer to this question involves communication between this stem cell and other cells in the system. And it involves a special type of cell called a stem cell niche cell. And so I'm going to tell you about a model, which is the stem cell niche model. And what the stem cell niche model is, is that basically there is a niche or compartment which promotes the self-renewal of cells in that compartment that makes them stem cells. So the stem cell niche you can think of as a compartment where signals, similar to the types of signals that we've been talking about over the past four or five lectures, regulate the behavior of the cell to ensure self-renewal and to suppress differentiation, so a compartment where signals promote stem cell renewal. I want to ask you one question before I move on, which is, how would you define-- how would you determine that there is a special type of cell that gives rise to all of the cells in an organ? If you were tasked with finding this and determining whether this was true or not, how might you do it? Does anyone have an idea of what experiment they would do, or what criteria they would have to determine whether or not this is a case? Let's say I gave you a cell, right? In a dish. And I asked you to tell me whether or not you would think this is a stem cell or not. Yes, miles? AUDIENCE: [INAUDIBLE] ADAM MARTIN: Mm-hmm. AUDIENCE: One cell is produced by the same cell, assuming that it's not a stem cell but if the cell [INAUDIBLE] it's a stem cell. ADAM MARTIN: Mm-hmm. So Miles suggested he would like to take the cell that I just gifted him, and let it grow up, and determine whether it gives rise to multiple cell types. And Miles said that, if it just gave rise to a single cell type, that would suggest that it's not a stem cell. But if it gave rise to multiple cell types, then it could be a stem cell, OK? And it's this type of experiment that-- this type of experiment has been done, where you can take an intestine from mice, and even you can take tissue from humans, and you can associate the cells from each other so that you're left with single cells that are separate. And you can then use some type of flow cytometry to separate cells. And you might be interested in a cell that maybe expresses some marker that you're interested in. And you can separate those cells and isolate them. And then you can take an individual cell and grow it in a dish, OK? And this has been done for intestinal stem cells. And if you take an intestinal stem cell and you grow it in a dish, it can grow up and form a bulk of tissue. But what's really remarkable about the result of this experiment is that, from a stem cell, you get this massive tissue. But it self--organizes into a structure that very much resembles a normal gut, meaning that there are crypts where the stem cells are localized. And then if you look at the different cell types in this, what's known as an organoid, you see all of the different cell types that are normally present in the gut, OK? And so this is an example of a type of experiment that's done to show whether or not a certain cell has the capability of functioning like a stem cell, OK? So if you start with a stem cell, you can regenerate the entire organ system essentially. You might also be familiar with bone marrow transplants. And so if you kill all the hematopoietic stem cells in, let's say, a mouse, then you can transplant in a single hematopoietic stem cell into that system. And it will regenerate all the blood cells in that system, OK? So those are just a few examples of how one might functionally define a stem cell in an experimental setting. All right. So now, I want to tell you about the types of signals and how these signals are promoting stem cell renewal, OK? So in this case, the niche, stem cell niche is going to be right here. And the types of signals involved-- I'll just draw some cells here. One of the types of cells that's part of the niche in the intestine is the cell type that I introduced you as the paneth cell. And these paneth cells localized to the base of the crypts together with the intestinal stem cells. And what paneth cells do is they send a signal to neighboring cells. And this signal is a Wnt signal, which I'll tell you about in just a minute. And so this is the intestinal stem cell here. And the paneth cell is signaling to that intestinal stem cell through a secreted ligand known as Wnt. So Wnt is a secreted ligand. And this signal is what's known as a juxtacrine signal, which just means that the cells have to be adjacent, or juxtaposed, to each other in order for the sending cell to signal to its neighbor, OK? So it's not signaling long range, but it's signaling to a neighboring cell, OK? So now, if we think about this intestinal stem cell, one thing intestinal stem cells do as they divide, right? So this intestinal stem cell could divide. Here is a cell in mitosis. It rounds up and divides. Here, you still have your paneth cell. And then when this cell divides, then it's going to be two cells, OK? So now, paneth cell, another cell. Here the two daughter cells. Here's the paneth cell. The paneth cell is still secreting this Wnt signal. So you have Wnt getting secreted. And it's going to signal to its neighbor. But this cell is starting to get farther and farther away from that signal, right? So you can imagine this is happening right here in the tissue, where this cell is at the boundary of the niche, OK? So this cell that gets pushed out of the niche is going to start to differentiate, OK? Because it's no longer seeing the signal, OK? So it's the lack of the Wnt signal which tells cells basically that they should start differentiating into the various cell types of the gut. But the cells that are stuck at the base of the crypt, down with the paneth cells, are still getting the Wnt signal. And so they remain an intestinal stem cell, OK? So here is just a diagram showing you part of that. So the niche is down here, where you have the stem cells. The paneth cells would be some of these blue cells at the base. But there are also other cells below the epithelial lining known as stromal cells that are also secreting Wnt. And so this compartment down here has high Wnt ligan activity. And that tells cells that remain here to stay stem cells. But the cells that are getting pushed up and moving up towards the lumen, they no longer receive this signal. And so they are going to start to differentiate, OK? So you can think of this system as almost a conveyor belt. There's a conveyor belt-like movement of cells from the base of the crypt up towards the tip of the villis, OK? And when the cells move away from the niche cells, then they no longer have the self-renewal signal. And therefore, they go on to differentiate. OK. We'll go back for now. All right. Now, I want to tell you a little bit about this signal, Wnt, because it's something that's come up before in the lecture, even though you might not know it. So Wnt is a ligand. So it functions much like a growth factor. This is a protein that is secreted from the cell, and then binds to receptors on other cells, and induces signaling events in those cells, OK? And Wnt stands for-- the W stands for wingless, OK? So you remember from earlier in the semester, wingless was identified as a mutant that disrupts the formation of wings in the fly, OK? So one of the places where these genes were discovered is in the fly. The nt of Wnt comes from int 1, which stands for the integration of mouse mammary virus 1. Sorry, that's a little bit more of a mouthful. The integration of mouse-- actually, sorry, mouse mammary tumor virus 1. I forgot the tumor part of it. OK. So the W in Wnt is from wingless. The nt in Wnt is from int 1, OK? And so as you can see, there are two different systems where this type of gene was discovered. They are very disparate from each other. One was in a developmental mutant, in the fruit fly. The other was in a mouse system, where the integration of the virus caused over-expression of Wnt, which caused it as tumor genesis, OK? So these disparate systems led to the identification of this Wnt molecule. And this is a defining member of a signaling pathway that regulates stem cell renewal. And I just want to briefly go through the logic of the signaling pathway, because I want you to get a sense that not all signaling pathways are like Ras-MAP kinase, but there can be different regulatory logic, OK? So what's the regulatory logic of this pathway? So the regulatory logic is shown in this cartoon. And I'm going to start with a cell that does not see Wnt ligand. So that would be the case on the left here. So if there's no Wnt ligand-- I want you to focus on what's going on here-- there's a complex that's present in the cytoplasm of the cell. And what it's doing is it's destroying this beta-catenin protein, OK? So what beta-catenin is-- among other things, it's a transcriptional coactivator. So it's basically a transcription factor. So it works with another protein to regulate the expression of certain genes, OK? And in the absence of Wnt, this beta-catenin transcription factor is destroyed and is not able to get into the nucleus, OK? So if there's no Wnt, then beta-catenin is destroyed. And it's destroyed using a system that I introduced in Monday's lecture, which is regulated proteolysis by polyubiquitination, OK? So the first real-- so you have regulated proteolysis by polyubiquitination. So the way this works, as seen here, is that beta-catenin is bound by this complex. And there's a kinase that phosphorylates beta-catenin. And then the phosphorylated beta-catenin recruits this E3 ubiquitin ligase, which polyubiquitinates it, OK? So that leads to the destruction of beta-catenin in the absence of a Wnt signal. But when Wnt ligand is around, this leads to the disassembly of this complex, which is known as the destruction complex. And that leads to beta-catenin accumulating. And once it accumulates, it goes into the nucleus and starts changing gene expression, OK? So in the presence of Wnt, beta-catenin is nuclear. And that's where it needs to be, if it's going to regulate gene expression, OK? So what you see is the logic of this pathway is you have a double negative, where you have a complex, which is known as the destruction complex. And it includes this gene, APC, which we're going to talk about on Friday. This destruction complex is inhibiting beta-catenin by destroying it. And the way that Wnt induces beta-catenin activation is by inhibiting the inhibitor. So Wnt inhibits the destruction complex, which then stabilizes beta-catenin and allows it to go to the nucleus, OK? So the other piece of the logic here is you have inhibition of an inhibitor to activate beta-catenin. Any questions on the pathway? OK. So now that we have our intestinal stem cells and we have a way, by increasing Wnt in this compartment, to maintain the intestinal stem cells through self-renewal, now we have to talk about the compensatory mechanism of death, which allows this tissue to maintain homeostasis. OK. So death-- in this case, death is going to be useful. And the process of death is called apoptosis. And apoptosis is Greek for falling off. And that's essentially what these cells are doing, because the cells, as they move from the base of the crypt up towards the lumen-- eventually, they're going to fall off into the lumen of the intestine. So again, I'll draw. Here's a villus. This is now a villus. And so what happens is, at the tip of the villus, cells are going to be shed from the lining of the epithelium into the lumen. The lumen is up here. This is the villus. And the cells are going to shed off into the lumen and be removed from the organ system, OK? So cells are shed into the lumen here, OK? And this is going to balance the renewal at the base of the crypts such that there's homeostasis. So I just-- the movie I was showing you at the beginning of class was a movie showing you what happens to a cell undergoing apoptosis. So in this case, the cell is binucleate and unhappy. And then you're going to see that it basically explodes, OK? So that's a cell undergoing apoptosis. But you see that there is a clear change in cell morphology and physiology associated with this. And I just want to point out that this is also something that we talked about earlier in the course. And we talked about experiments, a genetic screen that led to the identification of the pathway that regulates apoptosis. And that was done by Robert Horvitz. Much of it was done by Robert Horvitz in his lab here at MIT. And for that work, Robert Horvitz, in addition to his colleagues, won the 2002 Nobel Prize. So you'll recall this is something we talked about in the context of a genetic screen. But this is something that's-- this is what it's doing in your intestine system. It's balancing renewal so that you have homeostasis. So during apoptosis, a cell goes through a series of changes. First-- or what happens eventually is the nucleus becomes fragmented and chromosomal DNA even gets fragmented. It gets chopped up. And also, the plasma membrane also starts to bleb and fragment such that it breaks up into what are known as these apoptotic bodies, OK? And so you can think of these apoptotic bodies as bite-sized pieces of cell that neighboring phagocytic cells can eat up and remove them from the body, OK? So in this case, the cells are being shed into the lumen. So they don't need to get eaten up, because they're just going to go out of the digestive tract. So there are-- cells have numerous ways to activate this apoptosis process. I'm going to tell you about two types of signals that regulate whether or not a cell undergoes apoptosis. The first is that there can be a signal that basically tells the cell to kill itself, OK? So you can think of this as a kill signal. And one type of way to activate this signal is if the DNA is irreparably damaged. So if there is a high level of DNA damage, this induces a signaling process in the cell. And one of the results of that signal-- in addition to regulating the cell cycle, if the DNA damage is great enough, it will induce an apoptotic signal. And it will activate the pathway that the Horvitz lab elucidated in the worm, OK? So there's a signal. And that leads to apoptosis and death. Another type of signal that's critically important for tissue homeostasis and determining whether or not cells live or die is a survival signal. So there are cell survival signals. And many of the growth factors, such as EGF which you've heard about-- in addition to inducing proliferation, these growth factors also tell the cell not to die, OK? So these could be growth factors. And what these cell survival signals do is they repress apoptosis. And so you can think of it where you have a cell constantly needs to be communicated to. And it needs to be told don't die, don't die, don't die, don't die. And then if you remove that signal, it won't be getting that information anymore. And it can undergo apoptosis, OK? So if we were to remove this signal, you remove the brakes on apoptosis. And the cell will undergo cell death, OK? So this ensures that you don't have a cell just kind of like going on and doing its own thing, because cells, in order to live, often need to have some sort of signal from another cell that tells them to live. And so there's some coordination between cells such that you don't have cells rampantly dividing out of control. Now, the reason we're doing this right before we talk about cancer on Friday is because everything that I'm telling you is really essential to understand how a tumor is formed in an organ system, OK? And I want to end the lecture by just planting a seed of an idea in your heads before we move on to talk about cancer on Friday. And I want you to think about the organization of this system, where you have stem cells undergoing renewal. And then the stem cells are just a small fraction of the cells in the system. And they're dividing slowly, OK? So let's think about the stem cells. The DNA on the stem cells, these are dividing slowly. And because they're dividing slowly, they're not going to-- their DNA is not going to accumulate as many mutations. So there is going to be fewer mutations. But these are the cells, and this is the genomic DNA that is going to stay with the organ, OK? So the stem cells are like the crown jewels of the organ. This is the material the organ wants to protect, because it's what's going to be lasting in the organ the entire lifetime of the organism, OK? So you get slow division here and self-renewal. And this cell will stay with the organ. But then where most of the mitosis and cell division happens and replication happens, it leads to an expansion of cells that all differentiate. And because the cells all differentiate, they will all eventually die and get removed from the organ, OK? And this is termed transient amplification. So when there's transient amplification of one of the daughters of this stem cell, this is where there's rapid division. And where there is rapid replication and division, this is where you can get the most mutations. But from the standpoint of cancer, that doesn't really matter, right? Because in order to have a tumor, the cells have to stay in the body. And so all of these cells are going to undergo programmed cell death, and then be shed into the lumen of the intestine and removed from the organism entirely, OK? And so this is actually one important way that our organs and our bodies prevent tumors from happening, because the cell type that is going to remain in our body is the one that's protected from accumulating mutations, OK? And we'll come back to this on Friday. And so I will see you on Friday. And we'll talk about cancer 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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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: So welcome, everybody. My name is Yen-Jie Lee. I am a assistant professor of physics in the physics department, and I will be your instructor of this semester on 8.03. So of course, one first question you have is, why do we want to learn about vibrations and waves? Why do we learn about this? Why do we even care? The answer is really, simple. If you look at this slide, you can see that the reason you can follow this class is because I'm producing sound wave by oscillating the air, and you can receive those sound waves. And you can see me-- that's really pretty amazing by itself-- because there are a lot of photons or electromagnetic waves. They are bouncing around in this room, and your eye actually receive those electromagnetic waves. And that translates into your brain waves. You obviously, start to think about what this instructor is trying to tell you. And of course, all those things we learned from 8.03 is closely connected to probability density waves, which you will learn from 8.04, quantum physics. And finally, it's also, of course, related to a recent discovery of the gravitational waves. When we are sitting here, maybe there are already some space-time distortion already passing through our body and you don't feel it. When I'm moving around like this, I am creating also the gravitational waves, but it's so small to be detected. So that's actually really cool. So the take-home message is that we cannot even recognize the universe without using waves and the vibrations. So that's actually why we care about this subject. And the last is actually why this subject is so cool even without quantum, without any fancy names. So what is actually the relation of 8.03 to other class or other field of studies? It's closely related to classical mechanics, which I will use it immediately, and I hope you will still remember what you have learned from 8.01 and 8.02. Electromagnetic force is actually closely related also, and we are going to use a technique we learned from this class to understand optics, quantum mechanics, and also there are many practical applications, which you will learn from this class. This is the concrete goal. We care about the future of our space time. We would like to predict what is going to happen when we set up an experiment. We would like to design experiments which can improve our understanding of nature. But without using the most powerful tool is very, very difficult to make progress. So the most powerful tool we have is mathematics. You will see that it really works in this class. But the first thing we have to learn is how to translate physical situations into mathematics so that we can actually include this really wonderful tool to help us to solve problems. Once we have done that, we will start to look at single harmonic oscillator, then we try to couple all those oscillators together to see how they interact with each other. Finally, we go to an infinite number of oscillators. Sounds scary, but it's actually not scary after all. And we will see waves because waves are actually coming from an infinite number of oscillating particles, if you think about it. Then we would do Fourier decomposition of waves to see what we can learn about it. We learn how to put together physical systems. That brings us to the issue of boundary conditions, and we will also enjoy what we have learned by looking at the phenomenon related to electromagnetic waves and practical application and optics. Any questions? If you have any questions, please stop me any time. So if you don't stop me, I'm going to continue talking. So that gets started. So the first example, the concrete example I'm going to talk about is a spring block, a massive block system. So this is actually what I have on that table. So basically, I have a highly-idealized spring. This is ideal spring with spring constant, k, and the natural length L0. So that is actually what I have. And at t equal to 0, what I am going to do is I am going to-- I should remove this mass a little bit, and I hold this mass still and release that really carefully. So that is actually the experiment, which I am going to do. And we were wondering what is going to happen afterward. Well, the mass as you move, will it stay there or it just disappear, I don't know before I solved this question. Now I have put together a concrete question to you, but I don't know how to proceed because you say everything works. What I am going to do? I mean, I don't know. So as I mentioned before, there is a pretty powerful tool, mathematics. So I'm going to use that, even though I don't know why mathematics can work. Have you thought about it? So let's try it and see how we can make progress. So the first thing which you can do in order to make progress is to define a coordinate system. So here I define a coordinate system, which is in the horizontal direction. It's the x direction. And the x equal to 0, the origin, is the place which the spring is not stressed, is at its natural length. That is actually what I define as x equal to 0. And once I define this, I can now express what is actually the initial position of the mass by these coordinates is x0. It can be expressed as x initial. And also, initially, I said that this mass is not moving. Therefore, the velocity at 0 is 0. So now I can also formulate my question really concretely with some mathematics. Basically, you can see that at time equal to t, I was wondering where is this mass. So actually, the question is, what is actually x as a function of t? So you can see that once I have the mathematics to help me, everything becomes pretty simple. So once I have those defined, I would like to predict what is going to happen at time equal to t. Therefore, I would like to make use physical laws to actually help me to solve this problem. So apparently what we are going to use is Newton's law. And I am going to go through this example really slowly so that everybody is on the same page. So the first thing which I usually do is now I would like to do a force diagram analysis. So I have this mass. This setup is on Earth, and the question is, how many forces are acting on this mass? Can anybody answer my question. AUDIENCE: Two. We got the-- So acceleration and the spring force. YEN-JIE LEE: OK, so your answer is two. Any different? Three. Very good. So we have two and three. And the answer actually is three. So look at this scene. I am drawing in and I have product here. So this is actually the most difficult part of the question, actually. So once you pass this step, everything is straightforward. It's just mathematics. It's not my problem any more, but the math department, they have problem, OK? All right. So now let's look at this mass. There are three forces. The first one as you mentioned correctly is F spring. It's pulling the mass. And since we are working on Earth, we have not yet moved the whole class to the moon or somewhere else, but there would be gravitational force pointing downward. But this whole setup is on a table of friction, this table. Therefore, there will be no more force. So don't forget this one. There will be no more force. So the answer is that we have three forces. The normal force is, actually, a complicated subject, which you will need to understand that will quantum physics. So now I have three force, and now I can actually calculate the total force, the total force, F. F is equal to Fs plus Fn plus Fg. So since we know that the mass is moving in the horizontal direction, the mass didn't suddenly jump and disappear. So it is there. Therefore, we know that the normal force is actually equal to minus Fg, which is actually Ng in the y direction. And here I define y is actually pointing up, and the x is pointing to the right-hand side. Therefore, what is going to happen is that the total force is actually just Fs. And this is equal to minus k, which is the spring constant and x, which is the position of the little mass at time equal to t. So once we have those forces and the total force, actually, we can use Newton's law. So F is equal to m times a. And this is actually equal to m d squared xt dt squared in the x direction, and that is actually equal to mx double dot t x. So here is my notation. I'm going to use each of the dot is actually the differentiation with respect to t. So this is actually equal to minus kxt in the x direction. So you can see that here is actually what you already know about Newton's law. And that is actually coming from the force analysis. So in this example, it's simple enough such that you can write it down immediately, but in the later examples, things will become very complicated and things will be slightly more difficult. Therefore, you will really need help from the force diagram. So now we have everything in the x direction, therefore, I can drop the x hat. Therefore, finally, my equation of motion is x double dot t. And this is equal to minus k over n x of t. To make my life easier, I am going to define omega equal to square root of k over n. You will see why afterward. It looks really weird why professor Lee wants to do this, but afterward, you will see that omega really have a meaning, and that is equal to minus omega squared x. So we have solved this problem, actually, as a physicist. Now the problem is what is actually the solution to this differential second-order differential equation. And as I mentioned, this is actually not the content of 8.03, actually, it's a content of 18.03, maybe. How many of you actually have taken 18.03? Everybody knows the solution, so very good. I am safe. So what is the solution? The solution is x of t equal to a cosine of omega t plus b sine omega t. So my friends from the math department tell me secretly that this is actually the solution. And I trust him or her. So that's very nice. Now I have the solution, and how do I know this is the only solution? How do I know? Actually, there are two unknowns, just to remind you what you have learned. There are two unknowns. And if you plug this thing into this equation, you satisfy that equation. If you don't trust me, you can do it offline. It's always good to check to make sure I didn't make a mistake. But that's very good news. So that means we will have two unknowns, and those will satisfy the equation. So by uniqueness theorem, this is actually the one and the only one solution in my universe, also yours, which satisfy the equation because of the uniqueness theorem. So I hope I have convinced you that we have solved this equation. So now I take my physicist hat back and now it is actually my job again. So now we have the solution, and we need to determine what is actually these two unknown coefficients. So what I'm going to use is to use the two initial conditions. The first initial condition is x of 0 equal to x initial. The second one is that since I released this mass really carefully and the initial velocity is 0, therefore, I have x dot 0 equal to 0. From this, you can solve. Plug these two conditions into this equation. You can actually figure out that a is equal to x initial. And b is equal to 0. Any questions so far? Very good. So now we have the solution. So finally, what is actually the solution? The solution we get is x of t equal to x initial cosine omega t. So this is actually the amplitude of the oscillation, and this is actually the angular velocity. So you may be asking why angular? Where is the angular coming from? Because this is actually a one-dimensional motion. Where is the angular velocity coming from? And I will explain that in the later lecture. And also this is actually a harmonic oscillation. So what we are actually predicting is that this mass is going to do this, have a fixed amplitude and it's actually going to go back and forth with the angular frequency of omega. So we can now do an experiment to verify if this is actually really the case. So there's a small difference. There's another spring here, but essentially, the solution will be very similar. You may get this in a p-set or exam. So now I can turn on the air so that I make this surface frictionless. And you can see that now I actually move this thing slightly away from the equilibrium position, and I release that carefully. So you can see that really it's actually going back and forth harmonically. I can change the amplitude and see what will happen. The amplitude is becoming bigger, and you can see that the oscillation amplitude really depends on where you put that initially with respect to the equilibrium position. I can actually make a small amplitude oscillation also. Now you can see that now the amplitude is small but still oscillating back and forth. So that's very encouraging. Let's take another example, which I actually rotate the whole thing by 90 degrees. You are going to get a question about this system in your p-set. The amazing thing is that the solution is the same. What is that? And you don't believe me, let me do the experiment. I actually shifted the position. I changed the position, and I release that really carefully. You see that this mass is oscillating up and down. The amplitude did not change. The frequency did not change as a function of time. It really matched with the solution we found here. It's truly amazing. No? The problem is that we are so used to this already. You have seen this maybe 100 times before my lecture, so therefore, you got so used to this. Therefore, when I say, OK, I make a prediction. This is what happened, you are just so used to this or you don't feel the excitement. But for me, after I teach this class so many times, I still find this thing really amazing. Why is that? This means that actually, mathematics really works, first of all. That means we can use the same tool for the understanding of gravitational waves, for the prediction of the Higgs boson, for the calculation of the property of the quark-gluon plasma in the early universe, and also at the same time the motion of this spring-mass system. We actually use always the same tool, the mathematics, to understand this system. And nobody will understands why. If you understand why, please tell me. I would like to know. I will be very proud of you. Rene Descartes said once, "But in my opinion, all things in nature occur mathematically." Apparently, he's right. Albert Einstein also once said, "The most incomprehensible thing about the universe is that it is comprehensible." So I would say this is really something we need to appreciate the need to think about why this is the case. Any questions? So you may say, oh, come on. We just solved the problem of an ideal spring. Who cares? It's so simple, so easy, and you are making really a big thing out of this. But actually, what we have been solving is really much more than that. This equation is much more than just a spring-mass system. Actually, if you think about this question carefully, there's really no Hooke's law forever. Hooke's law will give you a potential proportional to x squared. And if you are so far away, you pull the spring so really hard, you can store the energy of the whole universe. Does that make sense? No. At some point, it should break down. So there's really no Hook's law. But there's also Hook's law everywhere. If you look at this system, it follows the harmonic oscillation. If you look at this system I perturb this, it goes back and forth. It's almost like everywhere. Why is this the case? I'm going to answer this question immediately. So let's take a look at an example. So if I consider a potential, this is an artificial potential, which you can find in Georgi's book, so v is equal to E times L over x plus x over L. And if you practice as a function of x, then basically you get this funny shape. It's not proportional to x squared. Therefore, you will see that, OK, the resulting motion for the particle in this potential, it's not going to be harmonic motion. But if I zoom in, zoom in, and zoom in and basically, you will see that if I am patient enough, I zoom in enough, you'll see that this is a parabola. Again, you follow Hooke's law. So that is actually really cool. So if I consider an arbitrary v of x, we can do a Taylor expansion to this potential. So basically v of x will be equal to v of 0 plus v prime 0 divided by 1 factorial times x plus v double prime 0 over 2 factorial x squared plus v triple prime 0 divided by 3 factorial x to the third plus infinite number of terms. v 0 is the position of where you have minimum potential. So that's actually where the equilibrium position is in my coordinate system. It's the standard, the coordinate system I used for the solving the spring-mass question. So if I calculate the force, the force, f of x, will be equal to minus d dx v of x. And that will be equal to minus v prime 0 minus v double prime 0 x minus 1 over 2 v triple prime 0 x squared plus many other terms. Since I have mentioned that v of 0-- this will be x. v of 0 is actually the position of the minima. Therefore, v prime of 0 will be equal to 0. Therefore. This term is gone. So what essentially is left over is the remaining terms here. Now, if I assume that x is very small, what is going to happen? Anybody know when x is very small, what is going to happen? Anybody have the answer? AUDIENCE: [INAUDIBLE]. YEN-JIE LEE: Exactly. So when x is very small, he said that the higher order terms all become negligible. OK? So that is essentially correct. So when x is very small, then I only need to consider the leading order term. But how small is the question. How small is small? Actually, what you can do is to take the ratio between these two terms. So if you take the ratio, then basically you would get a condition xv triple dot 0, which will be much smaller than v double prime 0. So that is essentially the condition which is required to satisfy it so that we actually can ignore all the higher-order terms. Then the whole question becomes f of x equal to minus v double prime 0 x. And that essentially, Hooke's law. So you can see that first of all, there's no Hooke's law in general. Secondly, Hook's law essentially applicable almost everywhere when you have a well-behaved potential and if you only perturb the system really slightly with very small amplitude, then it always works. So what I would like to say is that after we have done this exercise, you will see that, actually, we have solved all the possible systems, which have a well-behaved potential. It has a minima, and if I have the amplitude small enough, then the system is going to do simple harmonic oscillation. Any questions? No question, then we'll continue. So let's come back to this equation of motion. x double dot plus omega squared x, this is equal to 0. There are two important properties of this linear equation of motion. The first one is that if x1 of t and x2 of t are both solutions, then x12, which is the superposition of the first and second solution, is also a solution. The second thing, which is very interesting about this equation of motion, is that there's a time translation invariance. So this means that if x of t is a solution, then xt prime equal to xt plus a is also a solution. So that is really cool, because that means if I change t equal to 0, so I shift the 0-th time, the whole physics did not change. So this is actually because of the chain law. So if you have chain law dx t plus a dt, that is equal to d t plus a dt, dx t prime dt prime evaluated at t prime equal to t plus a. And that is equal to dx t prime dt, t prime equal to t plus a. So that means if I have changed the t equal to 0 to other place, the whole equation of motion is still the same. On the other hand, if the k, or say the potential, is time dependent, then that may break this symmetry. Any questions? So before we take a five minute break, I would like to discuss further about this point, this linear and nonlinear event. So you can see that the force is actually linearly dependent on x. But what will happen if I increase x more? Something will happen. That means the higher-ordered term should also be taken into account carefully. So that means the solution of this kind, x initial cosine omega t, will not work perfectly. In 8.03, we only consider the linear term most of the time. But actually, I would like to make sure that everybody can at this point, the higher-order contribution is actually visible in our daily life. So let me actually give you a concrete example. So here I have two pendulums. So I can now perturb this pendulum slightly. And you you'll see that it goes back and forth and following simple harmonic emotion. So if I have both things slightly oscillating with small amplitude, what is going to happen is that both pendulums reach maxima amplitude at the same time. You can see that very clearly. I don't need to do this carefully. You see that they always reach maxima at the same time when the amplitude is small. Why? That is because the higher-order terms are not important. So now let's do a experiment. And now I go crazy. I make the amplitude very large so that I break that approximation. So let's see what will happen. So now I do this then. I release at the same time and see what will happen. You see that originally they are in phase. They are reaching maxima at the same time. But if we are patient enough, you see that now? They are is oscillating, actually, at different frequencies. Originally, the solution, the omega, is really independent of the amplitude. So they should, actually, be isolating at the same frequency. But clearly you can see here, when you increase the amplitude, then you need to consider also the nonlinear effects. So any questions before we take a five-minute break. So if not, then we would take a five-minute break, and we come back at 25. So welcome back, everybody. So we will continue the discussion of this equation of motion, x double dot plus omega square x equal to 0. So there are three possible way to like the solution to this equation. So the first one as I mentioned before, x of t equal to a cosine omega t plus b sine omega t. So this is actually the functional form we have been using before. And we can actually also rewrite it in a different way. So x or t equal to capital A cosine omega t plus phi. You may say, wait a second. You just promised me that this is the first one, the one is the one and only one solution in the universe, which actually satisfy the equation of motion. Now you write another one. What is going on? Why? But actually, they are the same. This is actually A cosine phi cosine omega t minus A sine phi sine omega t. So the good thing is that A and phi are arbitrary constant so that it should be you can use two initial conditions to determine the arbitrary constant. So you can see that one and two are completely equivalent. So I hope that solves some of the questions because you really find it confusing why we have different presentations of the solution. So there's a third one, which is actually much more fancier. The third one is that I have x of t. This is actually a real part of A-- again, the amplitude-- exponential i omega t plus phi, where i is equal to the square root of minus 1. Wait a second. We will say, well, professor, why are you writing such a horrible solution? Right? Really strange. But that will explain you why. So three is actually a mathematical trick. I'm not going to prove anything here because I'm a physicist, but I would like to share with you what I think is going on. I think three is really just a mathematical trick from the math department. In principle, I can drive it an even more horrible way. x of t equal to a real part of A cosine omega t plus phi plus i f of t. And f of t is a real function. In principle, I can do that. It's even more horrible. Why is that? Because I now have this function. I take the real part, and I actually take the two out of this operation. So f of t is actually the real function. It can be something arbitrary. And i can now plot the locus of this function, the solution on the complex print. Now I'm plotting this solution on this complex print. What is going to happen is that you're going to have-- That is what you are going to get. If I am lucky, if this f of t is confined in some specific region, if I not lucky, then it goes out of the print there. I couldn't see it. Maybe it go to the moon or something. But if you are smart enough, and I'm sure you are, if I choose f of t equal to A sine omega t plus phi, can anybody tell me what is going to happen? AUDIENCE: [INAUDIBLE]. YEN-JIE LEE: Would you count a circle? Very good. If I plot the locus again of this function, the real axis, imaginary axis, then you should get a circle. Some miracle happened. If you choose the f of t correctly, wisely, then you can actually turn all this mess into order. Any questions? So I can now follow up about this. So now I have x of t is equal to the real part of A cosine omega t plus phi plus iA sine omega t plus phi. And just a reminder, exponential i theta is equal to cosine theta plus i sine theta. Therefore, I arrive this. This is a real part of A exponential i omega t plus phi. So if I do this really carefully, I look at this the position of the point at a specific time. So now time is equal to t. And this is the real axis, and this is the imaginary axis. So I have this circle here. So at time equal to t, what you are getting is that x is actually-- before taking the real part, A, exponential i omega t plus phi, it's actually here. And this vector actually shows the amplitude. Amplitude is A. And the angle between this vector pointing to the position of this function is omega t plus phi. So this is actually the angle between this vector and the real axis. So that's pretty cool. Why? Because now I understand why I call this omega angular velocity or angular frequency. Because the solution to the equation of motion, which we have actually derived before, is actually the real part of rotation in a complex print. If you think about it, that means now I see this particle going up and down. I see this particle going up and down. You can think about that, this is Earth. If there is an extra dimension mention, which you couldn't see. Actually, this particle in the dimension where we can see into the extra dimension, which is hidden is actually rotating. And while we see that reality, it's a projection to the real axis. You see? So in reality, this particle is actually rotating, if you add the image and the extra dimension. So that is actually pretty cool, but the purity artificial. So you can see that I can choose f of t to be a different function, and then this whole picture is different. But I also would create a lot of trouble because then the mathematics become complicated. I didn't gain anything. But by choosing this functional form, you actually write a very beautiful picture. Another thing, which is very cool about this is that if I write this thing in the exponential functional form, since we are dealing with differential equations, there is a very good property about exponential function. That is it is essentially a phoenix function. Do you know what is a phoenix? Phoenix is actually some kind of animal, a long-beaked bird, which is cyclically called the regenerated or reborn. So basically, when this phoenix die, you will lay the eggs in the fire and you were reborn. This is actually the same as this function. I can do differentiation, still an exponential function, and differentiate, differentiate, differentiate. Still exponential function. So that is very nice because when we deal with differential equation, then you can actually remove all those dots and make them become just exponential function. So essentially, a very nice property. So the first property, which is very nice is that it cannot be killed by differentiation. You will see how useful this is in the following lectures. The second thing, which is really nice is that it has a very nice property. So basically the exponential i theta 1 times exponential i theta 2, and that will give you exponential i theta 1 plus theta 2. So what does that mean? That means if I have a solution in this form, A exponential i omega t plus phi. And I do a times translation, t become t plus A. Then this become A exponential i omega t plus A plus phi. So this means that times translation in this rotation is just a rotation in complex print. You see? So now t becomes t plus A. Then you are actually just changing the angle between this vector and the x-axis. So as time goes on, what is going to happen is that this thing will go around and around and around and the physics is always the set, no matter when you start counting, and the translation is just the rotation in this print. Any questions? So I think this is actually a basic slide just to remind you about Euler's formula. So basically, the explanation i phi is equal to cosine phi plus i sine phi. And I think it will be useful if you are not familiar with this. It is useful to actually review a little bit about exponential function, which will be very useful for this class. So I'm running a bit faster today. So let's take a look at what we have learned today. We have analyzed the physics of a harmonic oscillator. So basically, we start by asking really just a verbal question, what is going to happen to this mass on the table attached to a spring. And what we have learned is that we actually use mathematics. Basically, we translate all what we have learned about this mass into mathematics by first define a coordinate system. Then I'd write everything using that coordinate system. Then I use Newton's law to help us to solve this question. And we have analyzed the physics of this harmonic oscillator. And Hooke's law, we found that he actually, not only works for this spring-mass system, it also works for all kinds of different small oscillations about a point of equilibrium. So basically, it's actually a universal solution what we have been doing. And we have found out a complex exponential function is actually a beautiful way to present the solution to the equation of motion we have been studying. So everything is nice and good. However, life is hard because there are many things which actually, we ignored in this example. One apparent thing, which we actually ignore, is the direct force. So you can see that before I was actually making this pendulum oscillate back and forth. What is happening now? There are not oscillating anymore. Why? Well, they stopped being. Apparently, something is missing. When I actually moved this system, if I turn off the air so that there's friction, then it doesn't really move. If I increase a bit, the air so that the slide have some slight freedom, then actually, you can see that you move a bit then you stop. If I increase this some more, you can see that the amplitude becomes smaller and smaller. So in the following lecture, what we are going to do is to study how to actually include a direct force into it again and of course, using the same machinery which we have learned from here and see if we can actually solve this problem. Thank you very much. We actually end up earlier today. Sorry for that. And maybe I will make the lecture longer next time. And if you have any questions about what we have covered today, I'm here available to help you.
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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: Where we'll spend the first part of today is finishing up where we left off last time with the experiments that were done to take a look at what types of polypeptides, the DNA, kDNA, J chaperone machinery interact with in E. coli. And once we finish up that, we'll transition into module 3, which is protein degradation. And most of today will just be some general background about proteases and protein degradation. And then on Wednesday, we'll begin to look at the macromolecular machines that are involved in those processes. OK. So last time we discussed DnaK, the chaperone, DnaJ, the co-chaperone. Right. So recall DnaJ went around and found some polypeptide in a non-native state and delivered it to DnaK, which can grip and hold on to the hydrophobic segments and somehow facilitate folding. And so where we left off were with experiments performed in a similar manner to what we saw for GroEL/GroES where pulse chase was done to label newly synthesized polypeptides. Right. So during the pulse period with this radio label methionine, newly synthesized polypeptides are labeled. And that allows us to see specifically what newly synthesized polypeptides DnaK, J are interacting with without the background of everything else in the cell. And why do we care about that? Imagine if we didn't somehow label to discriminate these newly synthesized polypeptides. Right. We can pull down many things in the precipitation, but we'd have no sense as to how long a given polypeptide existed in the cell. So maybe it was newly synthesized. Maybe it had been around a long time and something happened to it such that it wasn't in a native fold and DnaK interacted with it. OK. So that's the key point with these pulse chase and this labeling for a short time period. OK. So the researchers had an antibody to DnaK. They had to test its specificity as we discussed for GroEL/GroES. And then after immunoprecipitation, it's necessary to do the analysis. And so as I noted last time, in this particular study, the analysis were less sophisticated than what we saw for GroEL/GroES so they just used one dimensional SDF page, which we're all pretty familiar with. And they didn't extend it to mass spec. But with that said, there are a number of observations that are helpful that come from this study. So what we're going to do is examine their gels and see what conclusions we can come up with. OK. So this is their experiment number one. And what they did was look at the soluble crude cell extracts that were generated in this pulse chase experiment. OK. And so what do we see in terms of how the data is presented? Right. We have two lanes on the left, one and two, that are basically total cytoplasmic proteins. And then the lanes three through six on the right are four samples that were immunoprecipitated with this anti-DnaK antibody. OK. So something that was done in these experiments that was different than the GroEL/GroES work is that they use two different E. coli strains. So they used a wild type E. Coli strain. So that strain expresses DnaK. And they also used a mutant E. coli that is deficient in DnaK. So that's what this delta DnaK means. So they did some genetic manipulation and knocked out DnaK. OK. And as we learned in the introduction, these chaperone system is not essential. So what do we see if we work through this gel? And first, we just want to go over what the data show and then why that's important here. So if we compare lanes one and two, what do lanes one and two tell us? So these are the total cell lysates, soluble fraction, from either wild type E. Coli or delta DnaK. And why do we care to run these? So no immunoprecipitation. So I'll give you a start and then you all can contribute to the next ones. OK? So what I would say looking at these two lanes is first it looks like the total amount of protein and the distribution of these proteins is similar for both the wild type E. Coli and the delta DnaK knockout. OK. There's proteins that are distributed across a wide range of molecular weights, from below 14 kilodaltons to upwards of 100. OK. And why is this important to show? One, we want to see what the cell lysate looks like in the absence of immunoprecipitation. Right? And two, it's important to know whether or not knocking out DnaK has done anything to change those cells. And at least at the level here, it looks like in terms of the total cellular polypeptide pool, it's pretty much comparable in terms of total protein. So what happens now here where we have this immunoprecipitation? So we have four different lanes. We're going to focus on three, four, and five and ignore six. Basically, in three, we see immunoprecipitation from the wild type E. coli; in four, immunoprecipitation with wild type when the sample has been treated with SDS, so we need to think about why that was done and what that experiment shows; and then lane five in the delta DnaK knockout. So first of all, what does line three tell us? Kenny? AUDIENCE: Can you get enrichment of higher molecular weight proteins? And as the note says that 15 times more protein was loaded into this well, so I think that's just so you can see the signal. But I think that just shows to prove that the higher molecular weight proteins are more enriched in that lane. ELIZABETH NOLAN: Yeah. So are many proteins immunoprecipitated? AUDIENCE: Yes. ELIZABETH NOLAN: Yes. Right. We see many bands across a range of molecular weights. Right. And then as Kenny said, we're seeing much more intensity up here than down here. Right. So it looks like polypeptides in the range of about 20 to about 60 kilodaltons are enriched. So maybe there's some preference in that size range here. So that's a good observation. What do we see in lane four? What happened for this sample? So effectively, the crude cell lysate, those extracts were treated with SDS. AUDIENCE: They didn't bind anything? ELIZABETH NOLAN: Yeah. So nothing bound. Right. What do we see here? Just one band for DnaK. So why didn't DnaK bind anything? AUDIENCE: It's denatured by the SDS. ELIZABETH NOLAN: Yeah. Not the page... which is the sodium dodecyl sulfate. Right. It's a denaturant and it will denature things. They don't need to be in a gel. If you add that to a sample, you'll have denaturation, right? So when these samples were denatured, DnaK didn't bind. Do we think that the SDS denatured DnaK itself? Right. So we do see one band here. Is that surprising? AUDIENCE: No, because if the DnaK line is suggesting that that's the molecular weight of of DnaK, it's around there. And the antibody you're using to do immunoprecipitation... I mean, would likely probably sub the other bind. ELIZABETH NOLAN: Yeah. So the antibody was still able to bind. Right. What kind of gel is this? Well, it's SDS page, but how is it being monitored? AUDIENCE: It's a [INAUDIBLE] so it's a biodome radiogram. ELIZABETH NOLAN: Yeah. We're looking at radioactivity too so just to keep that in mind for the ban. Right. That's what's allowing us to see. What about lane five? AUDIENCE: It shows that in the delta DnaK line that nothing's pulled down by the immunoprecipitation. ELIZABETH NOLAN: Right. So no DnaK. Nothing's pulled down. This is a very helpful control, because imagine if you did see bands, that would indicate that there's some lack of selectivity with this immunoprecipitation step. So I think that's quite a nice experiment they added into this piece of work here. OK. So moving on to their next experiment, what happens during the chase if we look at different time points during the chase period? So this is similar again to what was done with the GroEL/GroES study. So what we're looking at is a one DSDS page. We have the molecular weight marker. All of these are with the immunoprecipitation. And we're looking at times from below one minute up to 10 minutes. So the question is what do we see in these data here. AUDIENCE: As time goes on, you see more and more concentrations of the smaller proteins. ELIZABETH NOLAN: Yes. So we're seeing fewer proteins as time passes. Right. Let's start with the first time point here. Does that look to be in pretty good agreement to what we saw on the prior slide? Right. We see that there's a number of polypeptides that DnaK is interacting with. Right. And they're over a range of molecular weights. Right. And then exactly as we just heard, as time progresses, what we see is, overall, there's fewer polypeptides. But it looks like there's fewer polypeptides of lower molecular weight here. So what does this suggest if you're going to interpret the data? AUDIENCE: Probably that lower molecular rate peptides are folded more quickly. ELIZABETH NOLAN: Yeah. So maybe the folding there is complete over this time or it's complete to a point that DnaK isn't needed anymore. Right here. What do you think about the DnaK band? AUDIENCE: Quite constant. AUDIENCE: Yeah. ELIZABETH NOLAN: Quite constant. So does that make sense? Yeah, why does that make sense? AUDIENCE: Because it's just the whole quantity is not changing. And it's not involving [INAUDIBLE].. Just like in the case of the less [INAUDIBLE] so when [INAUDIBLE]. ELIZABETH NOLAN: Yeah, right? So there was some newly-- what this indicates, right? There was some newly synthesized DnaK in those 15 seconds of the pulse. Right? And that has stuck around. And that's all precipitated to the same degree in each sample here. So what about this data here? And then, how helpful is this data? Right? So effectively, what was done is, radioactivity was measured by liquid scintillation counting. OK, so they measured the total radioactivity in each sample prior to separation. And then, they've converted that to some arbitrary scale of proteins bound to DnaK. Right? So we see that, over time, the total radioactivity decreases and effectively comes to some sort of plateau. So that's just some nice quantitation in terms of what we see here in the gel, right? It's quite easy to measure the total radioactivity in a sample. And you get a measure of that from liquid scintillation counting. And then, you can look at the gel, right? And they're in good agreement there. And as I said, this is completely arbitrary, what's on the y-axis. So do these experiments tell us much about the specifics of DnaK function? So we see polypeptides being bound. We see over time that fewer are bound. What's actually happening in the cell? AUDIENCE: I don't think we can conclude much from these. Only that we know that it interacts with the polypeptides for some amount of time. ELIZABETH NOLAN: Yeah. Right. I agree. So is it acting as a foldase? Is it acting as an holdase-- an un-foldase? That's not clear from the data presented in these experiments. And I'd say overall, there are studies that show different things, depending on the system there, for this. So that's where we're going to close with the chaperone systems. And where we're going to move into-- actually, one more comment, right, before closing on the chaperones. What happens if certain ones are deleted? So just to reiterate, not all of these systems are required for cell viability of E. coli. It's only GroEL/GroES. Right? So you might ask if trigger factor or DnaK or DnaJ is deleted, what happens in the cell to keep things functioning properly? And just one observation. If trigger factor is deleted, OK, there's no growth phenotype. Is that surprising? Right? That observation may depend on growth conditions. But say you're in some standard growth conditions. What's observed is that, in the absence of trigger factor, DnaK and J can basically compensate for that loss of function. OK? But then, if trigger factor and DnaK are deleted, at higher temperatures, that becomes lethal here. OK? The cells can't cope for that. But at lower temperature, GroEL/GroES can compensate for that loss of function. OK. So we're on to protein degradation. There's some incredible macromolecular machines involved in this unit here. And we'll move on to that one come Wednesday. Just if we think about where we're going with lifecycle of a protein, right, we've gone from synthesis to folding. We've learned that misfolding can occur. OK? And at some point these polypeptides, whether they're folded or unfolded, need to be degraded. So they have some lifetime in the cell. OK? And so we can think about proteases, so classical enzymes, like trypsin. And we can think about proteasomes, which are degradation chambers. And these players are really important because they have a role, big picture, in controlling the dynamics and lifetimes of all proteins and cells. So what are some of our questions for this module? Why are proteins degraded? We just said a little bit about that. How are proteins degraded? And what types of proteases exist? We'll briefly today touch upon the general catalytic mechanism because that's important to have this background for thinking about the degradation chambers. So what are the general mechanisms and what are the active site machineries? Protease inhibitors are really important at the lab bench, and they also have a big role in therapeutics. And so we'll talk about those a bit here. And then moving into protein degradation machines, we're going to look at ClpXP from E. coli as a case study. So we need to think about, what are the structures of these degradation machines, what are the mechanisms? How do they differ in prokaryotes and eukaryotes? So after spring break, Joanne will spend some time talking about the eukaryotic proteasome there. We won't talk about it as much immediately here, but we'll come back to that later. And how are proteins that are destined for degradation by a proteasome tagged to get to that destination? So here are our topics. An overview, which is where we'll focus today. And then, looking at ClpXP, and down the road, the 26S proteasome. So first, thinking about proteases. Some general points to get everyone up to speed. So because we all know proteases catalyze the hydrolysis of peptide bonds. So we can just think of some peptide and that peptide bond gets hydrolyzed to give us these products here. Why do we need a protease? The bottom line is just that spontaneous hydrolysis of peptide bonds is very slow, right? So we can leave a protein or a polypeptide on the bench top. And maybe it will unfold. Maybe it will precipitate. But it's not going to have the peptide bonds being broken unless something else has been done to it, right? So we can think about a half life on the order of seven years. And so proteases give tremendous rate accelerations on the order of 10 to the ninth. And we can just think about chemistry for a minute and what we might do as a chemist to hydrolyze a peptide bond. So hydrolysis is pH dependent. And so in chemistry we'll use acid or base to hydrolyze a peptide bond. And we can think about base catalyzed reactions, such as this one, where we have our OH-minus group attacking, or acid-catalyzed reactions, as this one here. OK? So effectively we can just think about pH dependence of hydrolysis. Just if we have rate and we have pH. Something on the order of this, right? Where we have enhancements at low and high pH and a relative minimum at neutral pH here for that. And so these types of chemistry is going to come up in the context of the protease enzymes, depending on the type, as we'll see in a few slides. So we can think about proteases as being irreversible biological switches, that these reactions are irreversible. And what does this mean from the standpoint of the cell? It means that the cell needs some way to handle and deal with these proteases, right, such that they don't cause unnecessary hydrolysis of polypeptides. That would be very deleterious to the cell, right, if a protease was running rampant and hydrolyzing proteins that it shouldn't here. So what are some strategies that the cell can use? One, cells are quite good at controlling protease activity, both in terms of space and time. And there's a variety of different strategies, depending on the locale and the protease. So regulation is really key here. And some examples are provided here. So one is that proteases will be stored as zymogens or inactivated precursors. And there'll have to be some event that activates this zymogen to give the active protease. Proteases can be stored in separate organelles here. So these might be zymogen granules or lysosomes. And sometimes they're stored with a protease inhibitor, as well. And another strategy, which is really the strategy we're going to focus on as we move forward in this module, is that degradation chambers are used, such that you have this huge macromolecular machine where all of the protease activity is in the inside. And what this means is that somehow a condemned protein that needs to be degraded by this machine needs to be tagged. And there needs to be some mechanism to get it in the inside of the chamber. So effectively, degradation will limit access of the active sites to the rest of the cellular environment. So that's what we see in ClpXP and this 26S proteasome. Just to note-- so just the other week in C&E News, there is a highlight of a pretty exciting paper. So I noted that proteases are of interest and important from therapeutic development. And here's a little excerpt about a molecule shown here that's found to hit the proteasome of malaria parasite. And so hopefully, by the end of this unit, if you go back and read this, you'll have some sense as to why is this a good inhibitor of the proteasome or a protease. And what's going on in terms of the proteasome machinery here. And how can we differentiate proteasomes from different organisms. Back to some of the strategies. Just an example is zymogen activation, and thinking a little bit from the perspective of the organism. So here, we can think about the gut. We're in the small intestine. So there's the epithelium, the cells, these are crypts. And here's the lumen, so the space where the food goes through, et cetera. What do we see? So inside the intestine, there's a protease named entarokinase. And it has a role of activating trypsinogen. So trypsinogen is a zymogen. It's produced by the pancreas. And the pancreas delivers trypsinogen and other things into the small intestine. And so once it reaches the small intestine where its activity is needed, it will be activated by the action of entarokinase to give trypsin. OK? And then what can happen? Trypsin can also activate trypsinogen, and it will also activate chymotrypsinogen to give chymotrypsin. Right? So the net result here is protease activity in the intestinal lumen, which is the extracellular space here. And so they travel from the pancreas in a form that's inactive and then become active in the intestinal lumen there. So as I said before, proteases are important. And if we think about this role in controlling dynamics and lifetimes of proteins and cells, what are some of those roles? And I guess I also point out this also-- they also can exist in the extracellular space. So if we think about homeostasis and how proteases can regulate homeostasis, just some examples. They can remove misfolded proteins or aggregated proteins. They can provide amino acids when needed, right? So after destruction of a polypeptide, you have small fragments or amino acid monomers. And they can modulate many cellular functions. So just some examples. And this is to show the broad range. We can think about blood clotting, the generation of hormones, just digestion and recycling of amino acids. So energy harvesting, the cell cycle, control of the cell cycle, and even cell deaths. So thinking about apoptosis here. And if we just select two of these cellular functions and how proteases play a role, what I have here is the maturation of insulin, a peptide hormone in the blood coagulation cascade. OK, so if we take a look, insulin is a really terrific molecule. And if you're looking from some trivia not shown here, it also binds zinc and forms an interesting oligomer. So if you're interested in metals, that's a good one. But what do we see? We see that insulin is synthesized as a prepropeptide. And so in blue, we have a signal sequence. And then we have these chains here. And look, there's a bunch of cysteines, right? So there's action of a protease. And what do we see? The signal sequence is cleaved and at some point in this process, there's formation of disulfide bonds, right, in some regiospecific manner. So this is pro insulin. And then what happens? There's another protease cleavage event that gives us the mature form of insulin. This grape chain here is removed. OK? So this is an example of a hormone being stored as an inactive precursor. And actually, there's many peptides that are stored as inactive precursors. And then some protease has to come and cleave a pro region. So in my group, we're interested in a family of antibacterial peptides called defensins that are in the intestine. And they have a pro region. And it's trypsin or another protease that comes along and has a cleavage event to release the active peptide. So not only limited to insulin here. If we look at the blood coagulation cascade, we can imagine that we don't want blood to coagulate on whim, right? That'd be a huge problem. So proteases are required to allow coagulation to occur. And what we can see here is that prothrombin is converted to thrombin by a protease. And thrombin is a serine protease. And we'll hear more about serine proteases in a little bit. That converts fibrinogen to fibrin. And as a result, coagulation occurs. And that's important for wounds. And there's many, many other examples. So if we think about types of proteases and mechanisms of catalysis, what I just would like you all to be aware of is that there's two general mechanisms. And we can think about four different mechanistic varieties within that. And so we can divide these up by proteases that are involved in covalent catalysis. So there's formation of a covalent acylenzyme intermediate. And this is what we'll see for serine proteases, cysteine proteases, and threonine proteases. So examples here are quite relevant to this module as ClpXP is a serine protease. And as you'll see later on, the eukaryotic proteasome is an end terminal threonine protease. The second general type are proteases that accelerate the direct attack of water on the substrate. OK, so this is non-covalent catalysis. And the types here are aspartyl proteases and zinc proteases. So there was a question a few lectures ago if there's metal-dependent proteases. And the answer is yes, zinc proteases. And we can also think about these from the standpoint of the acid and base catalyzed chemistry we saw before. So just for some trivia. If we think about the human proteome-- 533 proteases. And this is a count here. So on the order of 200 serine proteases, 140 cysteine, around 190 metalloproteases, and 21 aspartyl proteases. So we have many of these enzymes to act at different places and points. If we take a look at the active site machinery, what do we see? So here we have the serine proteases. They have a catalytic triad comprised of aspartate, a histidine and a serine here. Cysteine proteases-- we see a cysteine and a histidine. And so these are the ones involved in covalent catalysis. Here we have the non-covalent catalysis. So the aspartic acid or aspartyl protease. We have two asp residues. And here we have an example of a zinc protease, where we see a single zinc ion coordinated by two histidines, and in this case, a glutamate and a bound water. OK? So if we think about just the covalent versus non-covalent catalysis here. So when I get further along. So imagine we just have some dipeptide. What we find in these enzymes is that they have what's called an oxyanion hole here. And we can think about the enzyme allowing attack as such. So that or some nucleophile here. So what do we get? We get a covalent acylenzyme intermediate. OK, we have the oxyanion hole. And these are the serine and the cysteine proteases here. OK? And we'll go through in more detail the mechanism in a minute. If we think about non-covalent catalysis, and again, we have our dipeptide. We can just think about for a minute one of the metalloproteases, right? So in these cases, the protease is accelerating the direct attack by water. So I imagine we have some metal here that has water bound, right? What happens? Imagine we can de-proteinate the water molecule. And then there can be attack. OK, so why does the metalloprotease allow this to occur? So what's happening when the water binds to the metal that will facilitate this? So we can think about the pKa of a water molecule, right? And what happens when a water molecule is bound to a metal, right? Say zinc. So how do we think about a metal? AUDIENCE: A Lewis acid. ELIZABETH NOLAN: Yeah. Right? We have a general Lewis acid here. Here. So what's going to be effect of the pKa of the bound water relative to unbound water? AUDIENCE: It'll be more acidic. ELIZABETH NOLAN: Right. We're going to lower the pKa of the bound water, which is going to help generate the nucleophile. Right? So that's how it's facilitating the direct attack here. OK, so what we're going to do is look at the serine protease example in a bit more detail. AUDIENCE: Are the end termini de-pertinated? Or is it-- ELIZABETH NOLAN: I am just-- what would it be at physiological pH? AUDIENCE: NH3? ELIZABETH NOLAN: The embryote NH3 plus. Right. So these are just showing a simple dipeptide that we have NH3 plus and O minus in terms of the acid ends there. OK. So just thinking about here, this covalent catalysis. So here's each protease type, the active site, and the nucleophile. So in the case of the serine proteases, the nucleophile is the serine side chain. And I'm showing this because ClpXP-- ClpP protease-- uses serine protease chemistry here. So what we observe in this overview is a generally accepted mechanism. And we see formation and collapse of this covalent acylenzyme intermediate. So if we take a look here, we have a bound polypeptide. This is the oxyanion hole provided by these two NH. Here we see the aspartate, the histidine, and the serine. Right? So what do we see happening here? First, there's formation of a tetrahedral intermediate. So there's an attack. OK? And here we have loss of the RNH2. And here what do we see? We see, basically, the histidine working on this water molecule. We have collapse of this acylenzyme intermediate, another tetrahedral intermediate, and release of the acid product here. OK? So in thinking about this, we think about the histidine as being a general acid general base involved in general acid-base catalysis, a proton carrier. We see this oxyanion hole providing stabilization here and here. Right? We have this negative charge. And something that you need to think about are the pKas. If we think about just pKas of amino acids and how this chemistry is happening. Right? So what is a little bit mysterious here, based on our knowledge of pKas of the catalytic triad? And they give some approximate values just for serine, histidine, and aspartate here. AUDIENCE: Well, each proton abstraction is being done by something that should, theoretically, have a lower pKa. So you have an aspartate abstracting a proton from a histidine, which is abstracting a proton from serine. So there has to be a lot of perturbation of the system for that to happen. ELIZABETH NOLAN: Yeah. Right. There needs to be a lot of perturbation to pKas for this to work, right? How easy is it to de-proteinate the serine by a typical histidine? Is that going to happen based on pKa? AUDIENCE: No. ELIZABETH NOLAN: Right? So there's something about this active site and the environment that's going to give perturbation of these values. If we just move beyond this cartoon form for a minute, and just look at the catalytic triad from chymotrypsin from a crystal structure. This is the orientation of the serine histidine and aspartate. And something to keep in mind is that different serine proteases, or different proteases in general, have different substrate specificity's, which means they prefer to cut before or after a given type of amino acid, depending on the side chain. And this is just a cartoon depiction indicating that, here's the peptide, here's some side chain, and there's some recognition site here. So there's a degree of substrate discrimination. For instance, trypsin likes to cut after arginine and lysine. But it will cut at other places, as well. Right? Chymotrypsin likes aromatic hydrophobic residues. Elastase likes small and uncharged residues here for that. So you may have seen diagrams or cartoons of specificity pockets for thinking about substrate discrimination amongst these proteases. And I guess what I would just say is, it's not so simple as those types of cartoons. If we look at the structures of serine proteases, just to compare, what we see is that, for trypsin, elastase, and chymotrypsin, they have similar overall structure. So this is an overlay of the three enzymes. And the catalytic triad is shown in red. OK? So despite this similar overall structure, they have distinct substrate preferences. And it's just something to be aware of. So you're not responsible for the origins of this substrate discrimination. And here's just a view showing more about the secondary structure of trypsin shown here. And in this case, there's an inhibitor bound. And I'll just note, in terms of the substrate preference, and these are the three types of activity we'll end up seeing within the eukaryotic proteasome. So if we just have some polypeptide. OK? OK, and so imagine we're thinking about hydrolysis here. OK, so the C terminal end of this amino acid with our one side chain. We think about the enzyme and the identity of R1. For trypsin, it prefers to cut after arginine or lysine. OK, so a positive charge. For chymotrypsin, phenylalanine, tyrosine, and also other ones like valine, leucine, and isoleucine. OK, so aromatic plus hydrophobic. And then elastase here. And we find that elastase prefers to cut after small. And if I'm boxing it, these are the ones we think is most preferred, small and uncharged residues. OK? So some discrimination based on side chain identity. So where we'll close the general background is just with a note on proteases and disease and protease inhibition. So if we consider various human diseases and proteases, there's many, many links. And many proteases are implicated in a variety of diseases and pathologies. And so this is a table just to give you a sense of the breadth. What we see in terms of the class is that all of the classes are represented here. And that we see diseases ranging from cardiovascular problems, to cancer, et cetera, cystic fibrosis, inflammation here. OK? So as a result, there is quite a bit of interest in terms of the possibility of protease inhibitors as therapeutics. And beyond that, they're also widely used in the lab. So how do these inhibitors work? Just as a general rule of thumb to think about, generally they react to form a covalent bond with the catalytic nucleophile. So for instance, for the serine proteases, they'll form some covalent bond with the active site serine residue. And we can classify these inhibitors as being either reversible inhibitors or irreversible inhibitors. So as those names indicate, if it's a reversible inhibitor, that covalent linkage between the protease or the proteasome and the inhibitor can be broken down. And types of reversible inhibitors, for instance, use aldehydes as the reactive group. So in contrast, the irreversible inhibitors form a covalent linkage that is not readily broken down with the catalytic nucleophile. And so irreversible inhibitors include vinylsulfones. And if you go back and look at that little excerpt from C&E News about this molecule that's inhibiting the proteasome of malaria, you'll see that it has a vinylsulfone on its terminus here. Epoxides are also employed. OK, and generally, if we have inhibitors that block the function of a protease or a proteasome, they're going to interfere with many critical cellular functions right here. And just in terms of cancer, just some observations. So it's been found that proliferating cells are sensitive to proteasome inhibitors. And there's some proteasome inhibitors that can selectively induce apoptosis in proliferating cells. And so cancer cells are proliferating, and there's interest in the use of these as anti-cancer drugs. So what I've included in these slides are some examples of inhibitors of each class, and then the mechanisms. Here are just three molecules. So we have either reversible or irreversible inhibitors, right? And what is there to note looking at these molecules? They're all polypeptide-like. Right? So there's amino acids or moieties that, with a little imagination, we can think about as being somewhat similar. And then we see these reactive groups on the terminus. So the aldehyde, for instance, the vinylsulfone. And so if you look at the structure of this molecule being used to inhibit the malaria proteasome, there's some clear similarities to these here. In terms of mechanisms, we can think about these reversible inhibitors. So for instance, the chemistry with the peptide aldehyde. Here we're seeing the nucleophile of the eukaryotic proteasome, which is really interesting because it's an end terminal threonine. That's why we're seeing it drawn as such here. So we can have formation and collapse of this species here. Or in the case of the irreversible inhibitors, we have the vinylsulfone and the chemistry that happens here for that. And so if you're interested in these, I encourage you to look at the mechanisms a bit more. And we'll see a little bit more on inhibitors being used experimentally as we go through the rest of this module. So where we'll start on Wednesday is looking at the structure of E. coli ClpXP, which is a degradation machine used to degrade certain condemned proteins there. OK?
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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. JOHN ESSIGMANN: We're still on storyboard 7. We're on panel C. Panel C is where I introduced the TCA cycle. As I mentioned earlier, respiration consists of three stages. The first is the pyruvate dehydrogenase reaction, which we just covered, taking pyruvate to acetyl-CoA. The second is the TCA cycle, or tricarboxylic acid cycle taking that acetyl-CoA and basically oxidizing it in order to generate CO2 but also to produce more reducing equivalents in the form of mobile electron carriers. And then the third step of respiration is taking those reducing equivalents to the mitochondrial inner membrane, where the molecules containing those reducing equivalents are oxidized. And then, the electrons from that oxidation reaction are used to power proton pumps that ultimately will generate the proton gradient that can be used for generation of ATP or for generation of movement or for other things. The tricarboxylic acid cycle, which is sometimes called the Krebs cycle, takes acetyl-CoA from several different sources. One of those sources is glycolysis to pyruvate and pyruvate to acetyl-CoA, as we've just seen. And the second source of acetyl-CoA is from fatty acid oxidation. We'll come to that pathway somewhat later. Again, looking at panel C, the TCA cycle starts with the reaction of acetyl coenzyme A, a 2-carbon compound with oxaloacetate, a 4-carbon compound, to form the 6-carbon product citrate. Citrate will lose 2 carbons as carbon dioxide, and in the process, there'll be a series of oxidation steps that generate three NADHes one FADH2, and one either GTP or ATP by substrate-level phosphorylation. In terms of the banking system of the cell, the NADHes that are generated in the mitochondrion are exchangeable for about three ATPs. FADH2 is exchangeable for about two ATPs. So if you look up the total number of nucleotide triphosphate, or NTP, equivalents that can be produced in the TCA cycle, you'll get about 12 ATPs for each 2-carbon unit of acetyl-CoA that's oxidized. As a detailed point, I want to mention that the two carbons that entered the TCA cycle as acetyl-CoA are not exactly the same two carbons that come out as CO2 in that cycle. The carbon dioxides from the input acetyl-CoA will emerge in later turns of the TCA cycle. Now we're going to look at panel D. In panel D, we start looking at the details of the TCA cycle. JoAnne explained why nature uses thioesters. The sulfur allows enolization stabilization of a carbanion at carbon 2, the carbon that's distal to the coenzyme A functionality. The carbanion is then able to attack the number 2 carbon, carbonyl, of oxaloacetate in the reaction catalyzed by citrate synthase. An intermediate is formed, citroyl coenzyme A, which loses its coenzyme A moiety by hydrolysis in a very thermodynamically irreversible step, resulting in the product citrate. This step is at the top of the pathway, and as is usually the case, this highly exothermic step makes the pathway, overall, irreversible. Chemically, citrate synthase does a mixed aldol-Claisen ester condensation. The product of the citrate synthase reaction, citrate, is a tertiary alcohol, and tertiary alcohols are relatively difficult to oxidize. The next enzyme in the pathway, aconitase, which is shown in storyboard 8, panel A, removes a water molecule and then adds a different water molecule back to rearrange the hydroxyl group, making a secondary alcohol, which is much easier to oxidize. Looking again at panel A, the hydroxyl group of high isocitrate is oxidized to a ketone, with the transfer of hydride to NAD+ to make NADH. This reaction is catalyzed by the enzyme isocitrate dehydrogenase, ICDH. The intermediate in this reaction is oxalosuccinate, which is a beta-keto acid. As we know from what JoAnne taught us, beta-keto acids are prone to spontaneous decarboxylation. So the second half of the isocitrate dehydrogenase reaction involves the loss of carbon dioxide in what's usually considered to be an irreversible step. I do want to point out, however, that under certain circumstances, you can re-add the carbon dioxide in order to make the reaction go in the other direction. After the ICDH reactions, which generated NADH and resulted in loss of CO2, the product is alpha-ketoglutarate, a 5-carbon keto acid. Take a look at the structure of alpha-ketoglutarate in the upper right-hand portion of storyboard 8, panel A. If you hold your finger over the top 2 carbons of alpha-ketoglutarate, you'll notice that the residue is pyruvate. So pyruvate plus an acetyl functionality equals, effectively, alpha-ketoglutarate. Now take a look back at the mechanism by which pyruvate is oxidized by pyruvate dehydrogenase. It's going to be a very similar mechanism for the oxidation of alpha-ketoglutarate. The product of the alpha-ketoglutarate dehydrogenase reaction is succinyl-CoA. Again, if you look at the structure of this molecule, succinyl-CoA, it's actually an acetyl-CoA with an acetyl group put onto one end. Now let's go back and look at the alpha-ketoglutarate from a different perspective. I want to make a point here in that alpha-ketoglutarate is an alpha-keto acid, and if we were to replace its keto group with an amino group, you'd convert this keto acid into the amino acid glutamic acid. As with most enzymatic reactions involving such nitrogen functionalities, a pyridoxal pyridoxamine phosphate cofactor will be needed to interconvert the alpha-ketoglutarate and glutamic acid. So glutamic acid can serve as a source of alpha-ketoglutarate if the cell is starved for TCA cycle intermediates. Alternatively, alpha-ketoglutarate can be a source for glutamic acid when a cell may need amino acids for protein biosynthesis. Now, let's turn to panel B, where we'll start with succinyl coenzyme A. Looking at the molecule of succinyl-CoA, note that I've labeled each of the atoms with a symbol. The triangle and square were from the original acetyl-CoA molecule that came in at the top of the pathway. The enzyme that processes succinyl-CoA is succinyl coenzyme A synthetase. As JoAnne taught us, synthetases are enzymes that typically need a nucleotide. In this case, the nucleotide involved is GDP in mammalian systems or ADP in bacterial systems. In the traditional clockwise direction of the TCA cycle, they are phosphorylated to form GTP or ATP, respectively. The molecule that you get after the phosphorylation reaction in the hydrolysis of the coenzyme A is succinate, a 4-carbon compound, which is also a dicarboxylic acid. This molecule is perfectly symmetrical and can tumble in three-dimensional space. The next enzyme in the pathway, succinate dehydrogenase, cannot distinguish one arm of its substrate, succinate, from the other. So if there were a radio label in the acetyl-CoA at the beginning of the pathway, that radio label would become scrambled at this point, uniformly distributed among the two carbons of the two arms. From this point onward in the TCA cycle, you will note that the label denoted as the triangle and box is scrambled, as indicated by triangle divided by 2 or box divided by 2. The next enzyme in the pathway is succinate dehydrogenase. This is the only membrane-bound enzyme in the TCA cycle. It is a dehydrogenase, and it uses flavin as a cofactor to help remove electrons from the succinate substrate. Flavin picks up electrons from succinate, converting FAD to FADH2 in the mitochondrial membrane. The product of the reaction is the alkene fumarate. The enzyme fumarase adds water to fumarate to form the alcohol product malate. The hydroxyl group of the alcohol malate is primed for oxidation by the next enzyme in the pathway, malate dehydrogenase, or MDH. MDH oxidizes malate, which is an alcohol, to a ketone. The ketone product is oxaloacetate. The hydride removed from malate is transferred to NAD+ to form NADH. As I mentioned, the product of the overall reaction is oxaloacetate, and its ketone functionality is now primed for attack by the next molecule of acetyl-CoA entering the TCA cycle. As a final point, I've mentioned several times that oxaloacetate is present at a very low concentration, only in the micromolar range, inside the mitochondrion of a mammalian cell. So it's always at rather limiting concentration. The cell has to work very hard to preserve enough of the oxaloacetate to enable the next cycle of the TCA cycle, that is the acquisition of the next acetyl coenzyme A group. One of the ways that the cell can generate oxaloacetate is by deamination, or transamination of aspartic acid to the keto acid oxaloacetate. We typically have plenty of aspartic acid and this PLP-mediated reaction helps to maintain a critical level of oxaloacetate. I want to return to storyboard 7 to make some comments about the importance of prochirality in some enzymatic reactions. This short interlude will help explain how to track a radio label in a TCA cycle intermediate as that intermediate progresses through the TCA cycle. As you will note, at first glance, the label does some unexpected things. But at the end of the day, the fact that the label does surprising things helped early biochemists figure out mechanistically how several enzymes work in concert during the linear steps of a pathway. We're going to look at storyboards 7, 8, and 9, starting with the chemical reaction at the bottom right of storyboard 7, panel D. This is the chemical reaction that's catalyzed by citrate synthase. You'll notice that I've highlighted the two carbons with either a triangle or a box. As we have seen, the nucleophile on acetyl-CoA attacks the electropositive carbon of the carbonyl functionality of oxaloacetate. The carbonyl functionality is a flat sp2 hybridized center. So if this were a typical organic chemical reaction, the electrophile could come in from either the top or the bottom, and you'd get two different stereochemistries in the product. That is the citrate at the very bottom would have equally labeled acetyl arms at the top and bottom of the molecule as I've drawn it. You would have a delta divided by 2 for the blue methylene group, and a box or square divided by 2 for the red carboxylate group in each of the two acetyl arms. Again, these arms are at the bottom and top of the molecule as drawn in the lower right of panel D. You'll note, however, that only the top acetyl group of citrate has the labels. Initially, the observation that only the top arm acquired label was a puzzle to early biochemists. One way to think about the citrate synthase reaction is to think about the oxaloacetate laying on the surface of the citrate synthase enzyme. Now imagine that the enzyme precludes, or blocks, access to the carbonyl from the bottom and allows access only to the top, giving rise to only one stereochemical outcome, the one that I've shown in the citrate to the right. Now move ahead to the storyboard number 9, panel B. This panel shows a more cartoon-like representation of the molecule of citrate. You can see the acetyl arm on the top, the pro-S arm, as having the labels. And the pro-R arm, the one that came from oxaloacetate, at the bottom, is label free. So while the pro-R and pro-S arms are chemically identical, they're going to be handled by the next enzyme in the series, aconitase, as being chemically different from one another. All biochemistry is going to be occurring on the pro-R arm, that is the arm that came in from oxaloacetate and not the arm that came in for acid 2 with acetyl coenzyme A. In the bottom right of panel B, I've sketched out an imaginary active site for the aconitase enzyme. I show a base picking up a proton from the pro-R arm of the citrate molecule. And you can see the elimination of the water molecule from the 3 carbon. So despite the fact that the citrate molecule is chemically symmetrical, aconitase, the next enzyme in the reaction series, is able to distinguish between the pro-R and the pro-S arms. At this point, I want you to look back at storyboard 8, panel A. Look once again at the citrate that is at the upper left-hand corner of storyboard number 8, and let's imagine that the aconitase chemistry has happened on the pro-S arm, that is the one in the box. Keep in mind that these experiments have shown that this does not happen. This is just a hypothetical scenario. In this hypothetical case, the hydroxyl group would end up on the number 2 carbon, the one with the blue triangle. You have to draw it out, but if you traced this molecule, in which the chemistry happened on the pro-S arm, all the way around to alpha-ketoglutarate, you would find out that the alpha-ketoglutarate dehydrogenase would liberate CO2 from the carboxylate that has the red box on it. This is in contrast to the molecule succinate that appears on storyboard 8, panel B. Succinate is another symmetrical molecule, but for some reason, succinate dehydrogenase cannot distinguish between the two acetyl arms. Because the enzyme is unable to distinguish the two arms, in this case, unlike that of aconitase, the radio label would become scrambled. The reason that I'm belaboring this point is because one of the ways that biochemists work out the chemical reactions involved in a biochemical pathway is by putting in some kind of labeled molecule. It could be a radio label or a heavy isotope. And then they trace the position of the label in the molecule as you move from molecule to molecule along the pathway. So label tracer studies are ones that are absolutely central to all of biochemistry. And as I mentioned earlier, you'll get lots of experience in the problem sets in using labels to work out the details of a pathway. Before leaving the TCA cycle, there are a couple of big picture points that I want to make. You put in two carbons as acetyl-CoA, deposit them into oxaloacetate to form citrate, a 6-carbon compound, and then in the cycle, you lose 2 carbons as carbon dioxide. That means that there's no loss or gain of carbon in this cycle. If you had just one molecule of oxaloacetate, you'd be able to complete the TCA cycle. What happens, however, when the cell is in an, let's say, "energy needed" situation, where it needs to buff up this cycle, that is increase the number of molecules cycling to be able to accommodate the processing of more and more molecules of acetyl-CoA? Those molecules of acetyl-CoA might flood in from carbohydrate metabolism or, as we'll see later, from catabolism of lipids. Let's look at panel A of storyboard 11. As shown in this figure, one way to accomplish increasing the carbon content of the TCA cycle is to take an amino acid, such as glutamate, and remove its amino group to form alpha-ketoglutarate. Note that if you increase the concentration of any one molecule in the TCA cycle, for example, alpha-ketoglutarate, you're effectively increasing the concentrations of all molecules in the cycle. Because it's a cycle, many of the molecules are in equilibrium with one another. Going clockwise around the TCA cycle from alpha-ketoglutarate, we come to succinyl-CoA. Succinyl-CoA is an entry point into the TCA cycle from odd-chain fatty acids in certain amino acids, such as methionine. So these molecules can give rise to succinyl-CoA that itself will then increase the concentration of all molecules in the TCA cycle. Our third primary input point is at oxaloacetate. It involves aspartic acid being deaminated into oxaloacetate. This is a very common way to increase the amount of oxaloacetate available for the TCA cycle. This reaction can dramatically increase the rate of processing of molecules by the TCA cycle. Finally, there's an enzyme we'll look at later called pyruvate carboxylase, or PC, which can take pyruvate in the mitochondrion, add CO2 to it, and form oxaloacetate. We're going to come to this enzyme later when we talk about carboxylase enzymes as a class. This is an important enzyme in the pathway of gluconeogenesis. So overall, I think you can see that there are several ways that a cell can increase the overall concentration of the intermediates of the TCA cycle. Increasing any one intermediate increases all of them. And that will increase the rate by which acetyl-CoA molecules can be processed ultimately to generate energy. The general word describing this buffing up of the TCA cycle is anapleurosis, which comes from the Greek word "filling up."
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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, WKB approximation, or semiclassical approximation. So this is work due to three people-- Wentzel, Kramers, and Brillouin-- in that incredible year, 1926, where so much of quantum mechanics was figured out. As it turns with many of these discoveries, once the discoveries were made, people figured out that somebody did them before. And that person was a mathematician, Jeffreys, who did it three years earlier, in 1923. The work had not become very popular. So some people write JWKB. But we will not do that. We'll note Jeffreys, but we'll follow this more standard notation. So these people were dealing with differential equations with slowly-varying spatial coefficients. That was the main thing. So we're continuing our approximation methods. We've done perturbation theory. We will add little pieces to the Hamiltonian. Now we consider things that are slowly varying in space. You might have a very simple Hamiltonian where nothing varies in space-- a constant potential. But as soon as the potential starts varying slowly, you have approximation methods. Those are the methods we're considering now. We will also consider, after we've finished WKB, time-dependent perturbation theory, where still things start slowly varying in time. So we'll have many, many things to do still. So this is called, also, the semiclassical approximation. Because classical physics gives you intuition about the quantum wave function. So it is a lesson in which you want to learn something about the quantum wave function, and you learn it by using classical physics. A quantity that is relevant here is that the Broglie wavelength of a particle. This is the Broglie. And many people say that semiclassical approximation has to do with the fact that the quantum mechanic effects are not that important. That may happen. The Broglie wavelength is much smaller than the physically relevant sizes of your apparatus. So you have a particle like an electron. And if the Broglie wavelength is very small compared to the aperture in the screen, the letter will go almost like a classical particle. When the Broglie wavelength is comparable with the size of the aperture in the screen, you will get diffraction effects, and the electrons will do quantum mechanical things. So the semiclassical approximation has to do with lambda being smaller than the length scale, L, of your physical problem. We will refine this. In fact, the whole search of understanding the semiclassical approximation is all about understanding this better. Because it's a little subtle. We will end up deciding that what you need is that the Broglie wavelength, suitably generalized, varies very slowly. We'll have to generalize the concept of the Broglie wavelength. We might get to it today. Mathematically, you can say, OK, I want this to be sufficiently small. So semiclassical limit was just take h going to 0. You should complain, of course. h is a constant of nature. I cannot take it equal to 0. But on the other hand, they could imagine other universes, maybe, where h has different smaller and smaller values in which quantum mechanical effects don't set in until much smaller scales. But at the end of the day, I will try to consider h to be small as an idea underlying a semiclassical approximation. And the intuition is that for h small-- we cannot tune it, but we can say it-- h small lambda becomes small. The lambda, the Broglie, dB. You're taking a quantity with units-- h bar is units. And saying it's small. It goes against lots of things. Things with units are not supposed to be small. You should compare it with the situation we had before. We had a very nice and clean situation with perturbation theory, where we had a unit-free thing that we consider it to be small. This time, we're going to try to consider h to be small. And it's going to be more delicate because it has units. It's going to be a more complicated story. The physics is interesting, and that's why this approximation is harder in some ways to understand than the ones we've done in perturbation theory. So how does this begin? It begins by thinking of a particle in a potential, V of x. And the particle has some energy, E. And then, if it's classical, as we're imagining now, it could be a three-dimensional potential. My sketch of course is just for one-dimensional. This is E. And you can solve for p squared, 2m, E minus V. This is a notion of local momentum because it depends on x. It's the momentum the particle would have. When it is at some position, x, you will have momentum, p, of x. And now, nobody forbids you from declaring that you're going to define a position-dependent Broglie wavelength, which is going to be h over p of x is your definition, which is equal to 2 pi h bar over p of x. And it's going to be local the Broglie. You see that the Broglie wavelength, when you first started in 804, was considered for a free particle-- always the [INAUDIBLE]. You have a free particle with some momentum it has at the Broglie wavelength. Why was the Broglie wavelength important? Because when you write the wave function, it's a wave with wavelength equal to the Broglie wavelength. The wave function with the Broglie wavelength solves the Schrodinger equation. That's sort of how it all came about. But it was all defined for a free particle. The Broglie defined it for a free particle with some momentum, p. And it all made sense, because you could write a wave function using the Broglie momentum. It was in fact E to the ipx over h bar. That was your wave function. But here, you have a particle moving with varying momenta. And we don't know how to write the solution of the Schrodinger equation, but there's classically this concept, and we could define the local, the Broglie wavelength, and we will have to discover what it means or how it shows up. But it probably shows up in some way. So there's two cases that probably we should consider. But before that, we looked at the Schrodinger equation-- the time-independent Schrodinger equation. So it's minus h squared over 2m Laplacian of psi of x is equal to E minus v of x, psi of x. All these vectors-- that's why I'm Laplacian, not second derivatives. But the right-hand side-- well, if I put the 2m to the other side, I get minus h squared Laplacian of psi of x, is equal to 2m E minus vx. That's p squared of x, psi of x. That's the local momentum squared. Maybe it's a curiosity, but it's now nice that the left-hand side is the momentum operator squared-- so it's p hat squared on the wave function-- is equal to p squared of x times a wave function. Kind of a nice result. Nice-looking Schrodinger equation. I don't know if you've seen it like that before. It's an operator on the left. And on the right, almost like an eigenvalue. It's an illegal eigenvalue. If it were a real eigenvalue problem, there should be a number, not a function here. But it's a function that acts a little like an eigenvalue. It's a nice way of thinking of the Schrodinger equation in the semiclassical approximation. Well, no. No approximation here so far in the semiclassical language, in which you call this the local momentum. And thus defined, it certainly is an exact statement-- no approximation whatsoever. OK, now I want to know these two circumstances of course. If you have E greater than v, you're in an allowed region. And p squared is really 2m E minus v of x. And it's convenient to define h squared k squared of x, the wave number. You remember that p equal hk was good notation. We use the wave number sometimes. We'll have here a local wave number as well. And if you're in the forbidden region, which energy is less than v-- forbidden region-- then minus p squared is positive, is 2m v minus E. That is positive this time, because v is greater than E. And that, we always used to call the penetration constant, the kappa, for wave functions that decay exponentially. So we call this h squared kappa squared of x. A local decaying factor wave number. It can be thought as an imaginary wave number. But that's notation. So, so far, so good. We will need one more piece of intuition as we work with this semiclassical approximation. We have not done any approximation yet. The approximation will come soon, as we will begin solving the Schrodinger equation under those circumstances. And we will take h bar as an expansion parameter-- will be fine and correct, but a little subtle. So the thing that will help us many times to understand these things is to write the wave function as a complex number. So suppose you have a wave function. We will write it as a complex number in the polar form-- radius times face. So the radius is going to be rho of x. The probability density, I claim, is this. And there is a phase that I will write as E to the i over h bar, S of x and t. It looks a little like an action. And it's, to some degree, the beginning of the path into rule formulation. It has lots of connections with the action principle. So here, rho and s are going to be real. So this is truly scale factor in front that determines the value. This is a pure phase, because it's an imaginary number times a real thing. S has units of h bar, or angular momentum. And indeed, psi squared is equal to rho of x and t. If you loop psi squared. The reason we focus on this wave function is that our solutions of WKB are going to have exactly that form. So we need to have an intuition as to what the observables of this wave function are. So the other observable is the current density. If you remember, it's h bar over m, imaginary part of psi star gradient psi. So this must be calculated. So what is this? Gradient of psi-- we must take the gradient of this. Gradient's a derivative, so it acts on one, acts on the other. When it acts on the first, it first acts as the relative 1 over 2 squared of rho of x and t, times the gradient of rho, times the phase factor, plus now I have to take the gradient on this quantity, and this will bring down an i, an h bar, the gradient of S, and then multiplied by the whole wave function, because this factor remains and the exponential remains. Now we can multiply by psi star to form what we need to get for the current. So psi star [INAUDIBLE] psi. If I multiply by that, I'm multiplying by the top line. For the first factor, I get 1/2 gradient of rho. The exponentials cancel. And for the second part, we get plus i over h bar, gradient of S, times psi squared, which is rho. The imaginary part of this is equal 1 over h bar, rho gradient of s. So finally, the current, which is h bar over m, times that imaginary part, is rho gradient of s, over m. A very nice formula. Basically, it says that the phase factor in the wave function determines the probability current. And it also says that if you want to think of this, here are the surfaces of constant phase. Here is our space. And S constant. So here is one valley of the phase, another valley of the phase. Those are surfaces in space of constant phase. The current is orthogonal to that. So the current is proportional to the gradient. The gradient of a function is always proportional. It's a normal vector to the surfaces of constant values. So the current is orthogonal to the surfaces of constant phase. If you have a fluid mechanics interpretation, J is rho v in fluids. So, so far, everything I've said could have been said in 804. These are properties of a general wave function. This is how you compute the current. The useful thing is that our WKB wave functions are going to be presented in that language. So if you think of the analogy with fluid mechanics. The current is the charge density times the velocity, and therefore the velocity would be identified with gradient of S over m. Or the momentum would be identified with gradient of S. That's not a quantum mechanical rigorous identification. Because gradient of S is a function. And therefore, p there would be a function. And a function of momentum-- momentum in quantum mechanics is an operator, and it has eigenvalues, which are numbers. They're not functions. But we already have seen the beginning of some momentum function. So that analogy is actually quite nice. Let me give you an example, and conclude with that. If you have a free particle, you have a wave function psi of x and t, which is E, to the ipx over h bar, times minus iEt over h bar. So in this case, rho is equal to 1, an S-- remember, S is read by having an i over an h bar out. And that gives you p dot x minus Et. So for a free particle, the gradient of S is indeed just the momentum. You take this gradient, and it's that. And therefore, that's a rigorous interpretation when you have a free particle, that the gradient of S is going to be the momentum. Interestingly, the derivative of S with respect to time is minus the energy. What will happen in the semiclassical approximation is that this S over there will depend on x, and this p will depend on x, and it will be this p that depends on x here. And we will see how to solve this equation in an approximation scheme where the changes are a little slow. And the notation S here is also motivated because actions in classical mechanics actually have this property. The gradient of the action-- if you think of the action as a function of coordinates, which is something you don't usually do, but if you do, in somewhat advanced classical mechanics, you see that the derivatives of the action-- spatial derivatives are the momentum, and the time derivatives are the energy. It's a nice relation between classical mechanics, and justifies once more the name of semiclassical approximation, which we will continue to develop next time. See you then.
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https://ocw.mit.edu/courses/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2013/8.333-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. PROFESSOR: So we've been wondering how a gas, such as the one in this room, with particles following Newtonian equations of motion, comes to equilibrium. We decided to explain that by relying on the Boltzmann equation. Essentially, we said, let's look at the density, or probability, that we will find particles with momentum p at location q at time t, and we found that within some approximations, we could represent the evolution of this through a linear operator on the left-hand side equal to some collision second order operator on the right-hand side. The linear operator is a bunch of derivatives. There is the time derivative, then the coordinate moves according to the velocity, which is P over m. And summation over the index alpha running from 1 to 3, or x, y, z, is assumed. And at this stage, there is symmetry within the representation in terms of coordinates and momenta. There is also a force that causes changes of momentum. Actually, for most of today, we are going to be interested in something like the gas in this room far away from the walls, where essentially there is no external potential. And for the interest of thinking about modes such as sound, et cetera, we can remove that. The second order operator-- the collision operator-- however, explicitly because we were thinking about collisions among particles that are taking place when they are in close contact, breaks the symmetry between coordinate and momenta that were present here. So this collision operator was an integral over the momentum of another particle that would come at some impact parameter with relative velocity, and then you had a subtraction due to collisions, and then addition due to reverse collisions. Now, what we said is because of the way that this symmetry is broken, we can distinguish between averages in position and in momentum. For example, typically we are interested in variations of various quantities in space. And so what we can do is we can define the density, let's say at some particular point, by integrating over momentum. So again, I said I don't really need too much about the dependence on momentum, but I really am interested in how things vary from position to position. Once we have defined density, we could define various averages, where we would multiply this integral by some function of P and q, and the average was defined in this fashion. What we found was that really what happens through this collision operator-- that typically is much more important, as far as the inverse time scales are concerned, than the operators on the left, it has a much bigger magnitude-- is that momenta are very rapidly exchanged and randomized. And the things that are randomized most slowly are quantities that are conserved in collision. And so we focused on quantities that were collision conserved, and we found that for each one of these quantities, we could write down some kind of a hydrodynamic equation. In particular, if we looked at number conservation-- two particles come in, two particles go out-- we found that the equation that described that was that the time derivative of the density plus moving along the streamline had a special form. And here, I defined the quantity u alpha, which is simply the average of P alpha over m, defined in the way that averages defined over here. And that this is the number of particles, how it changes if I move along the streamline And variations of this, if you are incompressible, comes from the divergence of your flow velocity. And actually, this operator we call the total, or streamline, derivative. Now, the next thing that we said is that OK, number is conserved, but momentum is also conserved. So we can see what equation I get if I look at momentum. But in fact, we put the quantity that was momentum divided by mass to make it like a velocity, and how much you deviate from the average that we just calculated. And this quantity we also call c. And when we looked at the equation that corresponded to the conservation of this quantity in collisions, we found that we had something like mass times acceleration along the streamline. So basically this operator multiplied by mass acting on this was external force. Well, currently we've said this external force to be zero-- so when we are inside the box. Here there was an additional force that came from variations of pressure in the gas. And so we had a term here that was 1 over n d alpha of P alpha beta, and we needed to define a pressure tensor P alpha beta, which was nm expectation line of c alpha c. Finally, in collisions there is another quantity that is conserved, which is the kinetic energy. So here the third quantity could be the kinetic energy-- or actually, we chose the combination mc squared over 2, which is the additional kinetic energy on top of this. And it's very easy to check that if kinetic energy is conserved, this quantity is also conserved. We call the average of this quantity in the way that we have defined above as epsilon. And then the hydrodynamic equation is that as you move along the streams-- so you have this derivative acting on this quantity epsilon-- what we are going to get is something like minus 1 over n d alpha of a new vector, h alpha. h alpha, basically, tells me how this quantity is transported. So all I need to do is to have something like mc squared over 2 transported along direction alpha. And then there was another term, which was P alpha beta-- the P alpha beta that we defined above-- times u alpha beta. U alpha beta was simply the derivative of this quantity, symmetrized. So the statement is that something like a gas-- or any other fluid, in fact-- we can describe through these quantities that are varying from one location to another location. There is something like a density, how dense it is at this location. How fast particles are streaming from one location to another location. And the energy content that is ultimately related to something like the temperature, how hot it is locally. So you have these equations. Solving these equations, presumably, is equivalent, in some sense, to solving the Boltzmann equation. The Boltzmann equation, we know, ultimately reaches an equilibrium, so we should be able to figure out how the system, such as the gas in this room, is if disturbed, comes to equilibrium, if we follow the density, velocity, and temperature. Now, the problem with the equations as I have written is that they are not closed in terms of these three quantities, because I need to evaluate the pressure, I need to evaluate the heat transfer vector. And to calculate these quantities, I need to be able to evaluate these averages. In order to evaluate these averages, I need to know f. So how did we proceed? We said well, let's try to find approximate solutions for f. So the next task is to find maybe some f, which is a function of p, q, and t, and we can substitute over there. Now, the first thing that we said was, OK, maybe what I can do is I can look at the equation itself. Notice that this part of the equation is order of 1 over the time it takes for particles to move in the gas and find another particle to collide with, whereas the left-hand side is presumably something that is related to how far I go before I see some variation due to the external box. And we are really thinking about cases where the gas particles are not that dilute, in the sense that along the way to go from one side of the room to another side of the room, you encounter many, many collisions. So the term on the right-hand side is much larger. If that is the case, we said that maybe it's justifiable to just solve this equation on the right-hand side as a zeroth order. And to solve that, I really have to set, for example, the right-hand side that I'm integrating here, to 0. And I know how to do that. If log f involves collision conserved quantities, then ff before is the same as ff after. And the solution that I get by doing that has the form of exponential involving conserved quantities, which are the quantities that I have indicated over here-- let's say, such as c. And so log of that would be something that involves-- I can write as mc squared over 2 with some coefficient that, in principle, varies from location to another location. I want to integrate this and come up with the density, so I put the density out here, and I normalize the Gaussian. And so this is a reasonable solution. Indeed, this is the zeroth order solution for f-- I'll call that f0. So once you have the zeroth order solution, from that you can calculate these two quantities. For example, because the zeroth order solution is even in c, the heat vector will be 0, because it is symmetric in the different components. The pressure tensor will be proportional to delta alpha beta. OK? We started with that. Put those over here, and we found that we could get some results that were interesting. For example, we could see that the gas can have sound modes. We could calculate the speed of sound. But these sound modes were not damped. And there were other modes, such as sheer modes, that existed forever, confounding our expectation that these equations should eventually come to an equilibrium. So we said, OK, this was a good attempt. But what was not good enough to give us complete equilibrium, so let's try to find a better solution. So how did we find the better solution? We said that the better solution-- let's assume that the solution is like this, but is slightly changed by a correction. And the correction comes because of the effect of the left-hand side, which we had ignored so far. And since the left-hand side is smaller than the right-hand side by a factor involving tau x, presumably the correction will involve this tau x. OK. Good? We said that in order to see that as a correction, what I need to do is to essentially linearize this expression. So what we did was we replaced this f's with f0's 1 plus g, 1 plus g, and so forth. The zeroth term, by construction, is 0. And so if we ignore terms that are order of g squared, we get something that is linear in g. OK? Now, it's still an integral operator, but we said let's approximate that integration and let's do a linearized one collision time approximation. What that approximation amounted to was that, independent of what this deviation is, if we are relaxed to the zeroth order solution over time scale that is the same, and that times scale we'll call tau x. So essentially we wrote this as minus f0, essentially g, which is the difference between f and f0, divided by tau x. f0, this was g. No. I guess we don't have this. We wrote it in this fashion. It's just writing the way that I wrote before. We wrote this as g over tau x, where g is the correction that I have to write here. And you can see that the correction is obtained by multiplying minus tau x with l acting on f0 divided by f0, which is l acting on log of f0. So I would have here 1 minus tau x-- let's make this curly bracket-- and then in the bracket over here, I have to put the action of l on the log of f0. So I have to-- this is my f0. I take its log. So the log will have minus mc squared over 2 kt and the log of this combination, and I do d by dt plus P alpha over m acting on this log, and then a bunch of algebra will leave you to the following answer. Not surprisingly, you're going to get factors of mkT. So you will get m over kT. You get c alpha c beta minus delta alpha beta c squared over 3 acting on this rate of strength tensor that we have defined over here. And then there are derivatives that will act on temperature-- because temperature is allowed to vary from position to position-- so there will be a term that will involve the derivative of temperature. In fact, it will come in the form over T c alpha multiplying another combination which is mc squared over 2kT minus 5/2. Let me see if I got all of the factors of 1/2 correct. Yeah. OK. So this is the first order term. Presumably, there will be high order corrections, but this is the improved solution to the Boltzmann equation beyond the zeroth order approximation. It's a solution that involves both sides of the equation now. We relied heavily on the right-hand side to calculate f0, and we used the left-hand side-- through this log l acting on log of f0-- to get the correction that is order of tau x that is coming to this equation. So now, with this improved solution, we can go back and re-check some of the conclusions that we had before. So let's, for example, start by calculating this pressure tensor P alpha beta, and see how it was made different. So what I need to do is to calculate this average. How do I calculate that average? I essentially multiply this f, as I have indicated over there, by c alpha c beta, and then I integrate over all momenta. Essentially, I have to do integrations with this Gaussian weight. So when I do the average of c alpha c beta with this Gaussian weight, not surprisingly, I will get the delta alpha beta, and I will get from here kT over m. Multiplying by nm will give me mkT. So this is the diagonal form, where the diagonal elements are our familiar pressures. So that was the zeroth order term. Essentially, that's the 1 over here, multiplying c alpha c beta before I integrate. So that's the 1 in this bracket. But there will be order of tau x corrections, because I will have to multiply two more factors of c with the c's that I have over here. Now, none of these terms are important, because these are, again, odd terms. So when I multiply two c's with three or one c the average will be zero. So all of the averages are really coming from these terms. Now, these terms involve four factors of c's. Right? There's the two c's that I put out here-- and actually, I really have to think of these as different indices, let's say nu nu, nu nu. Summation convention again assumed. And then when I multiply by c alpha c beta, I will have, essentially, four indices to play with-- c alpha c beta, c nu c nu. But it's all done with the Gaussian weight out here. And we showed and discussed how there was this nice fixed theorem that enabled you to rapidly calculate these Gaussian weights with four factors of c, or more factors of c-- doesn't really matter. And so in principle you know how to do that, and I'll skip the corresponding algebra and write the answer. It is proportional to minus tau x. And what it gives you, once you do these calculations, is a factor of u alpha beta minus delta alpha beta over 3 nu gamma gamma. And again, let me make sure I did not miss out anything. And apparently I missed out the factor of 2. So really, the only thing that's happened, once we went and included this correction, we added this term. But the nice thing about this term is that potentially, it has off-diagonal terms. This matrix initially was completely diagonal. The corrections that we have calculated are potentially off-diagonal coming from this term that corrected the original Gaussian weight. So where is that useful? Well, one of the problems that we discussed was that I can imagine a configuration of velocities, let's say close to a wall-- but it does not have to be a wall, but something where I have a profile of velocities which exist with components only along the x direction, but vary along the y direction. So this ux is different from this ux, because they correspond to different y's. So essentially, this is a pattern of sheering a gas, if you like. And the question is, well, hopefully, this will come to relax and eventually give us, let's say, uniform velocity, or zero velocity, even better-- if there's a wall, and the wall velocity is zero. So how does that happen? Well, the equation that I have to satisfy is this one that involves u. So I have m. Well, what do I have to have? I have du by dt plus something that acts on ux. ux only has variations along the y direction. So this term, if it was there, had to involve a derivative along the y direction multiplying uy, but it's not present, because there's no uy. On the right-hand side of the equation, I have to put minus 1 over n. Again, the only variations that I'm allowed to have are along the y direction. So I have, when I do the summation over alpha, the only alpha that contributes is y. And so I need the pressure y, but I'm looking for velocities along the x direction, so the other index better be x. OK? So the time course of the velocity profile that I set up over here is determined by the y derivative of the yx component of the pressures tensor. Now, previously our problem was that we stopped at the zeroth order, and at the zeroth order, the pressure tensor was diagonal. It didn't have a yx component. So this profile would stay forever. But now we do have a yx component, and so what do I get? I will get here minus 1 over n dy dy. The yx component will come from minus 2 tau x multiplying nkT and then multiplying-- well, this term, again, is diagonal. I can forget about that term. So it comes from the uxy. What is uxy? It is 1/2 of the x derivative of uy that doesn't exist, and the y derivative of ux that does exist. OK? So what we have, once we divide by m, is that the time derivative of ux is given by a bunch of coefficients. The n I can cancel if it does not vary. What I have is the 2's cancel. The n's cancel. I will have tau x kT over m. tau x kT over m, that's fine. And then the second derivative along the y direction of ux. So suddenly, I have a different equation. Rather than having the time derivative does not change, I find that the time derivative of ux is proportional to Laplacian. It's a diffusion equation. And we know the solution to the diffusion equation, how it looks qualitative if I have a profile such as this. Because of the fusion, eventually it will become more and more uniform in time. And the characteristic time over which it does so, if I assume that, let's say, in the y direction, I have a pattern that has some characteristic size lambda, then the characteristic relaxation time for diffusion will be proportional to lambda squared. There's a proportionality here. The constant of proportionality is simply this diffusion coefficient. So it is inversely. So it is m kT tau x. Actually, we want to think about it further. kT over m is roughly the square of the terminal velocities of the particles. So lambda squared divided by v squared is roughly the time that you would have ballistically traveled over this line scale of the variation. The square of that time has to be provided by the characteristic collision time, and that tells you the time scale over which this kind of relaxation occurs. Yes? AUDIENCE: So if x depends linearly, why? Would still get 0 on the-- why can't-- PROFESSOR: OK. So what you are setting up is a variation where there is some kind of a sheer velocity that exists forever. So indeed, this kind of pattern will persist, unless you say that there's actually another wall at the other end. So then you will have some kind of a pattern that you would relax. So unless you're willing to send this all the way to infinity, is will eventually relax. Yes? AUDIENCE: Can you just say again the last term on the second line, the 1/2 term. Where did that come from? Can you just explain that? PROFESSOR: This term? AUDIENCE: Yeah. The last half of it. PROFESSOR: The last half of it. So what did we have here? So what I have is that over here, in calculating the pressure, when I'm looking at the xy component, I better find some element here that is off diagonal. What's the off diagonal element here? It is u xy What is u xy? u xy is this, calculated for. So it is 1/2 of dx uy, which is what I don't have, because my uy is 0. And the other half or it, of symmetrization, is 1/2 the value of x. Yes? AUDIENCE: [INAUDIBLE] derivative [INAUDIBLE] for ux. You're starting out the second term uy uy. Also, shouldn't there be a term ux dx? PROFESSOR: Yes. Yeah. So this is a-- yes. There is a term that is ux dx, but there is no variation along the x direction. I said that I set up a configuration where the only non-zero derivatives are along the y direction. But in general, yes. You are right. This is like a divergence. It has three terms, but the way that I set it up, only one term is non-zero. AUDIENCE: Also, if you have [INAUDIBLE] one layer moving fast at some point, the other layer moving slower than that. And you potentially can create some kind of curl? But as far as I understand, this would be an even higher order effect? Like turbulence. PROFESSOR: You said two things. One of them was you started saying viscosity. And indeed, what we've calculated here, this thing is the coefficient of viscosity. So this is really the viscosity of the material. So actually, once I include this term, I have the full the Navier-Stokes equations with viscosity. So all of the vortices, et cetera, should also be present and discussed, however way you want to do it with Navier-Stokes equation into this equations. AUDIENCE: So components of speed which are not just x components, but other components which are initially zero, will change because of this, right? PROFESSOR: Yes, that's right. That's right. AUDIENCE: You just haven't introduced them in the equation? PROFESSOR: Yes. So this is I'm looking at the initial profile in principle. I think here I have set up something that because of symmetry will always maintain this equation. But if you put a little bit bump, if you make some perturbation, you will certainly generate other components of velocity. OK? And so this resolved one of the modes that was not relaxing. There was another mode that we were looking at where I set up a situation where temperature and density were changing across the system, but their products-- that is, the pressure-- was uniform. So you had a system that was always there in the zeroth order, despite having different temperatures at two different points. Well, that was partly because the heat transport vector was zero. Now, if I want to calculate this with this more complicated equation, well, what I need-- the problem before with the zeroth order was that I had three factors of c, and that was odd. But now, I have a term in the equation this is also odd. So from here, I will get a term, and I will in fact find eventually that the flow of heat is proportional to gradient of temperature. And one can compute this coefficient, again, in terms of mass, density, et cetera, just like we calculated this coefficient over here. This will give you relaxation. We can also look at the sound modes including this, and you find that the wave equation that I had before for the sound modes will also get a second derivative term, and that will lead to damping of the modes of sound. So everything at this level, now we have some way of seeing how the gas will eventually come to equilibrium. And given some knowledge of rough parameters of the gas, like-- and most importantly-- what's the time between collisions, we can compute the typical relaxation time, and the relaxation manner of the gas. So we are now going to change directions, and forget about time dependents. So if you have questions about this section, now may be a good time. Yes? AUDIENCE: [INAUDIBLE] what is lambda? PROFESSOR: OK. So I assume that what I have to do is to solve the equation for some initial condition. Let's imagine that that initial condition, let's say, is a periodic pattern of some wavelength lambda. Or it could be any other shape that has some characteristic dimension. The important thing is that the diffusion constant has units of length squared over time. So eventually, you'll find that all times are proportional to some lengths where times the inverse of the diffusion constant. And so you have to look at your initial system that you want to relax, identify the longest length scale that is involved, and then your relaxation time would be roughly of that order. OK. So now we get to the fourth section of our course, that eventually has to do with statistical mechanics. And at the very, very first lecture, I wrote the definition for you that I will write again, that statistical mechanics is a probabilistic approach to equilibrium-- also microscopic-- properties of large numbers of degrees of freedom. So what we did to thermodynamics was to identify what equilibrium microscopic properties are. They are things such as identifying the energy, volume, number of particles of the gas. There could be other things, such as temperature, pressure, number, or other collections of variables that are independently sufficient to describe a macrostate. And what we're going to do is to indicate that macroscopic set of parameters that thermodynamically characterize the equilibrium system by big M. OK? Clearly, for a probabilistic approach, you want to know something about the probability. And we saw that probabilities involving large numbers-- and actually various things involving large numbers-- had some simplified character that we are going to exploit. And lastly, in the last section, we have been thinking about microscopic description. So these large number of degrees of freedom we said identify some point, let's say, for particles in a six-n dimensional phase space. So this would for gas particles be this collection p and q. And that these quantities we know are in fact subject to dynamics that is governed by some Hamiltonian. And we saw that if we sort of look at the entirety of the probability in this six n dimensional phase space, that it is subject to Liouville's question that said that dp by dt is a Poisson bracket of H and p. And if the only thing that we take from equilibrium is that things should not change as a function of time, then requiring this probability in phase space to be independent of time would then require us to have a p which is a function of H, which is defined in p and q and potentially other conserved quantities. So what we are going to do in statistical mechanics is to forget about how things eventually reach equilibrium. We spent a lot of time and energy thinking about how a gas reaches equilibrium. Having established what it requires to devise that solution in one particular case, and one of few cases where you can actually get far, you're going to ignore that. We say that somehow my system reached this state that is time independent and is equilibrium, and therefore the probability should somehow have this character. And it depends, however, on what choice I make for the macrostate. So what I need here-- this was a probability of a microstate. So what I want to do is to have a statement about the probability of a microstate given some specification of the macrostate that I'm interested in. So that's the task. And you're going to do that first in the ensemble-- and I'll tell you again what ensemble means shortly. That's called microcanonical. And if you recall, when we were constructing our approach to thermodynamics, one of the first things that we did was we said, there's a whole bunch of things that we don't know, so let's imagine that our box is as simple as possible. So the system that we are looking is completely isolated from the rest of the universe. It's a box. There's lots of things in the box. But the box has no contact with the rest of the universe. So that was the case where essentially there was no heat, no work that was going into the system and was being exchanged, and so clearly this system has a constant energy. And so you can certainly prescribe a particular energy content to whatever is in the box. And that's the chief identity of the microcanonical ensemble. It's basically a collection of boxes that represent the same equilibrium. So if there is essentially a gas, the volume is fixed, so that there would be no work that will be done. The number of particles is fixed, so that there is no chemical work that is being done. So essentially, in general, all of the quantities that we identified with displacements are held fixed, as well as the energy in this ensemble. So the microcanonical ensemble would be essentially E, x, and N would be the quantities that parametrize the equilibrium state. Of course, there's a whole collection of different microstates that would correspond to the same macrostate here. So presumably, this bunch of particles that are inside this explore a huge multidimensional microstate. And what I want to do is to assign a probability that, given that I have fixed E, x, and n, that a particular microstate occurs. OK? How do I do that? Well, I say that OK, if the energy of that microstate that I can calculate is not equal to the energy that I know I have in the box, then that's not one of the microstates that should be allowed in the box. But presumably there's a whole bunch of micro states whose energy is compatible with the energy that I put in the box. And then I say, OK, if there is no other conserved quantity-- and let's assume that there isn't-- I have no way a priori of distinguishing between them. So they are just like the faces of the dice, and I say they're all equally likely. For the dice, I would give 1/6 here. There's presumably some kind of a normalization that depends on E, x, and n, that I have to put in. And again, note that this is kind of like a delta function that I have. It's like a delta of H minus E, and it's therefore consistent with this Liouville equation. It's one of these functions that is related through H to the probability on the microstate. Now, this is an assumption. It is a way of assigning probabilities. It's called assumption of equal a priori probabilities. It's like the subjective assignment of probabilities and like the faces of the dice, it's essentially the best that you can do without any other information. Now, the statement is that once I have made this assumption, I can derive the three laws of thermodynamics. No, sorry. I can derive two of the three laws of thermodynamics. Actually, three of the four laws of thermodynamics, since we had the zeroth law. So let's proceed. OK? So we want to have a proof of thermodynamics. So the zeroth law had something to do with putting two systems in contact, and when they were in equilibrium, there was some empirical temperature from one that was the same as what you had for the other one. So basically, let's pick our two systems and put a wall between them that allows the exchange of energy. And so this is my system one, and they have an spontaneous energy bond. This is the part that is two, it has energy E2. So I start with an initial state where when I look at E2 and E2, I have some initial value of E1, 0, let's say, and E2, 0. So my initial state is here, in these two. Now, as I proceed in time, because the two systems can exchange energy, E1 and E2 can change. But certainly, what I have is that E1 plus E2 is something E total that is E1,0 plus E2,0. Which means that I'm always exploring the line that corresponds to E1 plus E2 is a constant, which it runs by 45-degree along this space. So once I remove the constraint that E1 is fixed and E2 is fixed, they can exchange. They explore a whole bunch of other states that is available to them. And I would probably say that the probability of the microstates is the same up to some normalization that comes from E1. So the normalization-- not the entirety 1 plus 2-- is a microcanonical ensemble but with energy E total. So there is a corresponding omega that is associated with the combined system. And to obtain that, all I need to do is to sum or integrate over the energy, let's say, that I have in the first one. The number of states that I would have, or the volume of phase space that I would have if I was at E1, and then simultaneously multiplying by how many states the second part can have at the energy that corresponds to E total minus E1. So what I need to do is to essentially multiply the number of states that I would encounter going along this axis, and the number of states that I would multiply going along the other axis. So let's try to sort of indicate those things with some kind of a color density. So let's say that the density is kind of low here, it comes kind of high here, and then goes low here. If I move along this axis, let's say that along this axis, I maybe become high here, and stay high, and then become low later. Some kind of thing. So all I need to do is to multiply these two colors and generate the color along this-- there is this direction, which I will plot going, hopefully, coming out of the board. And maybe that product looks something like this. Stating that somewhere there is a most probable state, and I can indicate the most probable state by E1 star and E2 star, let's say. So you may say, OK, it actually could be something, who says it should have one maximum. It could have multiple maxima, or things like that. You could certainly allow all kinds of things. Now, my claim is that when I go and rely on this particular limit, I can state that if I explore all of these states, I will find my system in the vicinity of these energies with probability that in the n goes to infinity, limit becomes 1. And that kind of relies on the fact that each one of these quantities is really an exponentially large quantity. Before I do that, I forgot to do something that I wanted to do over here. I have promised that as we go through the course, at each stage we will define for each section its own definition of entropy. Well, almost, but not quite that. Once I have a probability, I had told you how to define an entropy associated with a probability. So here, I can say that when I have a probability p, I can identify the average of log p, the factor of minus p log p, to be the entropy of that probability. Kind of linked to what we had for mixing entropy, except that in thermodynamics, entropy had some particular units. It was related to heat or temperature. So we multiplied by a quantity kb that has the right units of energy divided by degrees Kelvin. Now, for the choice of the probability that we have, it is kind of like a step function in energy. The probability is either 0 or 1 over omega. So when you're at 0, this p log p will give you a 0. When you're over here, p log p will give you log of omega. So this is going to give you kb log of omega. So we can identify in the macrocanonical ensemble, once we've stated what E, x, and n are, what the analog of the six that we have for the throwing of the dices, what's the number of microstates that are compatible with the energy. We will have to do a little bit of massaging that to understand what that means in the continuum limit. We'll fix that. But once we know that number, essentially the log of that number up to a factor would give something that would be the entropy of that probability that eventually we're going to identify with the thermodynamic entropy that corresponds to this system. But having done that definition, I can rewrite this quantity as E1 e to the 1 over kb S1 plus 1 over kb S2. This one is evaluated at E1. This one is evaluated at E2, which is E total minus E1. Now, the statement that we are going to gradually build upon is that these omegas are these types of quantities that I mentioned that depend exponentially on the number of particles. So let's say, if you just think about volume, one particle then, can be anywhere in this. So that's a factor of v. Two particles, v squared. Three particles, v cubed. n particles, v to the n. So these omegas have buried in them an exponential dependence on n. When you take the log, these are quantities that are extensing. They're proportionate to this number n that becomes very large. So this is one of those examples where to calculate this omega in total, I have to evaluate an integral where the quantities are exponentially large. And we saw that when that happens, I could replace this integral essentially, with its largest value. So I would have 1 over kb S1 of E1 star plus S2 of E2 star. Now, how do I identify where E1 star and E2 star are? Well, given that I scale along this axis or along that axis, essentially I want to find locations where the exponent is the largest. So how do I find those locations? I essentially set the derivative to 0. So if I take the derivative of this with respect to E1, what do I get? I get dS1 by dE1. And from here, I get dS2 with respect to an argument that is in fact has a minus E1, so I would have minus dS2 with respect to its own energy. With respect to its own energy argument, but evaluated with an energy argument that goes opposite way with E1. And all of these are calculated at conditions where the corresponding x's and n's are fixed. And this has to be 0, which means that at the maxima, essentially, I would have this condition. So again, because of the exponential character, there could be multiple of these maxima. But if one of them is slightly larger than the other in absolute terms, in terms of the intensive quantities, once I multiply by these n's, it will be exponentially larger than the others. We sort of discussed that when we were doing the saddle point approximation, how essentially the best maximum is exponentially larger than all the others. And so in that sense, it is exponentially much more likely that you would be here as opposed to here and anywhere else. So the statement, again, is that just a matter of probabilities. I don't say what's the dynamics by which the energies can explore these two axes. But I imagine like shuffling cards, the red and the black cards have been mixed sufficiently, and then you ask a question about the typical configuration. And in typical configurations, you don't expect a run of ten black cards, or whatever, because they're exponentially unlikely. So this is the same statement, that once you allow this system to equilibrate its energy, after a while you look at it, and with probability 1 you will find it at a location where these derivatives are the same. Now, each one of these derivatives is completely something that pertains to its own system, so one of them could be a gas. The other could be a spring. It could be anything. And these derivatives could be very different quantities. But this equality would hold. And that is what we have for the zeroth law. That when systems come into equilibrium, there is a function of parameters of one that has to be the same as the function of the parameters of the other, which we call the empirical temperature. So in principal, we could define any function of temperature, and in practice, for consistently, that function is one over the temperature. So the zeroth law established that exists this empirical function. This choice of 1 over T is so that we are aligned with everything else that we have done so far, as we will see shortly. OK? The next thing is the first law, which had something to do with the change in the energy of the system. When we go from one state to another state had to be made up by a combination of heat and work. So for this, let's expand our system to allow some kind of work. So let's imagine that I have something-- let's say if it is a gas-- and the quantity that would change if there's work done on the gas is the volume. So let's imagine that there is a piston that can slide. And this piston exerts for the gas a pressure, or in general, whatever the conjugate variable is to the displacement that's we now allow to change. So there's potentially a change in the displacement. So what we are doing is there's work that is done that corresponds to j delta x. So what happens if this j delta x amount of work is done on the system? Then the system goes from one configuration that is characterized by x to another configuration that is characterized by x plus delta x. And let's see what the change in entropy is when that happens. Change in entropy, or the log of the number of states that we have defined. And so this is essentially the change starting from E and x to going to the case where x changed by an amount dx. But through this process I did work, and so the amount of energy that is inside the system increases by an amount that is j delta x. So if all of these quantities are infinitesimal, I have two arguments of S now have changed infinitesimally, and I can make the corresponding expansion in derivatives. There is a dS with respect to dE at constant x. The corresponding change in E is j delta x. But then there's also a dS by dx at constant E with the same delta x, so I factored out the delta x between the two of them. Now, dS by dE at constant x is related to the empirical temperature, which we now set to be 1 over T. Now, the claim is that if you have a situation such as this, where you have a state that is sitting in equilibrium-- let's say a gas with a piston-- then by definition of equilibrium, the system does not spontaneously change its volume. But it will change its volume because the number of states that is availability to it increases, or S increases. So in order to make sure that you don't go to a state that is more probable, because there are more possibilities, there are more microstates, I better make sure that this first derivative is 0. Otherwise, depending on whether this is plus or minus, I could make a corresponding change in delta x that would increase delta S. So what happens here if I require this to be 0 is that I can identify now that the derivative of this quantity dS by dx at constant E has to be minus j over T. Currently, I have only these two variables, and quite generically, I can say that dS is dS by dE at constant x dE plus dS by dx at constant E dx. And now I have identified these two derivatives. dS by dE is 1 over T, so I have dE over T. dS by dx is minus j over T, so I have minus j dx over T. And I can rearrange this, and I see that dE is T dS plus j dx. Now, the j dx I recognize as before, it's the mechanical work. So I have identified that generically, when you make a transformation, in addition to mechanical work, there's a component that changes the energy that is the one that we can identify with the heat. Yes? AUDIENCE: Could you explain why you set that S to 0? PROFESSOR: Why did I set delta S to 0? So I have a box, and this box has a certain volume, and there's a bunch of particles here, but the piston that is holding this is allowed to slide to go up and down. OK? Now, I can ask what happens if the volume goes up and down. If the volume goes up and down, how many states are available? So basically, the statement has been that for each configuration, there is a number of states that is available. And if I were to change that configuration, the number of states will change. And hence the log of it, which is the entropy, will change. So how much does it change? Well, let's see what arguments changed. So certainly, the volume changed. So x went to x plus dx. And because I did some amount of work, pdv, the amount of energy that was in the box also changed. OK? So after this transformation with this thing going up or down, what is the new logarithm of the number of states? How much has it changed? And how much of the change is given by this? Now, suppose that I make this change, and I find that suddenly I have a thousand more states available to me. It's a thousand times more likely that I will accept this change. So in order for me to be penalized-- or actually, not penalized, or not gain-- because I make this transformation, this change better be 0. If it is not 0, if this quantity is, let's say, positive, then I will choose a delta v that is positive, and delta S will become positive. If this quantity happens to be negative, then I will choose a delta v that is negative, and then delta S will be positive again. So the statement that this thing originally was sitting there by itself in equilibrium and did not spontaneously go up or down is this statement this derivative 0. AUDIENCE: That's only 0 when the system is in equilibrium? PROFESSOR: Yes. Yes. So indeed, this is then a relationship that involves parameters of the system in equilibrium. Yeah? So there is thermodynamically, we said that once I specify what my energy and volume and number of particles are in equilibrium, there is a particular S. And what I want to know is if I go from one state to another state in equilibrium, what is the change in dS? That's what this statement is. Then I can rearrange it and see that when I go from one equilibrium state to another equilibrium state, I have to change internal energy, which I can do either by doing work or by doing heat. OK? Now, the second law is actually obvious. I have stated that I start from this configuration and go to that configuration simply because of probability. The probabilities are inverse of this omega, if you like. So I start with some location-- like the pack of cards, all of the black on one side, all of the red on the other side-- and I do some dynamics, and I will end up with some other state. I will, because why? Because that state has a much more volume of possibilities. There is a single state that we identify as all-black and all-red, and there's a myriad of states that we say they're all randomly mixed, and so we're much more likely to find from that subset of states. So I have certainly stated that S1 evaluated at E1 star, plus S2 evaluated at E2 star, essentially the peak of that object, is much larger than S1 plus E1 plus 0 plus S2 [INAUDIBLE]. And more specifically, if we sort of follow the course of the system as it starts from here, you will find that it will basically go in the direction always such that the derivatives that are related to temperature are such that the energy will flow from the hot air to the colder body, consistent with what we expect from thermodynamics. The one law of thermodynamic that I cannot prove from what I have given you so far is the third law. There is no reason why the entropy should go to 0 as you go to 0 temperature within this perspective. So it's good to look at our canonical example, which is the ideal gas. Again, for the ideal gas, if I am in a microcanonical ensemble, it means that I have told you what the energy of the box is, what the volume of the box is, and how many particles are in it. So clearly the microstate that corresponds to that is the collection of the 6N coordinates and momenta of the particles in the box. And the energy of the system is made up by the sum of the energies of all of the particles, and basically, ideal gas means that the energy that we write is the sum of the contributions that you would have from individual particles. So individual particles have a kinetic energy and some potential energy, and since we are stating that we have a box of volume v, this u essentially represents this box of volume. It's a potential that is 0 inside this volume and infinity outside of it. Fine. So you ask what's the probability of some particular microstate, given that I specified E, v, and n. I would say, well, OK. This is, by the construction that we had, 0 or 1 over some omega, depending on whether the energy is right. And since for the box, the energy is made up of the kinetic energy, if sum over i p i squared over 2m is not equal to E, or particle q is outside the box. And it is the same value if sum over i p i squared over 2m is equal to v and q i's are all inside the box. So a note about normalization. I kind of skipped on this over the last one. p is a probability in this 6N dimensional phase space. So the normalization is that the integral over all of the p i's and qi's should be equal to 1. And so this omega, more precisely, when we are not talking about discrete numbers, is the quantity that I have to put so that when I integrate this over phase space, I will get 1. So basically, if I write this in the form that I have written, you can see that omega is obtained by integrating over p i q i that correspond to this accessible states. It's kind of like a delta function. Basically, there's huge portions of phase space that are 0 probability, and then there's a surface that has the probability that is 1 over omega and it is the area of that surface that I have to ensure is appearing here, so that when I integrate over that area of one over that area, I will get one. Now, the part that corresponds to the q coordinates is actually very simple, because the q coordinates have to be inside the box. Each one of them has a volume v, so the coordinate part of the integral gives me v to the m. OK? Now, the momentum part, the momenta are constrained by something like this. I can rewrite that as sum over i p i squared equals to 2mE. And if I regard this 2mE as something like r squared, like a radius squared, you can see that this is like the equation that you would have for the generalization of the sphere in 3N dimensions. Because if I had just x squared, x squared plus y squared is r squared is a circle, x squared plus y squared plus z squared is r squared is a sphere, so this is the hypersphere in 3N dimensions. So when I do the integrations over q, I have to integrate over the surface of a 3N dimensional hypersphere of radius square root of 2m. OK? And there is a simple formula for that. This surface area, in general, is going to be proportional to r raised to the power of dimension minus 1. And there's the generalization of 2pi r over 4pi r squared that you would have in two or three dimensions, which we will discuss next time. It is 2 to the power of 3n over 2 divided by 3n over 2 factorial. So we have the formula for this omega. We will re-derive it and discuss it next time.
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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: Before we get started, let me ask you guys if you have any questions, pragmatic or otherwise, about the course so far. Seriously? To those of you reading newspapers, I encourage you to find a slightly different time to do so. I really encourage you find a slightly different time to do so, thanks. So far we've done basic rules of quantum mechanics. We've done solids. We understand a lot about electrons in atoms, the periodic table, and why diamond is transparent. One thing we did along the way is we talked about spin. We found that when we looked at the angular momentum commutation relations, these guys. We found that the commutators are the same. For these commutators, we can get total angular momentum, l l plus 1 times h bar squared for l squared. And h bar m for a constant and integer m for angular momentum in a particular direction, which we conventionally called z. But we also found that there were half integer values of the total spin and of the spin in a particular direction which, for example, with Spin s as 1/2, 3/2, 5/2, et cetera. And we discovered that riding a wave function on a sphere that interpreted these states as states with definite probability to be at a particular position on a sphere, a function of theta and phi, was inconsistent. In order for the wave function to satisfy those properties that have those eigenvalues and, in particular, half integer eigenvalues, we discovered that the wave function had to be doubly valued. And thus it would equal to minus itself, it equaled 0. So the rest of today and tomorrow is going to be an exploration of spin, or tomorrow-- next lecture, is going to be an exploration of spin and the consequences of these spin 1/2 states. Exactly what they are, we already saw that they're important for understanding the structure of the periodic table. So we know they're there. And they're present from the experiment, the Stern-Gerlach experiment that we've discussed many times. But before I get onto that, I want to improve on the last experiment we did. So in particular, in this last experiment we talked about the effective mass of an object interacting with a fluid or an object interacting with its environment. And why the mass of the object that's moving is not the same as the mass of the object when you put it on a balance. And that's this basic idea of renormalization. And we demonstrated that. I demonstrated that with a beaker of water. So I had a beaker of water and pulled a ping pong ball under water. We calculated that it should accelerate upward when released at 20 times the acceleration of gravity, depending on the numbers you use, very, very rapidly. And in fact it went glug, glug, glug, but it wasn't a terribly satisfying experiment because it's very hard to get the timing right. And the basic issue there is that the time scales involved were very short. How long did it take for a ping pong ball to drop from here to the surface? Not much time. And to rise through the water, not a whole lot of time. So I did that experiment, it was sort of comical. But I wanted to improve on it. So over the weekend, I went down to my basement and tweaked the experiment a little bit. And I called up a couple of my friends, and we did an improved version of this experiment. So this is a diver, I think this one is Kathy. Oh, shoot, we need to turn off the lights. Sorry. I totally forgot. good You'll never see this if we don't. You can do it. There we go. All right. So here you see my friend, I think this one's, actually, is it Kathy? I'm not sure. It's hard to tell when they have their marks on. And we're in the tank at the New England Aquarium, and she's going to perform this experiment for us. And we're going to film it, as you can see, the bubbles moving rather slowly, with a high speed camera filming at 1,200 frames per second with which we'll be able to analyze the data that results. The camera cost as much as a nice house. And it's not mine, but it's important to have friends who trust you. OK, so here we are. You might notice something in the background. Before we get started, I just want to emphasize that one should never take casually the dangers of doing an experiment. When you plan an experiment you must, ahead of time, design the experiment, design the experimental parameters. We designed the lighting. We designed everything. We set it up, but there are always variables you haven't accounted for. And a truly great experimentalist is one who has taken account of all the variables. And I just want to emphasize that I'm not a great experimentalist. So here, for example, is a moment. I probably should have thanked the sharks, it just occurred to me. Anyone who wants to take this experimental data, which, as you can probably guess, filmed for different purposes, I just manage to get the ping pong ball into the tank. Anyone who wants to get this and actually take the data, come to me, I'll give you the raw footage. And you can read off the positions, and hopefully for next lecture we'll have the actual plot of the acceleration as a function of time. So with that said and done, the moral of the story is you have to account for all variables. The other moral of the story is that you saw this very vividly. In the motion of the ping pong ball up, when it was released, there is that very rapid moment of acceleration when it bursts up very, very rapidly. But it quickly slows in its acceleration. Its acceleration slows down. In fact, a slightly funny thing happens. If you look carefully, and again, anyone who wants this can get access to the video, what you'll see is that the ping pong ball accelerates and then it sort of slows down. It literally decreases in velocity, accelerates and slows down. Can anyone think what's going on in that situation? AUDIENCE: Boundary layer formation. PROFESSOR: Sorry? AUDIENCE: Boundary layer formation. PROFESSOR: Good. Say that in slightly more words. AUDIENCE: It's starting to pick up more and more water [INAUDIBLE]. PROFESSOR: Yeah. Exactly. so what's going on is as this guy starts slowly moving along, it's pulling along more and more water, each bit of water around it is starting to drag along the layers of water nearby, and it builds up a sheath of water. Now that water starts accelerating, and the ping pong ball and the bubble of water that it's dragging along need to come to equilibrium with each other. They need to settle down smoothly to a nice uniform velocity. But it takes a while for that equilibrium to happen. And what actually happens is that the ping pong ball drives up. It pulls up the water. Which then drags with the ping pong ball. So you can see that in the acceleration, which is oscillatory with a slight little oscillation. So it's a damped but not overdamped harmonic oscillator motion. Any other questions about the effective mass of an electron and solid before moving-- Yeah? AUDIENCE: Why does it speed up after? That explains why it slows down because it's forming that sheath-- PROFESSOR: Right. Why it speeds back up is it's sort of like a slingshot. As this guy gets going a little ahead of the pack of water, the pack of water has an extra driving force on top of the buoyancy. It has the fact that it's a little bit displaced. So it catches up, but it's going slightly greater velocity than it would be if there were uniform velocity. So this guy catches up with the ping pong ball. OK so this is an of course of course in fluid mechanics, but I guess we don't actually need this anymore. Picking up on spin. So the commutation relations for spin are these. And as we saw last time, we have spin states. We have we can construct towers of states because from the sx and s1 we can build s plus-minus is equal to sx plus-minus sy. Sorry. i, thank you. So we can build towers of states using the raising and lowering operators as plus-minus. And those states need to end, they need to terminate. So we find that the spin can have totalling momentum of s squared h bar squared l l plus 1. And S in some particular direction, which we conventionally called z, is h bar m. I don't want to call this l. I want to call this s. For orbital angular momentum this would be l and this would be an integer. But for spinning angular momentum these are all the states we could build, all the towers we could build, which were 2n plus 1 over 2, which were not expressible in terms of wave functions, functions of a position on a sphere. These are all the 1/2 integer states. So s could be 1/2, 3/2, 5/2, and so on. And then m sub s is going to go from minus s to s in integer steps, just like m for the orbital angular momentum, l. So I want to talk about these states in some detail over the rest of this lecture and the next one. So the first thing to talk about is how we describe spin. In principle, this is easy. What we want, is we want to describe the state of a particle that carries this intrinsic angular momentum spin. So that's easy. The particle sits at some point, but the problem is it could be sitting at some point with angular momentum with spin in the z direction, say, plus h bar over 2. And let's focus on the case s is equal to 1/2. So I'll be focusing, for the lecture, for simplicity on the case, total spin is 1/2, which is the two-state ladder, but all of this generalizes naturally. In fact, that's a very good test of your understanding. So for s as 1/2, we have two states, which I will conventionally call the up in the z direction and the down in the z direction. And I will often omit the subscript. If I omit the subscript it's usually z unless from context you see that it's something else. So the wave function tells us the state of the system. But we need to know now for a spinning particle whether it's in the spin 1/2 up or spin 1/2 down state. And so we could write that as there's some amplitude that it's in the plus 1/2 state of x and at position x. And there's some amplitude that it's at position x and it's in the minus 1/2 state or the down state. And we again need that the total probability is one. Another way to say this is that the probability that we find the particle to be at x with plus or minus h bar upon 2 being the spin in the z direction. So at x, spin in the z direction is h bar upon 2 plus or minus is equal to norm squared of psi plus or minus of x squared. So this is one way you could talk about spin, and you could develop the theory of spin nicely here. But it's a somewhat cumbersome formalism. The formulation I want to introduce is one which involves matrices and which presages the study of matrix mechanics which you'll be using in 805. So instead, I want to take these two components, and what we see already is that we can't use a single wave function to describe a particle at spin 1/2. We need to use two functions. And I want to organize them in a nice way. I'm going to write them as psi is equal to-- and I'll call this capital psi of x-- is a two component vector, or so-called spinner, psi up of x and psi down of x. So it's a two component object. It's got a top and a bottom component. And notice that its conjugate, or its adjoint, psi dagger, is going to be equal to psi up, complex conjugate psi down, a row vector, or a row spinner. And for normalization we'll need that the total probability is 1 which says that psi capital dagger psi, psi capital with psi is equal to 1. But this is going to be equal to the integral dx. And now we have to take the inner product of the two vectors. So integral dx of psi up squared plus psi down squared. Cool? Yep? AUDIENCE: What's the coordinate x representing here? PROFESSOR: The coordinate x is just representing the position. So what I'm saying here is I have, again, we're in one dimension just for simplicity, it's saying, look if I have a particle that carries spin 1/2, it could be anywhere. Let's say it's at position x. So what's the amplitude at position x and spinning up, and I'm not going to indicate spinning down. OK? I like my coffee. So that's what the of x indicates, and I've just been dropping the of x. So there's some probability that it's at any given point and either spin up or spin down. Now, again, it's important, although I'm going to do this, and we conventionally do this spin up and down, this spin is pointing in a vector space that's two dimensional. It's either plus 1/2 or minus 1/2 h bar upon 2 h bar. So it's not like the spin is an arrow in three dimensional space that points. Rather, what it is, it's saying, if I measure the spin along some axis, it can take one of two values. And that was shown in the Stern-Gerlach experiment, where if we have a gradient of the magnetic field, dbz, in the z direction, and this is our Stern-Gerlach box, and we send an electron in, the electron always comes out in one of two positions. OK. Now, this is not saying there is a vector associated with this, that the electron has a angular momentum vector that points in some particular direction. Rather, it's saying that there are two possible values, and we're measuring along the z direction. Cool? So it's important not to make that mistake to think of this as some three dimensional vector. It's very explicitly a vector in a two dimensional vector space. It's not related to regular rotations. Yeah? AUDIENCE: Where you write psi of x equals psi plus 1 plus psi [INAUDIBLE]. PROFESSOR: Yes. AUDIENCE: Do we need a 1 over square root of 2 in front of that thing? PROFESSOR: Yeah, I haven't assumed their normalization, but each one could be independently normalized appropriately. AUDIENCE: So [INAUDIBLE]. PROFESSOR: Right. The whole thing has to be properly normalized, and writing it this way, this is just the [INAUDIBLE]. Good. So there's another nice bit of notation for this which is often used, which is probably the most common notation. Which is to write psi of x is equal to psi up at x times the vector 1, 0 plus psi down of x times the vector 0, 1. So this is what I'm going to refer to as the up vector in the z direction. And this is what I'm going to refer to as the down in the z direction vector. And that's going to allow me to write all the operations we're going to need to study spin in terms of simple two by two matrices. Yeah? AUDIENCE: Will those two psi's not be the same? PROFESSOR: Yeah, in general, they're not. So for example, here's a situation, a configuration, a quantum system could be in. The quantum system could be in the configuration, the particle is here and it's spinning up. And it could be in the configuration the particles over here, and it's spinning down. And given that it could be in those two configurations it could also be in the superposition over here and up and over here and down. Right? So that would be different spatial wave functions multiplying the different spin wave functions, spin part of the wave function. Make sense? OK. So these are not the same function. They could be the same function. It could be that you could be either spin up or spin down at any given point with some funny distribution, but they don't need to be the same. That's the crucial thing. Other questions? Yeah. AUDIENCE: I know that [INAUDIBLE] is a way to call them. So like [INAUDIBLE] something, are they anti-parallel, perpendicular, or are they something? PROFESSOR: Yeah. So here's what we can say. We know that an electron which carries sz is plus h bar upon 2, and a state corresponding to minus h bar upon 2 are orthogonal because those are two different eigenvectors of the same operator, sz. So these guys are orthogonal. Up in the z direction and down to the z directions are orthogonal. And thank you for this question, it's a good way to think about how wrong it is to think of up and down being up and down in the z direction. Are these guys orthogonal? These vectors in space? AUDIENCE: No. PROFESSOR: No, they happen to be parallel with a minus 1 in the overlap, right? So sz as up sc as down are orthogonal vectors, but this is clearly not sz as down, right? So the direction that they're pointing, the up and down, should not be thought of as a direction in three-dimensional space. AUDIENCE: It's just [INAUDIBLE]. PROFESSOR: It's just a different thing. What it does tell you, is if you rotate the system by a given amount, how does the phase of the wave function change. But what it does tell you is how the spin operations act on it. In particular, sz acts with a plus 1/2 or minus 1/2. They're just different states. Yeah? AUDIENCE: Is this similar to what happens when polarizations [INAUDIBLE]? PROFESSOR: It's similar to the story of polarizations except polarizations are vectors not spinners. It's similar in the sense that they look and smell like vectors in three-dimensional space, but they mean slightly-- technically-- slightly different things. In the case of polarizations of light, those really are honest vectors, and there's a sharp relationship between rotations in space. But that's a sort of quirk. They're both spin and vectors. These are not. OK so here's a notation I'm going to use. Just to alert you, a common notation that people use in Dirac notation is to say that the wave function is equal to psi up at x times the state up plus psi down at x times the state down. So for those of you who speak Dirac notation at this point, then this means the same thing as this. For those of you who don't, then this means this. What I want to do is I want to develop a theory of the spin operators, and I want to understand what it means to be a spin 1/2 state. Now, in particular, what I mean by develop a theory of the spin operators, if I was talking about four orbital angular momentum, say the orbital angular momentum in the z direction, I know what operator this is. If you hand me a wave function, I can act on it with lz and tell you what the result is. And that means that I can construct the eigenfunctions. And that means I can construct the allowed eigenvalues, and I can talk about probabilities. Right? But in order to do all that I need to know how the operator acts. And we know how this operator acts. It acts as h bar upon i dd phi where phi is the angular coordinate around the equator. And so given any wave function, a function of x, y, and z or r, theta, and phi, I can act with this operator, know how the operator acts. And it's true, that again, lx with lx is equal to i h bar lz. It satisfies the same time computation relation as the spin. But we know the spin operators cannot be expressed in terms of derivatives along a sphere. I've harped on that many times. So what I want to know is what's the analog of this equation for spin? What is a representation of the spin operators acting on the spinners, acting on states that carry half integer spin? We know it's not going to be derivatives. What is it going to be? Everyone cool with the goal? In order to do that, we need to first decide just some basic definition of spin in the z direction. So what do the angular momentum operators do? Well whatever else is true of the spin in the z direction operator, sz acting on a state up is equal to h bar upon 2 up. And sz actually on a state down is equal to h bar minus upon 2 down. And similarly s squared on up or down-- Oh, by the way, I'm going to relatively casually oscillate between the notations up and plus, up or down, and plus or minus. So sometimes I will write plus for up and minus for down. So I apologize for the sin of this. s squared on plus, this is the plus 1/2 state, is equal to h bar squared ss plus 1, but s is 1/2, so 1/2 times 1/2 plus 1 is 3/4 h bar squared times 3 over 4. And we get the same thing up, ditto down. Because s squared acts the same way on all states in a tower. Going up and down through a tower of angular momentum states, raising and lowering, does not change the total angular momentum because s plus and s minus commute with s squared because they have exactly the same commutation relations as the angular momentum. That's an awesome sound. So I want to know what these look like in terms of this vector space notation, up and down. And for the moment I'm going to dispense entirely with the spatial dependence. I'm going to treat the spatial dependence as an overall constant. So we're equally likely to be in all positions. So we can focus just on the spin part of the state. So again I want to replace up by 1, 0 and down by 0, 1. And I want to think about how this looks. So what this looks like is sz acting on [INAUDIBLE] 1, 0 is equal to h bar upon 2, 1, 0. And sz on 0, 1 is h bar upon 2, 0, 1. And s squared on 1, 0 is equal to 3 h bar squared on 4, 1, 0. And ditto for 0, 1. Yeah? AUDIENCE: That [INAUDIBLE] should have a line over it. PROFESSOR: Oh, thank you. Yes, it really should. AUDIENCE: So how do you get 3/4 there? PROFESSOR: 3/4, good. Where that came from is that remember that when we constructed the eigenfunctions of l squared, l squared acting on a state lm was equal to h bar squared l l plus 1. l plus 1 on [? file m. ?] Now if we do exactly the same logic, which we actually did at the time. We did in full generality whether the total angular momentum was an integer or a half integer. We found that if we took, I'm just going to use for the half integer states the symbol s, but it's exactly the same calculation. s squared on phi and again sm sub s is equal to h bar squared s s plus 1 phi s m sub s. OK and so for s equals 1/2 then s, s plus 1, is equal to 1/2 times 3/2, which is 3/4. Yeah? AUDIENCE: [INAUDIBLE] s equals [INAUDIBLE]. Isn't that [INAUDIBLE]? PROFESSOR: Ah, but remember, does l go negative? Great. Does s go negative? No. s is just labeling the tower. So s is 0, it's 1, it's 2. And so for example, here, these are states where the s is 1/2 and the s in the z direction can be plus 1/2 or minus 1/2. s is 3/2 and then s in the z direction can be m is 3/2, 1/2, minus 1/2, minus 3/2. OK. Other questions? OK, yeah. AUDIENCE: What about the lowering and raising of those? PROFESSOR: Good, we're going to have to construct them, because, what are they? Well, they lower and raise, so we're going to have to build the states that lower and raise. AUDIENCE: And would lowering on the up will give you down, but raising on up--? PROFESSOR: Awesome. So what did it mean that we had towers? Let me do that back here. So the question is, look, we're going to have to use the raising and lowering operators at the end of the day, but what happens if I raise the bottom state-- what if I raise down? I'll get up. And what happens if I raise up? You get 0. That's the statement that the tower ends. On the other hand, if s is 3/2, what happens if I raise 1/2? I get 3/2. And if I raise 3/2, I get nothing, I get 0, identically. And that's the statement that the tower ends. So for every tower labeled by s, we have a set of states labeled by m which goes from minus s to s in integer steps. The raising operator raises us by 1, the lowering operator lowers by 1. The lowering operator annihilates the bottom state, the raising operator kills the top state. Cool? Yeah. AUDIENCE: Do states like 3/2 and 5/2 have anything akin to up and down? PROFESSOR: Yeah, so-- do they have anything akin to up and down. Up and down is just a name. It doesn't really communicate anything other than it's shorthand for spin in the z direction 1/2. So the question could be translated as, are there convenient and illuminating names for the spin 3/2 states? And I don't really know. I don't know. I mean, the states exist. So we can build nuclear particles that have angular momentum 3/2, or 5/2, all sorts of things. But I don't know of a useful name. For the most part, we simplify our life by focusing on the 1/2 state. And as you'll discover in 8.05, the spin 1/2 states, if you know them very, very well, you can use everything you know about them to construct all of the spin 5/2 8/2-- well, 8/2 is stupid, but-- 9/2, all those from the spin 1/2. So it turns out spin 1/2 is sort of Ur-- in a way that can be made very precise, and that's the theory of Lie algebras. Yeah. AUDIENCE: Can you just elaborate on what you meant by, you can't really think of spin as an angular momentum vector? PROFESSOR: Yeah. So OK, good. So the question is, elaborate a little bit on what you mean by, spin can't be thought of as an angular momentum vector. Spin certainly can be thought of as an angular momentum, because the whole point here was that if you have a charged particle and it carries spin, then it has a magnetic moment. And a magnetic moment is the charge times the angular momentum. So if it carries spin, and it carries charge, and thus it carries magnetic moment-- that's pretty much what we mean by angular momentum. That's as good a diagnostic as any. Meanwhile, on top of satisfying that experimental property, this, just as a set of commutation relations, these commutation relations are the commutation relations of angular momentum. It just turns out that we can have states with total angular momentum little s, which is 1/2 integral-- 1/2, 3/2, et cetera. Now, the things that I want to emphasize are twofold. First off, something I've harped on over and over again, so I'll attempt to limit my uses of this phrase. But you cannot think of these states with s is 1/2 as wave functions determining position on a sphere. So that's the first sense in which you can't think of it as equivalent to orbital angular momentum. But there's a second sense, which is that up and down should not be thought of as spin in the z direction being up and spin in the z direction being down meaning a vector in three dimensions pointing up and a vector in three dimensions pointing down, because those states are orthogonal. Whereas these two three-dimensional vectors are not orthogonal-- they're parallel. They have a non-zero inner product. So up and down, the names we give these spin 1/2 states, should not be confused with pointing up in the z direction and down in the z direction. It's just a formal name we give to the plus 1/2 and minus 1/2 angular momentum in z direction states. Does that answer your question? AUDIENCE: Yeah, so, when you make a measurement of the value of spin-- so perhaps you do a Stern-Gerlach experiment-- and you get what the spin is, can you not then say, all right, this is spin plus 1/2, spin minus? It's as z is positive 1/2 as z is minus 1/2? PROFESSOR: Yeah, absolutely. So if you do a Stern-Gerlach experiment, you can identify those electrons that had spin plus 1/2 and those that had spin minus 1/2, and they come out in different places. That's absolutely true. I just want to emphasize that the up vector does not mean that they're somehow attached to the electronic vector that's pointing in the z direction. Good. Yeah. Go ahead, whichever. AUDIENCE: How do we verify that uncharged particles have spin? PROFESSOR: Yeah, that's an interesting question. So the question is, how do we know if an uncharged particle has spin? And there are many ways to answer this question, one of which we're going to come to later which has to do with Bell's inequality, which is a sort of slick way to do it. But a very coarse way is this way. We believe, in a deep and fundamental way, that the total angular momentum of the universe is conserved, in the following sense. There's no preferred axis in the universe. If you're a cosmologist, just stay out of the room for the next few minutes. So there's no preferred axis in the universe and the law of physics should be invariant under rotation. Now, if you take a system that has a bunch of particles with known angular momentum-- let me give you an example. Take a neutron. A neutron has spin 1/2. Wait, how did I know that? We can do that experiment by doing the following thing. We can take a neutron and bind it to a proton and see that the resulting object has spin 1. So let me try to think of a way that doesn't involve a neutron. Grant me for the moment that you know that a neutron has spin 1/2. So let's just imagine that we knew that, by hook or by crook. We then do the following experiment. We wait. Take a neutron, let it sit in empty space. When that neutron decays, it does a very cool thing. It decays relatively quickly into your proton and an electron that you see. You see them go flying away the proton has positive charge, and the electron has negative charge and it goes flying away. But you've got a problem. Because you knew that the neutron had spin 1/2, which is [INAUDIBLE]. And then you decay one into a proton and an electron. And the total angular momentum there is 1/2 plus 1/2 or 1/2 minus 1/2. It's either 1 or 0. And you've got a problem. Angular momentum hasn't been conserved. So what do you immediately deduce? That another particle must have also been emitted that had 1/2 integer angular momentum to conserve angular momentum. And it couldn't carry any charge because the electron and the proton were neutral, and the neutron is neutral. So things like this you can always deduce from conservation of angular momentum one way or the other. But the best way to do it is going to be some version of addition of angular momentum where you have some object like an electron and a proton and you allow them to stick together and you discover it has total spin 1. Yeah. We can talk about that in more detail afterwards. That's a particularly nice way to do the experiment. Yeah. AUDIENCE: Angular momentum [INAUDIBLE] weird vector since if you reflect your system through [INAUDIBLE]. How does that work? PROFESSOR: Yeah, OK, good. I don't want to get into this in too much detail, but it's a really good question, so come to my office hours and ask or go to recitations and ask. It's a really good question. The question is this-- angular momentum has a funny property under parity, under reflection. So if you look in a mirror this way, here's angular momentum and it's got-- right-hand rule, it's got angular momentum up-- if I look in the mirror, it's going this way. So it would appear to have right angular momentum down. That's what it looks like if you reflect in a mirror. Other direction. So that's a funny property of angular momentum. It's also a true property of angular momentum. It's fine. And what about spin, is the question. Does spin also have this funny property under parity, is that basically the question? Yeah, and it does. And working out exactly how to show that is a sort of entertaining exercise. So again, it's beyond the scope of the lecture, so come ask me in office hours and we can talk about that. Yeah, one more. AUDIENCE: For orbital angular momentum, say for l equals 1, we had states like m equals plus 1 and minus 1. PROFESSOR: Yes. AUDIENCE: And we did think of those as angular momentum vectors. PROFESSOR: Absolutely. AUDIENCE: But those states are also orthogonal, are they not? PROFESSOR: Yeah, those states are also orthogonal. AUDIENCE: So even though the angular momentum vectors aren't orthogonal, they're still-- it's just a different sense. PROFESSOR: That's exactly right. So again, even in the case of integer angular momentum, you've got to be careful about talking about the top state and the bottom state corresponding to pointing in some direction, because they're orthogonal states. However, they do correspond to a particular angular momentum vector in three dimensional space. They correspond to a distribution on the sphere. So there's a sense in which they do correspond to real rotations, real eigenfunctions on a sphere, and there's also a sense in which they don't, because they're still orthogonal. That's exactly right. So let me move on. I'm going to stop questions at this point. So good. So these are the properties that need to be satisfied by our operators. And it's pretty easy to see in this basis what these operators must be. Sz has eigenvectors 1, 0 and 0, 1. So Sz should be equal to h bar upon 2 1, 0, 0, minus 1. So let's just check this on 1, 0 gives me 1, 0, so it gives me the same thing back times h bar upon 2. Cool. And acting on 0, 1, or the down state, we get h bar upon 2 times 0 minus 1. That could be minus 1. Oh, sorry, 1, 0 gives me 0 and 0 minus 1 on 1 gives me minus 1, which is 1 with a minus sign. That's a minus sign. So this works out like a champ. And S squared, meanwhile, is equal to-- well, it's got to give me h bar squared times 3/4 for both of these vectors. So h bar squared-- and meanwhile, these are eigenstates-- h bar squared times 3/4 times 1, 0, 0, 1. So we know one other fact, which was brought up just a minute ago, which was that if we take S plus and you act on the state 0, 1, what should you get? If you raise your 1-- 1, 0. Great. So we also worked out the normalization coefficient on the problem set. And that normalization coefficient turns out to be 1. And let's be careful-- we've got an h bar, for dimensional analysis reasons. So meanwhile, S minus, similarly, on 0, 1, is equal to 0. And S plus on 0, 1, is equal to-- oh, sorry, we already did that. We want S minus on 1, 0. Let's see-- S minus on-- we want S plus on 1, 0 is equal to-- right, 0. And S minus on 1, 0 is equal to h bar 0, 1. OK, so putting all this together, you can pretty quickly get that S plus is equal to-- we need an h bar and we need it to raise the lower one and kill the top state. So on 1, 0, what does S plus do? That gives us 0 that gives us 0. Good. And on the lowered state, 0, 1, that gives me a 1 up top and that gives me a 0 downstairs, so it works out like this. So we've got h bar. Similarly, S minus is equal to h bar times 0, 0, 1, 0. So we've got these guys-- so much from just the definitions of raising and lowering. And by taking inner products, you can just derive those two lines from these. But notice that Sx is equal to S plus plus S minus upon 2, and Sy is equal to S plus minus S minus upon 2i. So this tells us that Sx is equal to h bar upon 2 times S plus-- we're going to get a 1 here-- plus S minus-- we're going to get a 1 here-- 0, 1, 1, 0. And Sy is equal to, again, upon 2i times h bar, h bar upon 2i, times S plus, which is going to give me 1 and minus S minus which is going to give me minus 1, 0, 0. But we can pull this i in, so 1/i is like minus i. So minus i times minus 1 is going to give me i and minus i times 1 is going to give me i. So-- AUDIENCE: Shouldn't it be minus i? PROFESSOR: Sorry? Yeah, did I write i? That should've been minus i. Thank you. So now we have a nice representation of these spin operations, of the spin operators. And explicitly we have that Sx is equal to h bar upon 2 0, 1, 1, 0, Sy is equal to h bar upon 2 0, minus i, i, 0. And Sz is equal to h bar upon 2 1, 0, 0, minus 1. So why is Sz the only one that's diagonal? Is it something special about Sz? AUDIENCE: I mean, we've chosen z as the axis along which to project S squared. PROFESSOR: Exactly. So the thing that's special about Sz is that at the very beginning of this, we decided to work in a basis of eigenstates of Sz, with definite values of Sz. So if they have definite values, then acting with Sz is just going to give you a number. That's what it is to be a diagonal matrix. You act on a basis vector, you just get a number out. So we started out by working in the eigenbasis of Sz. And as a consequence, we find that Sz is diagonal. And this is a general truth that you'll discover in matrix mechanics when you work in the eigenbasis of an operator, that operator is represented by a diagonal matrix. And so we often say, rather than to work in an eigenbasis, we often say, to diagonalize. Yeah. AUDIENCE: Are the signs right for Sy? Because if we had h bar over 2i, and as we initially had a 1 in the top right and minus 1 in the bottom left, shouldn't we just multiply by i? PROFESSOR: I'm pretty sure-- so originally, we had a downstairs i, right? So let's think about what this looked like. This was 1 and minus 1, right? Agreed? So this is minus 1 over i. So we pull in the i. So that we go from minus 1 to minus 1 over i. And we go from 1 to 1 over i. And I claim that one over i is minus i. AUDIENCE: Oh, OK. PROFESSOR: OK? And minus 1 over i is i. That cool? Good. OK, so this is i and minus i. I always get that screwy, but it's useful to memorize these matrices. You might think it's a little silly to memorize matrices. But these turn out to be ridiculously useful and they come up all the time. This is called sigma x. This is called sigma y. And this is called sigma z. And different people decide whether to put the 1/2 in there or not, the h bar does not go in there. Some people put in the 2, some people don't put in the 1/2, it's a matter of taste. Just be careful and be consistent, as usual. And these are called the Pauli matrices because A, we really like Pauli and B, Pauli introduced them. Although he didn't actually introduce them in some sense-- this mathematical structure was introduced ages and ages and ages ago. But physicists cite the physicist, not the mathematician. OK I'm not saying that's good. I'm just saying it happens. So notice a consequence of these. An important consequence of these-- the whole point here was to build a representation of the spin operators. Now whatever else the spin operators do, they had better satisfy that computation relation, otherwise they're not really spin operators. That's what we mean by being spin operators. So let's check. Is it true that Sx commutator with Sy is equal to i h bar Sz? So this is a question mark. And let's check. Let's do the commutator. From the Sx, we're going to get an h bar upon 2. From the Sy, from each Sy, we're going to get a factor of h bar upon 2, so I'll just write that as h bar upon 2 squared-- times the commutator of two matrices-- 0, 1, 1, 0 commutator with 0, minus i, i, 0. This is equal to h bar squared upon 4 times-- let's write this out. The first term is going to be this matrix times this matrix. That's going to be, again, a matrix-- 0, 1, 0, 1. So that first one is a 1. 0, 1, minus i-- oh, sorry, that's an i. 0, 1, that's an i, that's a minus i. So 0, 1, 0, i gives me an i. 0, 1, minus i, 0 gives me a 0. Second row-- 1, 0, 0, i gives me 0. And 1, 0, minus i, 0 gives me minus i. And then the second term is the flipped order, right? The commutator term. So we get minus the commutator term, which is going to be 0, minus i, 0, 1. That gives me minus i. 0, minus i, 1, 0, that gives me 0. Bottom row-- i, 0, 0, 1-- 0. And i, 0, 1, 0 give me i. OK, so notice what we get this is equal to h bar squared upon 4. And both of those matrices are the same thing. Those matrices are both i, 0, 0, minus i with minus i, 0, 0, i, giving us i, 0, 0, minus i times 2 from the two terms. The 2's cancel, and this gives me h bar squared upon 2 times i, 0, 0, minus i. But this is also known as-- pulling out an i and an h bar-- times h bar upon 2 1, 0, 0, minus 1. This is equal to i h bar Sz. So these matrices represent the angular momentum commutators quite nicely. And in fact, if you check, all the commutators work out beautifully. So quickly, just as a reminder, what are the possible measurable values of Sz for the spin 1/2 system? What possible values could you get if you measured Sz, spin in the z direction? Yeah, plus or minus h bar upon 2. Now what about the eigenvectors? What are they-- of Sz? In this notation, there are these states. There's one eigenvector, there's the other. But let's ask the same question about Sx. So for Sx, what are the allowed eigenvalues? Well, we can answer this in two ways. The first way we can answer this is by saying look, there's nothing deep about z. It was just the stupid direction we started with. We could have started by working with the eigenbasis of Sx and we would've found exactly the same story. So it must be plus or minus h bar upon 2. But the reason you make that argument is A, it's slick and B, it's the only one you can make without knowing something else about how Sx acts. But now we know what Sx is. Sx is that operator. So now we can ask, what are the eigenvalues of that operator? And if you compute the eigenvalues of that operator, you find that there are two eigenvalues Sx is equal to h bar upon 2 and Sx is equal to minus h bar upon 2. And now I can ask, well, what are the eigenvectors? Now, we know what the eigenvectors are because we can just construct the eigenvectors of Sx plus. And if you construct the eigenvectors of Sx plus-- should I take the time? How many people want me to do the eigenvectors of Sx explicitly? Yeah, that's kind of what I figured. OK, good. So the eigenvectors of Sx are, for example, on 1, 1 is equal to-- well, Sx on 1, 1, the first term, that 0, 1, is going to give me a 1, the 1, 0 is going to give me a 1. So this is h bar upon 2 coefficient of Sx on 1, 1. And Sx on 1, minus 1 is going to give me h bar upon 2 minus 1, minus 1, because all Sx does is swap the first and second components. So it gives me minus 1, takes it to the top, but that's just an overall minus sign. So again, we have the correct eigenvalues, plus and minus h bar upon 2, and now we know the eigenvector. So what does this tell us? What does it tell us that up in the x direction is equal to 1 over root 2 if I normalize things properly. Up in the z direction plus down in the z direction. That's what this is telling me. This vector is equal to up in the z direction-- that's this guy-- plus down in the z direction. But this isn't properly normalized, and properly normalizing it gives us this expression. So what does this tell us? This tells us that if we happen to know that the system is in the state with angular momentum, or spin, angular momentum in the x direction being plus 1/2, then the probability to measure up in the z direction in a subsequent measurement is 1/2. And the probability to measure down is 1/2. If you know it's up in the x direction, the probability of measuring up in the z or down in the z are equal. You're at chance. You're at even odds. Everyone agree with that? That's the meaning of this expression. And similarly, down in the x direction is equal to 1 over root 2, up in the z direction minus down in the z direction. And we get that from here. This state is explicitly, by construction, the eigenstate of Sx as we've constructed Sx. And we have a natural expression in terms of the z eigenvectors up and down. That's what this expression is giving us. It gives us an expression of the Sx eigenvector in the basis of Sz eigenvectors. So for example, this tells you that the probability to measure up in the z direction, given that we measured down in the x direction first-- so this is the conditional probability. Suppose I first measured down in the x direction, what's the probability that I subsequently measure up in the z direction? This is equal to-- well, it's the norm squared of the expansion coefficient. So first down in the x direction and the probability that we're up in the z direction is 1 upon root 2 squared. Usual rules of quantum mechanics-- take the expansion coefficient, take its norm squared, that's the probability-- 1/2. And we can do exactly the same thing for Sy. So let's do the same thing for Sy without actually working out all the details. So doing the same thing for Sy, up in the y direction is equal to 1 upon root 2 times up in the z direction plus i down in the z direction. And down in the y direction is equal to 1 upon root 2 up in the z direction minus i down in the z direction. And I encourage you to check your knowledge by deriving these eigenvectors, which you can do given our representations of Sy. Now here's a nice thing that we're going to use later. Consider the following. Consider the operator S theta, which I'm going to define as cosine theta Sz plus-- oh, sorry. I'm not even going to write it that way. So what I mean by S theta is equal to-- take our spherical directions and consider an angle in the-- this is x, let's say, y, x, z-- consider an angle theta down in the zx plane, and we can ask, what is the spin operator along the direction theta? What's the angular momentum in the direction theta? If theta, for example, is equal to pi/2, this is Sx. So I'm just defining a spin operator, which is the angular momentum along a particular direction theta in the xz plane. Everyone cool with that? This is going to turn out to be very useful for us, and I encourage you to derive the following. And if you don't derive the following, then hopefully it will be done in your recitations. Well, I will chat with your recitation instructors. And if you do this, then what are we going to get for the eigenvalues of S theta? What possible eigenvalues could S theta have? [? AUDIENCE: None. ?] PROFESSOR: Good. Why? AUDIENCE: Because it can't have any other redirection [INAUDIBLE] no matter where you start. PROFESSOR: Fabulous. OK. So the answer that was given is that it's the same, plus or minus h bar upon 2 as Sz, or indeed as Sx or Sy, because it can't possibly matter what direction you chose at the beginning. I could have called this direction theta z. How do you stop me? We could have done that. We didn't. That was our first wave answering. Second wave answering is what? Well, construct the operator S theta and find its eigenvectors and eigenvalues. So I encourage you to do that. And what you find is that, of course, the eigenvalues are plus or minus h bar upon 2 and up at the angle theta is equal to cosine of theta upon 2 up in the z direction plus sine of theta upon 2 down in the z direction. And down theta is equal to cosine theta upon 2 up in the z direction minus sine of theta over 2 down in the z direction. Oops, no. I got that wrong. This is sine, and this is minus cosine. Good. That makes more sense. OK. So let's just sanity check. These guys should be properly normalized. So if we take the norm squared of this guy, the cross terms vanish because up z and down z are orthogonal states. So we're going to get a co squared plus the sine squared. That's 1. So that's properly normalized. Same thing for this guy. The minus doesn't change anything because we norm squared. Now, I'll check that they're orthogonal. If we take this guy dotted into this guy, everything's real. So we get a cosine from up up, we're going to get a cosine sine. And from down down, we're going to get a minus cosine sine. So that gives us 0. So these guys are orthogonal, and they satisfy all the nice properties we want. So this is a good check-- do this-- of your knowledge. And if we had problem sets allowed this week, I would give you this in your problem set, but we don't. Yeah, OK. Suppose that I measure. So I want to use these states for something. Suppose that I measure Sz and find Sz is equal to 1/2 h bar plus h bar upon 2, OK, at some moment in time. First question is easy. What's the state of the system subsequent to that measurement? AUDIENCE: [INAUDIBLE] PROFESSOR: Man, you all are so quiet today. What's the state of the system subsequent to measurement that Sz is plus h bar upon 2? AUDIENCE: Up z. PROFESSOR: Up z. Good. So our state psi is up z upon measurement, OK, after measurement. And I need new chalk. OK. Now, if I measure Sx, so 1, 2, measure Sx, what will I get? AUDIENCE: [INAUDIBLE] PROFESSOR: OK. What values will I observe with what probabilities? Well, first off, what are the possible values that you can measure? AUDIENCE: Plus or minus h bar over 2. PROFESSOR: Right, the possible eigenvalues, so which is plus or minus h bar upon 2, but with what probability? AUDIENCE: [INAUDIBLE] PROFESSOR: Exactly. So we know this from the eigenstate of Sx. If we know that we're in the state up z, we can take linear combinations of this guy to show that up z is equal to 1 over root 2. So let's just check. If we add these two together, up x and down x, if we add them together, we'll get 1/2, 1/2, 2/2, a root 2 times up z, is up x plus down x, up x plus down x, dividing through by the 1/2. So here we've expressed up in the z direction in a basis of x and y, which is what we're supposed to do. So the probability that I measure a plus 1/2 as x is equal to plus 1/2 h bar upon 2 is equal to 1 over root 2 squared, so 1/2. And ditto, Sx is equal to minus h bar upon 2. Same probability. OK? What about Sy? Again, we get one of two values. Oops, plus or minus h bar upon 2. But the probability of measuring plus is equal to 1/2, and the probability that we measure minus 1/2 is 1/2. That's h bar upon 2. OK? So this should look familiar. Going back to the very first lecture, hardness was spin in the z direction, and color was spin in the x direction, I guess. And we added, at one point, a third one, which I think I called whimsy, which is equal to spin in the y direction. OK? And now all of the box operation that we used in that very first lecture, you can understand is nothing other than stringing together chains of Stern-Gerlach experiments doing Sx, Sy, and Sz. So let's be explicit about that. Let's make that concrete. Actually, let's do that here. So for example, suppose we put a random electron into an Sz color box. OK. Some are going to come out up, and some are going to come out down in the z direction. And if we send this into now an Sx color box, this is going to give us either up in the x direction or down in the x direction. And what we'll get out is 50-50, right? OK. So let's take the ones that came out down. And if we send those back into an Sz, what do we get? AUDIENCE: 50-50. PROFESSOR: Yeah, 50-50, because down x is 1 over 2 up z plus 1 over 2 down z. What was going on in that very first experiment, where we did hardness, color, hardness? Superposition. And what superposition was a hard electron? It was 1 over a 2, white plus black. We know precisely which superposition, and here it is. OK? And now, let's do that last experiment, where we take these guys, Sx down, and I'm turning this upside down, beam joiners. We take the up in the x direction and the down in the x direction, and we combine them together, and we put them back into Sz, what do we get? Well, what's the state? 1 over root 2 down x plus 1 over root 2 up x? AUDIENCE: [INAUDIBLE] PROFESSOR: It's up z. Up z into an Sz box, what do we get? Up z with 100%. That's white with 100%, or I'm sorry, hard with 100%. AUDIENCE: [INAUDIBLE] down z with a down z. PROFESSOR: Sorry. AUDIENCE: You put in down. PROFESSOR: Oh, I put in down. Shoot, I'm sorry. Well, yeah, indeed, I meant to put in up. Yes, down z with 100% confidence. If we remove the mirror, 50-50. If we add in the mirror, 100%. And the difference is whether we're taking one component of the wave function, or whether we're superposing them back together. All right? Imagine we know we have a system in this state, and I say, look, this component is also coincidentally very far away, and I'm going to not look at them. So of the ones that I look at, I have 1 over root 2 up x. But if I look at the full system, the superimposed system, that adds together to be an eigenstate of Sz. These are our color boxes. Yeah? AUDIENCE: But how do you know, when you put the two beams to the beam joiner, it serves to add their two states together? PROFESSOR: Yeah, excellent. This is a very subtle point. So here we have to decide what we mean by the beam splitter. And what I'm going to mean by the beam splitter, by the mirrors and beam joiners-- so the question is, how do we know that it does this without changing the superposition? And what I want to do is define this thing as the object that takes the two incident wave functions, and it just adds them together. It should give me the direct superposition with the appropriate phases and coefficients. So if this was plus up x, then it stays plus up x. If it's minus up x, it's going to be minus up x. Whatever the phase is of this state, when it gets here it just adds together the two components. That's my definition of that adding box. AUDIENCE: So realistically, what does that look like? PROFESSOR: Oh, what? You think I'm an experimentalist? [LAUGHTER] Look, every time I try an experiment, I get hit by a shark. OK? [LAUGHTER] Yeah, no. How you actually implement that in real systems is a more complicated story. So you should direct that question to Matt, who's a very good experimentalist. OK. So finally, let's go back to the Stern-Gerlach experiment, and let's actually run the Stern-Gerlach experiment. I guess I'll do that here. So let's think about what the Stern-Gerlach experiment looks like in this notation, and not just in this notation, in the honest language of spin. And I'm going to do a slightly abbreviated version of this because you guys can fill in the details with your knowledge of 802 and 803. OK. So here's the Stern-Gerlach experiment. We have a gradient in the magnetic field. This is the z direction. And we have a gradient where B in the z direction has some B0, a constant plus beta z. OK? So it's got a constant piece and a small gradient. Everyone cool with that? It's just the magnetic field gets stronger and stronger in the z direction, there's a constant, and then there's a rate of increase in the z direction. And I'm going to send my electron through. Now remember, my electron has a wave function, psi electron, is equal to-- well, it's got some amplitude to be up, a up z, plus some amplitude to be down, b down z. And if this is a random electron, then its state is going to be random, and a and b are going to be random numbers whose norm squared add up to 1, proper normalization. Cool? So here's our random initial state, and we send it into this region where we've got a magnetic field gradient. And what happens? Well, we know that the energy of an electron that carries some angular momentum is a constant, mu naught, times its angular momentum dotted into any ambient magnetic field. Whoops, sorry, with a minus sign. This is saying that magnets want to anti-align. Now, in particular, here we've got a magnetic field in the z direction. So this is minus mu 0 Sz Bz. And Bz was a constant, beta 0-- sorry, B0 plus beta z. So the energy has two terms. It has a constant term, which just depends on Sz, and then there's a term that depends on z as well as depending on Sz [INAUDIBLE]. So we can write this as a-- so Sz, remember what Sz is. Sz is equal to h bar upon 2, 1, 0, 0, minus 1. So this is a matrix with some coefficient up a-- do I want to write this out? Yeah, I guess I don't really need to write this out. But this is a matrix, and this is our energy operator. And it acts on any given state to give us another state back. It's an operator. OK. And importantly, I want this to be only-- this is in some region where we're doing the experiment, where we have a magnetic field gradient. Then outside of this region, we have no magnetic field and no magnetic field gradient. So it's 0 to the left and 0 to the right. So as we've talked about before, the electron feels a force due to this gradient to the magnetic field. The energy depends on z, so the derivative of the energy with respect to z, which is the force in the z direction, is non-zero. But for the moment that's a bit of a red herring. Instead of worrying about the center of mass motion, let's just focus on the overall phase. So let's take our initial electron with this initial wave function a up z plus b down z, and let's note that in this time-- so what does this matrix look like? OK. So fine, let's actually look at this matrix. So Sz is the matrix 1, 0, 0, minus 1 times h bar upon 2. So we have h bar upon 2 minus mu 0. And then B0 plus beta z, which I'll just write as B, which is equal to some constant C, 0, 0, some constant minus C. Everyone agree with that? Where a constant, I just mean it's a number, but it does depend on z, because the z is in here. Now, here's the nice thing. When we expand the energy on up z, this is equal to C of z up z, because there's our energy operator. And energy on down z is equal to minus Cz of z down to z. So up and down in the z direction are still eigenfunctions of the energy operator. We've chosen an interaction, we've chosen a potential which is already diagonal, so the energy is diagonal. It's already in its eigenbasis. So as a consequence, this is the energy, energy of the up state is equal to, in the z direction, is equal to plus C. And energy in the down state, energy of the down state, is equal to minus C of z. Everyone agree with that? So what this magnetic field does it splits the degeneracy of the up and down states. The up and down states originally had no energy splitting. They were both zero energy. We turn on this magnetic field, and one state has positive energy, and the other state has negative energy. So that degeneracy has been split. Where did that original degeneracy come from? Why did we have a degeneracy in the first place? AUDIENCE: [INAUDIBLE] PROFESSOR: Spherical symmetry. And we turn on the magnetic field, which picks out a direction, and we break that degeneracy, and we lift the splitting. OK? So here we see that the splitting is lifted. And now we want to ask, how does the wave function evolve in time? So this was our initial wave function. But we know from the Schrodinger equation that psi of t is going to be equal to-- well, those are already eigenvectors-- so a e to the minus i e up t upon h bar up z plus b e to the minus i e down t upon h bar down z. All right? But we can write this as equals a e to the i mu 0 b0 t over 2 plus i mu 0 beta t z upon 2 up z plus b same e to the minus i--oops, that should be, oh, no, that's plus-- that's e to the i mu 0 b0 t over 2. And if we write this as two exponentials, e to the minus i mu 0 beta t z upon 2-- oh no, that's a minus, good, OK-- down z. OK. So what is this telling us? So what this tells us is that we start off in a state, which has some amplitude to be up and some amplitude to be down, a and b. And at a later time t, sine of t, what we find after we run it through this apparatus is that this is the amplitude. What do you notice? Well, we notice two things. The first is that the system evolves with some overall energy. So the phase rotates as usual. These are energy eigenstates, but the amount of phase rotation depends on z. So in particular, this is e to the i some number times z. And what is e to the i some number times z? If you know you have a state that's of the form e to the i some number, which I will, I don't know, call kz times z, what is this telling you about the system? This is an eigenstate of what operator? AUDIENCE: Momentum. PROFESSOR: Momentum in the z direction. It carries what momentum? AUDIENCE: h bar kz. PROFESSOR: h bar kz, exactly. So what about this? If I have a system in this state, what can you say about its momentum in the z direction? AUDIENCE: [INAUDIBLE] PROFESSOR: It's non-zero. Right here's a z. These are a bunch of constants. It's beta, t, mu 0, 2, i. So it's non-zero. In fact, it's got minus some constant times z. Right? So this is a state with negative momentum. Everyone agree? It's got momentum z down in the z direction. What about this guy? Momentum up. So the states that are up in the z direction get a kick up. And the states that are down in the z direction get a kick down. They pick up some momentum down in the z direction. Yeah? AUDIENCE: Where did that big T come from? PROFESSOR: Big T should be little t, sorry. Sorry, just bad handwriting there. That's just t. Yeah? AUDIENCE: So in this case, the eigenvalues-- it's a function of z? PROFESSOR: Mhm. AUDIENCE: Is that allowable? PROFESSOR: Yeah. So that seems bad, but remember what I started out doing. Oop, did I erase it? Shoot, we erased it. So we started out saying, look, the wave function is up in the z direction, or really rather here, up in the z direction times some constant. Now, in general, this shouldn't be a constant. It should be some function. So this should be a function of position times up in the z direction. This is a function position times down in the z direction. So what we've done here is we said, under time evolution, that function changes in time, but it stays some linear combination up and down. So you can reorganize this as-- the time evolution equation here is an equation for the coefficients of up z and the coefficient of down z. AUDIENCE: OK. PROFESSOR: Other questions? OK. So the upshot of all this is that we run this experiment, and what we discover is that this component of the state that was up z gets a kick in the plus z direction. And any electron that came from this term in the superposition will be kicked up up z. And any electron that came from the superposition down z will have down. Now, what we really mean is not that an electron does this or does that, but rather that the initial stage of an electron that's here, with the superposition of z up and down, ends up in the state as a superposition of up z being up here and down z being down here. OK? An electron didn't do one, it didn't do the other. It ends up in a superposition, so the state at the end. So what the Stern-Gerlach experiment has done, apparatus has done, is it's correlated the position of the electron with its spin. So if you find the amplitude, to find it up here and down is very small. The amplitude to find it up here and up is very large. Similarly, the amplitude, to find it down here and up is 0. And the amplitude, to find it down here and down is large. Cool? So this is exactly what we wanted from the boxes. We wanted not to do something funny to the spins, we just wanted to correlate the position with the spin. And so the final state is a superposition of these guys. And which superposition? It's exactly this superposition. So these calculations are gone through in the notes that are going to be posted. OK. So questions at this point? Yeah? AUDIENCE: And so when you put the electron through the Stern-Gerlach device, does that count as a measurement of the particle's angular momentum? PROFESSOR: Excellent. When I put the electrons through the Stern-Gerlach device, do they come out with a definite position? AUDIENCE: No. PROFESSOR: No. At the end of the experiment, they're in a superposition of either being here or being here. And they're in a superposition of either being up spin or down spin. Have we determined the spin through putting it through this apparatus? No. We haven't done any measurement. The measurement comes when we now do the following. We put a detector here that absorbs an electron. We say, ah, yeah, it got hit. And then you've measured the angular momentum by measuring where it came out. If it comes out down here, it will be in the positive. So this is a nice example of something called entanglement. And this is where we're going to pick up next time. Entanglement says the following. Suppose I know one property of a particle, for example, that it's up here, or suppose I'm in the state psi is equal to up, so up here and up in the z direction plus down there and down in the z direction. OK? This means that, at the moment, initially, if I said, look, is it going to be up or down with equal probabilities, 1 over root 2 with equal amplitudes, if I measure spin in the z direction, what value will I get? What values could I get if I measure spin in the z direction? Plus or minus 1/2. And what are the probabilities that I measure plus 1/2 or minus 1/2? AUDIENCE: [INAUDIBLE] PROFESSOR: Even odds, right? We've done that experiment. On the other hand, if I tell you that I've measured it to be up here, what will you then deduce about what its spin is? AUDIENCE: Plus. PROFESSOR: Always plus 1/2. Did I have to do the measurement to determine that it's plus 1/2. No, because I already know the state, so I know exactly what I will get if I do the experiment. I measure up here, and then the wave function is this without the 1 over root 2, upon measurement. But as a consequence, if I subsequently measure up z, the only possible value is up in the z direction. Yeah? And this is called entanglement. And here it's entanglement of the position with the spin. And next time, what we'll do is we'll study the EPR experiment, which says the following thing. Suppose I take two electrons, OK, take two electrons, and I put them in the state one is up and the other is down, or the first is up and the second is down. All right? So up, down plus down, up. OK? I now take my two electrons, and I send them to distant places. Suppose I measure one of them to be up in the x direction. Yeah? Then I know that the other one is down in that direction. Sorry. If I measure up in the z direction, I determine the other one is down in the z direction. But suppose someone over here who's causally disconnected measures not spin in the z direction, but spin in the x direction. They'll measure one of two things, either plus or minus. Now, knowing what we know, that it's in the z eigenstate, it will be either plus 1/2 in the x direction or minus 1/2 in the x direction. Suppose I get both measurements. The distant person over here does the measurement of z and says, aha, mine is up, so the other one must be down. But the person over here doesn't measure Sz, they measure Sx, and they get that it's plus Sx. And what they've done as a result of these two experiments, Einstein, Podolsky, and Rosen say, is this electron has been measured by the distant guy to be spin z down, but by this guy to be measured spin x up. So I know Sx and Sz definitely. But that flies in the face of the uncertainty relation, which tells us we can't have spin z and spin x definitely. Einstein and Podolsky and Rosen say there's something missing in quantum mechanics, because I can do these experiments and determine that this particle is Sz down and Sx up. But what quantum mechanics says is that I may have done the measurement of Sz, but that hasn't determined anything about the state over here. There is no predetermined value. What we need to do is we need to tease out, we need an experimental version of this tension. The experimental version of this tension, fleshed out in the EPR experiment, is called Bell's inequality. We studied it in the very first lecture, and we're going to show it's a violation in the next lecture. See you guys next time. [APPLAUSE]
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https://ocw.mit.edu/courses/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2013/8.333-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. PROFESSOR: Let's say that I tell you that I'm interested in a gas that has some temperature. I specify, let's say, temperature. Pressure is room temperature, room pressure. And I tell you how many particles I have. So that's [INAUDIBLE] for you what the macro state is. And I want to then see what the corresponding micro state is. So I take one of these boxes, whereas this is a box that I draw in three dimensions, I can make a correspondence and draw this is 6N-dimensional coordinate space, which would be hard for me to draw. But basically, a space of six n dimensions. I figure out where the position, and the particles, and the momenta are. And I sort of find that there is a corresponding micro state that corresponds to this macro state. OK, that's fine. I made the correspondence. But the thing is that I can imagine lots and lots of boxes that have exactly the same macroscopic properties. That is, I can imagine putting here side by side a huge number of these boxes. All of them are described by exactly the same volume, pressure, temperature, for example. The same macro state. But for each one of them, when I go and find the macro state, I find that it is something else. So I will be having different micro states. So this correspondence is certainly something where there should be many, many points here that correspond to the same thermodynamic representation. So faced with that, maybe it makes sense to follow Gibbs and define an ensemble. So what we say is, we are interested in some particular macro state. We know that they correspond to many, many, many different potential micro states. Let's try to make a map of as many micro states that correspond to the same macro state. So consider n copies of the same macro state. And this would correspond to n different points that I put in this 6N-dimensional phase space. And what I can do is I can define an ensemble density. I go to a particular point in this space. So let's say I pick some point that corresponds to some set of p's and q's here. And what I do is I draw a box that is 6N-dimensional around this point. And I define a density in the vicinity of that point, as follows. Actually, yeah. What I will do is I will count how many of these points that correspond to micro states fall within this box. So at the end is the number of mu points in this box. And what I do is I divide by the total number. I expect that the result will be proportional to the volume of the box. So if I make the box bigger, I will have more. So I divide by the volume of the box. So this is, let's call d gamma is the volume of box. Of course, I have to do this in order to get a nice result by taking the limit where the number of members of the ensemble becomes quite large. And then presumably, this will give me a well-behaved density. In this limit, I guess I want to also have the size of the box go to 0. OK? Now, clearly, with the definitions that I have made, if I were to integrate this quantity against the volume d gamma, what I would get is the integral dN over N. N is, of course, a constant. And the integral of dN is the total number. So this is 1. So we find that this quantity rho that I have constructed satisfies two properties. Certainly, it is positive, because I'm counting the number of points. Secondly, it's normalized to 1. So this is a nice probability density. So this ensemble density is a probability density function in this phase space that I have defined. OK? All right. So once I have a probability, then I can calculate various things according to the rules of probability that we defined before. So for example, I can define an ensemble average. Maybe I'm interested in the kinetic energy of the particles of the gas. So there is a function O that depends on the sum of all of the p squareds. In general, I have some function of O that depends on p and q. And what I defined the ensemble average would be the average that I would calculate with this probability. Because I go over all of the points in phase space. And let me again emphasize that what I call d gamma then really is the product over all points. For each point, I have to make a volume in both momentum and in coordinate. It's a 6N-dimensional volume element. I have to multiply the probability, which is a function of p and q against this O, which is another function of p and q. Yes? AUDIENCE: Is the division by M necessary to make it into a probability density? PROFESSOR: Yes. AUDIENCE: Otherwise, you would still have a density. PROFESSOR: Yeah. When I would integrate then, I would get the total number. But the total number is up to me, how many members of the ensemble I took, it's not a very well-defined quantity. It's an arbitrary quantity. If I set it to become very large and divide by it, then I will get something that is nicely a probability. And we've developed all of these tools for dealing with probabilities. So that would go to waste if I don't divide by. Yes? AUDIENCE: Question. When you say that you have set numbers, do you assume that you have any more informations than just the microscopic variables GP and-- PROFESSOR: No. AUDIENCE: So how can we put a micro state in correspondence with a macro state if there is-- on the-- like, with a few variables? And do you need to-- from-- there's like five variables, defined down to 22 variables for all the particles? PROFESSOR: So that's what I was saying. It is not a one-to-one correspondence. That is, once I specify temperature, pressure, and the number of particles. OK? Yes? AUDIENCE: My question is, if you generate identical macro states, and create-- which macro states-- PROFESSOR: Yes. AUDIENCE: Depending on some kind of a rule on how you make this correspondence, you can get different ensemble densities, right? PROFESSOR: No. That is, if I, in principle and theoretically, go over the entirety of all possible macroscopic boxes that have these properties, I will be putting infinite number of points in this. And I will get some kind of a density. AUDIENCE: What if you, say, generate infinite number of points, but all in the case when, like, all molecules of gas are in right half of the box? PROFESSOR: OK. Is that a thermodynamically equilibrium state? AUDIENCE: Did you mention it needed to be? PROFESSOR: Yes. I said that-- I'm talking about things that can be described macroscopically. Now, the thing that you mentioned is actually something that I would like to work with, because ultimately, my goal is not only to describe equilibrium, but how to reach equilibrium. That is, I would like precisely to answer the question, what happens if you start in a situation where all of the gas is initially in one half of the room? And as long as there is a partition, that's a well-defined macroscopic state. And then I remove the partition. And suddenly, it is a non-equilibrium state. And presumably, over time, this gas will occupy that. So there is a physical process that we know happens in nature. And what I would like eventually to do is to also describe that physical process. So what I will do is I will start with the initial configuration with everybody in the half space. And I will calculate the ensemble that corresponds to that. And that's unique. Then I remove the partition. Then each member of the ensemble will follow some trajectory as it occupies eventually the entire box. And we would like to follow how that evolution takes place and hopefully show that you will always have, eventually, at the end of the day, the gas occupying the system uniform. AUDIENCE: Yeah, but just would it be more correct to generate many, many different micro states as the macro states which correspond to them? And how many different-- PROFESSOR: What rule do you use for generating many, many micro states? AUDIENCE: Like, all uniformly arbitrary perturbations of particles always to put them in phase space. And look to-- like, how many different micro states give rise to the same macro state? PROFESSOR: Oh, but you are already then talking about the macro state? AUDIENCE: A portion-- which description do you use as the first one to generate the second one? So in your point of view, let's do-- [INAUDIBLE] to macro states, and go to micro state. But can you reverse? PROFESSOR: OK. I know that the way that I'm presenting things will lead ultimately to a useful description of this procedure. You are welcome to try to come up with a different prescription. But the thing that I want to ensure that you agree is that the procedure that I'm describing here has no logical inconsistencies. I want to convince you of that. I am not saying that this is necessarily the only one. As far as I know, this is the only one that people have worked with. But maybe somebody can come up with a different prescription. So maybe there is another one. Maybe you can work on it. But I want you to, at this point, be convinced that this is a well-defined procedure. OK? AUDIENCE: But because it's a well-defined procedure, if you did exist on another planet or in some universe where the physics were different, the point is, you can use this. But can't you use this for information in general when you want to-- like, if you have-- the only requirement is that at a fine scale, you have a consistent way of describing things; and at a large scale, you have a way of making sense of generalizing. PROFESSOR: Right. AUDIENCE: So it's sort of like a compression of data, or I use [INAUDIBLE]. PROFESSOR: Yeah. Except that part of this was starting with some physics that we know. So indeed, if you were in a different universe-- and later on in the course, we will be in a different universe where the rules are not classical or quantum-mechanical. And you have to throw away this description of what a micro state is. And you can still go through the entire procedure. But I want to do is to follow these set of equations of motion and this description of micro state, and see where it leads us. And for the gas in this room, it is a perfectly good description of what's happened. Yes? AUDIENCE: Maybe a simpler question. Is big rho defined only in spaces where there are micro states? Like, is there anywhere where there isn't a micro state? PROFESSOR: Yes, of course. So if I have-- thinking about a box, and if I ask what is rho out here, I would say the answer is rho is 0. But if you like, you can say rho is defined only within this space of the box. So the description of the macro state which has something to do with the box, over which I am considering, will also limit what I can describe. Yes. And certainly, as far as if I were to change p with velocity, let's say, then you would say, a space where V is greater than speed of light is not possible. That's the point. So your rules of physics will also define implicitly the domain over which this is there. But that's all part of mechanics. So I'm going to assume that the mechanics part of it is you are comfortable. Yes? AUDIENCE: In your definition of the ensemble average, are you integrating over all 6N dimensions of phase space? PROFESSOR: Yes. AUDIENCE: So why would your average depend on p and q? If you integrate? PROFESSOR: The average is of a function of p and q. So in the same sense that, let's say, I have a particle of gas that is moving on, and I can write the symbol, p squared over 2m, what is this average? The answer will be KT over 2, for example. It will not depend on p. But the quantity that I'm averaging inside the triangle is a function of p and q. Yes? AUDIENCE: So if it's an integration, or basically the--? PROFESSOR: The physical limit of the problem. AUDIENCE: Given a macro state? PROFESSOR: Yes. So typically, we will be integrating q over the volume of a box, and p from minus infinity to infinity, because classically, without relativity, this is a lot. Yes? AUDIENCE: Sorry. So why is the [INAUDIBLE] from one end for every particle, instead of just scattering space, would you have a [INAUDIBLE]? Or is that the same thing? PROFESSOR: I am not sure I understand the question. So if I want to, let's say, find out just one particle that is somewhere in this box, so there is a probability that it is here, there is a probability that it is there, there is a probability that it is there. The integral of that probability over the volume of the room is one. So how do I do that? I have to do an integral over dx, dy, dz, probability as a function of x, y, and z. Now I just repeat that 6N times. OK? All right. So that's the description. But the first question to sort of consider is, what is equilibrium in this perspective? Now, we can even be generous, although it's a very questionable thing, to say that, really, when I sort of talk about the kinetic energy of the gas, maybe I can replace that by this ensemble average. Now, if I'm in equilibrium, the results should not depend as a function of time. So I expect that if I'm calculating things in equilibrium, the result of equations such as this should not depend on time, which is actually a problem. Because we know that if I take a picture of all of these things that I am constructing my ensemble with and this picture is at time t, at time t plus dt, all of the particles have moved around. And so the point that was here, the next instant of time is going to be somewhere else. This is going to be somewhere else. Each one of these is flowing around as a function of time. OK? So the picture that I would like you to imagine is you have a box, and there's a huge number of bees or flies or whatever your preferred insect is, are just moving around. OK? Now, you can sort of then take pictures of this cluster. And it's changing potentially as a function of time. And therefore, this density should potentially change as a function of time. And then this answer could potentially depend on time. So let's figure out how is this density changing as a function of time. And hope that ultimately, we can construct a solution for the equation that governs the change in density as a function of time that is in fact invariant in time. It is going back to my flies or bees or whatever, you can imagine a circumstance in which the bees are constantly moving around. Each individual bee is now here, then somewhere else. But all of your pictures have the same density of bees, because for every bee that left the box, there was another bee that came in its place. So one can imagine a kind of situation where all of these points are moving around, yet the density is left invariant. And in order to find whether such a density is possible, we have to first know what is the equation that governs the evolution of that density. And that is given by the Liouville's equation. So this governs evolution of rho with time. OK. So let's kind of blow off the picture that we had over there. Previously, they're all of these coordinates. There is some point in coordinate space that I am looking at. Let's say that the point that I am looking at is here. And I have constructed a box around it like this in the 6N-dimensional space. But just to be precise, I will be looking at some particular coordinate q alpha and the conjugate momentum p alpha. So this is my original point corresponds to, say, some specific version of q alpha p alpha. And that I have, in this pair of dimensions, created a box that in this direction has size dq alpha, in this direction has size dp alpha. OK? And this is the picture that I have at some time t. OK? Then I look at an instant of time that is slightly later. So I go to a time that is t plus dt. I find that the point that I had initially over here as the center of this box has moved around to some other location that I will call q alpha prime, p alpha prime. If you ask me what is q alpha prime and p alpha prime, I say, OK, I know that, because I know with equations of motion. If I was running this on a computer, I would say that q alpha prime is q alpha plus the velocity q alpha dot dt, plus order of dt squared, which hopefully, I will choose a sufficient and small time interval I can ignore. And similarly, p alpha prime would be p alpha plus p alpha dot dt order of dt squared. OK? So any point that was in this box will also move. And presumably, close-by points will be moving to close-by points. And overall, anything that was originally in this square actually projected from a larger dimensional cube will be part of a slightly distorted entity here. So everything that was here is now somewhere here. OK? I can ask, well, how wide is this new distance that I have? So originally, the two endpoints of the square were a distance dq alpha apart. Now they're going to be a distance dq alpha prime apart. What is dq alpha prime? I claim that dq alpha prime is whatever I had originally, but then the two N's are moving at slightly different velocities, because the velocity depends on where you are in phase space. And so the difference in velocity between these two points is really the derivative of the velocity with respect to the separation that I have between those two points, which is dq times dq alpha. And this is how much I would have expanded plus higher order. And I can apply the same thing in the momentum direction. The new vertical separation dp alpha prime is different from what it was originally, because the two endpoints got stretched. The reason they got stretched was because their velocities were different. And their difference is just the derivative of velocity with respect to separation times their small separation. And if I make everything small, I can, in principle, write higher order terms. But I don't have to worry about that. OK? So I can ask, what is the area of this slightly distorted square? As long as the dt is sufficiently small, all of the distortions, et cetera, will be small enough. And you can convince yourself of this. And what you will find is that the product of dq alpha prime, dp alpha prime, if I were to multiply these things, you can see that dq alpha and dp alpha is common to the two of them. So I have dq alpha, dp alpha. From multiplying these two terms, I will get one. And then I will get two terms that are order of dt, that I will get from dq alpha dot with respect to dq alpha, plus dp alpha dot with respect to dp alpha. And then there will be terms that are order of dt squared and higher order types of things. OK? So the distortion in the area of this particle, or cross section, is governed by something that is proportional to dt. And dq alpha dot over dq alpha plus dp alpha dot dp alpha. But I have formally here for what P i dot and qi dot are. So this is the dot notation for the time derivative other than the one that I was using before. So q alpha dot, what do I have for it? It is dh by dp alpha. So this is d by dq alpha of the H by dp alpha. Whereas p alpha dot, from which I have to evaluate d by dp alpha, p alpha dot is minus dH by dq alpha. So what do I have? I have two second derivatives that appear with opposite sign and hence cancel each other out. OK? So essentially, what we find is that the volume element is preserved under this process. And I can apply this to all of my directions. And hence, conclude that the initial volume that was surrounding my point is going to be preserved. OK? So what that means is that these classical equations of motion, the Hamiltonian equations, for this description of micro state that involves the coordinates and momenta have this nice property that they preserve volume of phase space as they move around. Yes? AUDIENCE: If the Hamiltonian has expontential time dependence, that doesn't work anymore. PROFESSOR: No. So that's why I did not put that over there. Yes. And actually, this is sometimes referred to as being something like an incompressible fluid. Because if you kind of deliver hydrodynamics for something like water if you regard it as incompressible, the velocity field has the condition that the divergence of the velocity is 0. Here, we are looking at a 6N-dimensional mention velocity field that is composed of q alpha dot and p alpha dot. And this being 0 is really the same thing as the divergence in this 6N-dimensional space is 0. And that's a property of the Hamiltonian dynamics that one has. Yes? AUDIENCE: Could you briefly go over why you have to divide by the separation when you expand the times between the displacement? PROFESSOR: Why do I have to multiply by the separation? AUDIENCE: Divide by. PROFESSOR: Where do I divide? AUDIENCE: dq alpha by-- PROFESSOR: Oh, this. Why do I have to take a derivative. So I have two points here. All of my points are moving in time. So if these things were moving with uniform velocity, one second later, this would have moved here, this would have moved the same distance, so that the separation between them would have been maintained if they were moving with the same velocity. So if you are following somebody and you are moving with the same velocity as them, thus, your separation does not change. But if one of you is going faster than the other one, then the difference in velocity will determine how you separate. And what is the difference in velocity? The difference in velocity depends, in this case, to how far apart the points are. So the difference between velocity here and velocity here is the derivative of velocity as a function of this coordinate. Derivative of velocity as a function of that coordinate multiplied by the separation. OK? So what does this incompressibility condition mean? It means that however many points I had over here, they end up in a box that has exactly the same volume, which means that the density is going to be the same around here and around this new point. So essentially, what we have is that the rho at the new point, p prime, q prime, and time, t, plus dt, is the same thing as the rho at the old point, p, q, at time, t. Again, this is the incompressibility condition. Now we do mathematics. So let's write it again. So I've said, in other words, that rho p, q, t is the same as the rho at the new point. What's the momentum at the new point? It is p plus. For each component, it is p alpha plus p alpha dot. Let's put it this way. p plus p dot dt, q plus q dot dt, and t plus dt. That is, if I look at the new location compared to the old location, the time changed, the position changed, the momentum changed. They all changed-- in each arguments changed infinitesimally by an amount that is proportional to dt. And so what I can do is I can expand this function to order of dt. So I have rho at the original point. So this is all mathematics. I just look at variation with respect to each one of these arguments. So I have a sum over alpha, p alpha dot d rho by dp alpha plus q alpha dot d rho by dq alpha plus the explicit derivative, d rho by dt. This entirety is going to be multiplied by dt. And then, in principle, the expansion would have higher order terms. OK? Now, of course, the first term vanishes. It is the same on both times. So the thing that I will have to set to 0 is this entity over here. Now, quite generally, if you have a function of p, q, and t, you evaluate it at the old point and then evaluate at the new point. One can define what is called a total derivative. And just like here, the total derivative will come from variations of all of these arguments. We'll have a partial derivative with respect to time and partial derivatives with respect to any of the other arguments. So I wrote this to sort of make a distinction between the symbol that is commonly used, sometimes d by dt, which is straight, sometimes Df by Dt. And this is either a total derivative or a streamline derivative. That is, you are taking derivatives as you are moving along with the flow. And that is to be distinguished from these partial derivatives, which is really sitting at some point in space and following how, from one time instant to another time instant, let's say the density changes. So Df by Dt with a partial really means sit at the same point. Whereas this big Df by Dt means, go along with the flow and look at the changes. Now, what we have established here is that for the density, the density has some special character because of this Liouville's theorem, that the streamlined derivative is 0. So what we have is that d rho by dt is 0. And this d rho by dt I can also write down as d rho by dt plus sum over all 6N directions, d rho by dp alpha. But then I substitute for p alpha dot from here. p dot is minus dH by dq. And then I add d rho by dq alpha. q alpha dot is dH by dp r. OK? So then I can take this combination with the minus sign to the other side and write it as d rho by dt is something that I will call the Poisson bracket of H and rho, where, quite generally, if I have two functions in phase space that is defending on B and q, this scalary derivative the Poisson bracket is defined as the sum over all 6N possible variation. The first one with respect to q, the second one with respect to p. And then the whole thing with the opposite sign. dA, dP, dB, dq. So this is the Poisson bracket. And again, from the definition, you should be able to see immediately that Poisson bracket of A and B is minus the Poisson bracket of B and A. OK? Again, we ask the question that in general, I can construct in principle a rho of p, q, let's say, for an equilibrium ensemble. But then I did something, like I removed a partition in the middle of the gas, and the gas is expanding. And then presumably, this becomes a function of time. And since I know exactly how each one of the particles, and hence each one of the micro states is evolving as a function of time, I should be able to tell how this density in phase space is changing. So this perspective is, again, this perspective of looking at all of these bees that are buzzing around in this 6N-dimensional space, and asking the question, if I look at the particular point in this 6N-dimensional space, what is the density of bees? And the answer is that it is given by the Poisson bracket of the Hamiltonian that governs the evolution of each micro state and the density function. All right. So what does it mean? What can we do with this? Let's play around with it and look at some consequences. But before that, does anybody have any questions? OK. All right. We had something that I just erased. That is, if I have a function of p and q, let's say it's not a function of time. Let's say it's the kinetic energy for this system where, at t equals to 0, I remove the partition, and the particles are expanding. And let's say the other place you have a potential, so your kinetic energy on average is going to change. You want to know what's happening to that. So you calculate at each instant of time an ensemble average of the kinetic energy or any other quantity that is interesting to you. And your prescription for calculating an ensemble average is that you integrate against the density the function that you are proposing to look at. Now, in principle, we said that there could be situations where this is dependent on time, in which case, your average will also depend on time. And maybe you want to know how this time dependence occurs. How does the kinetic energy of a gas that is expanding into some potential change on average? OK. So let's take a look. This is a function of time, because, as we said, these go inside the average. So really, the only explicit variable that we have here is time. And you can ask, what is the time dependence of this quantity? OK? So the time dependence is obtained by doing this, because essentially, you would be adding things at different points in p, q. And at each point, there is a time dependence. And you take the derivative in time with respect to the contribution of that point. So we get something like this. Now you say, OK, I know what d rho by dt is. So this is my integration over all of the phase space. d rho by dt is this Poisson bracket of H and rho. And then I have O. OK? Let's write this explicitly. This is an integral over a whole bunch of coordinates and momenta. There is, for the Poisson bracket, a sum over derivatives. So I have a sum over alpha-- dH by dq alpha, d rho by dp alpha minus dH by dp alpha, d rho by dq alpha. And that Poisson bracket in its entirety then gets multiplied by this function of phase space. OK. Now, one of the mathematical manipulations that we will do a lot in this class. And that's why I do this particular step, although it's not really necessary to the logical progression that I'm following, is to remind you how you would do an integration by parts when you're faced with something like this. An integration by parts is applicable when you have variables that you are integrating that are also appearing as derivatives. And whenever you are integrating Poisson brackets, you will have derivatives for the Poisson bracket. And the integration would allow you to use integration by parts. And in particular, what I would like to do is to remove the derivative that acts on the densities. So I'm going to essentially rewrite that as minus an integral that involves-- again. I don't want to keep rewriting that thing. I want to basically take the density out and then have the derivative, which is this d by dp in the first term and d by dq in the second term, act on everything else. So in the first case, d by dp alpha will act on O and dH by dq alpha. And in the second case, d by dq alpha will act on O and dH by dp alpha. Again, there is a sum over alpha that is implicit. OK? Again, there is a minus sign. So every time you do this procedure, there is this. But every time, you also have to worry about surface terms. So on the surface, you would potentially have to evaluate things that involve rho, O, and these d by d derivatives. But let's say we are integrating over momentum from minus infinity to infinity. Then the density evaluated at infinity momenta would be 0. So practicality, in all cases that I can think of, you don't have to worry about the boundary terms. So then when you look at these kinds of terms, this d by dp alpha can either act on O. So I will get dO by dp alpha, dH by dq alpha. Or it can act on dH by d alpha. So I will get plus O d2 H, dp alpha dq alpha. And similarly, in this term, either I will have dO by dq alpha, dH by dp alpha, or O d2 H, dq alpha dp alpha. Once more, the second derivative terms of the Hamiltonian, the order is not important. And what is left here is this set of objects, which is none other than the Poisson bracket. So I can rewrite the whole thing as d by dt of the expectation value of this quantity, which potentially is a function of time because of the time dependence of my density is the same thing as minus an integration over the entire phase space of rho against this entity, which is none other than the Poisson bracket of H with O. And this integration over density of this entire space is just our definition of the expectation value. So we get that the time derivative of any quantity is related to the average of its Poisson bracket with the Hamiltonian, which is the quantity that is really governing time dependences. Yes? AUDIENCE: Could you explain again why the time derivative when N is the integral, it's rho as a partial derivative? PROFESSOR: OK. So suppose I'm doing a two-dimensional integral over p and q. So I have some contribution from each point in this p and q. And so my integral is an integral dpdq, something evaluated at each point in p, q that could potentially depend on time. Imagine that I discretize this. So I really-- if you are more comfortable, you can think of this as a limit of a sum. So this is my integral. If I'm interested in the time dependence of this quantity-- and I really depends only on time, because I integrated over p and q. So if I'm interested in something that is a sum of various terms, each term is a function of time. Where do I put the time dependence? For each term in this sum, I look at how it depends on time. I don't care on its points to the left and to the right. OK? Because the big D by Dt involves moving with the streamline. I'm not doing any moving with the streamline. I'm looking at each point in this two-dimensional space. Each point gives a contribution at that point that is time-dependent. And I take the derivative with respect to time at that point. Yes? AUDIENCE: Couldn't you say that you have function O as just some function of p and q, and its time derivative would be Poisson bracket? PROFESSOR: Yes. AUDIENCE: And does the average of the time derivative would be the average of Poisson bracket, and you don't have to go through all the-- PROFESSOR: No. But you can see the sign doesn't work. AUDIENCE: How come? PROFESSOR: [LAUGHS] Because of all of these manipulations, et cetera. So the statement that you made is manifestly incorrect. You can't say that the time dependence of this thing is the-- whatever you were saying. [LAUGHS] AUDIENCE: [INAUDIBLE]. PROFESSOR: OK. Let's see what you are saying. AUDIENCE: So Poisson bracket only counts for in place for averages, right? PROFESSOR: OK. So what do we have here? We have that dp by dt is the Poisson bracket of H and rho. OK? And we have that O is an integral of rho O. Now, from where do you conclude from this set of results that d O by dt is the average of a Poisson bracket that involves O and H, irrespective of what we do with the sign? AUDIENCE: Or if you look not at the average failure of O but at the value of O and point, and take-- I guess it would be streamline derivative of it. So that's assuming that you're just like assigning value of O to each point, and making power changes with time as this point moves across the phase space. PROFESSOR: OK. But you still have to do some bit of derivatives, et cetera, because-- AUDIENCE: But if you know that the volume of the in phase space is conserved, then we basically don't care much that the function O is any much different from probability density. PROFESSOR: OK. If I understand correctly, this is what you are saying. Is that for each representative point, I have an O alpha, which is a function of time. And then you want to say that the average of O is the same thing as the sum over alpha of O alpha of t's divided by N, something like this. AUDIENCE: Eh. Uh, I want to first calculate what does time derivative of O? O remains in a function of time and q and p. So I can calculate-- PROFESSOR: Yes. So this O alpha is a function of AUDIENCE: So if I said-- PROFESSOR: Yes. OK. Fine. AUDIENCE: O is a function of q and p and t, and I take a streamline derivative of it. So filter it with respect to t. And then I average that thing over phase space. And then I should get the same version-- PROFESSOR: You should. AUDIENCE: --perfectly. But-- PROFESSOR: Very quickly, I don't think so. Because you are already explaining things a bit longer than I think I went through my derivation. But that's fine. [LAUGHS] AUDIENCE: Is there any special-- PROFESSOR: But I agree in spirit, yes. That each one of these will go along its streamline, and you can calculate the change for each one of them. And then you have to do an average of this variety. Yes. AUDIENCE: [INAUDIBLE] when you talk about time derivative of probability density, it's Poisson bracket of H and rho. But when you talk about time derivative of average, you have to add the minus sign. PROFESSOR: And if you do this correctly here, you should get the same result. AUDIENCE: Oh, OK. PROFESSOR: Yes. AUDIENCE: Well, along that line, though, are you using the fact that phase space volume is incompressible to then argue that the total time derivative of the ensemble average is the same as the ensemble average of the total time derivative of O, or not? PROFESSOR: Could you repeat that? Mathematically, you want me to show that the time derivative of what quantity? AUDIENCE: Of the average of O. PROFESSOR: Of the average of O. Yes. AUDIENCE: Is it in any way related to the average of dO over dt? PROFESSOR: No, it's not. Because O-- I mean, so what do you mean by that? You have to be careful, because the way that I'm defining this here, O is a function of p and q. And what you want to do is to write something that is a sum over all representative points, divided by the total number, some kind of an average like this. And then I can define a time derivative here. Is that what you are-- AUDIENCE: Well, I mean, I was thinking that even if you start out with your observable being defined for every point in phase space, then if you were to, before doing any ensemble averaging, if you were to take the total time derivative of that, then you would be accounted for p dot and q dot as well, right? And then if you were to take the ensemble average of that quantity, could you arrive at the same result for that? PROFESSOR: I'm pretty sure that if you do things consistently, yes. That is, what we have done is essentially we started with a collection of trajectories in phase space and recast the result in terms of density and variables that are defined only as positions in phase space. The two descriptions completely are equivalent. And as long as one doesn't make a mistake, one can get one or the other. This is actually a kind of a well-known thing in hydrodynamics, because typically, you write down, in hydrodynamics, equations for density and velocity at each point in phase space. But there is an alternative description which we can say that there's essentially particles that are flowing. And particle that was here at this location is now somewhere else at some later time. And people have tried hard. And there is a consistent definition of hydrodynamics that follows the second perspective. But I haven't seen it as being practical. So I'm sure that everything that you guys say is correct. But from the experience of what I know in hydrodynamics, I think this is the more practical description that people have been using. OK? So where were we? OK. So back to our buzzing bees. We now have a way of looking at how densities and various quantities that you can calculate, like ensemble averages, are changing as a function of time. But the question that I had before is, what about equilibrium? Because the thermodynamic definition of equilibrium, and my whole ensemble idea, was that essentially, I have all of these boxes and pressure, volume, everything that I can think of, as long as I'm not doing something that's like opening the box, is perfectly independent of time. So how can I ensure that various things that I calculate are independent of time? Clearly, I can do that by having this density not really depend on time. OK? Now, of course, each representative point is moving around. Each micro state is moving around. Each bee is moving around. But I want the density that is characteristic of equilibrium be something that does not change in time. It's 0. And so if I posit that this particular function of p and q is something that is not changing as a function of time, I have to require that the Poisson bracket of that function of p and q with the Hamiltonian, which is another function of p and q, is 0. OK? So in principle, I have to go back to this equation over here, which is a partial differential equation in 6N-dimensional space and solve with equal to 0. Of course, rather than doing that, we will guess the answer. And the guess is clear from here, because all I need to do is to make rho equilibrium depend on coordinates in phase space through some functional dependence on Hamiltonian, which depends on the coordinates in phase space. OK? Why does this work? Because then when I take the Poisson bracket of, let's say, this function of H with H, what do I have to do? I have to do a sum over alpha d rho with respect to, let's say-- actually, let's write it in this way. I will have the dH by dp alpha from here. I have to multiply by d rho by dq alpha. But rho is a function of only H. So I have to take a derivative of rho with respect to its argument H. And I'll call that rho prime. And then the derivative of the argument, which is H, with respect to q alpha. That would be the first term. The next term would be, with the minus sign, from the H here, dH by dq alpha, the derivative of rho with respect to p alpha. But rho only depends on H, so I will get the derivative of rho with respect to its one argument. The derivative of that argument with respect to p alpha. OK? So you can see that up to the order of the terms that are multiplying here, this is 0. OK? So any function I choose of H, in principle, satisfies this. And this is what we will use consistently all the time in statistical mechanics, in depending on the ensemble that we have. Like you probably know that when we are in the micro canonical ensemble, we look at-- in the micro canonical ensemble, we'd say what the energy is. And then we say that the density is a delta function-- essentially zero. Except places, it's the surface that corresponds to having the right energy. So you sort of construct in phase space the surface that has the right energy. So it's a function of these four. So this is the micro canonical. When you are in the canonical, I use a rho this is proportional to e to the minus beta H and other functional features. So it's, again, the same idea. OK? That's almost but not entirely true, because sometimes there are also other conserved quantities. Let's say that, for example, we have a collection of particles in space in a cavity that has the shape of a sphere. Because of the symmetry of the problem, angular momentum is going to be a conserved quantity. Angular momentum you can also write as some complicated function of p and q. For example, p cross q summed over all of the particles. But it could be some other conserved quantity. So what does this conservation law mean? It means that if you evaluate L for some time, it is going to be the same L for the coordinates and momenta at some other time. Or in other words, dL by dt, which you obtained by summing over all coordinates dL by dp alpha, p alpha dot, plus dL by dq alpha q alpha dot. Essentially, taking time derivatives of all of the arguments. And I did not put any explicit time dependence here. And this is again sum over alpha dL by dp alpha. p alpha dot is minus dH by dq alpha. And dL by dq alpha, q alpha dot is dH by dp alpha. So you are seeing that this is the same thing as the Poisson bracket of L and H. So essentially, conserved quantities which are essentially functions of coordinates and momenta that you calculate that don't change as a function of time are also quantities that have zero Poisson brackets with the Hamiltonian. So if I have a conserved quantity, then I have a more general solution. To my d rho by dt equals to 0 requirement. I could make a rho equilibrium which is a function of H of p and q, as well as, say, L of p and q. And when you go through the Poisson bracket process, you will either be taking derivatives with respect to the first argument here. So you would get rho prime with respect to the first argument. And then you would get the Poisson bracket of H and H. Or you would be getting derivatives with respect to the second argument. And then you would be getting the Poisson bracket of L and H. And both of them are 0. By definition, L is a conserved quantity. So any solution that's a function of Hamiltonian, the energy of the system is conserved as a function of time, as well as any other conserved quantities, such as angular momentum, et cetera, is certainly a valid thing. So indeed, when I drew here, in the micro canonical ensemble, a surface that corresponds to a constant energy, well, if I am in a spherical cavity, only part of that surface that corresponds to the right angular momentum is going to be accessible. So essentially, what I know is that if I have conserved quantities, my trajectories will explore the subspace that is consistent with those conservation laws. And this statement here is that ultimately, those spaces that correspond to the appropriate conservation law are equally populated. Rho is constant around. So in some sense, we started with the definition of rho by putting these points around and calculating probability that way, which was my objective definition of probability. And through this Liouville theorem, we have arrived at something that is more consistent with the subjective assignment of probability. That is, the only thing that I know, forgetting about the dynamics, is that there are some conserved quantities, such as H, angular momentum, et cetera. And I say that any point in phase space that does not violate those constants in this, say, micro canonical ensemble would be equally populated. There was a question somewhere. Yes? AUDIENCE: So I almost feel like the statement that the rho has to not change in time is too strong, because if you go over to the equation that says the rate of change is observable is equal to the integral that was with a Poisson bracket of rho and H, then it means that for any observable, it's constant in time, right? PROFESSOR: Yes. AUDIENCE: So rho of q means any observable we can think of, its function of p and q is constant? PROFESSOR: Yep. Yep. Because-- and that's the best thing that we can think of in terms of-- because if there is some observable that is time-dependent-- let's say 99 observables are time-independent, but one is time-dependent, and you can measure that, would you say your system is in equilibrium? Probably not. OK? AUDIENCE: It seemed like the same method that you did to show that the density is a function of [INAUDIBLE], or [INAUDIBLE] that it's the function of any observable that's a function of q. Right? PROFESSOR: Observables? No. I mean, here, if this answer is 0, it states something about this quantity averaged. So if this quantity does not change as a function of time, it is not a statement that H and 0 is 0. A statement that H and O is 0 is different from its ensemble average being 0. What you can show-- and I think I have a problem set for that-- is that if this statement is correct for every O that the average is 0, then your rho has to satisfy this theorem equals to-- Poisson bracket of rho and H is equal to 0. OK. So now the big question is the following. We arrived that the way of thinking about equilibrium in a system of particles and things that are this many-to-one mapping, et cetera, in terms of the densities-- we arrived that the definition of what the density is going to be in equilibrium. But the thermodynamic statement is much, much more severe. The statement, again, is that if I have a box and I open the door of the box, the gas expands to fill the empty space or the other part of the box. And it will do so all the time. Yet the equations of motion that we have over here are time reversal invariant. And we did not manage to remove that. We can show that this Liouville equation, et cetera, is also time reversal invariant. So for every case, if you succeed to show that there is a density that is in half of the box and it expands to fill the entire box, there will be a density that presumably goes the other way. Because that will be also a solution of this equation. So when we sort of go back from the statement of what is the equilibrium solution and ask, do I know that I will eventually reach this equilibrium solution as a function of time, we have not shown that. And we will attempt to do so 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. PROFESSOR: All right, a couple of announcements. Next week two minor celebrations, maybe receptions we would call them. Quiz 2 Tuesday based on homework 2, and periodic table quiz Thursday. I provide the numbers, you provide the letters. I guess that's how it works. And the contest ends Friday five PM. Last day we looked at the Bohr Model and we developed equations for the radius of the electronic and the orbit of the one electron atom. The energy of the electron and the velocity of the electron. And we found that for all of these they were a function of n. Quantum number. n takes on discrete values. One, two, three, and so on. We say that these energies radii velocities are quantized. They take discrete values. And then later in the lecture we started looking for evidence. And we found ourselves in an exercise of reconciliation with data taken by Angstrom about 50 years earlier and fit to an equation by J.J. Balmer. And we were part way through that and adjourned. So I'd like to pick up the discussion at that point. I've done a different drawing of what's going on inside the gas discharge tube. Last day I had the ballistic electronic here, and this is boiling off the cathodes. The cathode is inside the gas discharge tube and this electron, if the voltage is high enough, will leave the cathode and shoot across this low pressure gas, which contains, among other things, atomic hydrogen. And I'm trying to depict the atomic hydrogen atom here. Here's the proton, which is the nucleus-- that's the sum total of the contents of the nucleus-- and here's the lone electron that is orbiting the nucleus at some initial value. n sub i. Could be ground state. Doesn't necessarily have to be. With some thermal energy this could be n greater than one. Then we reason that if the electron, ballistic electron, and it's trajectory across the gas discharge tube over to the anode, which is charged positively. If it collided with this electron it could impart some of its energy, thereby promoting the electron from ni up to nf, the final level. And the electron would be up here. And for this transition there would be an energy cost. That energy cost is delta e. Delta e is the energy to go from ni to nf. And so the kinetic energy-- half mv squared of the incident electron-- is diminished by this amount. And the electronic continues on it's merry way at a slower speed. We assume it's mass doesn't change. The only way we can change its energy is to slow it down. And there's a conservation of energy, so the sum of the energy of the scattered electron and the transition energy of the electron within hydrogen must equal the incident kinetic energy of the ballistic electron. But there's more. This is a-- no pun intended-- a one shot deal. This is ballistics, and so the electron is not sustainably promoted. It falls back down. And when it falls back down we have the transition energy now given off. Here, to promote we had to call for energy. When the electron falls down it gives off that energy. And that energy is given off in the form of an emitted photon. And it's that emitted photon and it's that emitted photon that ultimately gives rise to the lines. The lines in the spectrum are generated by the emitted photon here. Everything else is preamble to this event, and this event gives rise to the emitted photons. And I think that's about where we got last day with the reconciliation. So let's look carefully here. We recognize that there needs to be conservation of energy again. In other words, the energy of the emitted photon, the energy of the photon, which we know from Planck is h times nu. I'm trying to distinguish. This, I'm making nu, the Greek symbol nu. And I put a little descender on it. It looks like a v, but I put a little ascender here to distinguish it. This is lowercase v, as in mv squared. This is nu. So h nu, or it could be hc over lambda. Or it could be hc nu bar. Three ways of writing the energy of the photon. And that must equal delta e of the transition. So let's keep going. We know that the delta e, the transition is given by the Bohr Model. Delta e transition will equal e final minus e initial, which will be minus kz squared-- I'm writing this generally, in this case with atomic hydrogen z as 1-- with minus k times 1 over n final squared minus one over n initial squared. What we can do then is equate these and roll them around to isolate nu bar. nu bar, then, equals minus kz squared over the product of the Planck constant, the speed of light, 1 over nf squared minus 1 over ni squared. Now for the Balmer series, that is to say the Balmer series of lines, that turns out to be a series where all of the transitions end up on n equals 2. We said nf equals 2. z equals 1 because we're talking about atomic hydrogen. Then what we have is a set of translations that go 1 over 2 squared minus 1 over ni squared. ni must be greater than nf, so ni must come from the set 3, 4, 5, et cetera. Furthermore, I'm going to put z equals 1. Let's evaluate k. We know that's 2.18 times 10 to the minus 18 joules, or 13.6 electron volts. And we know the Planck constant, 6.6 times 10 to the minus 36. And this is 3 times 18 of the 8th all in SI units. So this gives me 1.1 times 10 to the 7th in reciprocal meters. And if I put all this together I end up with exactly they equation that was published by Balmer. Exactly Balmer's equation in 1885, rewritten to express it in SI units. 1 over 2 squared minus 1 over-- I'm just going to put ni squared-- or ni equals 3, 4, 5, 6. This is Balmer exactly. Balmer exactly. So the assumption of this planetary model, with all of the restrictions that Bohr placed on it in order to get this set of equations, reconciled with laboratory data. Very, very significant. So here we are. Those four lines all can be derived from the Bohr model. And here's another cartoon from your book and showing what I'm trying to depict here, namely it's the falling down, the return to the state from which the electron was promoted that generates the photon. And the set of those lines is what gives you this. There so there's the validation of the sixth piece. So Bohr Model agrees with Angstrom's data, but it also suggests other experiments. Let's think about this for a second. OK, here's another cartoons from your thing. You know, I told you that this thing is unstable. And in the Balmer series it goes from n equals 2 up. But there's a ground state, n equals 1. What was wrong with those electronics in Sweden in 1853 than Angstrom could never find any electron that would fall all the way down to the ground state? What's wrong with it? Well, here's the answer. It has to do with instrumentation. So this is an example where science goes further thanks for the advent of new instrumentation that allows this to make measurements that previous people couldn't make, even though they were very competent experimentalists. Angstrom could have found n equals 1 series, but couldn't see them because he was using a photographic plate. This shows you the range of sensitivity for photographic plates. Here's the electromagnetic spectrum. Out here you have low energy radio waves and up here you have x-rays and gamma rays and so on. And the visible spectrum is parked right here in the middle, and here it is unpacked for you. And it roughly runs from 400 to 700 nanometers. That's the invisible spectrum. So wavelength increasing from left to right, which means energy frequency in wave number increase from right to left. They're complimentary, right? e nu, nu bar on the top, lambda is on the bottom. Some spectra are plotted in lambda, some are plotted in wave number, whatever. And by the way, I want to show you the power of knowing a few things. I don't expect you to know a lot of facts, but I expect you to know a few things. Every educated person ought to know that the visible spectrum runs round numbers 400 to 700 nanometers. But look, I can take 700 nanometers and use this formula and convert it to energy. And I'm going to get something like 3 times 10 to the minus 19 joules. Yuck. Instead, I go in electron volts. 1.8 ev. Over here, 400 nanometers is 3.1 ev. So round numbers, the visible spectrum spans 2 to 3 electron volts. Our eyes are photo detectors that operate on the band width 2 to 3 electron volts. that's easy to remember. Those are good numbers. So where does that leave us? It leaves us here. We go back and we see these numbers, 656, 486, 434, they're all in the visible spectrum. So I went and I did a little calculation. I said, well what would I have if we'd gotten the wave length for the transition from 2 down to 1? This is n equals 2 down to ground state. If you plug in the numbers to the Bohr model, you'd find that that would give you 122 nanometers. 122 nanometers? Well, 122 nanometers is going to put you way over. It's too high energy, right? 122 nanometers is going to put you off to the left there into the ultraviolet, where the photographic film was not sensitive. So he couldn't measure those lines. So now I'm going to end by putting the master equation that captures all of this. And the master equation that captures of all of this is here. It's that nu bar goes as r times z squared, 1 over nf squared minus 1 over ni squared. So this is the most general form for all 1 electron atoms. That's why I've got z squared in there for all 1 electron atoms. And this is called the Rydberg equation. Named after another Swedish spectroscopist at the University of Lund. I think a Swede would probably pronounce this something Rydberg, but you don't have to say that. You can just say Rydberg and it'll be fine. And in honor of Rydberg, the constant here is given the symbol, capital R. The capital R as the Rydberg constant and it has a value of 1.1 times 10 to the 7th reciprocal meters in good ST units. Well, there was more evidence for the support of Bohr's Model. More evidence for the support of Bohr's Model. By the way, as the detectors got better and better we could get more and more lines. You see these, as you get higher and higher series ending on higher and higher end numbers, you move off into the infrared. Because this is not to scale. These n equal 4, n equal 5 are closer and closer to closer together in terms of energy. They're farther and farther apart in terms of spacing, but they're closer and closer together in terms of energy. Because they're farther from the nucleus. You say, gee, shouldn't it cost more energy to go farther? Uh-uh. Because you're farther from the positive nucleus. Be careful. Don't let your intuition send you in the wrong direction. It's all about Coulombics. Anyway, so the Lyman series ends at n equals 1. And these are different scientists. Paschen, Bracket, Pfund, Humphreys, and so on. So maybe, I don't know, if somebody hasn't claimed n equals 214, all the lines that end there, you know, maybe that could be your name on the series. As if anybody cares. Looks like this quantum condition is validated. See this is really important because this was the big break away from classical theory. That the motion of a body, something with mass, could be quantized in its behavior shook this physics community. But this reconciliation of the data says that assumption is valid. There's more that happens. So in 1913 in Berlin-- remember 1913 is when Bohr published the paper-- 1913 in Berlin there was James Franck and Gustav Hertz. James Franck and Gustav Hertz. And they were conducting experiments on gas discharge tubes. Only they filled a gas discharge tube, instead of with hydrogen, they filled it with mercury vapor. So gas discharge tube-- GDT-- gas discharge tube containing mercury vapor. The same thing. Put the electrodes, connect on a power supply, and started varying the potential. So I'm going to show you what they found. This is the Planck voltage, and this is the current between electrodes, or if you like, across the tube. Between the electrodes, or through the tube. Or if you like, tube current. Meaning from one electrode to the other. The tube current. Well, low voltage, low current. High voltage, high currently. They get up to a certain value of voltage, all of sudden the tube starts glowing blindly and the current falls to 0. Then they continued to raise the voltage. More voltage, more current, up, up, up, up, up. And then they get to another critical value of voltage, even more intensity. And then the current falls to 0. So you look at those data and say, well, what's that got to do with the Bohr Model? Because mercury is not a 1 electron atom. It's got a boat load of electrons. This is not a 1 electron atom. So you say, I know what it is. It's ionization energy. Must be ionizing the mercury. So you go to the periodic table and you look up the ionization energy of mercury and you discover that that's 10.4 volts. 10.4 electron volts is the ionization energy. And this first null is at 4.9 volts. Well, 4.9 is a long way from 10.4. And this second null occurs at 6.7 volts. 6.7 volts. So what's this telling us? What this is telling us is that when you get to a value of 4.9 volts, you've hit a certain value that allows you to promote electrons within mercury between one level and the next level. And those electrons are being promoted and then cascading down. And they're cascading down and they're emitting in the visible and it's blinding you. Say, OK, so what does that mean? Well, it means that the Bohr Model, which is for a 1 electron atom, assumes that energy levels within it are quantized. These data indicate that on the basis the behavior of this gas discharge tube, there must be quantized energy levels inside of mercury, which means all atoms have quantized energy levels. You understand? Everything is quantized. That's really powerful. It starts off with this nerdy little 1 electron atom, and now he's applying it across matter. And this is gas, this is more elaborate gas. Heaven forbid, it might exist in liquids and solids. So that's the Franck, Hertz experiment. So his stock goes way, way up as a result of that. And they win a Nobel Prize. Here's James Franck. Here's Gustav Hertz. You know what the Hertz is. 200 kilohertz, so on. That's Hertz. James Franck was at Gottingen when he won this, but he ultimately came to the United States when the political changes started occurring in the thirties in Germany. Franck decided to seek safer surroundings and ended up at the University of Chicago, where there is to this day the James Franck Institute of Physics. Very, very high-end physics institution. So this is good. But all good things come to an end. So 1913 was a bittersweet year for Bohr. Because he got some good news, but he also got some bad news. So now I want to move over to limitations of the Bohr Model. Limitations of the Bohr Model. So I know what you're going to say-- well, it only talks about 1 electron atoms, so that's a limitation. No, there's more to it than that. Even the 1 electron atom model doesn't capture everything. I'm going to summarize the limitations. I'm going to show you three, and they all fall under the general umbrella of fine structure. Fine structure. In other words, the Bohr Model is good, give us the big lines, but when you start looking more carefully it fails to capture some of the physics. So first of all, let's go back to some earlier data. 1887. 1887, there were already data out there that were going to give heartburn to the Bohr Model. And those data were taken by Michelson and Morley. Michelson and Morley. Everything I've taught you so far, with one exception, has been European science. Americans were not active in science because this was a young country. We were really good engineers because we were blockaded by the rest of the world. We had to live by our wits-- that's where you get the term "yankee ingenuity." Science was hifalutin stuff. We didn't have time for it. But towards the latter half of the 19th century, we started moving into fundamental science. The first American to win the Nobel Prize was Michelson. Michelson was doing work at Case in Cleveland, which eventually became Case Western Reserve University. So he was at Case in Cleveland and he was studying optics. And he was a brilliant experimentalist. In fact, he made the first reliable measure of the speed of light. Back before 1900. What they were doing is they were looking at Angstrom's lines and they noticed something peculiar. If you take a look at even this drawing, you notice the red line is a little bit fatter than the others. Now you might just say, well, that's just the artist taking liberties and somebody didn't catch it in proof reading. But in point of fact, what he found was that if you look at that line, which is really the line for the 3-2 transition-- the 3-2 transition in the Balmer series-- what you find is that if you look at the photographic plate more carefully, you find that this thing in fact is a pair of lines, but very, very closely spaced. This is known as a doublet. Two lines very closely spaced, centered at 656 nanometers. And with his interferometer he gets super, super good data. And he could split the doublet. Well, what's that mean for Bohr? Bohr has no way of explaining this. If you look at the Bohr Model, you've got n equals 2, you've got n equals 3, alright? So this is energy 2, energy 3, right? And so when the electron falls from 3 to 2, we get a photon of a certain value. It's going to be nu 3 to 2. That's the frequency or wave number, what have you. Now, the fact that you've got a doublet here means that there must be two transitions, but darn close. There's either a 3 and a 3 primed, or there's a 2 and a 2 primed, but it's not simply 3 and 2. So that piece of information runs counter to the Bohr Model. Bohr Model is silent about it. It gets the big picture, but if you look more carefully it can't capture the doublet. And Michelson ultimately gets the Nobel Price. And I think I've got him here. There he is. The Nobel Prize. By the time he got the Nobel Prize he was at the University of Chicago, but he did the work that won the Nobel Prize for him at Case. So sometimes when you see even Millikan, Millikan did his work at University of Chicago, but eventually took a position at Caltech. So the Nobel Prize says, Robert Millikan, Caltech. But he didn't do that work at Caltech. He did it at Chicago. Anyways, you can go to the Nobel website. You can read about these people. And what's really cool is when you win the Nobel Prize-- you notice I didn't say if-- I say, when you win the Nobel Prize, what you do is you get on an airplane, you go to Stockholm, and then you go and you have dinner in this beautiful hall. I've been there and it's gorgeous, gilded and so on. Very nice kitchen, excellent wine list. And-- yes-- and you can go there and they serve meals. the menu is taken from previous Nobel Prize dinners. So you can sit and-- whatever it is, it could be the Nobel Prizes of 1927 and that's what's going to be on the menu today-- and after the dinner they have a presentation ceremony with the King of Sweden. You get your Nobel Prize, and then people listen to your lecture. And those Nobel lectures are really, really expository. So if you want to go and read the Nobel lecture that Michelson gave on the occasion of winning the Nobel Prize, you'll probably learn all of about this and more. It's really, really good, so go there and read. Now back to the story. Second problem with the Bohr Model. 1896-- see, all this data had been accumulating-- 1896, there was a postdoc by the name of Zeeman. Piet Zeeman. He was a postdoc at Leiden. Leiden in Holland under Lorentz. You'll learn about the Lorentz force when you study 802. What he was doing-- again, gas discharge tube. So this was gas discharge tube, and what Zeeman was doing on his postdoc was in a magnetic field. These people were doing all sorts of experiments. They were trying to block out the whole experimental space. So one guy, his specialty is high energy. One guy's specialty is low pressure. These people are taking a gas discharge tube and putting in the jaws of a powerful, permanent magnet and then measuring the spectrum. And what he found was that for certain lines, this was the rest-- b, I'm going to use as magnetic field-- in the absence of applied magnetic field you have a line. And this is not a doublet, triplet-- it's just a plain old line. Well behaved line. But when they take that gas discharge tube and put it into a magnetic field, they see a plurality of lines. And furthermore, the spacing-- I'm going to use c here-- the spacing in the lines is proportional to the intensity of the magnetic field. No magnetic field, single line. Modest magnetic field, modest amount of what is called line splitting. So a modest amount of applied magnetic field, modest splitting. Intense magnetic field, intense splitting. Bohr Model is silent about that. Because you know, if you've got different lines, it means you must have different energy levels. It's as though the energy level diagrams opens up in a magnetic field. The Bohr Model can't account for that. And parenthetically, they got the Nobel Prize, too. So there's Piet Zeeman. Got his PhD in 1896. He's got his Noble Prize, 1902. He's off to a good start, I'd say. And there's Lorentz. Two of them. We'll get to him in a second. So third piece of bad news for the Bohr Model. And that comes, again, in 1913 in November. In November of 1913 there was a man by the name of Stark in Germany. And Stark was doing analogous experiments. He was studying gas discharge tube in electric fields. Obviously, you've got an electric field across the electrodes to excite the electrons. But he's taking a whole gas discharge tube and putting it between flights and then applying an electric field. And what did he find? He found the same sort of thing. He got line splitting in an E field, and furthermore that extent of splitting, extent dependent upon the intensity. E intensity. So no field, no splitting. Modest field, modest splitting. Intense field, intense splitting. Well, again, that's a headache for the Bohr Model. So this is all three problems, and it's all under aegis of fine structure. So we know the Bohr Model has its limitations. OK, Stark. I know he's got his Nobel Prize. There he is. So 1913 ends on a sour note. But people don't give up. 1916, Arnold Sommerfeld in Munich. He was a professor of physics and he proposed modifications. Modifications to Bohr Model. It's a patch, we would call it a patch. going? To put a patch on the Bohr Model. And what's he going to do? What's the gist of his idea? Well, he retains the planetary structure. He liked that idea-- nice orbits, so on. But he took a page out of Kepler's book. The planets in the Kepler model, when they revolve around the sun their orbit is not circular. It's elliptical. So Sommerfeld said, why don't we give that a try? What if we said the electronic orbit can be elliptical or circular? And he was quite specific. He said, suppose-- and this, again, is not to scale, but to emphasize this is going to be elliptical or circular, but very, very mild eccentricity. What I'm going to draw for you is extreme eccentricity to make a point. But suppose we had the circular orbit as I'm drawing it now, and then we had an elliptical orbit that is centered on that circle. So it's mild eccentricity. We might have another one-- let's do one more. This is good enough. The gist here is that we have a circular orbit and an elliptical orbit, but the bandwidth here is very, very narrow. So this is very, very thin. And it's sort of like an egg shell. So if I asked you, what's the dimension of an egg? You'd say, well, it's dimension of the surface of the egg. Then I'd say, but the egg shell has some thickness, right? But that thickness is relatively small in comparison to the total dimension of the egg. So an analogy. He said that the range of distance from the nucleus, whether it's circular or elliptical, is very, very narrow. So we can say the set of circular and elliptical orbits lie within a shell, as in egg shell. So this is a shell model. It's a shell model. So now how do you designate the different orbits? You've got some that are circular, some that are elliptical. He needs to distinguish them and he needs to be able to label them. So he introduces new quantum numbers to allow us to name them. So let's go and take a look at the quantum numbers that Sommerfeld gave us. So he starts off with n. He retains that from the Bohr Model and he calls that the principal quantum number. And it's primary attribute is size. It captures the distance, the principal r from the nucleus. And it takes values 1, 2, 3, all the way up to infinity. So n equals 1, small radius. n equals 10, large radius. Oh, by the way, there's another numbering system. This is what we use, but the spectroscopists use letters. The spectroscopists use letters-- why? Because remember the Balmer series? Everybody was hooked on the Balmer series and it ended up being n equals 2. And then later with better detectors we find there's an n equals 1. So the spectroscopists said, we're going to get fooled again. So we're going use letters. And we're going to start with the letter k. It's in the middle of the alphabet. That way if we discover even lower energies, we've got some head room here, we can label those. But we never found any. So if you go over to Building Thirteen and you do some x-ray refraction and you use the line that emanates from a copper target, n equals 1-- it's called the k alpha line of copper. To this day. So k, l, m, and so on. You can't get to infinity, obviously. You know, I didn't think this thing through. Now the l. l is, what's his name? Sommerfeld. And it's called the orbital quantum number. Why? Because he said that the electron is in an orbital instead of an orbit. Orbit is Bohr, orbital is Bohr-Sommerfeld. And it speaks to the shape. Somehow, I've got to distinguish between elliptical and circular. And it takes values 0, 1, up to n minus 1. So the n number controls the range of l. And again, the spectroscopists, they're real number weenies, they're afraid, so they use s, lowercase. See this is uppercase, this is lowercase. s, p, d, f. For sharp, this is the sharpest line from the l equals 0-- then the principal because as you go to z they all seem to converge and look like hydrogen-- d is diffuse, f is fine, and then after that they ran out of ideas so g and h. So you'll talk about the one s-orbital, meaning n equals 1, l equals 0. And there are some values here for shapes. I'm going to put that right above it. When l equals 0, l equals 0 means you have a circular orbit. And when l equals 1 it's elliptical. And when l equals 2 it's much more complex, and we'll just leave it at that. 1, 2, and 3. So there's l values. And then m is the magnetic quantum number. And it talks about orientation. I'll show you what I mean by that in a second. The values are governed by l, which is governed by n. Starts at l, l minus 1, goes down through 0, goes to minus values and ends at minus l. So for example, we could do something like this-- when n equals 1, then l most equal 0, so therefore m must equals 0. So this means for n equals 1, it's only a circular orbit and this thing is going to be immune to line splitting in a magnetic field. When n equals 2, l can equals 0 or l can equal 1. When l equals 0, m equals 0. That's boring, that's circular. But here's another possibility. And that is, when l equals 1 then m can equal 1, 0, and minus 1. Now I said it has something to do with orientation. Most of quantum mechanics doesn't translate into the Cartesian world, but this one does, mercifully, and I think it's a cute analogy. If I were to tell you that I've got three different quantum numbers and I've got an elliptical thing-- and one way to think, see, the number 0 looks like a circle and the number 1 has some asperity associated with it, so you can think of that as the ellipse-- so I know that I can, with no prior knowledge of where the true origin of the universe it is, I can arbitrarily define a set of rectangular coordinates, orthogonal coordinates, x, y, and z. And that means I could put one orbital here, one orbital here, and one orbital here. So those are three orthogonal orientations, which I think is consistent with the fact that m takes on three values. OK, that's cute. So that's as far as Sommerfeld went. I'm going to go and do something as a retronym. I want to get the fourth quantum number up here now, but we're going to pause the story. We're going to fast forward to 1925 so I can get the last quantum number up here. And that's called the spin quantum number. And it takes values plus or minus a half. Where did that come from? Well, in 1922-- oh, you know everybody's getting Nobel Prizes and I didn't give Niels Bohr his proper recognition. He gets the Noble Prize, as well. Oh, when Sommerfeld turned 80, they had a symposium in his honor and they published a book. And the book had papers and well-wishes, papers that were given at the symposium. And in the front they had Sommerfeld's picture and they also had this twin picture, this diptych. So on the right is Sommerfeld, and on the left is the same picture but they've morphed it. Now remember, there's no Photoshop. Horrors, there's no Photoshop. Can you imagine? So how could they do this? They had to take the negative, which was a photographic plate, and when they were printing the negative using a light box they had to hold the negative on an angle to get the distortion. And in holding it on an angle to get the distortion, they turned this image into something that was a little more spread out. And the caption that went with this, "To Arnold Sommerfeld, who taught us that the circle is the degenerate form of the ellipse." Now that's geek humor. I mean, they laughed. They thought that was so funny, hahaha. You know. They were having a great time. It was Germany, and nineteen twenties. And there he is. OK, so now let's go to 1922. This is the Stern-Gerlach experiment. Very interesting experiment. This is really physical vapor deposition. Over here I've got a crucible and it's full of molten silver. So Stern and Gerlach were studying the magnetic behavior of liquid metals. So what they were doing is they had this-- over here you see it's red even though it's silver, because this is that about 1,000 centigrade. Silver melts at about 960. Everything, I don't care what it's color is at room temperature, at 1,000 degrees it's red. It's called red hot. All right. So this is red hot silver and there's a vapor here and there's a slit and the silver atoms come out of the slit. And they go across over here to a substrate and then they pause it on the substrate. So making little band of silver on the substrate. And furthermore, he sometimes put them through a magnetic field. So he's got a slit here that narrows the beam, and then he sends it through a magnetic field that is asymmetric. It's divergent. Can you see here? Look at the end. The south pole as a tip and the north pole is this arc, this cup. So the field lines don't go just directly from tip to tip, they go from tip off to the side. So you can see the divergence of the magnetic field. And so he looked at what kind of deposits he got as a function of the magnetic field. Here's what the observed. Very puzzling. The whole thing was about Maxwell's equation. So he's got a silver beam. And when it went directly from the furnace to the substrate he just got the shadow of the slit. And they have the split crosswise with respect to the divergent magnetic field. So if you look at the substrate you just see a band. So this is a band of silver and you can imagine there was a slit out here and it just cast a shadow and that's the band of silver. This is PVD, physical vapor deposition of silver. Now when b is not equal 0, what would you expect? You'd think the beam would bend, right? So what do you think happens? The beam bends up? The beam bends down? Or beam bends to the right or to the left? Think about it. I don't want to hear your answer. Think about it. What do they observe? What they observe is, if this is where the original one is. Two. The beam splits in two. And it gets two deposits, one above, one below, of equal intensity. That's a problem. Beam splitting. But now it's a beam of matter. Beam splitting. Boy, they had them scratching their heads on that one. No way to explain that. So along come a couple of graduate students. 1925, couple of graduates. So this is 1992 in Frankfurt. 1925, two graduate students in Leiden, again. Gaudsmit and Uhlenbeck. They're just like my TA's. Grad students. And they looked at this thing, I don't know, maybe sitting around over a beer one night, and they said, you know, so far what we've been saying is the electron revolves around the nucleus. And sometimes it revolves in a circular orbit, and sometimes it revolves in an elliptical orbit, but here's the electron revolving. And they said, what if in addition to revolve, the electron rotated so that it's going like this? [GESTURES] But there's two choices. It can be going like this, or can be going like this. [GESTURES] Now, it's a charged species and it's rotating, which means that it's going to have a magnetic moment depending on rotation. And now I'm going to send it through a divergent magnetic field. Doesn't it follow to reason that if I put it through a magnetic field and I've got some of them doing this and some of them doing that, they're going to go in different directions? Opposite directions? And what do you think the numbers are? If I give you Avogadro's number of silvers, you think I'm going to get a dominant clockwise and a minority anti-clockwise? No. We're going to get equal numbers. Some are going to spin like this, some are going to spin like that. And you're going say, but electrons don't spin, they're not doing this. But if you model them as though they are doing this, you get those results. Those results make sense. And so they introduced the spin quantum numbers. And I think these are the ones that have been erased. Such is education. OK. So you know it. S plus or minus a half. By the way, Gaudsmit and Uhlenbeck were here during World War Two. They worked in Building Four. You go down the corridor, Building Four just off the Infinite Corridor, there's a plaque there for the Radiation Laboratory. That's where they worked, in the Rad Lab. That's where radar was first engineered. There was work in the UK, there was work in other places. But this is the Radiation Laboratory, started here and they were both here at the time. OK, well, I think that's a-- So this is a plate from the paper. No magnetic field with the magnetic field. And by the way, why did they choose silver? They chose silver because it's atomic number is 47-- it has an odd number of electrons. You're going to learn later that you get two electrons in an orbital, and if you have two electrons, one will be spin up, one will be spin down. There's no magnetic moment. So they were clever about choosing an element that had an odd number of electrons so that there would, at the end, be an unpaired electron. And there's Otto Stern with his Nobel Prize. And he came to the United States, as well. And you're going to see the ascendancy of American Science as people flee Europe up in the nineteen thirties. And America is the beneficiary, and then you see American science rise. But for now it's European science. OK, so I'm going to talk a little bit about hydrogen and transportation. And we're going to talk about the Hindenburg because it was full of hydrogen. And to give you a sense of scale, this is what a 747 would look like and is what the Titanic would look like. It was almost as long as the Titanic. It was built in Germany by the Zeppelin company. And the Titanic, the Hindenburg rather, was LZ129 serial number. That's Luftschiff Zeppelin. Airship Zeppelin. 135 feet in diameter. 804 feet long. How long is a football field? So that's a big boat. Seven million cubic feet of gas, giving you 112 tons of useful lift. You ever have to lift something very heavy, there's your sky crane. So why are they using hydrogen? Well, when the Nazis came to power in Germany, Congress passed the Helium Control Act. The dominant supplier of helium to the world was the United States. Helium comes from helium wells in the earth. And so as of 1933 the United States refused to sell Helium to Germany, so the engineers were forced to use hydrogen. Next best thing. Here are some posters. "Only 2 1/2 half days to Europe." And here one, a German one. "And now over the North Atlantic." That's Manhattan. That's the lower tip of Manhattan. There's the Chrysler Building, look at that. Now look at that picture, isn't that magnificent? 10 transatlantic flights, 1936. 1002 passengers. Cruising speed, 78 miles an hour. Took two and a half days. By the way, 100 feet in diameter-- when people traveled, they traveled in style. They had a ballroom there and a grand piano. People didn't sit like this. [CROUCHES] With a plastic knife and fork. That's progress, right? Two and a half dancing, tux, tails, champagne. Now, like this. [CROUCHES] May sixth, 1937. Arrival of first flight to the U.S. while docking in Lakehurst, New Jersey. Why were they docking in Lakehurst, New Jersey? If you go to the top of the Empire State Building, look and you will see at the corners moorings. Moorings sticking out. The plan was to dock airships at the Empire State Building. So you'd come in from Europe, you'd dock at Fifth Avenue, get on the elevator, and there you were. When they tried to dock the air currents were so violent that they couldn't safely dock the ship. So then they moved across to the fair grounds at Lakehurst, New Jersey, where obviously this mooring is much closer to the ground than the top of the Empire State Building. The wild currents, they're bad, but they're manageably bad. At the top of the Empire State Building, impossible. There's another image. So what happened? It did not explode. It did not explode. It couldn't explode. Seven million cubic feet of hydrogen to explode requires seven million cubic feet of oxygen instantaneously. And air is 20% oxygen. So it was a very violent fire, roman candle from the point of egress of the hydrogen. Most of the people on board walked off the Hindenburg. Most of the people walked off the Hindenburg uninjured. They think it was electrical discharge in the vicinity of a hydrogen leak. Recent research has indicated the skin was made of resin finished with a lacquer dope. And then to make it shiny they put aluminum powder. And why they put iron oxide on the inside I don't know, but this is what NASA uses for solid rocket motor grains. So when this thing catches fire, this is a thermite reaction and could be very violent. And this spelled the end of rigid error ships in commercial air transportation. Now this a U.S. Navy airship filled with helium. And there was a small gasoline fire, look what happened. Again, it was the skin. That's a blow up of that one. So I looked at that and I thought, geez, that looks Lichtenstein, doesn't it? You know this one? This one. Look at that. Look at that. So you know, I can be an artist, too, right? OK, I'm going to tell you one more story. Another Niels Bohr story. So in 1896 there was a guy, and astronomer at Harvard called Pickering. Pickering at Harvard, 1896, and he was studying the lines in star light. And he attributed to some of the spectra that he was getting, he said he was seeing atomic hydrogen in star light. And then there was a fellow in London called Fowler. And Fowler, in 1912, reproduced the experiments in the laboratory. He put gas in a tube and got the same thing in the lab. So this guy's at Harvard and the other guy is at London. Well, Bohr looks at this stuff and he says, you guys are wrong. You guys are wrong-- your lines are off by a factor of 4x. You've got the right series, but you got the wrong element. What you guys are looking at is helium plus. You're not looking at hydrogen, and you know, from-- it goes to z squared. So the lines are going to be shifted by factor of four because this z is equal to 2. So Fowler was a pompous ass and he didn't like being called on his bad science. So he does a calculation and he looks more carefully and he says, Bohr, you're wrong. In fact, our lines are off by 4.0016. Now don't laugh. The reason is the spectroscopy was so precise that they could go to five significant figures. So Bohr says, hmm. And he goes back and he says, you know, we've been doing all these calculations with a one electron atom just neglecting the center. So he redoes the calculations for the entire Bohr model, including considerations of the mass of the nucleus and the mass of the electron in the form of the reduced mass. The reduced mass is-- you're learning this in-- the reciprocal of the sum of the reciprocals. And when he does that, he gets that the value of the line shift should be 4.00163. So he says, you guys are wrong. It should be 4.0016. You got 4.0016, you idiots. You're looking at helium plus. That was Bohr. Did not want to get into an argument with Bohr. All right. Have a nice weekend.
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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, we continue our discussion of quantum states of light, and we focus on single photons as qubits today. But before I get into this, I want to ask you if you have any questions about the last unit we be completed on Wednesday-- namely, the discussion or non classical light-- in particular, squeezing light-- and we also discussed a lot of exciting quantum mechanical effects related to beam splitters. So, any questions? Anything you would like to see discussed? So, as I pointed out, we want to now talk about realization of quantum logic and quantum gates with single photons, but I also said that we are actually using the language of quantum communications-- a kind of information processing to describe general physics in a very nice way. So, first-- when we talk about photons, we should talk about our qubits, and right now, you may think that having a single photon and having no photon in a mode are two possibilities. And you could use them as qubits. However, I will tell you today that it is better, actually, to use always one photon, but in two different wave guides-- two different modes. But, we'll get there in a moment. So, the first task at hand is we want to manipulate photons. And for that, I want to introduce phase shifters and beam splitters. And just to give you the punchline right away, what I want to show you is that those two simple devices -- beam splitter, a half-silvered mirror, a phase shifter, which is just a piece of glass which you can put into one of the laser beams-- that those two elements and those two basic operations, if they are now put together-- if you do a beam splitter, a phase shifter, another beam splitter and such-- you can realize any single qubit operation. Therefore, if you have our qubit cubic state with single photons, any possible state of those qubit can be realized using these optical elements. Therefore, when I say we want to realize how we can manipulate photons with optical elements and such, I'm not going for a zoo of optical elements. Those two will do it all. So, with the phase shifter, that's what we discussed at the end of last class. If you have two modes, and we shift the phase in one of the modes, just simply put a piece of glass into it, then there is a phase shift. And why do we need two modes? Well, phases are usually-- at least in experiments-- not absolutely defined. You need a phase reference, and here, the mode a acts as a phase reference. Let me address one question which a student asked me after class, and this was related to what phases appear in more than one place. The student's question was motivated that if you have one photon, it's a flux state. A flux state is a circle in the causal probability, and therefore, the electric field has no phase. Well, this is the phase of the electric field, but here, we're talking about the phase of the wave function. And a flux state within photons is an eigenstate of the harmonic oscillator. It has a time dependence, e to the i omega t, and we can change its phase. So, this is a kind of phase shift I've introduced here. So, the second element we need to realize-- arbitrary single qubit operation is the beam splitter, and I introduced a beam splitter by just saying, hey look, I think that's a good Hamiltonian. Let's see what this Hamiltonian does to the two modes, and then we realize-- yes, it takes those two modes, and it mixes the modes with cosine theta sine theta reading factors. That's exactly what you expect a beam splitter to do. I'm not sure we need here now because it's part of your homework assignment. You will show that if you have a coherent state, the coherent state is split by a ratio which is cosine square theta sine square theta. So, it's exactly what you'd expect from a beam splitter. So from that, you realize now what this angle theta is, which I just put into the Hamiltonian. The cosine square and sine square of it is the reflection and the transmission of the beam splitter. Any questions at that point? Let me now introduce matrix representation, which will come in handy. You remember that you transform an operator-- the mode operator, the annihilation operator a and b. We multiply with the beam splitter operator on the left and on the right, b dega-- b ab dega. But there is often a simpler way how the right inside can be written, and this is by using a matrix representation, which goes as follows. We can say that the two operators are transformed by the following matrix, and this matrix represents the beam splitter. So that's what the beam splitter does to the mode operators. Now, we want to talk about what does the beam splitter do to quantum states, to single photon states. So now, we want to transform the single photon states. It's a little bit like, well-- we know what to do in the Heisenberg picture to the operator, but now, we want to see what happens to the wave functions. So, now-- I want to say it very slowly, because it shouldn't lead to any confusion later. We have single photons-- we can have the single photon in mode b, or we can have it in mode a. These are the two possibilities. And right now, I label it like this-- where the first number is your occupation number of mode a, and the second number is your occupation in mode b. You have to be a little bit careful-- I'm just saying that keep your alert level high. Later, I will use qubit representation, where the qubit one means the photon is in mode a, and the qubit zero means the photon is in mode b. So when we talked about logical states of our two levels used, and the photon can be here or the photon can be there, zero means the photon is there. It doesn't mean that we have zero photons. But sometimes, I have to talk about the photons-- and that's what I'm doing right now-- and zero means no photon in this mode. But I will remind you of that as we go along. So we have the two quantum states-- one zero, and zero one. And let's just see what the beam splitter is doing to one photon in mode b. I'm changing notation from one line to the next in my notes. Just give me a split second. Yes. So, let's use this convention. So, just to elaborate what I just said is-- this convention one zero is the direct product of the Hilbert space for mode b, with one photon with the Hilbert space of mode a, and we have zero photons there. And sometimes, you want to denote the state by putting a comma in between. That's all the same. So, how do we transform the quantum state with the beam splitter operator? Well, we could now apply to the quantum stage, but we just learned how to apply to operators, so let's just use what we already know, because we can write the state like this. And now, we want to insert the unit. We want to insert the unit operator b dega b, and now, we can use our knowledge of the operator. We know how this transforms. We had this above the transformation of the operators b and b dega, and when we apply b to nothing-- to no photons, we get no photons. So now, by taking from the page above the transformation or the operator b dega, it's a linear combination of a dega and b dega. We find what we expect-- the photon can now be in the same mode, or it can appear in the other mode. And the coefficients from the transformation of cosine theta minus sine theta. And if you would send the other state through the beam splitter, we find the cosine theta sine theta component. So, since the total probability to have a photon is cosine square plus sine square is unity, we find what we expected, but it's nice to see it. Namely, that b conserves the photon number-- of course, this should have been obvious from the outset, because we used a Hermitian operator, a unitary time evolution, and that is energy- conserving. What happens if we use a state which has one photon in each mode, and we act on it with the beam splitter? Well, it gives a superposition of the two photons can now be in either of the two modes, or they can be distributed one in each mode. And the coefficients are because we have two transformations, products of sine and cosine. Here is cosine square plus sine square theta. Here is a plus sine, square root two plus square root two. So that means if we allow one photon to be in each mode, it leads us actually out of the Hilbert space of single photon states, because we have a certain probability now that we have two photons in each mode. So, we don't want that. So if you want to deal with only one photon, we should restrict our attention, our formalism to the states which have nor more than one photon. And now, if we act on those states with beam splitters, we don't get out of this subspace or fill that space. However, we also want to omit this state because it's more elegant to do it, but also, if you have a single photon state and we have a detector which has not 100% efficiency and we do a measurement-- well, we don't know if we have the state zero one, but we just didn't detect it, or we had the state zero zero. Whereas if you deal with two states where you always have a photon-- exactly one photon-- and you read out your system and you detect nothing, well you discard the measurement. And then, you only take the measurements which have detected one photon. So you have the ability to deal with finite efficiencies and losses in a very straightforward way. Yes? AUDIENCE: The operation of dega 1, 1 doesn't look so unitary to me. PROFESSOR: Yes. And it is in my lecture notes. Minus. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yes. And it's right in my lecture notes. Sorry. Sometimes, when you're talking, explain it, build it up-- yes. Thank you. Other question? Good. So what I just said, that we want to restrict the Hilbert space to exactly one photon, this is so important for implementations and discussions of quantum information protocols that it has its own name. It's called the dual-rail photon state. The dual-rail photon state space. Exactly one photon, but the photon can either be in mode a or in mode b. So this part-- this dual-rail photon state space-- is a Hilbert space which has a basis which is zero one and one zero, and this is of course a two-level system. In this so-defined Hilbert space, we can have an arbitrary state, [? psi ?] which is-- as in any two-dimensional Hilbert space-- it's a linear superposition of your two base states with coefficients alpha and beta. And the few of them for which we have prepared right now is that any possible state of this Hilbert space can be created from any of the base state, that is from zero one simply by beam splitters and phase shifters. When I taught this unit the last time, I went through the proof, but I want to make room for more in-class discussion like we had on Wednesday with the clicker question, so what I decided is I give you the idea behind the proof, and the formal parts you can simply fill in by reading about it on the Wiki page. So, let me just focus on the idea. So the proof is that if I use some column vector for alpha beta-- that we can identify with the beam splitter and the phase shifter-- we can identify them as rotation operators. This is immediate obvious if I use the transformation on psi by beam splitter, we learned just a few minutes ago how the beam splitter operator acts on the base state. So now, we know how it acts on an arbitrary linear combination. And the matrix here is nothing else than the rotation matrix with the minus sign around the y-axis. So that's a beam splitter. The phase shifter is not changing any state. It shifts one of the states by e to the minus i phi, but I can sort of summarize it by taking half of it out as a prefector, and then adding here e to the i phi over two. And this here is immediately recognized as the rotation matrix around the z-axis. Now, you may ask what about this? Well, this is an irrelevant global phase factor. If you have any state in Hilbert space and you change the global phase, it doesn't matter. Therefore, we are realizing that the beam splitter is a rotation around the y-axis. It's now a definitional thing, but we put a factor of two here by an angle two theta, and the phase shifter is a rotation around the z-axis by an angle phi. Now, with that, we can formulate what is called Bloch's theorem, that any unitary operation of a two-dimensional Hilbert space can be written as a product of rotations. Now, let's first count an arbitrary unitary operation while a two by two matrix in complex space has eight numbers-- four real parts, four imaginary parts. But if the matrix is unitary, they are only four independent elements-- alpha, beta, gamma, delta. And if you parametrize the unitary matrix with alpha, beta, gamma, delta which is shown in the weekly notes, then we can write this arbitrary unitary matrix by a combination of an overall phase factor. Which is irrelevant because it's a global phase factor-- and then a rotation around z, a rotation around y, and a rotation around the z-axis. So, if I rephrase this theorem in modern language, I would say we have a qubit which is a two-level system characterized by two numbers alpha beta. And we can do an arbitrary operation in the Hilbert space of the qubit, which is called an arbitrary single qubit operation. So, can be performed by phase shifters and beam splitters. Just to make sure that this doesn't lead to any confusion, the phase shift which is relevant for this Hilbert space is a phase shift which is relative between the two states, and can be detected by an interference experiment. Those global phase shifts cannot be detected. There is no observable, no procedure to observe those-- unless you would have a a third mode. But then, we would expand the Hilbert space to three levels and then, of course, it's the global phase shift in the two-dimension Hilbert space becomes a relative phase shift within the three-dimension Hilbert space. And in modern language, those single qubit operations are called quantum gates. So, all gates which act on a single qubit can be realized with phase shifters and beam splitters. Let me give you one example for that. In quantum computation, quantum information science, there is a very important gate-- the Hadamard gate, which is described by a transformation which has 1 minus 111. And I said it can be realized with a beam splitter and phase shifter. So let me just show it to you in using our symbolic language. So if you look at the matrix we had derived for the beam splitter, you realize that we get the Hadamard gate by aiding the phase shifter by pi. So, in that sense, single qubit operations are checked off. We know how to deal with that. We want to now move towards two qubit operations-- and then, it's getting really interesting. As you may know, you can do any arbitrary operation in many qubits by just having single and two qubit operations. Therefore, once we know how two qubits interact-- how photons in two different qubits interact, we are done-- we have universal quantum gates. The element I need for two qubit operation is the Mach-Zehnder interferometer. Just a side remark, I introduce a Mach-Zehnder interferometer here as a way to manipulate qubits. On the other hand, most of you know that interferometers for light and interferometers for atoms are used for some of the most precise measurements ever done. So, in that sense, this shows the dual use of the formalism I'm introducing here. We really use some of the formalism and the language we develop here for qubit operation, and discuss next week the ultimate accuracy-- the fundamental limits to the accuracy you can get in precision measurements based on interferometry. Let me just write down the sentence because it has a lot of key words in it. So the dual-rail photon presentation of a qubit-- this is what we have achieved so far. So this allows us now to discuss interferometers simply as [INAUDIBLE] gates. And what isn't interferometer-- an interferometer is, in its basic [INAUDIBLE], you have a beam splitter, you have two paths, and then you use another beam splitter which recombines the two beams. So, let me draw two beam splitters. These are two beam splitters. We have an input [INAUDIBLE] b, an input mode a. They are sort of mixed. Here are the two arms of the interferometer. And now, we create output modes b prime and a prime. What we have introduced here is the beam splitter b, and by putting the dot on the other side, this is b dega. Since b dega times b equals one equals the identity, this interferometer is doing nothing. If the photon starts in mode b, it comes out in the upper mode b. If it starts in a, it comes out in the mode a. But of course, this is just the beginning, because we can now introduce here an arbitrary phase shift phi. And now, it is an interferometer. Now, we can read out with the output the phase shift, and ultimately, with very high precision. I what you to appreciate that. If we had used single qubits in one mode-- no photon, one photon. I gave you many reasons why we shouldn't do that, but then, we would be dealing with one mode, and this would actually be something which involves two qubits. But since our single qubit system is a two-level system where the photon is here and there, we can now send single photons with the interferometer-- describe the interferometer as a single qubit operation. At least it's nice. It allows us to keep things simple, and discuss things at a very fundamental level. So, what we want to figure out-- what is this interferometer doing? So, we want to know if we have an input state-- what is the output state? And the input state can now be any arbitrary wave function in our dual-rail Hilbert space. So, it's always one photon, but it can be in a random superposition in an arbitrary superposition of states a and b. Well, with the formalism we have developed now, it becomes very simple. All we have to do is we have to act with the first beam splitter with a phase shifter, and with a second beam splitter onto the state. And this is nothing else than taking two rotation matrices and multiplying them. The beam splitter we have chosen, I said there was sort of a phase convention when I peeked the beam splitter. It is a rotation by pi over 2. This is minus pi over 2, and the z-rotation is by minus phi. I'm not sure if I manage to do that, but if this is the y-axis and we start with the qubit, the first beam splitter rotates it down. Then, the rotation by z-- thus, that-- and the second beam splitter rotates by y. Therefore, the qubit is like that. Therefore, if you count all x's with the things I did is-- I started with a qubit here. It went down like this, like this. So what I did is-- in the end, I simply rotated it around the x-axis. If you want, you can multiply the matrices, or you can draw block sphere visualization. That you say we have an x-axis, y-axis, and z-axis. So the z is vertical. So, we started out with the qubit along the z-axis. The first one was rotation around the y-axis. So then, we were here. The second operation was rotation around the z-axis. And then, we have to rotate back around the y-axis, which eventually gives this as the final state. So ultimately, the product of all this is a rotation around the x-axis. Our interferometer-- I'm not sure if you've ever heard the language, but it's really elegant and powerful. An interferometer with an arbitrary phase in this two-level Hilbert space with this sort of geometric interpretation we have given, is nothing else than a rotation around the x-axis. Therefore, we know immediately the two limiting cases when we rotate by zero, it's sort of balanced. Output and input are the same, because we're not rotating at all. Whereas if we have a phase shift of 180 degrees, we swap the two qubits. So, we swap the two modes, and swapping modes is an inversion of the qubit. So that's almost trivial, but we need this definition-- we need this knowledge to take it to the next step. Remember, we want to do something more interesting than rotating two level systems. We want to get two qubits, combine them, entangle them, and all that. So what we need for that is-- we have to expand the Hilbert space. I introduced the Mach-Zehnder interferometer as the device to do it. But now, we need a way how a second photon from another qubit can interact with the photon of our first qubit. And we want to do that by introducing first the non-linear Mach-Zehnder interferometer. I want to throw in a non-linear element, and then we are ready to allow the second qubit through the non-linear element to manipulate the first qubit. And then, we have interactions between qubit. We have two qubit gates. Let me maybe just deviate for my notes. I really want to show you what I'm doing, and that you have it clearly in mind where I'm aiming it. So what we want to accomplish is-- instead of having a phase shifter with an externally controlled phase, I want to put in some non-linear crystal-- which has an index of refraction which has a phase shift. But now, the face that the non-linear crystal has the property, that its index of refraction changes when I take another laser beam and send it through. So, the laser beam through-- because the index of refraction is intensity-dependent, the green laser beam changes index of refraction of the non-linear crystal-- also called Kerr medium. And that means now, because the blue beam goes through the same crystal, that the green beam controls now the phase shift of the blue beam. And now, we have interacting photons-- one photon interacts with the second photon. Let's work it out. So our goal is to develop a description for non-linear Mach-Zehnder interferometer. What we need is a non-linear medium. So, let me just introduce for two or three minutes non-linear medium, how we describe it, and then we put in into our interferometer. So, in linear optics-- just to remind you, we have a polarization on electric dipole moment, which is proportional to the electric field of light, and x i is the polarizability. The most complicated case, it can be polarizability tensor. Now, this is linear. If you want two non-linear, then we have the linear relationship simply as you can say the first terminal [INAUDIBLE] expansion. So we expand the response of the medium in powers of the electric field, and we have the susceptibilities chi 1, chi 2, chi 3, and so on. We have already encountered a non-linear element which had a chi 2 susceptibility, and these where the crystals we put in our OPOs into our optical parametric oscillators. Remember, the optical parametric oscillator involved-- how many modes? Three. We had one photon, which was broken down into two photons. But that also means that the reverse process is you drive the beam-- you drive the OPO with omega, but the e square term creates an electric polarization as 2 omega. And if you have an oscillating polarization as 2 omega, you create light at 2 omega. So, I just want to remind you that the chi square term is what we have already encountered. That's not what we need for the non-linear Mach-Zehnder interferometer. What we need now is a chi 3 interaction, which is the Kerr medium. Because we want to realize the following Hamiltonian. The Hamiltonian, the Kerr medium is also called cross-- that's cx -- cross-phase modulation. One beam can affect the phase of another beam, and this is called cross-phase modulation. There's also self-phase modulation-- that one laser beam changes the index of refraction for itself-- but here, we want the cross-phase modulation. So, we know that if we have a mode, and we shift its phase, we're not changing any photons. So if we're in an eigenstate of mode a with one, two, or three photons and we simply shift its phase, we are diagonal in a dega a, and the phase shift is proportional to prefector here. So this Hamiltonian is simply a phase shift for photons in mode a, but if you now multiply that with b dega b, the phase shift is now proportional to the number of photons in mode b. Now, we have a situation that the phase shift from mode a proportional to the intensity in beam b, and vice versa. So, I think it's self-evident that this Hamiltonian is sequential description of the process we want to introduce know. This is the cross-phase modulation Hamiltonian. Isn't it much more elegant to describe non-linear optics with Hamiltonians than with parametrized susceptibilities and all that? It's just amazing-- you write in a few letters and it has all the power of the classical description-- plus extra, because it includes quantum fluctuations and everything. I hope you appreciate the power of the language we are developing. This is the cross-phase modulation Hamiltonian. And now, we want to use this Hamiltonian in a crystal of lengths L Well, you know if you have a time-propagation operator, you put the Hamiltonian in the exponent and multiply with t, but for propagating laser beam, t and L are related, so this is now the unitary transformation if you propagate through that crystal. We are now defining that by saying we have this non-linear susceptibility, things will be proportional to the links L, and then we have the operator a dega a, b dega b. And let me choose for the following discussion that the links are chosen, such that-- when multiplied with a non-linear susceptibility, we get a phase shift of pi. Now, we are ready to describe what happens when we have a Kerr medium, and we have two input modes. So, let's write down the table of possible combinations and operations. Well, when we act on the vacuum, we get nothing. What happens when we have one photon in one mode? What happens to that photon? Well, the extra phase shift coming from cross-phase modulation is zero, because we have no photon in one of the modes, and a dega a or b dega b acting on this mode gives zero. Therefore, as long as we have only one photon, we have no phase shift. The exponential factor is one, and we simply reproduce the original state. So, the only non-trivial situation by construction is when we have one photon in each mode, and then we get the matrix element e to the ikl. And we adjusted the lengths of the crystal that this is just minus one. In other words, in this Hilbert space, we get a phase shift of pi-- we get a minus sign when we have one photon in each mode. That's all what this medium does. And if you ever thought about it-- how to deal with this situation-- that when I said there is one photon which provides a phase shift to the other photon. But the other photon is also providing a phase shift to the first photon. So you may wonder. When you have a combined state of the two, how do you deal that one photon shifts the other photon, and vice versa? But when you apply the operator, you don't even have to think about it. You see that the operator for the combined Hilbert space of the two photons gives you a phase shift of pi. It's not pi for one photon times pi for the other one-- it's pi for the whole quantum state. Any questions? We have a Kerr medium, which means we are ready now to put it into our interferometer. I'm doing at this time-- insert picture. This is what you want to do now. We have now three modes of light. You would say, hey-- a qubit, each qubit is two. What is it-- is it one and a half qubit? Yes, it is-- but we extended two cubits in a moment. So what we simply have right now is we have the mode c-- which controls the non-linear medium, and we have what we have already discussed. Our interferometer with a phase shifter, which is now the Kerr medium. So I think even without doing any math, we know that by construction, when we have no photons in c, that the modes a prime b prime transform-- become a and b, because we have a balance interferometer which is the identity transformation, unless we have a phase shift. But when we have one photon in state c, then we actually swap. So, that's pretty cool. A single photon in mode c will redirect the photon from ab to ba. So that's how one photon in mode c controls what happens to the photon in ab. If you want a general description of that, we obtain the output state from the input state by using the matrix b. This was our interferometer based on two beam splitters. It involved mode a and b. But now, we throw in the phase shifter, which is our Kerr operator-- our non-linear phase shifter with b and c. This new addition to the interferometer is described by e to the i, non-linear susceptibility L, b dega b, c dega c. Let me give you one intermediate result. We have now beam splitters on either side. And that leads us to an expression. While the beam splitters do nothing to the mode c, but the beam splitters transform the modes in linear combinations. And if you do some tedious rearrangement of terms-- more details are given in the Wiki-- you find that the essential result is now the operator, which we get is a dega b minus b dega a over 2. I'm not giving you the non-essential terms, which are just phase shifts-- global phase shifts. So, let me write it down in blue here. This is just the result of the operator manipulation. If I would cover that-- what is that? Do you remember a dega b minus b dega a in the exponent? It's a beam splitter. And the beam splitter-- the beam splitter matrix was sine theta cosine theta. So we had a theta here. Now, you realize that this non-linear Mach-Zehnder interferometer, the three optical elements can be replaced by simply a single-beam splitter for mode a and b, but the angle theta of the beam splitter is now controlled by the photon field in state c. In other words, we have now a beam splitter. I told you that a beam splitter is nothing else than a rotation around the y-axis. So now, this non-linear Mach-Zehnder interferometer is simply a beam splitter with a rotation angle given by this term. Any questions? This is a situation we just had, and we can now get to two qubits by simply adding one more rail. Remember, two rails with exactly one photon in it is a qubit. So now, we have a qubit here, and we have a qubit there. And you see immediately one way how the upper qubit x on the lower qubit-- if the lower qubit has a photon in state c, it flips the other qubit. But if the upper qubit has a photon in d, it doesn't do anything to the lower qubit. Now, we have a two-qubit operation. This is shown here. But what I want to discuss now is the situation when we throw in one more beam splitter, and this beam splitter acts on the upper qubit. Well-- isn't it great how quickly we go from simple elements to something which looks quite sophisticated? So, this device now is a single optical device which can entangle qubits, which leads to entanglement. And actually, entanglement is our next big topic we want to talk about-- entanglement between particles, entanglement between photons. And by introducing the Mach-Zehnder interferometer, I was actually building up the situation, which is extremely simple elements which lead to entanglement. We have everything for the description-- each of those elements is described by a matrix which we have developed. Therefore, by just multiplying those matrices, you find immediately how this whole device is now manipulated into qubits. So, if you want, you can just take the crank, use what we have already done, and crank out the result for it. Now I wish I had a bigger screen, because I want to describe what's going on for you. What I want to discuss with you now is what happens when we take this device, and we use exactly the states 0,1 0,1 at the input as I indicated. In terms of qubit language, I say that the first qubit is in spin-up, and the lower qubit is also in spin-up. Spin-down would mean that the single photon is in the other state. Our two-level system-- our dual-rail single qubit two-level system is the photon can be either in one of those states. This one, we call spin-up-- this one, we call spin-down. What I want to discuss with you now, what happens when we start with this symmetric input up-up, which means 0,1 0,1. And since I don't want to scroll up and down so often, everything is trivial beam splitters, except for the situation when we have one photon each here, and then the Kerr medium gives us a minus sign. So that's what I want to put in. Remember, I want to develop the physics in one, two, three temporal steps. The first temporal step is simply two beam splitters, and this is shown here. We start out-- and now we have beam splitter for the upper qubit, beam splitter for the lower qubit. And well, by just multiplying it out, we are now obtaining this summation state for the four different modes. And I'm dropping here factors of square root-- or you can collect them at the end, if you want. And I said the only interesting situation is where we have two photons-- one each in the two middle modes, because now, the Kerr medium kicks in and gives us a minus sign. Now, you can show by inspection that if you now apply the beam splitter to modes a and b, you take 1, 0 into a linear combination of 1 and 1, but the beam splitter is only acting on the last two numbers here. Then you obtain a state which looks very simple. So this is the output state of the device. Any questions? And this output state is-- if I use the spin language, it's an entangled state. So now, you have one simple example how just using linear optics, very simple elements you can start with one photon here, one photon here, a simple product state, and what comes out is an entangled state. Questions? Let me tell you what I'm not telling you-- namely, the Wiki developed by a professor [INAUDIBLE] has now a wonderful section on a famous quantum algorithm-- the Deutsch-Jozsa algorithm. Pretty much using the elements we discuss, you can now realize the Deutsch-Jozsa algorithm, which is one of the famous algorithms where quantum logic-- quantum computation is faster and more efficient than classical computation. I decided not to present it to you, because it just leads to an even more complicated diagram and more formulae. You should just sit back and slowly read it by yourself. There is no new idea introduced-- it's just the concepts we have discussed can lead to very powerful algorithms. What I want to rather do is-- I want to continue our discussion and now talk about the object we have encountered here-- namely, entangled states. But if you have any questions, if you want me to explain anything more about the single qubit manipulation, the dual-rail photon state, I would be happy to do that. Our next chapter is on entangled photons, and what we can accomplish in the next 10 or 15 minutes is mainly a definition and some of the properties. What we will do next week is we will talk about how we measure entanglement-- you already had one homework assignment where you discussed one way to measure entanglement, but there are others which we want to discuss next week. I also want to show you next week how entanglement leads to the Einstein-Podolsky-Rosen paradox and Bell's inequalities. And eventually, I want to show you how you can entangle not only photons, but how you can entangle atoms. But that's an outlook. I think, first of all, we want to understand what is entanglement. Entanglement is-- let me first [? motivate ?] it. Many people regard entanglement as the most quantum essence you can find. Well, there's always a discussion, but it's really quantum. Some people would say waves are quantum. But often, we find quantum systems-- very famously, Bose-Einstein condensates, which really behave like electromagnetic waves. They can be split. They can interfere. But Bose-Einstein condensates are so big, have so many atoms that you hardly ever encounter quantum fluctuations. You're really in the classical limit of matter waves, and I would actually say this is maybe not at the heart of quantum mechanics. While it's nice, it's powerful, it's important-- but this is not really the new feature which quantum mechanics has shown us in nature. You would say-- well, what else is quantum? Maybe certain quantum fluctuations going down to single photons, and I would agree that's a quantization of the electromagnetic field. That's much more quantum than having millions of condensates, millions of condensed atoms in one single wave. But what I would say is even more quantum than the single photon is the entanglement. Entanglement is sort of pure quantum mechanics. With single photons, you can still use intuition. But with entanglement, that is really bizarre. This is really- I hate the word, but let me use it-- this is sort of pure quantum weirdness. And there are people who have spent most of their career until now to just show the weirdness of quantum mechanics, and showing to what peculiar phenomena entanglement leads. So entanglement has really lead to a lot of-- I should be careful with the word surprising-- surprising always means you didn't have enough imagination to think about it, but yes, I would say-- really truly unanticipated, and in that sense, surprising developments. Entanglement is, therefore, for me, the most quantum aspect of quantum mechanics. It was through the Einstein-Podolsky-Rosen argument-- it was actually Einstein-Podolsky-Rosen in the '30s who argued-- Einstein-Podolsky-Rosen were really the first-- this was almost 5 or 10 years after the development of quantum mechanics. But it was Einstein-Podolsky-Rosen who looked at the properties of entangled states. This famous experiment where you have something entangled in position and momentum space. And then, Einstein-Podolsky-Rosen found a paradox, and their conclusion was that the properties of entangled states implies that quantum mechanic is not compete-- is not a complete description of the word. Well, we don't any longer share this opinion. It was John Bell and others who said that entanglement really exemplifies that quantum mechanical correlations go beyond classical mechanics-- go beyond the classic picture. Therefore, you can say if you have a quantum system like a Bose-Einstein condensate, where you have end particles doing in unison what one particle does. This is sort of like the laser where many photons do what one photon does-- but it fits very, very well into the concept of classical probabilities and an intuitive understanding of why many particles do what one particle does. But if the particles are entangled, that's almost like you turbo-charge your system. Your system has now more oomph, more power in it. And this is shown that it has extra correlation-- there is something extra in it, which you would never get from any classical limit. And this extra oomph, this extra power turns out to be a real resource. So, entanglement-- this extra correlation-- is a resource. And resource means it's good for something. It enables teleportation. Remember when we talked about the teleportation scheme-- Alice and Bob could only teleport a quantum state because they shared entangled photons. It was this entangled state which Alice manipulated with a measurement, and Bob used this half of the entangled state to recreate the original state. So, it's the engine [INAUDIBLE] up behind teleportation in the world of quantum computation. The exponential speedup of quantum computers versus classical computers in quantum algorithms is due to the entanglement, which we can put into a quantum system. And eventually, next week, you will see that if you have an atom interferometer with entangled states, we can operate it at a precision which is better than short noise. So in other words, you use laser light to measure something, and you're limited by the fundamental noise limit of classical physics. Now, you entangle your laser beam and you get higher precision out of it. So this shows that entanglement is a very special resource. It's very precious, very powerful and can extend what physics can do. So, let's define it. I'm not describing to you entanglement in the most generous situation. If you talk about many modes-- if you talk about not just pure states, but statistical operators-- it can become quite involved. I'd rather start with the simple situation, which for conceptual discussions is also the most important one. I restrict our focus now on two modes. I want to ask if you have two modes a-- actually, two modes means two subsystems here. If you have two subsystems a and b-- let me tell you what entanglement is not, and then I make it the definition. There wouldn't be anything special if you have a system which has two parts a and b. The global system is psi ab. But if your system would simply factorize into a wave function of subsystem a, direct product with subsystem b, and this here is the tensor product. Then, there would be nothing special going on between subsystem a and subsystem b-- and this is non-entangled. Everything else is entangled. So, this is our definition. When we have two separate systems a and b, we call this situation the total system bipartite. So, a bipartite state which has sort of half of it in system a, half of it in system b. We need those two different systems-- it's a composite system a plus b. And this state is entangled-- if and only if you cannot find two states psi a and psi b, such that the state can be factorized. We need a few examples to fully comprehend this discussion-- what systems a and b qualify. What does it mean if there does not exist any combination? So, let me give you some examples now, and then I think we stop. If you have the 0, 0 state, photons in two mode, we can factorize it-- and this is not entangled. If you have a state which is 0, 0 plus 1, 1-- well, you cannot write it as a product of one state times another state. This is entangled. Well, if you take the following state-- Is that entangled or not? I've written it as a sum of four states, but if you just stare at it for a split second, you realize you can just write it as a product of two states. It's just a product of 0, 1 with 0, 1. So, a state is not entangled if you can write it as a linear superposition of those states. The definition of entanglement is-- if you try hard and hard and hard, and there doesn't exist any product state decomposition, then it is entangled. So, that was easy. If you use a state like-- let me just leave the 1, 1 out. What about this one? Well, you can try as hard as you want-- you will not find the decomposition. So eventually, when I say you have to try hard to find decomposition, maybe what we want is we want to apply some operator or some procedure to this state, and get the answer, yes-no. This naturally asks for question-- how in general can we measure if a state is entangled or not? But this is something we'll discuss next week. So let me just conclude with the following. I'm focusing here and for pretty much all of this cause when we talk about entanglement, about pure states. If you have decoherence-- if you have the transition from pure states to density matrices-- it gets much more involved to talk about entanglement and measure entanglement. But at least the basic definition can still be maintained for statistical operators. If you describe the system-- the total system by one statistical operator-- the system is not entangled. If the statistical operator can be broken up into a direct tensor product of statistical operator for system a and system b. Otherwise, the system is entangled. So the basic definition can be generalized to density matrices, but a lot of things become much, much more messy if you don't have pure states. Any last-minute questions? See you on Monday.
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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. On Monday, we went through and looked at the functional forms for sp two hybrid orbitals, as found in the case of the BH three molecule. Now, you should recognize that there are other hybridization schemes that go along with different geometries. So, BH three was trigonal planar. And we are going to talk some more about BH three today. If you had been considering last time instead a tetrahedral carbon atom, then the hybridization scheme we would have needed to develop would have been the sp three hybridization scheme. That would be associated with a tetrahedral carbon atom. On the other hand, if we had a linearly coordinated carbon atom or some other main group atom, boron or beryllium, for example, then we would have had to develop the sp hybridization scheme. You should not miss the forest for the trees in putting into context from my previous lecture on the sp two specific case for that type of geometry. But now, rather than going through more examples of valance bond theory treatments of molecules, instead, it is time for me to introduce you to the tenants of the molecular orbital theory for treatment of molecular electronic structure. I mentioned valance bond theory as having been introduced by Linus Pauling. And then, I also named Robert S. Mulliken, MIT undergraduate, as the father of molecular orbital theory. And these are both two different perspectives on viewing electron structure of molecules that arises from the results of quantum mechanics. In the case of valance bond theory, we have a situation where you have localized electron pairs. In the case of molecular orbital theory, electron pairs are still going to be important, but they are going to be able to be delocalized over the entire molecule in some cases. So, this is very different. This is a major difference between MO theory and valance bond theory. In the case of the BH three molecule that we considered last time, you had four positively charged nuclei. And you arrange those four plus charges at the points in space that correspond to the equilibrium geometry of the molecule. And then you sprinkle in your six valance electrons, and you want to understand how those six electrons become organized in space in response to that electric field set up by those four positively charged nuclei. And we took the approach last time that we are going to localize electron pairs in between nuclei. And, because structurally the BH three molecule is symmetric and the three hydrogens are indistinguishable from one another, we decided that we were going to make hybrids. And so we talk about hybridization. The idea behind hybridization was to change the atomic orbitals of boron by mixing them, so that we would have one that would be able to point at each of the three hydrogens in space to form three nicely, perfectly directed electron sigma bonds between boron and hydrogen. That was the scheme that we adopted, this hybridization scheme, but in MO theory we are not going to do that. We are not going to interfere with the intrinsic atomic orbital structure of a boron atom in order to make bonds. And we are going to see that there are some predictions that come out of the MO treatment for the molecule that differ from those that came out of the valance bond treatment for the molecule. And so here, no hybrids. In terms of just the vernacular of chemical structure, you will hear sp three as being used interchangeably with the notion of a tetrahedron. But in valance bond theory, it refers to a particular hybridization scheme in which we actually mix s and p as a preparative to bond formation in a molecule. And we mix the atomic orbitals on that central atom in a hybridization scheme. In MO theory, we are not going to do that. It is very important that you keep these theories and the language associated with the theories separate in your minds so you can see the difference between these theories. And then one of the consequences of this valance bond theory and the hybridization scheme is that it is not so good for excited states. And what that means is that we were developing a scheme to describe the bonding in the molecule in its ground electronic state. Molecules can have excited states, just like atoms can have electronic states. And over here, in molecular orbital theory, we are going to find that we do a much better job with excited states. And that is important for understanding the range of properties associated with molecular systems. And you are going to see, indeed, that the energy level scheme for the six valance electrons in the BH three molecule is different, depending on whether you use valence bond theory or molecular orbital theory. Now, for molecular orbital theory, we are going to need, again, like we do for valance bonds, to have some kind of a procedure for forming molecular orbitals, conceptually. And the first step in such a procedure is that you are going to want to analyze the three-dimensional shape of the molecule. And we do this, of course, when we talk about the valance-shell electron-pair repulsion theory for predicting molecular structure. We are going to look at the structure. And we want to identify, usually by inspection, sets of symmetry-related -- -- atoms or orbitals. And here, I am talking about atomic orbitals. I will come back in a moment and talk about what I mean by symmetry-related. This is a concept that can be put on a very nice, firm, mathematical footing. And, in fact, if you find this type of analysis interesting and would like to see more of the math that can help you to organize the results of quantum mechanics in terms of symmetry, then you will want to put the 5.04 subject on your calendar for the future. That subject is devoted, in large part, to the applications of group theory to chemistry and chemical problems. And symmetry plays a big role in that. And then, two, we are going to form combinations. And let me just further quantify this by saying we are going to form linear combinations -- -- of symmetry-related orbitals. One of the big approximations that we usually kind of take for granted in the molecular orbital theory of electronic structure is the LCAO approximation. I am just going to mention that parenthetically, here. This is that we can form molecular orbitals that will be linear combinations of atomic orbitals. This LCAO approximation arises from the fact that we can solve the Schrˆdinger equation exactly for the hydrogen atom, but for big molecules and many electrons systems, we cannot. And what we like to do is to take the atomic orbital wave functions, that is, atomic orbital here in LCAO, and use those wave functions in our approximation of molecular orbitals. We are saying that we can combine these atomic orbitals on the atoms that are in a molecule to form the molecular orbitals that will be able to spread out and delocalize over the entire molecule. This is inherent in the development of the theory that I am working on here. And you should see that this will pop up in a number of cases, but it is an inherent approximation that we are accepting. And then, three, we are going to combine the linear combinations from part two -- -- with central atom atomic orbitals. And that is what we will do. And if we have done that properly, we will have arrived at molecular orbitals for the system in question. Now, I want to take a little bit of time to go through each one of these steps in order to define the problem that we have for this alternative way of viewing the electronic structure of the BH three molecule. Now, why am I choosing the BH three molecule for this? Well, I am choosing it because it is an easy problem and one that is illustrative of the steps that go into forming molecular orbitals for a system. In your textbook, you will see that before you get to something like BH three, first diatomic molecules are considered. Those, I will show you on Friday, are actually a little trickier to understand than is the case with BH three. That is because in molecular orbital theory we have the same number of MOs as AOs. If we start out with a certain number of atomic orbitals that come into play by virtue of the atoms that are in the molecule, then that number of orbitals will be the same as the number of molecular orbitals that we will get at the end of the problem. They won't all be filled. We will have some that are empty. But we are going to have the same number of molecular orbitals as atomic orbitals that go into the problem. And so what that means is that the complexity of the problem is related to the number of MOs and, hence, the number of AOs. The complexity of the problem scales with the number of atomic orbitals in the problem. We actually call these our basis functions. And one of the things that John Pople talks about, if you have gone and looked at his video that I pointed you to in the problem set, he talks about the wonderful fact that our computational power has gotten so great, and will grow so much greater in the future, that we are able to computationally handle problems of the calculations of the properties of enormous molecules that bring into a problem an enormous number, a vast number of atomic orbitals. And so, a great computational power is necessary to apply molecular orbital theory, or even the more recent density functional theory, to problems of electronic structure that we need to grapple with in order to predict properties of molecular systems. And in the BH three molecule, which is, as I said, a simple problem of electronic structure and a nice illustrative one for today's purposes, we have, this is a seven orbital problem. Why is this a seven orbital problem? It is a seven orbital problem because boron has four valance orbitals, an s, a px, a py, a pz. And we have three hydrogens, each with their 1s orbital. We have our valance orbitals on boron. We don't count the 1s orbital on boron because that is filled and is a core orbital, and it doesn't get involved in chemical bonding. The valance shell for boron is the n equals two shell. We have 2s, 2px, 2py and 2pz. And we have three hydrogens, each with their 1s orbital. If we were to tackle right now the problem of the molecular orbital energy level diagram for a diatomic molecule like O two, dioxygen, we would find immediately that this analysis of how many atomic orbitals we have available would give us the value eight. And so it is a more complicated problem. It has one more atomic orbital than this problem, even though this one has four atoms because three of these atoms happen to be hydrogens, which only bring in one orbital to the problem. That is why I am choosing this. And then, in addition, I am choosing this because the consequences of the symmetry of the BH three molecule are that the orbital interactions that we are going to identify occur in nice pair-wise sets. And that makes things especially easy to see in terms of how chemical bonds arise in the context of MO theory for a problem like this. And so, to now go ahead and carry out step one for this, we are going to draw the molecule and try to identify sets of symmetry-related atoms/orbitals. And, before you do that, I just would like to show you an example of a highly symmetric molecule, because this notion of symmetry is something that, at this point in time, I really only want you to gain an intuitive grasp of, I am not going to quantify it. But, if you look at a molecule like the one I have placed on the left and right-hand screens, you will see that it is a large, round molecule. Actually, round, spherical things are the most symmetric things that we can think of. Here, you can imagine that this is a big ball with atoms located at various points on the surface of the ball. This is just a ball and stick representation of the C60 molecule, also known as Buckminsterfullerene. It is a geodesic dome-type of molecule. This molecule has the chemical formula C60. There are no hydrogens in this molecule. All the atoms are displayed. And each carbon atom sits in a position where it is adjacent to one five-membered ring and two six-membered rings on the surface of this spherical molecule. And that is true of every carbon atom on the surface of this whole molecule. As you go on in chemistry, if you go into the analysis of molecules using nuclear magnetic resonance spectroscopy, you will find that it is really important to be able to identify the symmetry of a molecule. And you will realize that the symmetry of the molecule is manifest in the nuclear magnetic resonance spectrum of a molecule. If you take the carbon-13 nuclear magnetic resonance spectrum of this molecule, a sample composed of this molecule, you will see that there is only a single C13 signal in the spectrum. And that is because all 60 carbon atoms are in an identical environment in this molecule. Every one of them feels exactly like everyone else. They are all equidistant from the center of this molecule. And they are all equidistant from their set of neighbors. And so, in that way of looking at a molecule, you would see that they are all equivalent. You may have encountered this molecule before, but this molecule's discovery -- -- was actually predicted initially from analysis of mass spectrometry data by Professor Smalley, a Nobel Prize winner whose efforts in this area have spawned off chemistries involving not only these nano-sized balls of matter, but also nanotubes made of carbon, and this whole area that we think of nano technology. Really a lot of it is dominated by the chemistry of new forms of carbon that arose with the discovery of Buckminsterfullerene. And that is just one example of a new allotrope of carbon that was discovered in recent years. This gives you an idea for symmetry. I could show you pictures, actually, of enormous biomolecules. There are large viruses that are composed of biopolymers, macromolecules that pack in a way that is symmetrical, so that you can see these things. If you view them in the right kind of representation, you will be able to see the symmetry in them. And one definition of symmetry that I would like you to take away from this picture is just that each of the atoms that are symmetry-related is indistinguishable. If you turn the molecule around and look at the atoms, those that are symmetry-related are indistinguishable from one another. And so, when we have a trigonal planer BH three molecule, is the boron symmetry related to any of the other atoms? No, it is not. What about this hydrogen? I labeled them last time A, B, and C. This hydrogen labeled C, is it symmetry-related to other hydrogens? Yes. And that is because of the symmetry of this molecule with these 120 degree bond angles on the planarity of the molecule. So, you have a set of three hydrogens. And their 1s orbitals are in space indistinguishable from one another. They are related by symmetry. And so, in this problem here, we have four orbitals here on the boron that are not symmetry-related. And then also incidentally let me point out that the boron atom is located at the center of gravity of this system. If atoms are going to be symmetry-related, they must not be located at the center of gravity of the system. Let's go over here and expand on these ideas. The method that I am going to develop here for forming the linear combinations has to do with thinking ahead, in this problem, to the fact that we are going to want to make linear combinations that have the correct symmetry to bond to atomic orbitals on the central boron atom. Let me draw the ones that are going to be relevant to this part of the problem. This one over here would be the boron pz orbital, so it has one positive lobe coming out of the board, negative lobe going back. Over here we would have the boron py orbital using the coordinate system that I had chosen last time, which is x up and y to the left. And then here we have the boron's px orbital. And then over here we have the boron's 2s orbital. And so, our challenge now will be to construct linear combinations, we are at part two, of the set of three hydrogen 1s orbitals that can match in symmetry the boron central atom's atomic orbitals. Last time, remember, for hybridization, we were making these orbitals mix with each other in order to point at the hydrogens. Now what we are doing is kind of an inverse concept. We are going to mix the hydrogen orbitals so that they have the right symmetry to interact with the central atom atomic orbitals. There is a nice parallelism here. Here is going to be our LCs. They key feature of the boron's 2s orbital is that it does not have any nodes. Remember nodes are surfaces. When you pass from one side of a node to the other, you get a change in sign of the wave function. And we indicate that change in sign by differential shading. The 2s orbital has no nodes whatsoever. And a way that we can construct a linear combination that has the same spatial, nodal properties as that boron 2s atomic orbital is as follows. We can involve contributions from each of the three hydrogen 1s orbitals. Remember this one is A, this one is B, and this one is C. This is going to be a linear combination of the three hydrogen 1s orbitals. I will write this one as follows. This one will be written as A plus B plus C. That indicates the 1s orbital on A, plus the 1s orbital on B, plus the 1s orbital on C. And, as we talked about last time, wave functions that we write should be normalized. And they should satisfy the unit orbital contribution rule. For normalization, here, I give a factor of one over root three for this linear combination formed as a symmetry match with the boron 2s orbital. You can see, I hope, what we are doing. We are projecting the nodal properties of the central atom atomic orbitals onto the linear combinations that we are forming. And we are going to form a complete set of three linear combinations in this way. Let's make one that has symmetry properties that remind us of this px orbital. On HA we are going to have a positive contribution to match the positive contribution of this lobe of the px orbital that points along the plus x axis. And then down here we are going to have contributions from hydrogens B and C. And they are going to be smaller, and they are going to be opposite in-phase. They are going to be opposite in-phase because we are building a linear combination that has a node approximately at the center of the system here, so as you go from positive x into the negative x region of space, the wave function changes sign to match the change in sign associated with the px orbital. We are projecting the nodal properties of px onto the linear combination of hydrogen orbitals that we are forming here. And this one, written in normalized fashion, will be root two over three, A minus one-half B minus one-half C. And, although what we are working with here are linear combinations of these symmetry-related hydrogen 1s wave functions, you are going to find that these coefficients on the atomic orbitals, that contribute eventually to the molecular orbitals, are going to come out as normalized and as unit orbital contributions. So that if we started out this problem with a single 1s orbital on HA, that will be entirely accounted for among these linear combinations and the molecular orbitals that we are going to make with them. Now, let's generate a linear combination having the nodal properties of py. And, in order to do that, we need to have a negative coefficient out here in the minus y direction. We are going to put in a contribution from HC as negative, like that, to match that. And then over here, we are going to make a contribution for HB that is positive. And then, noting that there is a nodal plane along the y,z-plane, which comes out of the board like this, so that we are always negative along minus y and we are positive along plus y. We match that here. And the coincidence of that nodal plane with the location of HA dictates no contribution from HA to this orbital, for reasons that actually we looked at last time, -- -- namely, that we cannot bring a hydrogen 1s orbital in here and also change sign on going half-way through that hydrogen 1s orbital, because s orbitals have to have the same sign everywhere. It does not contribute to this linear combination. And our normalized form for this will be one over root two B minus C. And then, if I were to ask the question, can I make a linear combination of the three hydrogen 1s orbitals that has the same nodal properties as pz, the answer would be no because they all lie in the x,y-plane and they are just s orbitals and cannot change sign as you go through the x,y-plane from plus z to minus z. And so, we are done here. And what you are going to find is that we have created these three linear combination according to step two. And taking into account both the symmetry properties of the molecule to identify a set of three symmetry-related atoms and orbitals. And then, taking into account an analysis of the nodal properties of the central atom atomic orbitals, so that we could project those out to help us find appropriate linear combinations for mixing with the central atom orbitals. And, when we do that mixing, we are going to find out that there are three ways that we can do it. We are about to move onto step three of this problem. We are going to need to combine these linear combinations with the central atom atomic orbitals according to the rules of MO theory to generate, first, bonding molecular orbitals. And the bonding molecular orbitals that we will get will be an in-phase combination -- -- of our LCs, our linear combinations of atomic orbitals, with our boron atomic orbitals. That will describe the chemical bonding in our system. And we will see that it contrasts in a very interesting way with the hybridization scheme developed last time. And the key phrase to underline, here, is in-phase. And what that in-phase means is that when two positive lobes of two orbitals centered on two different atoms are juxtaposed and neighbor one another and can have good overlap of their atomic orbital wave functions, that leads to in-phase constructive interference and stabilization of the electrons associated with that newly formed bonding molecular orbital. And that stabilization is what we call the chemical bond. The analogy to that in valence bond theory is the idea that a pair of electrons associated with two nuclei in a sigma bond is more stable because it experiences, simultaneously, two positive charges. And here we are generalizing that and allowing electrons to flow over the molecule as a whole. But now we have a new concept, and that is antibonding. Antibonding molecular orbitals will be out-of-phase combinations that are repulsive and lead to high energy interactions. And when, as in the case of the problem we are considering here, the interactions occur in pair-wise sets, we will find that we get very nicely, for every bonding molecular orbital, a corresponding anti-bonding molecular orbital. Also, one of the interesting things is that if you start putting electrons into anti-bonding orbitals, if your system just happens to be so constructed as to have many electrons, such that you fill up not only the bonding molecular orbitals of electrons to make the chemical bonds, but you continue on and you have enough electrons to keep going and put them into anti-bonding orbitals, those antibonds start to cancel the bonds. And so, you will see a very nice progression of this as we study the MO theory of the homonuclear diatomic molecules, starting on Friday. And then, here is another concept that arises from the MO analysis of a molecule, and that is that certain orbitals can be non-bonding. And an example of this would be a lone pair of electrons. And it happens when an orbital or a linear combination of orbitals finds no counterpart -- -- of like nodal symmetry. These nodal properties of orbitals are very important, and I will show you later how the nodal properties are related to the energies of the orbitals, as we consider them. Having given you this preview of how orbitals are going to be able to combine in the MO theory, let's see how it actually takes place in the case of BH three. Now we are drawing another example of an energy-level diagram, where the energy is low at the bottom and rises as it goes up. And these energy-level diagrams that we are now developing for molecules are analogous to those that you studied earlier in the semester for atoms. And we are just generalizing this notion to the atoms. And what I am drawing over here is our boron 2s orbital and here is our boron 2px, 2py, 2pz orbital. Those horizontal bars just represent the energy of these orbitals in the molecule. Here I am just redrawing the boron atom. And then, over here on the right, we are going to see that we have our linear combinations that we developed. These are our three H 1s linear combinations. We found that the three hydrogens in BH three were symmetry equivalent, so we generated linear combinations. The pictures of them are over there. I can give them names. Why don't I call them D, E, and F. We have D constructed to match the nodal properties of the boron's 2s orbital. E and F were respectively constructed to match the nodal properties of the boron's 2s, 2px, and 2py orbitals. And I am showing you their relative energies. And now we need to do this issue referenced here as point three. We need to combine these things. I have these on the one hand and these on the other. Our seven atomic orbitals have been changed into four atomic orbitals and three linear combinations, so I still have seven orbitals total. And now I am going to combine these with these according to their nodal properties to generate seven molecular orbitals. And let's do it this way. We are going to take linear combination D that has been constructed to match the boron 2s orbital in terms of nodal symmetry properties, and we are going to make a bonding combination. I am going to draw these pictorially in a moment. And, in this case, those are the only two orbitals in my seven orbital system here that have that set of nodal symmetry properties. And for every bonding MO, I must have an antibonding MO. This is one of our molecular orbitals. And there is going to be a corresponding molecular orbital up here, high in energy. And this will be an antibonding molecular orbital that will be the out-of-phase combination of the boron 2s with this linear combination that I labeled D. Antibonding molecular orbitals are usually denoted with a star. We have a bonding combination and an anti-bonding combination. Now, we can form two more bonds that will spread out over the molecule because, if you recall, we had our px and py pair that served as the nodal template for our construction of linear combinations E and F. We are going to be able to match up those to form two more bonding molecular orbitals. And these will be found to be higher in energy than the first one that we formed from D. Here is a pair of bonding molecular orbitals that derive from linear combinations E and F combining in-phase with boron's px and py atomic orbitals. And there will be a corresponding antibonding combination, where we allow those orbitals to interact in an out-of-phase manner. And you will see what that means shortly. Let me put that up there and add a star to indicate that this high energy pair of molecular orbitals is an antibonding pair of orbitals. I have six orbitals now in my molecular orbital energy-level diagram for BH three. And that means I am not done because I have to have seven, and I started out with seven atomic orbitals. Look over here. 2pz was an atomic orbital on boron that did not find any way of serving as a template for making a linear combination involving the three hydrogens. And so, it comes over here as nonbonding. It has no counterpart of like nodal symmetry because of the location of those three hydrogens in the x,y-plane, which is a nodal plane for the boron pz orbital. And so, this one is nonbonding. These three orbitals up here are anti-bonding. And the ones down at the bottom, which are the lowest in energy, corresponding to being able to most tightly hold onto electrons in them are bonding molecular orbitals. And so our electrons can fill into this MO energy-level diagram in that way. We have our six electrons that come into this problem. We have boron bringing in three valance electrons and three hydrogens each bringing in one valance electron. There are six electrons to put into the diagram, filling up three of the molecular orbitals and then leaving empty pz. Let me introduce a little bit more MO language right now. This one here will be called the highest occupied molecular orbital, and this one here will be called the lowest unoccupied molecular orbital. The reason why I am drawing attention to these orbitals is that in chemistry, the chemical properties derive oftentimes from those orbitals that are in what is called the frontier orbital region. And the frontier orbitals are those close in energy to the HOMO-LUMO gap. And I will come back to this. But those highest energetically lying electrons are going to be the ones responsible for nucleophilic properties of the molecule and basic properties of the molecule and reducing properties of the molecule. Whereas, low-lying empty orbitals are going to be the ones responsible for acidic properties of the molecule or oxidizing properties of the molecule. And we will come back to that in a moment. But that is something pretty general and very useful that comes out of studying molecular orbital energy-level diagrams. Now that we have the diagram, let's see what the orbitals in the diagram look like. And I will try to do this relatively quickly. Here, let's start with the lowest lying molecular orbital in the system. This is a representation of an in-phase combination of the boron 2s plus D, where D is defined up here as that linear combination. This will be a molecular orbital having the nodal properties of a 2s orbital centered on that central atom. That is where our lowest-lying two electrons reside. Now, if you look at the LUMO over there and then go up one orbital in energy, you will be looking at this orbital, which is the boron 2s orbital minus D. And the thing that makes this out-of-phase linear combination so much higher in energy than its in-phase counterpart is the appearance, now, of a nodal surface. And this node is between the nuclei. It goes all the way around and is between the central atom s orbital and those peripheral hydrogens. And I will show you a picture of it. This is our BH three LUMO plus one molecular orbital. You can see that we have a wave function in the center of one sign. And then, as we go along any one of the B-H bond vectors from boron to hydrogen, we change phase midway along the bond from positive to negative. And that is true, no matter which of the three B-H bonds we pick to traverse along. That one has the nodal properties as drawn down there for the boron 2s interacting in an out-of-phase manner with linear combination letter D. And so, next time, I will finish up and show you what these other orbitals look like as calculated. I hope you have enjoyed this. We will see more MO theory on Friday.
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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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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 PROFESSOR: The final quiz is scheduled for a week from today, on December 8. Following that weekend, we'll have one class on the 13th. And the last days of classes are Monday, Tuesday, and Wednesday, the 12th, 13th, and 14th. So I know a number of you spoke up last time and said that you have a conflict on Thursday, the eighth, with a quiz in another class. So could I see a show of hands again of how many of you have this conflict? So three, six people, seven people, you're going to enroll in the other course just so you can put off taking the quiz. That's unfortunate, I don't think anybody should have to take two quizzes in one day. We can't move it up. We'll have to move it back. So I don't know if I am violating any Institute rule, but I know that it is strictly illegal to give assignments that are due after the last day of classes, let alone have a quiz after the last day of classes, which has not been previously scheduled as a final examination. So we can't give it after Wednesday, the 14th, the last day of classes. So, do any of the six impacted people have a strong preference for when we should schedule the quiz? AUDIENCE: Monday? PROFESSOR: Due it Monday? It doesn't have to be that soon. Monday? AUDIENCE: Why not Tuesday? PROFESSOR: Because Tuesday we have a class. AUDIENCE: Or Wednesday? Well we would have a Thursday class instead. Correct? PROFESSOR: No. If we're doing it after the quiz next week, there's a class. We have a last meeting here on the 13th. And I'm not going to shift it for other people, just for the impacted. So of the six who are entitled to vote, how many would prefer to have it Monday, the 12th? Three and I know how it's going to turn out. And for Wednesday, the 14th? Three. AUDIENCE: I can't take the test because-- not because of another test-- I'm out of town. And I might come back Saturday. And Monday is just a little quick to take it after we get back. PROFESSOR: I think I'm going to have to make an executive-- yes AUDIENCE: Professor, when you mean by conflict, do you mean having an exam at the exact time? PROFESSOR: No, it's on the exact same day. Considering this is a two hour examination, to have another examination exactly the same day, if you're not brain dead after an hour or two, you will be pretty close to it. So I think that is an unfair penalty to pay. I think, to keep it as close as possible to the quiz that the rest of the people will be taking, why don't we make it-- since it's a tie vote-- on Monday, the 12th. Do it sooner, rather than later. And I'll let you know next time of where we will hold it. I'll have to arrange a room for it. So that will not be a make-up quiz. It would be a made-up-- invented-- quiz. And, as I say on Monday, I should have all of the homework and the previous quiz to return to you. Reason you don't have it now is all the little crabbed handwriting that you see before you in the form of these notes, which takes forever. All right, so since we've satisfied the unpleasant aspects of the end of the term, let's get back to discussing some of the other piezoelectric effects that we have defined. And then also ask the question rhetorically, is there anything like a representation surface for a third-ranked tensor property? And it doesn't look promising. But there are some things that we can do to discuss variation of properties with direction, and we'll see directly what those are. But first let's look at the converse piezoelectric effect. And this again is a third-ranked tensor property, but what we do is to have the elements of strain, epsilon ij, a second-ranked tensor. And to take advantage of this curious relation between the directed converse effects, we define the converse piezoelectric effect as giving you nine elements of strain in terms of a third-ranked tensor dijk times e sub i. So what is not standard is our convention for the order of the subscripts on the moduli. And we make up the rules, we can do it any way we like. And the advantage of defining it this way, in nonstandard tensor notation, is that we can use the same coefficients for both the direct and the converse effects. So let me-- to illustrate what these equations look like-- write out a few examples. Epsilon 11 would be d1jk times e sub 1. I'm writing it in simple fashion, and not expanding fully, just to save space and time. Epsilon 22 would be d2jk times e2. Epsilon 33 will be d3jk times epsilon 3. And let me stop after these six terms, and write, at least for these, an expansion. Because this gives us some interesting information. I'm not doing the equal signs in here. So this would say that the element of strain epsilon 11 is d111 times e1 plus d112 times e2 plus d113 times e3. And I want to say this is 11k. And I want to say that this is 22k. The next term would be epsilon 22, and that would be d122 times epsilon 2 plus d212. AUDIENCE: Wouldn't the 2's be balanced? PROFESSOR: Hm? AUDIENCE: Wouldn't that be d2's? PROFESSOR: Yeah, you're right. This should be epsilon jk equals dijk times e sub i. So this is all e1. This is e1. And this one goes with this one. This 2 goes with this one. You're right. d2 and this is the d311 times e11. This would be 122 times e1 plus 222 times e2 plus d322 times e3. And the fourth one would be e33 equals d133 times e1 plus d233 times e2 plus d333 times e3. OK, these elements here are the three that appear in the box that I had indicated for the terms djk. When we write the direct piezoelectric effect, this would be the box of coefficients. When we write the converse effect, this is the box of coefficients. And now what I wanted to point out is that delta v over v is equal to episilon 11 plus epsilon 22 plus epsilon 33, which is the trace of the strain tensor. And we're going to get one set of terms which depends on the x1 component of the field, another set of three terms that depend on the x2 component of the field, and another one on the x3 component. If these expressions sum to 0, then there would be no volume change. So this first set of 3-- 3 of the first 6 equations-- give you an indication of when the application of a field will result in no volume change in the sample. And that again, I remind you, can be for two reasons. It could be because all of these six of the nine piezoelectric moduli are 0. Or alternatively, if you examined the form of the piezoelectric moduli matrices that is required by symmetry constraints, you'll find that there are a substantial number of point groups for which some terms are 0. But then there is, in addition, an equality between some of the other terms, which make the volume change zero. Even though, not all of the elements within this 1/2 box of moduli are zero. So that is an interesting effect. And it turns out that the majority of the non-centrosymmetric point groups do not have a volume change, when you apply in a field in any way you choose. A few do, but it's a minority. Alright, but this is not the main point of writing this. I want to-- at this point-- point out that this tells you about the volume change. And then we would have additional terms, we would have a term of the form e1 something like e123 or e132. And this would be e14. This would also be e14 since we replace both of those subscripts by a single subscript. And the tensor elements that would go in here would be d123. We just want one of the them. d123 times e1, and then we want d223 times e2, and then d323 times e3. We're writing one of the specific equations for the shear strings. If we write the expression for d132, this is going to be d132 times e1 plus d232 times e2 plus d332 times e3. There are two interesting consequences of this. The strain tensor is symmetric. And this element of strain-- why have I got three subscripts in here? Don't want that one in there. These two strains are equal. And therefore, if we would apply just an e1 for example, let e be equal to just a component of field along x1. The strains have to be equal. But we have two different tensor elements here. And the only way that strain can be symmetric, and it's defined as such, is that d123 be identical to d132. And that resolves the issue that came up in connection with the direct case electric effect. We said the direct piezoelectric effect depends just on the sum of the elements dijk and dikj. And since they're lumped together, all we can measure is the sum. And, so, we'll just have to call that a single matrix element. The converse piezoelectric effect tells us these tensor elements have to be equal, if the strain tensor is to be symmetric. So that says that, since we defined d16 as the sum of d123 plus d132, this says that d132 is equal to d123 is equal to 1/2 of d16, just making the equality in the reverse direction. The converse effect let's us say that 123 has to be 132, that any ijk has to be equal to a dijk. So if I tried now to write this first expression in the reduced subscript notation, e23 is what we let e5 be. And now we have this equal to and in our reduced subscripts, we have this as 1/2 of d16. And the field that's multiplying this is piezoelectric modulus is e1. And that messy factor of 2 has come back to haunt us again. It's like trying to stuff a jack-in-the box back in the box. It keeps popping up. We ate the factor of 2 in defining the matrix representation of the piezoelectric electric modulus. And now when we try to go to a reduced subscript notation for the converse piezoelectric effect, we've got a 1/2 in there. And similarly, the second equation would be e5-- same result epsilon 5-- and it's 132, but 132 is 1/2 of d16 times e1. And we have a similar mess for the other coefficients here. So what do we do? Do we say that the relation between strain epsilon j equals dij e sub j has 1/2 in front of several of the coefficients and not in others? Well, we can't really absorb the factor of 2 in the definition of the piezoelectric moduli, because we've already done that. So the only thing we can do is to say that we will have to take the off diagonal strains, and define them as having a 1/2 in front, and we add these up. So we will have to write matrix strain in this reduced subscript notation. We'll have to take e11, epsilon 12, epsilon 13, epsilon 21, epsilon 22, epsilon 23, epsilon 31, epsilon 32, and epsilon 33. And in converting this to matrix form, we'll call this epsilon 1, analogous to what we did for the tensile stresses. We'll call this epsilon 2 and this epsilon 3. And then for all of the off-diagonal elements of strain, in order to avoid the factor of 2 popping up in front of the matrix representation of the piezoelectric moduli, we're going to have to put in here 1/2 of epsilon 4, 1/2 of epsilon 5, and 1/2 of epsilon 6, and same for the off diagonal terms 1/2 of epsilon 5, 1/2 of epsilon 4, and 1/2 of epsilon 6. So the moral of this story is that you can't win, but if you play it right, you can come out even. So only if we define the reduced subscript strains in this fashion, can we write an expression of this form. So this algebra is carried through for you for the other elements in the notes. But this is the way we are forced, unless we want to have a factor of 2 in some terms, and not in others, is the way we have to define matrix strain. All this is formalism and definition, but I'd like to now do two things. First of all, give you some examples of real numbers for piezoelectric moduli, and then ask the question about representation surfaces. Once again, these numbers are in the handout for you, so you don't have to make note of them. But one of the very important piezoelectric materials is the quartz form of SiO2. SiO2 has many polymorphic forms. Quartz is the form that's stable at room temperature, and it has point group 32 asymmetric. And there are higher temperature polymorphs of SiO2. There's a phase transition in quartz to a more symmetric form, and then there are cubic forms at the highest temperatures. Now, quartz is not the material that displays the largest piezoelectric moduli. But it has the following advantages. One is it is a naturally occurring material that is very inexpensive. So it's not an exotic expensive material. Very stable. It's not water soluble. Extremely tough. You can take a thin wafer of quartz, for example, if you want to make a monochromator for a diffraction experiment. You can take a thin wafer of quartz, and bend it like this, and it does not break, very elastic. And if you want a material that's going to earn its living by being squished, you want something that doesn't plastically deform and something that is very hard and resistant to stress. So quartz, even though the moduli are not the largest, is a very attractive material, and is used in a variety of devices. There was a time when CB radios were very, very popular. Everybody had to have one in their car. I guess so they could pretend that they were truck drivers. But anyway, you don't have them anymore now that cellphones have come in in existence. But for your CB radio, you needed something called a crystal. And they were fairly expensive. And the number of channels on which you could communicate dependent on the number of crystals that you could plug into your CB radio. The so-called crystal was exactly that. It was a little black box that looked almost like a transistor. And there were two leads coming out of it. If you ever got curious and broke this thing open, what you found was a nice wafer of quartz. And on the wafer of quartz-- brazed onto it-- were two wires. And that's all there was in the box. The crystal really was a crystal. And the crystals had been very precisely ground to thicknesses such that when a field caused these wafers to hit a resonance, that resonance would be at exactly a particular frequency. And that was the frequency of that channel. One of the crises, during the Second World War, is that the highest quality natural crystals of quartz come from Brazil, and during the conflict the sea channels were essentially blocked. And so, people-- in order to make all these communication devices-- had to learn how to synthesize crystals of quartz synthetically. And there are a number of companies, such as Sylvania up on the North Shore, that developed entire buildings devoted to growing single crystals of quartz. And they're big tanks like something out of the aquarium. And in the center of the tank is a rod, and seeds of quartz are placed on the rod. And the thing very slowly rotates around in this solution. And on the rod, eventually, are single crystals of quartz that are this size. And its a very spectacular thing to see. In any case, symmetry 3 2, and the moduli that are 0 and non-zero. If we refer the reference axes to a set of coordinates with x1 in this direction, and x2-- since it has to be orthogonal to x1-- in between the two-fold axes, and x3. And for all materials of commerce that are anisotropic, there has to be some standard for defining the choice of axes. For example, crystallographers would say the unique axis should be along the z direction, the x3 direction. But why isn't x1 and x2 in between the two-fold axis? There's some professional society that is responsible for giving standards for representing property measurements in some mutually agreed upon form. And this is the standard set of axes for the quartz and symmetry 3, 2. The moduli, dij, not dijk, but dij, these two are constrained to be equal. This one is 0. This is minus 0.67, 0, 0, 0, 0, 0, 0, 0.67, 4.6, 0, 0, 0, 0, 0, 0, 0. One of the strange materials for which no field can create a strain-- field along x3 cannot create a strain. And these are all in units of 10 to the minus 12 coulombs per Newton. An example they give you here, if you apply a field of 100 volts per centimeter, which is not terribly large, but would be comparable to what you have in some electronic device, perhaps. So this, since our units are MKS, this would correspond to 10 to the 4 volts per meter. The strain epsilon 1, which is d11 times e1. It turns out to be minus 2.3, which means it contracts. That's the significance of the negative times 10 to the 4. And that turns out to be 10 to the minus 12 times 10 to the fourth. That turns out to be a strain of minus 2.3 times 10 to the minus eighth. 10 to the minus eighth is not exactly a large point strain. You're not going to see the crystal wafer twitch and jump, if you apply a field of 100 volts on it. But yet, even a strain of this sort is more than enough to be useful. But this is just to illustrate that quartz is not the most sensitive of piezoelectric materials. Another one that I give you data for is so-called ADP. And this is a widely used material. This is ammonium dihydrogen phosphate. And this is one of the family of salts that have very large piezoelectric responses. The nice part about it is that it's water soluble, so you can grow very, very large crystals easily from solution. The nasty part about it is that it is water soluble, so you have to be careful to protect this material from moisture if you're going to use it in any sort of device. But if we look at the moduli, relative to the standard axes, and this has point group 4-bar 2m. So x3 is taken along the 4-bar access. Then there are two-fold axes and orientations like this. And since they are orthogonal, you can take both x1 and x2 along the two-fold axes. And the numbers here for dij, are 0, 0, 0, 1.7, 0, 0, 0, 0, 0, 0, 1.7, 0. This is one of the interesting tensors where there's a diagonal row of non-zero terms off on the right hand side. And finally, the big surprise is the third modulus this is 51.7. So you can see this has a very strong effect. This is over 10 times the maximum piezoelectric modulus in quartz. So this is a material that's very commonly used in transducers. This is again times ten to the minus 12 coulombs per Newton. One of the very, very exciting developments in recent years is a class of materials that are perovskites. And they are very, very new. I gave you the reference to the first one that was reported, and that was just in the spring of 2000. And these are perovskites. Perovskites are materials that in the type form are cubic. But depending on composition and temperature, they can transform to a distorted version of this very simple cubic structure that is tetragonal. And this material can exhibit piezoelectric effects. And whether it distorts or not depends on the relative sizes of what goes into the perovskite. Perovskite has a composition ABO3 like barium titanate is one example. And the material has two different cations. And they have different valences so they have different sizes. And I won't bother to describe the structure, but it is only a very restricted locus in the field RA versus RB, where both the A and the B can remain in contact with the oxygen without distortion. And it turns out to be a line that does something like that. Any other perovskite-- and there are lots of them in this field of radii-- has to have one of the ion sort of flopping around. And if that gets too serious, the structure distorts so that all these ions can remain in contact with the oxygen. OK in order to do that, many of them distort to tetragonal forms. Others distort to super structures, which have very, very large unit cells. But in any case, when you're right at the phase boundary between the distorted structure and the true perovskite structure, these materials sometimes have very, very complicated x solutions of the two phases, on a very sort of fine scale. And it's not known exactly why they have this property. But they have absolutely enormous piezoelectric moduli, very close to this phase boundary. And the references that I give you-- here-- is a compound that is a lead titanium zinc niobate. And the complicated composition is to get you close to this phase boundary. This has a d333 that is greater than 2,000 picocuries per Newton. And pico is 10 to the minus 12. So this is a piezoelectric electric modulus that is 10 to the 3 times d11 for quartz. So 3 orders of magnitude stronger than this very commonly used piezoelectric material. Some of these materials have strains getting close to 1%. So this is something that will actually twitch on the lab bench, when you apply a field to it. So these are entirely new. People still don't know the origin of this behavior. And it's still under considerable study. So this is a new family of materials. It's very exciting, and undergoing a lot of investigation and development work at the moment. Alright, we are almost out of time. Time goes fast when you're having fun. Let me raise the question that we'll consider next time, which will be one of our last lectures. And that is, is it possible to create representation surfaces that tell you how the piezoelectric properties of a particular material will vary with direction? Well, vary with what? Well, we talk about the direct piezoelectric effect. This gives us components of-- well let's look at the simpler one, in terms of what we apply. We have the converse piezoelectric effect that says that epsilon ijk is going to be dijk times e sub i. So we've got a piece of material, and we apply a field, e sub i. So we can vary this in space relative to a coordinate system x1, x2, x3. But how in the world are we going to show what happens? So I always do an extra thing in here. There are 9 components to the strain tensor, which is symmetric. So they're really six responses that are unique. So yes, we can define the direction of the applied field. But they're going to be 6 different strains. So we're going to need six representation surfaces. One for each of the three tensile strains, and one for each of the three shear strains. So you can't do it with a single surface. So you can't do much other than say, there are certain responses which are intended to emphasize one particular sort of strain or one particular sort of polarization. So one of the things we might do is to cut a very thin plate of something like quartz, and subject it to a uniaxial stress. So let's say sigma along the x3 axis. So we're looking at a very restricted strain tensor. That's 0, 0, 0, 0, 0, 0, 0, 0, sigma 3. And in response to that strain, there are going to be three different components of the polarization. Polarization is manifested as a charge per unit area. So if we make a very thin plate-- to be sure there will be charges induced on these thin edges-- but if it's got a surface area that's a large compared to the area of these thin edges, we are going to be measuring primarily p3-- the component of polarization that's normal to this surface-- and that might have a charge per unit area that is comparable to these other two charged surfaces. But because the area by design of our specimen is so large, the response that would be most easy to detect, and which would be the largest response, by design, would be p3, which is the charge per unit area on this surface. So this is an effect we can define for a particular sample, and for a particular special form of the generalized force. And we then can ask, what is the value of the single modulus that relates p3 to sigma 3? And that's a question we can ask. And we can plot that response as a function of direction of a plate that we consider as being cut out of a single crystal, and different orientations, and then ask how this modulus-- which connects the two-- changes with the orientation of x3. So that is a question we can ask. And these surfaces are absolutely wild, highly anisotropic, can be identically zero in certain special directions, and they are very interesting, and a lot of fun to look at. So we'll take a quick look at a couple of those, which won't come as a surprise because they're already worked out for you in the notes. So we'll take a look at one of those. And in the problem set, I invite you to amuse yourself by looking at such representation surfaces for two other point groups. And with that, having kept you til five after the hour I will quit.
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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: Levinson's theorem, in terms of derivations, that we do in this course, this is probably the most subtle derivation of the semester. It's not difficult, but it's kind of interesting and a little subtle. And it's curious, because it relates to things that seem to be fairly unrelated. But the key thing that one has to do is you have to use something. How all of the sudden are you going to relate phase shifts to bound states? The one thing you have to imagine is that if you have a potential and you have states of a potential, if the potential changes, the states change. But here comes something very nice. They never appear or disappear. And this is something probably you haven't thought about this all that much. Because you had, for example, a square, finite square well. If you made this more deep, you've got more bound states. If you made it less deep, the bound states disappear. What does it mean the bound states disappear? Nothing, really can disappear. What really is happening, if you have the bounds, the square well. You have a couple of bound states, say. But then you have an infinity of scatterings states. And as you make this potential less shallow, the last bound state is approaching here. And at one state, it changes identity and becomes a scatterings state, but it never gets lost. And how, when you make this deeper and deeper you get more state, is the scattering state suddenly borrowing, lending you a state that goes down? The states never get lost or disappear. And you will, say, how could you demonstrate that? That sounds like science fiction, because, well, there's infinitely many states here. How do you know it borrowed one? Well, you can do it by putting it in a very large box. And then the states here are going to be finitely countable and discrete. And then you can track and see indeed how the states become bound. So you'd never lose or gain states. And that's a very, very powerful statement about quantum states in a system. So this is what we're going to need to prove Levinson's theorem. So Levinson theorem theorem-- so it relates a number of bound states of the potential to the excursion of the phase shift. So let's state it completely. It relates the number N of bound states of the potential to the excursion, excursion of the phase from E equals 0 to E equals infinity. So in other words, it says that N is 1 over pi delta of 0 minus delta of infinity, a number of bound states of your potential is predicted by the behavior of the phase shift of scattering. So how do we do this? This is what we want to prove. So consider, again, our potential of range R and 0 here and here is x. And I want you to be able to count states, but discovering states are a continuous set of states. So in order to count states, we're going to put a wall here, as well, at some big distance L much bigger than A than R. And we're going, therefore-- now the states are going to be quantized. They're never going be quite scatterings states. They're going to look like scatterings states. But they're precisely in the way that they vanished at this point. Now you would say, OK, that's already a little dangerous to be. Oh, you've changed the problem a little, yes. But we're going to do the analysis and see if the result depends on L. If it doesn't depend on L and L is very large, we'll take the limit this L goes to infinity. And we claim, we have answer. So we argue that L is a regulator, regulator to avoid a continuum, continuum of states to avoid that continuum of states. All right, so let's begin to count. To count this thing, we start with the case where there is no potential again. And why is that? Because we're going to try to compare the situation with no potential to the situation with potential. So imagine let V identical is 0, no potential and consider positive energy states. These are the only states that exist. There are no bound states, because the potential is 0. Well, the solutions were found before, we mentioned that these are what we call phi effects or sine of kx. But now we require that phi of L is equal to 0, because we do have the wall there. And therefore, si of kl must be 0 and kl must be n pi and n is 1, 2 to infinity. You know we manage with the wall to discretize this state, because the whole world is now an infinite, a very big box, not infinite, but very big. So you've discrete the state. The separations are microelectron volts, but they are discrete. You can count them. And then with this state over here, we think of counting them. And you say, well, I can count them with n. So if I imagine the k line from 0 to 50, the other states are over, all the values of k. And they could say, well, I even want to see how many states there are in a little element dk. And for the that you would have that dk taken a differential here is dn times pi. So of the number of states that there are in dk, dn-- let me right here-- dn equals l over pi dk is the number of positive energy states in dk. In the range dk, in the range of momentum, dk that little interval, there are that many positive energy eigenstates. So far, so good. So now let's consider the real case. Repeat for the case there is some potential. Well, you would say, well, I don't know how to count. I have to solve the problem of when the potential makes a difference. But no, you do know how to count. So repeat for V different from 0. This time we have a solution for x greater than R. We know the solution. This is our universal solution with the phase. So there you have the si effect is e to i delta sine kx plus delta of k. That's a solution. Yes, you have the solution always. You just don't know what delta is. But you don't know what delta-- you don't need to know what delta is to prove the theorem, You just need to know it's there. So here it goes. And this time the wall will also do the same thing. We'll demand the si of x vanishes for x equal L. So this time we get that kx-- no kL plus delta of k is equal to some other number n prime times pi multiple of pi. This phase-- this total phase has to be a multiple of pi. And what is n prime? I don't know what is n prime? It is some integers. I don't know whether it starts from 1, 2, 3 or from 100 or whatever. The only thing we care is that, again, taking a little differential, you have dk times L plus d delta, dk, times dk is equal to the dn prime times pi. We take an infinitesimal version of this equation, which again tells me how many positive-- all these states are positive energy states. They have k. So all these are positive energy states. So from this equation, we get that dn prime is equal to L over pi dk plus 1 over pi d delta dk times dk, which gives me if I know the dk, again, the number of states that you have in that range of k, You see it's like momentum is now quantized. So for any little range of momentum, you can tell how many states there are. And here it is how many states there are, positive energy states, positive. This is the number of positive energy states in dk with V different from 0, here is with V equal 0.
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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When we analyzed how the position vector changed, we know that the velocity for circular motion is given by the radius times the rate that the angle is changing. And it points tangential to the circle. So let's draw a few characteristic arrows to show that. At this point, we'll draw these pictures with d theta dt positive. So the velocity points like that. It points like this. It points like that. And these are all the velocity vectors at different times. Notice that if we make-- consider the special case in which d theta dt is a constant, in that instance, the magnitude of the velocity, v, is given by r magnitude of d theta dt. And that is also a constant. But the velocity vector is changing direction. And we know by definition that the acceleration is the derivative of velocity. And so what we see here is where we have a vector that's constant in magnitude but changing direction. And we now want to calculate the derivative in this special case. We refer to this case as uniform circular motion. So this special case is often called uniform circular motion. OK. How do we calculate the derivative of the velocity? Well, recall that the velocity vector, r d theta dt-- those are all constants-- because it's in the theta hat direction, once again, will decompose theta hat into its Cartesian components. You see it has a minus i hat component and a plus j hat component. The i hat component is opposite the angle. So we have minus sine theta of t i hat plus cosine theta of t j hat. So when I differentiate the velocity in time, this piece is constant, so I'm only again applying the chain rule to these two functions. So I have r, d theta dt. And I differentiate sine. I get cosine with a minus sign. So I have minus cosine theta. I'll keep the function of t, just so that you can see that-- d theta dt i hat. Over here, the derivative of cosine is minus sine d theta dt. That's the chain rule-- sign of theta dt, d theta dt, j hat. And now I have this common d theta dt term, and I can pull it out. And I'll square it. Now whether do you think that dt is positive or negative, the square is always positive, so this quantity is always positive. And inside I have-- I'm also going to pull the minus sign out. And I have cosine theta of t i hat plus sine theta of t j hat. Now what we have here is the unit vector r hat t. r hat has a cosine adjacent in the i hat direction and a sine component in the H Hut direction. So our acceleration--
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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: So, first step, we need to cut our DNA. Step one, cut, which is going to be DNA restriction enzymes. It turns out, quite remarkably, that if I have a sequence of DNA, five prime A G C T A G A A T T C T T A C C three prime, and we'll come backwards filling in the sequence. It turns out that molecular biologists are so cool that they invented a protein that is able to recognize that six letter sequence, G-A-A-T-T-C. And it is able-- actually it's G-A-A-T-T-C on this strand, but what is it coming back? It's G-A-A-T-T-C. It's the same thing. So actually this is a palindrome. That's kind of nice. It's a palindrome-- it's a reverse palindrome. It's the same word spelled backwards on the other strand. So what it does is it cuts it like that. And what it produces is a DNA molecule like this and a DNA molecule like that, that's mostly double stranded, but has a four base pair overhang. The overhang reads T-T-A-A here. It reads A-A-T-T there. And remember this is five prime to three prime, five prime to three prime. And there you go. This guy has its little phosphate at the end there. This guy has his little hydroxyl over there. And it cuts it. Now that is an incredible piece of engineering. To come up with a protein, to devise a protein, that is able to recognize those six bases and cut at those six bases. And cut in just this way making a really clean overhang here. It's this cool five prime overhang. Who do you think invented this cool protein? What engineer came up with this cool protein? AUDIENCE: MIT engineers. PROFESSOR: MIT engineers, yeah. Not a chance. Not a chance. This is a really tough feat. This is something that can only be done by the smartest engineers on the planet. And MIT engineers are unfortunately only the smartest human engineers on the planet. Who came up with this is E. coli. AUDIENCE: So you found it somewhere in nature? PROFESSOR: Sorry? AUDIENCE: You found it somewhere in nature? PROFESSOR: Of course you find it somewhere in nature. Almost everything important that we say molecular biologists have come up with, it means molecular sat at the feet of the true masters, bacteria, and learned from the true masters. This protein is found in nature. And it's found in E. coli. In fact, it's found in E. coli strain R. And it was the first such protein found in E. coli strain R, so it gets the name EcoR1. And it cuts the DNA like this. Pretty cool. Pretty cool. Now, it turns out that E. coli has this EcoR1. How often does E. coli-- so whenever I take EcoR1, this protein, purified from E. coli, and I add it to DNA it always cuts at this site, which we call an EcoR1 site. How frequently do we expect, what's the frequency of EcoR1 sites? G-A-A-T-T-C, how often will that occur at random? One in-- AUDIENCE: Two to the sixth? PROFESSOR: One in two to the sixth? How many letters do I have? AUDIENCE: Four, oh, four to the sixth. PROFESSOR: One in four to the sixth. My frequency should be about one in four to the sixth, which is about what? What's four to the sixth? It's two to the 12th. It's about 4,000. It's about one in 4,000 letters. One in 4,000 bases. So it's very convenient. One in every 4,000 bases it'll roughly cut. It'll cut at roughly one 4,000 bases. Why doesn't E. coli cut its own DNA? If it's got this protein floating around in its cell, why isn't it chopping up its own DNA? Doesn't have G-A-A-T-T-C? Yeah, the problem is it's so frequent. That'd be really hard to make sure-- I mean, E. coli has 4 million letters in its genome. This should cut every 4,000 bases. You expect about 1,000 such sequences. It might be hard to arrange not to cut-- not to have any such sequences. It's a good idea, is not to have any, but an alternative-- AUDIENCE: [INAUDIBLE]. PROFESSOR: It protects them. It turns out E. coli, instead of avoiding the sequence altogether, has another trick up its sleeve. E. coli protects this sequence whenever it occurs. So it turns out that whenever you have a stretch of the E. coli genome that has this G-A-A-T-T-C in it, what E. coli does is it puts-- I'm just writing M-E here for a methyl group. Right, C-H three up there. It puts a methyl group-- I'll write C-H three. There we go. It puts a methyl group on the A, that middle A. Well, that is a cute trick that E. coli uses, putting a methyl group there. Because what happens is, when there's a methyl group, right at that position, the enzyme no longer recognizes and no longer cuts there. So that's kind of clever. E. coli makes this protein that can recognize G-A-A-T-T-C, but it has a second protein that puts methyl groups there. And this protein happens by accident to be called a methylase. It has a methylase. And the methylase protects that sequence. So now, this is really cool engineering, but kind of dumb. What's it doing there? It has something that cuts the sequence and it protects the sequence, why bother having this? Yeah? AUDIENCE: You can use it to cut it at places to unwrap the true strands. PROFESSOR: That's an interesting idea. We could use it to cut our DNA and open it up to unravel our true strands. It's a thought. Yes? AUDIENCE: To protect the bacteria from viruses? PROFESSOR: Protect the bacteria from viruses. How do you protect yourself from viruses? Well, you have an immune system with immune cells and antibodies and all that. Does E. coli have an immune system? Why doesn't it have immune cells? Because it's like one cell. How's it going to have an immune system, right? So suppose E. coli gets a cold. Suppose it gets infected by a virus. How's it going to protect itself? Cut at a frequently occurring DNA sequence. Now the virus, of course, isn't methylated there, bingo. That's how it tells its own-- you can tell cell from an invader. E. coli can tell cell from an invader because it's methylated its own G-A-A-T-T-C sites, but the virus isn't methylated there. Way cool. This is an immune system for E. coli. Now, it turns out-- so this is protection. These restriction enzymes protect E. coli from viruses. It turns out that E. coli is not alone in this clever trick. It turns out that other bacteria have also thought of this trick. So it turns out that there is another restriction enzyme that cuts at G-G-A-T-C-C. And on the other strand it goes G-G-A-T-C-C. It, again, cuts in that distinctive pattern. And it's called BamH1. And there's another guy. And he cuts at A-A-G-C-T-T, A-A-G-C-T-T. And it also cuts like that. And it's called HindIII. And there's some that cut at G-A-T-C, just the four letter word. And they cut like that. And there's some that cut at C-A-G-C-T-G, C-A-G-C-T-G, and this, cuts smack in the middle. In other words, there's a wide number of different tricks. Some cut at six bases. Some cut at four bases. Some cut at eight bases. Some cut leaving an overhang. Some cut smack in the middle. Some cut leaving the overhang in the other direction. Some allow a degenerate base in the middle. It doesn't care which base is in the middle. There's a zillion different solutions that bacteria have come up with for their immune system. And so, if I want to cut up some human DNA all I need is say, this protein EcoR1 or BamH1 or HindIII or MVL 1 or PVU 2 or et cetera. And I can do that by growing up E. coli and purifying the protein. And if I wanted HindIII, I would grow up haemophilus influenza and purify the protein. So in a molecular biology lab, today, if you want to cut up human DNA, you could grow up some E. coli and purify EcoR1 or haemophilus influenza. And that is indeed what ancient molecular biologists did in prehistoric days in the 1970s and 1980s. They would purify their own restriction enzymes. They're still alive today. You can talk to them. There are many of them on the faculty. And they'll tell you how it put hair on their chest to be able to purify their own restriction enzymes. What do you do today? Order it online from the catalog, right? You know, there's the catalog, the New England Bio Labs catalog. Let's see what we got here. Restriction enzymes, modifying enzymes, polymerases, all right, EcoR1, sale on EcoR1 right now. $100 buys you 10,000 units of EcoR1. It's in the catalog. You can go online. You can order it. You can have it tomorrow by FedEx. So, but that's how it works. It's in the catalog. So you can get any restriction enzyme you want to cut DNA anywhere you want to.
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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: OK, OK, Settle down. Settle down. It's 11:05. It's time to learn. All right. Last day we were looking at the emissions spectrum of the target in the x-ray tube. The target is the anode, and it's being a bombarded by electrons that have been accelerated across a potential difference of some tens of thousands of volts, and this was what we saw as the output, intensity versus wavelength. We see this whale shaped curve which we have attributed to bremsstrahlung, which is the breaking radiation. And then if the voltage is high enough, high enough to do what? High enough to eject inner shell electrons, we will get the cascade, and the cascade will give rise to specific values of wavelength associated with transitions within the target atom. There's too much talking. I want it absolutely silent or else somebody is going to leave. That's the only way it works here. I talk. You listen. And if you have a problem with that, there's a door here. There's a door there. And there's a door at the back. So let's get it straight right now. It's the only way it works here. Now what happens is that these characteristic lines are calculable in part. The K alpha line, the L alpha line we saw from Moseley's law, which is up here on the slide. and the leading edge of the whale shape curve, the bremsstrahlung is calculable by the Duane-Hunt Law. The rest of this is not calculable. So there it is. What I want to do today is harness this radiation. The reason we were studying x-rays in first place was that we said they had a length scale that was comparable to that of atomic dimensions. And so now we want to close the circle here and probe atomic arrangement by x-ray diffraction. And it goes by this three-letter initialization XRD. And in order to use x-ray diffraction to probe atomic structure, we're going to make another model. And this model is going to be simple. And it's going to be inaccurate, but it's going to be good enough to explain the phenomena that we're going to study. So the first thing I want to do in constructing a model that will allow to use x-ray diffraction, is to model the atoms as mirrors. That's the first assumption that we make. We model the atoms as mirrors. And when we model the atoms as mirrors, this means that we will invoke the laws of specular reflection. So if this table top is a plane of atoms, and I have a beam coming down at an angle, the laws of specular reflection say that the angle of incidence will equal the angle of reflection. So that's all we're doing by modeling the atoms as mirrors. Laws of speculative reflection apply. So that simply means that the angle of incidence, theta incidence equals theta reflection. And you'll see how that comes into play in a minute. And then the second thing we're going to do in order to get intensities is to use the concept of constructive and destructive interference. OK. So we apply interference criteria. And I will explain what that means in a second with a diagram. But in dense text, it simply means that in phase rays-- it sounds like this rain in Spain falls mainly in the plain-- in-phase rays amplify. And out-of-phase rays dampen. So at some point, you'll study this in physics, but we need it right now, so I'll give you a little bit of foreshadowing, and then you'll be ahead of the game when you meet this in physics. So what I'm going to do is I'm going to show you two rays, and they both have the same wavelength. And I want to show you in phase. And so I want to illustrate this concept here, in-phase rays amplify. So let's give one ray here. I'm going to depict it as a wave. All right, so that's ray number 1. And beneath it, is ray number 2. It's got the same wavelength. So that means crest-to-crest distance is the same, and if it's in phase, in phase means that crest lines up with crest, trough lines up with trough. So I'm going to make another one seemingly identical. It's supposed to be identical to the capacity of my ability to draw. And so what this means is that these will amplify. So 1 plus 2, 1 plus 2 will give me something that looks like this. I'm going to line it up again. Only now the amplitude is double the amplitude of a single ray. That's what I'm trying to depict here. OK. So this is amplification. Now over here, I want to do the same thing. Only I'm going to have two rays, same wavelength, but out of phase. OK, so in phase is here, out of phase is to the right. So we'll have ray number 3, which does this. And ray number 4, I'm going to line up with the crest of ray number 3, the trough of ray number 4, but it's going to be the same wavelength. So the distance between successive crests is the same. Only they're going to be offset by exactly half wavelength. So the crest here of wave 3 lines up with the trough of wave 4. The trough of wave 3 lines up with the crest of wave 4, and so on, and so on, and so on. And if I drew this thing accurately, it would be obvious to you. But since I can't draw very well, then we're going to take 3 plus 4. And the sum of the crest and the trough is 0. The sum of the trough and the crest is 0. And so the sum here is 0. All right this is total destruction. This is cancellation. So in the extreme, you can have destructive interference that dampens to the point of obliteration. OK, so that's the two extremes, full amplification and complete cancellation. So now, I want to take this concept of planes as mirrors and rules of interference criteria, and now what I want to do is illuminate. And I want to use the full board here. So here's what we're going to do. We're going to take in phase, coherent-- that's what coherent means. Coherent means that the radiation is in phase and monochromatic. What's that mean? Monochromatic means one color, which means single wavelength. So I have something of a single wavelength in phase. Coherent, monochromatic, incident radiation, so I've got a plurality of rays. I have this thing flooded by a beam. So what I'm going to do, is I'm going to draw some atoms here. And I'll put a second batch on top. And now I'm going to model these as mirrors. So I'm going to have the incident beam come in like so. The incident beam comes in like so at an angle theta. So this is theta incident. This is the angle of incidence. And according to our model, the beam is reflected at the same angle. So this is theta, reflection of theta. Reflection equals theta incidence. So that's the first thing that we do. And then the second thing we do, is we use the laws of interference. So I'm going to take a second beam, and it's going to go down to this atom here. So I'm going to bring it down like so. And it's going to come in at the same angle. And it's going to leave at the same angle. And if you'll excuse the drawing, these lines should be parallel. They should be coming in parallel, and they should be leaving parallel where these angles theta i and theta r are the same. Now how do we invoke the interference criteria? Well what I'm going to do is put a marker here. I'll put a marker. And I'll call the first one ray number 1, and the second one ray number 2. And what I'm going to show is if I start from ray number 1-- maybe it's time for colored chalk-- so if I start from ray number 1, to show that it's in phase with ray number 2, I'm going to draw this little waveform, and they're both lined up. So you can see crest lines up with crest. Trough lines up with trough, and they both have the same wavelength. So that makes the point coherent, monochromatic. But now things get interesting, because you can see that ray number 1 travels a shorter distance than ray number 2, and we can put on some coordinates here and mark the geometry, so we can say that the point at which the ray number 2 has to start traveling a longer distance, I'll drop a normal down, and I'm going to call this point A. This is point A. The bottom here is point B. And then up here where it catches up with ray number 1, the reflected ray number 1, I'm going to call this point C. So you can you can see that ray number 2 has to travel this extra distance, AB plus BC. And now, if I want to get out to here where I'm now in the reflected zone, and I want ray number 1 reflected and ray number 2 reflected to continue to be in phase, there's a geometric constraint on the dimension of AB plus BC, isn't there? That length, AB plus BC must be a whole number of wavelengths. It must be an integer number of wavelengths of whatever this thing is, otherwise, these two will not being in phase. So that's the interference criterion. And n can equal 1, 2, 3, and so on, so this is called the order of reflection. And then we can go through the trigonometry. I'm not going to do it in class. It's just a waste of time. It'll bore everybody to tears. But if you want to do that at home at some point when you've got nothing to do, you can go through. You've got all the numbers here. You know the value of lambda. You know the distance AB, and you know what this distance is. This distance between successive planes is your dhkl, isn't it? So if you go through all of this analysis, you will show that n lambda is equal to 2 times the d of the hkl spacing times the sign of theta. And this is called Bragg's Law. And Bragg's Law governs the reflection of incident radiation by a crystal. Now you'll notice I didn't put an apostrophe here. Some people put the apostrophe here, because the man's name was Bragg. Some people put the apostrophe here because, in fact, it was father and son. And they both together won the Nobel Prize in 1915. Now there have been people who lived to see, as Nobel Prize winners, one of their children win the Nobel Prize. but this is the only time in history a father and son team together won the Nobel Prize for the same work. And there are undoubtedly people sitting in this room who think the fact that the father and son could work together in physics for an extended period of time alone is deserving of the Nobel Prize. Now in 3.091, I'm going to keep it simple, always choose first order reflection, always n equals 1 in Bragg's Law. So therefore, we will write Bragg's Law as lambda equals 2d sin theta. And to make the point, the d is specific to a particular set of planes. So it's a d spacing of the hkl planes. And it's the theta associated with the correct reflection of the hkl planes. Now how does destructive interference come into play? Destructive interference comes into play should there be a situation where I have a third ray. I'm going to bring a third ray in also at the same angle of incidence. And that ray is halfway between these two rays. And what situation could that be? That could be a situation in which the crystal structure has in a plane out from the board, an atom that sits halfway between these other two atoms. And if that happens, you can see. We can go through the derivation. But there's going to be a path length difference that's exactly 1/2 wavelength different. That's bad. It's not just there's going to be some measure of reduction. It's going to be total cancellation. And so by going through the set of crystal structures and recognizing that when you have 1/2 wavelength difference, you get destructive interference. And therefore, you will see no line. The incident radiation will come in at this angle, and at the reflection angle for that particular plane, there will be nothing detected. So you can say that you have a combination of selection rules that involve the integral of both the rules of interference and interaction with the crystal structure. So let's generalize this and say if we take interference criteria plus the crystal structure, that is to say, the instant relationship of atom positions for a particular specimen, the combination of that will give rise to the set of expected reflections. So only when you satisfy the Bragg criterion do you get reflection. So you move the specimen. And only at that special angle will you get the reflections, will you get constructive interference. So this, in fact, is a fingerprint. This is a fingerprint of the crystal. In this case, we're not getting the chemical identity. We're getting the structural identity. So we can determine if something is BCC, FCC, and so on. So for example in BCC, that's exactly the case that I just illustrated. So in BCC we have atoms at the eight corners. And we have an atom dead center. There's a central atom. And the central atom lies in 0 0 2 plane, doesn't it? The base of this is 0 0 1. And the top is an 0 0 1. But the central atom is an 0 0 2. And 0 0 2 is halfway between successive 0 0 1. So can you see that light that comes in? And it's going to be constructively reinforced. Off of 0 0 1 is going to be destructively reinforced off of 0 0 2, and the result is that in BCC, no 0 0 1 reflection observed. The same thing happens in face centered cubic, right? What's the face of face centered cubic look like? It looks like this, doesn't it? So there's an 0 0 1. Here's an 0 0 1. What about this space centered atom? Where is it? That's an 0 0 2. It's halfway between 0 0 1. So at the angle where you would've expected, if you use the Bragg Law and calculate the angle at which for your particular crystal, because you know the d spacing. You fix the wavelength of your x-radiation coming out of the generator. At the angle where you would expect to see reflection off of 0 0 1, you will see nothing, because 0 0 2 is canceling 0 0 1. So you don't have to worry about all this. This has all been tabulated for you. Somebody has gone through and done all of this. Well, this is just making the point. See, there's a simple cubic. All the planes are reflecting. There's body-centered cubic. That's a/2. There's face-centered cubic a/2. 0 0 2 is going to cancel 0 0 1. You're not going to see anything there. So people have gone through. And they've made this set of rules for reflection. So simple cubic, you get reflection from all the planes. Body-centered cubic, you just get these planes here. And it turns out that there's a simple rule that compactly represents which planes are going to reflect in body-centered cubic. And in BCC, the hkl, it's the planes for which h plus k plus l is an even number. h plus k plus l is even. And 0 was counted as even. So for example, 0 0 1 h plus k plus l is 1. It doesn't reflect. 0 0 2, 0 plus 0 plus 2 is 2. It does reflect. And so on. You just go down the whole line. And so these are in ascending order of h squared plus k squared plus l squared, because that's a nice way of deciding how to add them up. And now in FCC the selection rule is a little bit different. In FCC you get reflection from planes when you write the HKL such that h, k, and l must be either all even numbers, or all odd numbers, or some people like to say unmixed. Some crystallographers say unmixed, meaning you can't have a combination of even numbers and odd number. So for example, 0 0 1 won't work because 0 is a zero. That's even. 1 is odd. But 0 0 2, h plus k plus l all even or all odd, there you go all even or all odd unmixed. So this will work. And then so on. So you can go through and see which ones work. All right. So the next one here is in the sequence. This sums to 1. And what would sum to 2? It would be 0 1 1. And 0 1 1 doesn't work either because zero is even. And 1 is odd. The next one of the sequence is 1 1 1. These are all odd. So that one reflects. So in FCC, you don't see 0 0 1, 0 1 1. But 1 1 1 does reflect. And then 0 0 2, which is the next one, because 0 squared plus 0 squared plus 2 squared is 4. This is even even. So that one works. So there's the sequence. And so now what we can do is use this technique in order to make measurements. But before we do so, I want to show you the experimental measurement, one way. There's several ways of conducting the measurement. And so the first way I'm going to show you is called diffractometry. And you can do this over in Building 13. If you get yourself a UROP, you might be assigned to make some measurements. So diffractometry is a form of x-ray diffraction. It's one of the techniques. And the way it works is you fix the wavelength and vary the angle of incidence, you don't rotate the x-ray generator. You rotate specimen and present continually varying angles to the thing. So this shows the technique in operation. So you have a specimen sitting here. The specimen can either be a thin film, or in other instances, we have a finely ground powder. So each of the powder grains presents a different angle. And coming out of the collimator is the monochromatic. We want a single wavelength. So this collimator means it's monochromatic and coherent. So that beam comes and strikes the specimen. And for historical reasons, instead of calling this the angle of incidence, and this the angle of reflection, they call the angle that goes to the detector 2 theta. In other words, the projection of the beam, this is theta incident equals theta r. So this is really 2 theta incident, or it's 2 theta reflected. It's the same thing. So you'll see x-ray data usually reported in units of 2 theta. And here's the detector. And then all you do is you rotate the specimen. And by rotating the specimen with a detector, you're able to get the entire set of reflections. And whenever you move through an angle that satisfies the Bragg Law, you get a peak in intensity. And here's what the output would look like. So you're plotting intensity as a function of 2 theta. So somewhere along here at around, it looks like about 38 degrees, you have satisfied the Bragg angle, and you get constructive interference. When you move off of 38 degrees, destructive interference reigns supreme, and you get almost no reflection. Then you get to the right angle for 2 0 0, you get reflection. Well, all you get is these lines. You don't get these numbers. These numbers you don't get for free. So here's the experiment that we're going to do. We're going to run our x-ray generator with a copper target. Why do I tell you the copper target? Because you're going to fix the wavelength. And you fix the wavelength by choosing the target. So we've got copper target in our x-ray generator, OK? So remember, the sample is not the target. The sample is what is being irradiated. This is being bombed by the electrons in the tube, copper target in x-ray tube, and that means that the lambda of the radiation is lambda copper. And I'm going to use lambda copper K alpha because I know it's wavelength to five significant figures. It's 1.5418 angstroms. And here's the data set. I looked this up in the literature. Here's the data set from the experiment. And that's all you get, a set of 2 thetas. So here's my challenge to you. I'm telling you you've got an unknown sample that's cubic, and there's your data set. The task for you is determine two parts to the question. And we're going to do it together. First part, determine the crystal structure. So it's either FCC, BCC, or simple cubic. And the second part is determine the lattice constant, the value of the lattice constant. So you can get quantitative measurements. So I'm going to show you how to do it. So how are we going to do it? We're going to use this technique. This is my self-help book for you. And here's the key. Here's how I came up with this. There's a way to unravel this. And the way you unravel this is to take these two relationships. You have lambda equals 2d sin theta is 1. And you also know that d-- in fact, I'm going to keep writing hkl subscripts here-- you know the dhkl is equal to a. That's this lattice constant. Be sure this is lattice constant. This is the cube edge measurement lattice constant divided by the square root of h squared plus k squared plus l squared. So what I do is I combine the two. If you combine these two, you can end up with this relationship. If you combine the two, you get lambda squared over 4a squared equals sin squared theta over h squared plus k squared plus l squared. And this is the key. Why is it the key? The value of lambda is set by you. You chose the target. The value of a is set by the sample. That's the lattice constant. So the ratio of two constants is a constant. Agreed? So this is a constant. So if the left side of the equation is a constant, the right side of the equation must be a constant. But h, k, and l vary. And theta varies. But the ratio of the variation, when mapped into sin square theta over h squared plus k squared plus l squared, this must be a constant. And that's going to be the way I work through this delightful problem. So let's see. Start with 2 theta values, and generate a set of sin squared theta values. That's what Sadoway says first. So I took 2 theta. And all I did was make 1 theta, and then took the sin of it, and then took the square the sin. So this is the data set transmogrified into sin squared theta. All right, what's the next thing he says? Normalize by dividing through with the first value. So I'm just going to take this whole series and divide it by 0.143. And now I end up, instead of 0.143, 0.191. I have 1, 1.34, and so on. That's what I mean by normalize. Normalize to the first entry. OK, What's next thing he says? Clear fractions. Clear fractions from the normalized column. So I'm going to multiply this by a common number. Because 1.34, remember this is experimental data. It's noisy. 1.34 looks like 4/3. Doesn't it? It's fuzzy logic. And 2.67 looks like 2 and 2/3, 3 and 2/3. That's roughly 4. 5 and 1/3. The 3 looks like a magic number. So if a multiply this column by 3, I get 3, 4, 8, 11, 12. What does it say next? Speculate on the h, k, l values, that if expressed as h squared plus k squared plus l squared would generate the sequence of the clear fractions column. And then that's going to take me to the selection rule. So I say, well how do I get 3? It's 1 plus 1 plus 1. 4 is 2 squared plus 0 squared plus zero squared. 8 is 2 squared plus 2 squared plus 0 squared. So I'm generating this thing. And now it's pretty obvious, right? Because now I've got to use these. And what do I see here? Well, 1 1 1 are all odd. 2 0 0, all even. 2 2 0, all even. Gee, this looks like it conforms to the selection rules for Bragg reflection from an FCC crystal And then to make sure that I'm on to something, what I do is this. Compute for each theta the value of sin square theta over h squared plus k squared plus l squared on the basis of those assumed values. What I'm doing is I'm saying if my hunch is right, whatever I choose for the theta and the assumed value h squared plus k squared plus l squared, that should be a constant that doesn't change. So let's test it. And sure enough, look. 0.0477, 0.0478, so it looks like I'm on to something. And furthermore, I know what this is. That 0.0477 is this. This is 0.0477. And I know my lambda. That's 1.5418. So now I can calculate my a. So now I've determined that it's FCC. Plus if I go ahead and I calculate the a value, I get 3.53 angstroms. And if it's FCC and it's 3.53 angstroms, I bet it's nickel. So that's how you index this stuff. So there we are. There are the selection rules. Now here's a little trick. Let me show you. Because when I go to index this, the first thing I do is I go for low hanging fruit. If you start looking for whether it's simple cubic, face-centered cubic, or body-centered cubic, let's look at the sum. So I've got simple cubic, body-centered cubic, and face-centered cubic. So the first plane on simple cubic, it's going to give me everything. It's going to give me 1, 2, 3, 4, 5, 6. And FCC, what does it give us? If you start looking down there, it gives you 3, 4, 8, 11, 12. Well, this is so different. 3, 4, 8, 11 is so different from 1, 2, 3, 4. What does BCC give you? It gives you 2, 4, 6, 8, 10, 12. Can you see you have a problem here? It's pretty easy to pull out FCC. But look at these two. Since you don't know, you're just normalizing. You can't tell. If I give you a number sequence that is in the order 2, 4, 6, 8, 10, how do you know that that couldn't be 1, 2, 3, 4, 5? There's a hook here, though. Look at the sequence of h plus h squared plus k squared plus l squared. Do you notice that there's no combination of h squared plus k squared plus l squared that gives you 7? You get 6. And then the next one is 8. But there is a sequence that gives you 14. So if you have a seventh line, if the seventh line to the first line, the ratio of h squared plus k squared plus l squared for angle number 7 to h squared plus k squared plus l squared. For angle number 1, if it's equal to 7-- and point of fact it's not 7:1, it's 14:2-- and if it's equal to 8 for the seventh angle-- then it must be simple cubic. And now you've sorted it all out. So you're an expert now. Now you can do it. OK, so you're going to get some practice on homework. So you can index crystals. You can determine a crystal structure. All right, a couple of other things we can do. There is a second technique. And it's called Laue diffraction after von Laue, who got the Nobel Prize in 1914. He beat the Braggs by one year for this technique. So he shines. In this case for Laue, it's a slightly different technique. What Laue does, is he uses light x-rays. That means variable lambda. So how would you get a fixed value of lambda? Well, you would pick off one of those lines, like maybe the K alpha line because it's a nice, singular line. And repress the bremsstrahlung and so on. But if I want a variable lambda, where do I go for variable lambda? I repress the lines. And I go for the bremsstrahlung. So now I've got lines varying all across the x region of the spectrum. So I use variable lambda. And I fixed the angle of incidence, whereas with diffractometry, I use a single value of lambda and vary. So here's the cartoon which shows what's going on. I'll do it one more time. So this is called camera obscura, which is darkened room. That's all this means is dark room. So let's say I've got the specimen on the back wall of this thing, and what I've got is white x-rays coming in. So the white x-ray enters through the face. I've got the plate here, and I've got the specimen sitting somewhere in between. And what happens is I get a spot pattern. I get a spot pattern. And the spot pattern is imitative of the symmetry of atomic arrangement. So I have to say a little bit about symmetry so we know what that means. So let's take a look at symmetry. So I'm going to talk a little bit about rotational symmetries. So let's look at that. So first one I'm going to look at is an 0 0 1 plane. So if I just take a projection of an 0 0 1 plane, it looks like this. Doesn't it? It's just the cube edge, or the cube bottom. So it's got a and a as the edge. And now, the rotational axis, it's about a normal rotational axis. A normal rotational axis. So what I'm going to ask is how many degrees do I have to go through before I get this same image back. You can see that you go 90 degrees, it'll come back. If I stop at 45 degrees, it's going to look like this. You're going to say I know he moved the specimen. If I go 90 degrees, you can't tell it apart. So we define fold. The fold is equal to 360 divided by the basic angle of rotation. So we would say that this plane here exhibits 4-fold symmetry, OK? Let's do a second one. The second one is 0 1 1. So if I look at 0 1 1, 0 1 1 at this plane. 0 1 1 goes across the diagonal. So I'm going to take the diagonal here and plot it like so. So it's going to have an edge of a. And it's going to have the diagonal, which is root 2 times a. And if I put a rotational axis in the center of that, clearly if I go only 90 degrees, I'm going to tell that the thing is rotated. I have to go 180 degrees before I can't tell that there's been any disturbance. So this one here has 2-fold symmetry. We'll do one more. Because there's only three major ones that we have to deal with. And that is going to be the 1 1 1 plane. And the 1 1 1 plane. And the 1 1 1 plane, I think I've got an image of it here. You can see the 1 1 1 plane drawn in this cube. So what's the face of 1 1 1 look like? The face of 1 1 1 looks like this. That's face of 1 1 1. And if you go through this analysis, this is root 3a because it's a diagonal of a difference persuasion. So this is 3-fold symmetry. This has 3-fold rotational symmetry. And that means if I take a crystal such as this, for example. I've told you before, this is my silicon wafer, and this silicon wafer has been cut from a single crystal. So I'm looking at the edge of an atomic plane. The question is, which one is it? The crystal didn't come with a label on the side. I don't know what the crystal growth axis was. I started with a single crystal, and I dipped it into molten silicon. And just like rock mountain candy, I drop the heating coils and cause the liquid to solidify around the seed crystal. And I grow this salami that's about 8 inches in diameter, about 2 meters long. So now I go and I cut these things with the diamond wheel. So I'm cutting them normal to the growth axis, but I still don't know what plane I'm looking at. So if I use the Laue technique and put the crystal like so, bombard it with white x-rays, and I look at the spot pattern. The spot pattern is going to give me one of those symmetries. And if I get a 3-fold symmetry, I know I'm looking at 1 1 1. I mean, it's not going to be some wacko plane. It's going to be either face, edge, or diagonal. And on the basis of the symmetry in the Laue pattern, I can tell which plane I'm looking at. They use this stuff. They use it to make the devices that are in your cell phone and your computer. I'm not just telling you a story. All right. So let's look at some others. We'll go back to the Escher prints. Everything has symmetry. OK, so what's this one? See this? What's the symmetry? 4-fold. How about this one? What do you think? 3-fold. I even went into Photoshop just to show you how dedicated I am. So I took this image. And I told Photoshop to give me 120 degrees rotation. And I got this. You can even see the fold in the book there. So I'm really doing it. It's coming back. It's 120 degrees rotation. How about this one? That's 3. Now these are real Laue patterns. This is 4-fold. This is obviously 3-fold. This one is hard to see. But it turns out it's 2-fold. The one axis is just a little bit longer than the other axis. What about this one? That's our simple cubic puppy. Yes, 1-fold. Translational symmetry without rotational symmetry. You can make this by just moving them sideways. But you have to go all the way around. So that's 1-fold symmetry translation. Look at this one? What's going on here? Is that one of the Bravais lattices? This is called Penrose tile, again rotational symmetry without translational symmetry. You can take a patch of this and move it rotationally. But you can't take and cover a wall with it. So that's rotational without translational. See? You can go around this way. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and reproduce this. But you can't use this as a unit cell and keep it hopping laterally and tile a wall with it. Here's another one, not a Bravais lattice. All right. So I told you at the beginning of this unit that there are ordered solids and disordered solids. We've been focusing on ordered solids. They have a unit cell. They're periodic. And we call them crystals. And next week we're going to start looking at disordered solid. So they have no building block, no long range order. And we call them glasses. For a long time, people thought that solis have to fall into one box or the other box. And then the best thing that can happen in science is not, oh, yeah. That's exactly what I expected. It's, I wonder what that means. And there was one of these moments in 1982. In 1982, a man by the name of Danny Shechtman from Technion in Israel, was over here in the United States working at the National Institute of Standards and Technology, which used to be National Bureau of Standards down in Gaithersburg, Maryland. And he was looking at a set of aluminum-manganese alloys. So aluminum is a metal. Manganese is a metal. You can make a solid solution which we call an alloy. And these are highly ordered. And what he found in his Laue measurements, were rotational symmetries that are impossible in a crystal. He was getting 5-fold symmetry. You can't get 5-fold symmetry. There is no set of planes here that will give you a 5-fold symmetry. They lack translational symmetry. They were called aperiodic. And the popular name for them was quasicrystals. So here's the Laue pattern of one of Danny Shechtman's aluminum-manganese specimens. I think this is 25% manganese and aluminum if I'm not mistaken. So let's count, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. You can't have 10 or 5. Atoms don't work that way. But he got it. He got it, 5- fold, 72 degrees. Now this is really, really exciting. This is a shot of his lab book. You're looking over his shoulder. These are the specimen numbers. And this is aluminum 25% manganese, April 8, 1982. So it's just another day at work. You could be doing this. And he can't believe what he's getting. He's wondering what's going on. You can't contribute this to a calibration error on the instrument. This thing is on unassailably 5-fold symmetry. The joke is that this is not far from the Pentagon. Only within shouting distance of the Pentagon would you discover 5-fold symmetry. So there it is. So where have we come with all of this? We've come with the ability to start with some very simple ideas about x-ray generation. And we've come to the point where we can characterize crystals, get quantitative measurements of their lattice constants, and by using Laue techniques, investigate symmetry. OK. Let's adjourn. We'll see you on Friday.
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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 do we state the issue of normalization? See, the spherical harmonics are functions of theta and phi. So it makes sense that you would integrate over theta and phi-- solid angle. The solid angle is the natural integration. And it's a helpful integration, because if you have solid angle integrals and then radial integrals, you will have integrated over all volume. So for this spherical harmonic, solid angle is the right variable. And you may remember, if you have solid angle, you have to integrate over theta and phi. Solid angle, you think of it as a radius of one. Here is sine theta. So what is solid angle? It's really the area on a sphere of radius one. The definition of solid angle is area over radius squared, the part of the solid angle that you have. If you're working with a sphere of radius one, it's the area element is the solid angle. So the area element in here would be, or the integral over solid angle-- this is solid angle, d omega-- you would integrate from theta equals 0 to theta equals pi of sine theta, d theta. And then you would integrate from 0 to 2 pi of d phi. Now, you've seen that this equation that began as a differential equation for functions of theta ending up being the differential equation for a function of cosine theta. Cosine theta was the right variable. Well, here it is as well, and you should always recognize that. This is minus d of cosine theta. And here you would be degrading from cosine theta equals 1 to minus 1. But the minus and the order of integration can be reversed, so you have the integral from minus 1 to 1 of d cos theta and then the integral from 0 to 2 pi of alpha. So this is the solid angle integral. You integrate d of cosine theta from minus 1 to 1 and 0 to 2 pi of d phi. And we will many times use this notation, the omega, to represent that integral so that we don't have to write it. But when we have to write it, we technically prefer to write it this way, so that the integrals should be doable in terms of cosine theta. So what does this all mean for our spherical harmonics? Well, our spherical harmonics turned out to be eigenfunctions or Hermitian operators. And if they have different l's and m's, they are having different eigenvalues. So eigenfunctions of Hermitian operators with different eigenvalues have to be orthogonal. So we'll write the main property, which is the integral of 2l, say l prime, m prime, of theta and phi. And he put the star here. I'll put the star just at the Y, and he's complex conjugated the whole thing. Remember that in our problem, one wave function was complex conjugated. The other wave function is not complex conjugated. They're different ones because l and m and l prime and m prime could be different. So orthogonality is warranted. Two different ones with different l's and different m's should give you different values of this integral. So at this moment, you should get a delta l prime l delta m prime m. And if l and m are the same as l prime and m prime, you have the same spherical harmonic. And all this tremendous formula over there, with 2l plus 1 and all these figures, you're guaranteed that in that case you get 1 here. So this formula is correct as written. That is the orthonormality of this solution. Probably, this stage might be a little vague for you. We saw this a long time ago. We may want to review why eigenfunctions of Hermitian operators with different eigenvalues are orthogonal and see if you could prove. And you do it, or is it kind of a little fuzzy already? We saw it over a month ago. So time to go back to the Schrodinger equation. So for that, we remember what we have. We have minus h squared over 2m Laplacian of psi plus V of r psi equals E psi. And the Laplacian has this form, so that we can write it the following way-- minus h squared over 2m 1 over r d second dr squared r psi-- I won't close the brackets here-- minus this term. So I'll write it minus one over h squared r squared l squared psi plus V of r psi is equal to E psi. Now, you could be a little concerned doing operators and say, well, am I sure this l squared is to the right of the r squared? r and l-- l has momentum. Momentum [INAUDIBLE] with r. Maybe there's a problem there. But rest assured, there is no problem whatsoever. You realize that l squared was all these things with dd phetas and dd phis. There was no r in there. It commutes with it. There is no ambiguity. We can prove directly that l squared commutes with r, and it takes a little more work. But you've seen what l squared is. It's the dd thetas and dd phis. It just doesn't have anything to do with it. So now for the great simplification. You don't want any of your variables in this equation. You want to elicit to a radial equation. So we try a factorized solution. Psi is going to be-- of all the correlates-- is going to be a product a purely radial wave function of some energy E times a Ylm of theta and phi. And we can declare success if we can get from this differential equation now a radial differential equation, just for r. Forget thetas and phi. All that must have been taken care of by the angular momentum operators. And we have hoped for that. In fact, if you look at it, you realize that we've succeeded. Why? The right-hand side will have a factor of y and m untouched. V of r times psi will have a factor of Ylm untouched. This term, having just r derivatives will have some things acting on this capital R and Ylm untouched. The only problem is this one. But l squared on Ylm is a number times Ylm. It's one of our eigenstates. Therefore, the Y's and m's drop out completely from this equation. And what do we get? Well, you get minus h squared over 2m 1 over r, d second, dr squared, r capital RE minus l squared on psi lm-- or Ylm now-- is h squared times l times l plus 1. So the h squared cancels. You get l times l plus 1 r squared, and then we get the RE of r times the psi lm that has already-- I started to cancel it from the whole equation. So I use here that L squared from the top blackboard over there, has that eigenvalue, and the psi lm has dropped out. Then I have the V of r RE equals E time RE of r. So this is great. We have a simplified equation, all the angular dependencies gone. I now have to solve this equation for the radial wave function and then multiply it by a spherical harmonic. And I got a solution that represents a state of the system with angular momentum l and with z component of angular momentum m. The only thing I have to do, however, is to clean up this equation a little bit. And the way to clean it up is to admit that, probably, it's better as an equation for this product. So let's clean it up by multiplying everything by r. dr squared of little r RE. If I multiplied by r here, I will have plus h squared over 2m l times l plus 1 over r squared rRE plus V of r times r times RE equals E times r times RE of r. So we'll call u of r r times RE of r, and look what we've got. We've got something that has been adjusted, but things worked out to look just right. Minus h squared over 2m d second, dr squared u of r plus-- let me open a parentheses V of r plus h squared over 2m r squared, l times l plus one u of r is equal to E times u of r. Here it is. It's just a nice form of a one-dimensional Schrodinger equation. The radial equation for the wave function dependents, a long r, has become a radial one-dimensional particle in that potential, in which you should remember two things. That this u is not quite the full radial dependent. The radial dependent is RE, which is u over r. But this equation is just very nice. And what you see is another important thing. If you look at the given particle in a potential, you have many options. You can look first for the states that have 0 angular momentum-- l equals zero-- and you must solve this equation. Then you must look at l equals 1. There can be states with l equals one. And then you must solve it again. And then you must solve for l equals 2 and for l equals 3 and for all values of l. So actually, yes, the three-dimensional problem is more complicated than the one-dimensional problem, but only because, in fact, solving a problem means learning how to solve it for all values of l. Now, you will imagine that if you learn how to solve for one value of l, solving for another is not that different. And that's roughly true, but there's still differences. l equals 0 is the easiest thing. So if the particle is in three dimensions but has no angular momentum-- and remember, l equals 0 means no angular momentum-- it's this case. l equals 0 means m equals 0. l squared is 0. lz is 0. This is 0 angular momentum.
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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 to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: Last time I talked about LCAO-MO for diatomic molecules. And I didn't finish, but the important point was that this is a toy model. This is based on a little bit of extension from something which is not really a toy model, H2 plus, to, basically, an interpretive framework that can be applied to, basically, all diatomic molecules. And the logic is relatively simple. We know this one. We get the molecular orbitals for H2 plus. There are basically two. There's a binding one and the anti-binding one. Of course, there's many more, but we don't care about them. And then we go to H2. And again, there's basically only two orbitals that we care about because the next higher principle quantum number has such high energy that we can just forget about them, because those states that derive from the higher principle quantum number are Rydberg states or complicated things, because they're at such high energy. And then it's a very small step to go from hydrogen to the AH molecules, because, well, we've got the electronic structure for the A atom, which is complicated-- more complicated than hydrogen. But because hydrogen only can make sigma bonds, because the p orbitals in hydrogen are so high that it's a simpler picture and can easily be understood. This is in all the textbooks, and it's more or less given to you as something to memorize. But there is a lot more to it than just memorization. And the important thing is that everything in LCAO-MO for diatomics is based on the periodic table. And the periodic table tells you about the ionization energies-- the periodicites of ionization energies. And so we can say for any non-integer-- Well, if we know what the ionization energy from a particular state is, we can use that ionization energy to derive a non-integer principle quantum number. This is all empirical. And then use that empirical principle quantum number to get the size of the orbital. And basically, the size is everything. Because everything is based on overlap, And the internuclear distance molecule is based on the relative sizes of the different atoms. And for different electronic states of the atoms, these sizes are different, because they have the ionization energy from that level implicitly expressed. So because you know about orbital sizes, and that the different atomic orbitals have different sizes, you can do an enormous amount as far as understanding the electronic structure of all diatomic molecules. Now, this-- when you have two states which have different energies-- We do something like this to describe the molecular orbitals that arise from them. That that's perturbation theory. So the solid line is the dominant character. The dotted line is the admix character of the other orbital. And this is all qualitative, but you have these different atomic orbital energies. You know them from the periodic table, basically. And so you can say, yes. There's going to be two orbitals arriving, derived from these two states. And one is polarized towards this atom. The other is polarized towards that atom. And so you get the shape of the orbital. And it's actually useful for saying, well, if I were to do chemistry, which end of this is electronegative, and which end is positive? Or which end of this would attach to a metal surface? Would it attach pointing into the surface or lying down? And there's all sorts of insights, if you can draw these sorts of pictures. And this is what we do as physical chemists. We take the crudest model, and we say, OK. We understand the important features. And as we discover new, important features, we build them in, or we forget about them because they're too subtle. And we're always looking for something where the crude picture doesn't work. And we then find the important thing that is needed. And for example, if you were to look at the molecular orbital diagram for C2, which is a perfectly legitimate gaseous molecule, you'll see that there's a little bit of ambiguity about which is the ground state. And this is an important thing. And there was a big controversy about this that was settled by spectroscopy. OK. So if you can build an intuitive picture for all diatomic molecules, you can also build an intuitive picture for what we could call chromophores. So there are a lot of molecules that are larger than diatomic molecules. But the electronic structure is mostly nothing, except a few atoms that are close to each other, where there's a double bond or there's something special. And so if you can do LCAO-MO for diatomic molecules, you can also address electronic structure in much larger molecules, because it's really due to a few important atoms or several different groups of important atoms. And so the idea is, in here, enable you to talk about the electronic structure of almost anything. And one of the things that we care about is spectroscopy. How do we learn about the structure of a molecule? And mostly, you learn about it from doing electronic spectroscopy. You do transitions between the ground state and some higher states. And you want to be able to predict what you are going to see. And for example-- and I think you may hear about this a little bit more-- if you know the spectrum of nitrogen, it doesn't start absorbing until far into the vacuum UV. And then go one atom over-- the spectrum of oxygen. Well, that defines the vacuum UV. It starts to absorb around 200 nanometers. And 2 and 1/2 billion years ago, oxygen started to enter into the atmosphere and profoundly changed life on Earth. Because with the vacuum UV radiation, nothing could live on the surface of land, because it would all be killed by this hard UV. And so everything was underwater at the bottom of the ocean. But all of a sudden, when oxygen appeared, there was protection. And then there's additional protection farther out towards the visible from ozone. It's not such a good shield, but it absorbs between 200 and 350 nanometers. And it protects us, too. But the big thing is oxygen. And why should oxygen be so different from nitrogen? And maybe you should think about that. OK. I should have mentioned the schedule up there. I'm going to give another lecture on perturbation theory-- one of my favorites, which I thought I wasn't going to get to give on Friday. And then Troy is going to give two lectures on quantum chemistry. And there's going to be significant lab experience. And you'll hear about that from the TAs. And so you will do real calculations. So LCAO-MO could be the organization structure for quantitative calculations. But quantitative calculations are usually huge basis sets, and you lose the LCAO completely. One could take the results of such a calculation, and then project it onto a LCAO-MO picture. And that's generally what we do. We reduce experiments, we reduce theory, to something that we can intuit. And it's really important to have intuition. OK. Huckel theory-- Huckel theory is another kind of non-rigorous theory. In fact, it's laughable in its simplicity. And the idea is that even such a crude theory can make sense out of huge families of molecules and enable you to be quantitative about things without having to do a huge calculation. Now, you almost always have to diagonize a matrix. And because usually, in Huckel theory, you're dealing with more than two or three relevant orbitals, it's a big deal. I don't know whether you have experience with matrix diagonalization routines, but there are many of them. And you're going to have experience with them. But the things you put into them have absolutely nothing to do with a real Hamiltonian. It's a make up picture. It's a picture based on, basically, two parameters-- alpha and beta. And I'll talk about this. And these are parameters that are agreed upon by international committee-- It's not a committee. But people say, OK. This is the value of alpha and beta that we're going to use to describe many problems. But I just want to make sure you understand that Huckel theory is only a little bit more ridiculous than LCAO-MO theory. Because in LCAO-MO, you really have some semi-empirical sense of how big orbitals are. And once you know how big orbitals are, you know what's going on. In this, you just have a couple of parameters that says, OK. These are the rules. Let's now apply that. OK. I'm carrying around too much stuff. So organic chemists are really wonderful, because they give you an abbreviated way of drawing a molecular structure. And almost everyone in this room is gifted in being able to visualize three-dimensional structures. Physical chemists tend not to be so gifted. But here is something where we don't have any hydrogens-- I mean, there are hydrogens, but they're implied. And we don't put carbons here. We just say, at every vertex, there is a carbon atom. And so we consider these things. And every organic chemist knows, if I draw something like this, that either I was stupid and I drew something that was impossible or what the structure is. OK. Now, this is a conjugated system. We have double bonds and single bonds. And that kind of thing is known to be unusually stable. And in order for it to work, it has to be planar. If you have a conjugated system which is not planar, then it's not as stable. So Huckel theory is based on the existence of unusual stability of conjugated systems. And it can be extended to non-conjugated systems-- to non-planar systems-- and we have rules for how to do that. But the simple rules are, you start with a picture, and you can write down a Hamiltonian. And it's a toy model, but it's still something that the computer has to diagonalize. So when we're doing a variational calculation expressed in matrix language-- We have a Hamiltonian matrix. We have eigenvectors, and we have eigenvalues. And we have the overlap matrix. And we have something where this is the alpha-- where, here, we can have non-orthonormal basis functions. And this makes them normalized. So these are the orthonormalized basis functions or basis vectors. And this is the kind of equation we have to solve. It's the generalized eigenvalue equation. And we don't like this, because it doesn't have the simplicity of just a Hamiltonian, where we have matrix elements along the diagonal and somewhere else. And the idea is, we want to take the secular determinate and make it 0 by adjusting values of the energy differences along the diagonal. But when we have this overlap matrix, it's complicated. Now, there's a way of dealing with the generalized eigenvalue equations, but one way to deal with it is to say s is equal to the unit matrix. You can do anything you want. And it's basically saying, there is no overlap between orbitals on adjacent atoms. We're going to neglect it, and then we're going to bring it back if we need it. But it's a wonderful simplification, because it enables you to write a simple, effective Hamiltonian, which looks just like h c alpha is equal to e alpha c alpha. OK. This, we know how to solve. And we can use the same procedure. OK. So these are the rules. If we have a planar molecule, we can say there are p orbitals-- one on each carbon atom or one on each atom that's not hydrogen, which are perpendicular to the plane of the molecule. So easy orbitals. And those are special. They give rise to pi bonds. Pi bonds are bonds where there is one plane of symmetry containing the bond, And then there are p x, and p y, and s. And these give rise to sigma bonds. So we have pi bonds and sigma bonds. Never the twain shall meet. And we don't care about the sigma bonds, because anybody can make sigma bonds. But only the special, perpendicular-to-the-plane guys, which are responsible for the fact that the molecule likes to be planar. And so we're only going to consider these p z orbitals-- one on each atom. So we don't care about no pi-sigma interactions. And we're going to neglect the sigma orbitals, because they take care of themselves. And we can do anything we want. It's just a question of, if we make too many ridiculous assumptions, we'll get ridiculous results. And this has been time tested, and it gives pretty useful stuff. And it provides a framework for making arguments about molecular structure and molecular reactivity. In organic chemistry you learn about resonance forms. And this is compatible with generating the resonance forms and saying, what is the relative importance? And what is the charge distribution, and bond strength, and everything like that? So it's really useful using the most primitive tools that organic chemists introduce at an early stage in your education. And that's one of the reasons why a lot of people become organic chemists, because it's so beautiful. OK. So our h i j matrix-- So we have a bunch of matrix elements, and we say, OK. h i i is equal to alpha. So every carbon atom has its alpha value. It's the same for all carbon atoms, regardless of who is nearby. And we have h I i plus or minus 1 adjacent atoms, and it's beta. That's it. That's the whole ball game. And it's really a simple-- There's far less here than in LCAO-MO. But it's still a toy. Both are toy models, and they're both very useful OK. And everything else which is not diagonal or off diagonal by 1-- 0. That's really convenient. And so you can draw the h matrix, regardless of what it looks like, as alpha, alpha, alpha, alpha, et cetera, along the diagonal, and beta, beta, beta, et cetera, beta, along the near diagonal. And if it's a ring, you have a beta here, here, and here. So that's pretty simple. So we have 0s, 0s-- so that it has a tri-diagonal structure with something up here and there. Never forget this thing here and there, which is present when you have a ring, and it's not present when you don't. That's it. That's Huckel theory. It's just there. OK. And so now, things can be more complicated. So if we're not content with what we get from the really primitive theory, we can do something like saying, well, we can make beta be dependent on internuclear distance. If the molecule, for some reason, is constrained to have not equal bonds lengths. So we can add an additional parameter-- some kind of reason for this beta to be dependent on r. But that's already getting sophisticated. And heteroatoms-- In other words, if you have a nitrogen instead of a carbon in a benzene-type ring, you can have-- So, well, nitrogen is different from carbon. It has a different-- In LCAO theory, the distance-- The ionization energy for nitrogen is different from carbon. It's larger. And so heteroatoms can be included if you use a different value of alpha. And now, alpha and beta are both negative. Now, this is a little bit fraudulent. Yeah. AUDIENCE: Do you also have to pick a new beta? Or-- ROBERT FIELD: Yes. But the main thing is alpha. AUDIENCE: I see. Why is that? ROBERT FIELD: Why is that? Because beta comes from overlap, even though we're neglecting overlap. And so the bond distances between normal and heteroatoms are not usually that different. But the thing is, you put what you need into the model. And the first thing you do is, you solve the most simple model. And you say, this is not quite what I wanted. And so I allow a couple of extra degrees of freedom. And it's really instructive how these things work. But the thing is, you're not calculating the matrix element of a real operator. It's all make believe. But it's really powerful, because what you're comparing is families of molecules. And the reality might be really complicated, but the complexity in each member of the family is pretty much the same. And what it's allowing you to see is, what are the differences? It allows you to see the big picture. I love this. And in fact, at an early stage of my education, I thought Huckel theory was wonderful. And it was what got me interested in quantum mechanics. Because you normally see Huckel theory before you know anything about quantum mechanics, because it's just a game. OK So heteroatoms-- You can fiddle with alpha. Now, the ionization energies for carbon and nitrogen are not that different. I'm sorry. They're very different, but the effect on the alpha value in Huckel theory is very small. Well, so it is. But the more electronegative or the higher the ionization energy, the alpha value becomes increasingly negative. Now, I was starting to say something is fraudulent, and I was distracted by a really good question. Alpha is on the diagonal. And we know we can determine the sign of a diagonal element. And we know we can't determine the sign of an off-diagonal element. But we say that alpha and beta are both less than 0. And the reason for this is that, when you saw the secular equation, especially-- You get two eigenvalues-- one where you have minus beta and one where you have plus beta. And so in a sense, beta is present, but it's just a sign choice. And since alpha and beta have to do with stability, we just say alpha is negative. We know that. And beta is chosen to be always negative. There's no, you could have had beta be positive. And you could do all the theory. It's just a lot more complicated explaining all the cases. All right. So let's continue with this. And so we can do heteroatoms. So if we have a non-planar system, we can say that beta is a function of the dihedral angle. We can put that in. We can do anything we want. We have a molecule. We're trying to describe its properties relative to the normal members of the group, which are planar and no heteroatoms. And we can do stuff that will accommodate these interesting differences, which you can impose by you putting the molecule in a constrained environment or doing stuff that distorts the geometry. So we do this. So when we do this, we get a Hamiltonian. We get the energies of the orbitals, and we get the eigenvector that corresponds to each energy. And the total energy involves the sum over the energies of each of the occupied orbitals-- the number of electrons in that. And we can also get bond order and charge. So sometimes, we want to know, what is the charge on each atom if the charge is going to be different from 0? Because that also controls chemistry. Negatively charged atoms are sought out by certain classes of reactants. Anyway, so you get all this stuff. And of course, some of it will require a little bit of patchwork, but you do this for your career. And you discover that there are certain things that I know how to handle. And I use my favorite parameters for it. And you get closer to the truth. And you always want to be surprised, because when the crude theory cannot be made consistent with observations, you know you did something special. You have a molecule, which has a property which is unexpected, and we like that. OK. So when you're doing quantum chemistry, there are five steps-- or when you're doing molecular orbital theory. And this is one of Troy's rules. We have a five-step procedure. And we define the atomic orbital basis set. And the basis set is one p z orbital per atom. And so we can have a molecular orbital, which is the sum i equals 1 to n c i u p z i. So this is the p z orbital on the i-th atom. This is the coefficient for this particular linear combination. OK. So we have a bunch of molecular orbitals. Next step-- compute h and s. But s is 0 because we made this ridiculous assumption. It's convenient. This is called the complete neglect of overlap. And we can do that. It's wrong. But when you include overlap, it leads to greater complexity of the calculation, and not much improvement in the results. And so we normally don't worry about the overlap. So all we care about is this. And as I said, h has this structure of alpha, beta, beta-- tri-diagonal structure and maybe something up here. That's it. Then we diagonalize the Hamiltonian. And so this gives us the eigenvectors or eigenvalues-- the energies and eigenvectors associated with each orbital energy. So we fill electrons into orbitals. Now, for benzene, what you would get-- Now, this is important. How many carbon atoms in benzene? Right. And how many orbitals will you get from six primitive orbitals? Right. And so there are six energy levels. And it happens that when you solve the secular equation, you get this pattern. This is a problem where the number of nodal planes determines the order of energy. And so if you have benzene, you can imagine that the orbitals that you can have will be no nodes-- nodal planes-- one nodal plane. And you could have the nodal planes between the atoms or through atoms-- and two nodal planes or three. You can't have any more with six atoms. And so you almost don't have to solve this secular equation at all. You can anticipate what the structure is going to be just by saying, no nodes, one node, two node, three node. And you can even anticipate double degeneracies, because if you have two nodal planes, you can put them through opposite bonds or through opposite atoms. And those are examples of the forms that you would deal with. The only thing you can't do is to know what the order the energies are. But there are tricks for that, too. The tricks for that include-- The sum of the eigenvalues is equal to sum of the diagonal nodes. And so we know that the six orbital energies will sum to 0-- will sum to 6 alpha. The betas go away. And there are other tricks that you can do, but generally, you solve the equation. You don't want to push your requirements for symmetry too far. So we end up doing that. And then stick diagrams-- We fill electrons into orbitals in energy order. And for benzene, there are six of them. And so this is, then, the lowest energy state of benzene. This is called the independent electron approximation. They don't know about each other. This is illegal. But it's legal in a sense, if you have two electrons in every orbital, you have nothing but singlet states. The ground state is always going to be a singlet state unless you have something really weird going on-- like in O2-- But that's not Huckel theory. But it's examinable. OK. Why does oxygen have a triplet ground state? That's something that every textbook says. You got to know that. And of course, they don't really tell you anything more except something to memorize. OK. So generally, you have two electrons in each orbital in singlets. Now, if you're going to do spectroscopy you would perhaps promote one of these guys to a higher state. And so you'll have singlets and triplets. But of course the only transition you would see would be a singlet-to-singlet transition. So you might as well forget about triplets. And you might as well forget about having to add [INAUDIBLE]. You can get away with murder, especially with Huckel theory. So we have a stick diagram. We fill electrons into the thing. And then we compute the energy of the many electron problem. And so let's just do this. And you've all seen this. So we number the atoms. We have our symmetric structure. And psi mu is going to be sum from i equals 1 to 6. C i mu p z i. And c mu is c 1 mu c 2 mu to c 6 mu. These are the mixing coefficients. Well, for benzene, it's pretty simple because you know that if you have no nodes and symmetry, all of these are going to be the same. And so if you put 1s here, you put a 1 over square root of 6 out in front for normalization. And you can figure out the eigenvectors for benzene-- all of them-- just by counting the number of nodes. And that's useful, because if you know the eigenvectors, then you can show what the eigenvalue is by multiplying the original Hamiltonian by an eigenvalue. So anyway, we have this. And then the Hamiltonian-- we have alpha, beta, and then 0s. And beta, alpha, beta, and then 0s, et cetera. And so we have this tri-diagonal structure. And because it's a ring, we have a beta here and here. OK. So then, we ask our computer to diagonalize this, or we use clever tricks from linear algebra to find the eigenvalues, but I don't recommend it. I mean, the general problem is going to be something where you have to use a computer. So don't develop tricks unless you want to check to see whether you programmed the computer correctly. And so when you do this, you get-- E 1, the lowest energy, is alpha plus 2 beta. E 2 is alpha plus beta. And E 3 is alpha minus beta. I'm sorry. E 2 and e 3 are both this. And E 4 and e 5 are this. And e 6 is alpha minus 2 beta. Now, we don't really care about these orbitals, because they don't put electrons in them. And so when you put the electrons in the orbitals, you get 2 for this, 2 for this, 2 for that. OK? And so we just calculate the sum of the energies. And we end up getting the energy levels for benzene. The ground state is going to be 2 alpha, plus 2 beta, plus 2 alpha, plus beta, plus 2 alpha, plus beta. Right? So this gives you 6 alpha plus 8 beta. Now, this-- You might have guessed it, but you didn't guess it. Your computer told you that. And the computer actually likes numbers rather than symbols. And so you actually obtain this simple structure from the computer-- requires a little bit of manipulation. But you still get 6 alpha plus 8 beta. And you also get the eigenvectors. And that's the stuff that you know. So c 1 is 1 over square root of 6. 1. 1. All of those. And c 2 is going to be something a little bit more complicated. We can have the nodal plane going through atoms. And so if it goes through atom one, it's also going to go through atom four, and so we have 0 and 0, 1, 1, 1, 1. And now, to figure out how to normalize that, we just-- 1 over square root of 4, and so on. We can figure these things out. C 3 is going to be an eigenvector. Now, instead of having the nodal plane going through atoms, it's going between atoms. And so instead of having any 0s, you're going to have something more complicated. And now I can see that there is something in my notes which is subject to how you'd actually impose the symmetry. But suppose you have something like this-- 2, 1, minus 1, minus 2, minus 1, 1. So why did I use these numbers? Well, I had to have a nodal plane here between atoms two and three. And the corresponding guy will be-- Well, that should have been 2 at the bottom. Let me just make sure I'm doing this right. No. I had a 2. OK. It's a 1. OK. And so the last 1 is here. So the sign change occurs twice. And those correspond to opposite bonds. Why were there 2s? Well, the 2s are between two atoms that have 1s, and so it's going to have a larger eigenvector. And you can figure it out lots of ways. You can also say, well, every atom has to be used up. Yes. AUDIENCE: So should there also be a sign change? ROBERT FIELD: I can't-- AUDIENCE: Should there also be a sign change in c 2? AUDIENCE: Yeah. In c 2 eigen c 2 6 should be [INAUDIBLE].. ROBERT FIELD: Did I screw up? AUDIENCE: Yeah. No, not-- in c 2. Not c 3. ROBERT FIELD: I'm sorry? AUDIENCE: The previous one. ROBERT FIELD: Oh, yeah. Yeah. So opposite sides. OK? All right. It doesn't matter. The computer tells you. But sometimes you can approach the problem very quickly, and just say, I know what the eigenvalues are going to be, and I have to normalize them. But one of the things that you often do if you're trying to be smart and skip steps is to say, OK. We have six eigenvectors. And each of the atomic orbitals gets used up completely among the six eigenvectors. And I do that all the time, because it's a very useful way of making sure I haven't screwed up. OK. So we get this, and that's perfectly OK. We don't know how significant it is, but say we had three of these-- three ethylenes-- I guess an organic chemist would just do that, right? But anyway, if we had three of these, when you do this you get 2 alpha plus 2 theta. OK? So we get 6 alpha plus 6 beta. And alpha and beta are both negative. And so the energy for three ethylenes-- did I screw up again? AUDIENCE: How are those ethylenes arranged with one another? ROBERT FIELD: They're separate. We could draw benzene. And so we could say, we have three isolated bonds. And since we don't care about the sigma structure, and no bonds are adjacent, you know that they're additive. And so we represent benzene. The primitive structure for benzene is three ethylenes. And benzene is 2 beta better. AUDIENCE: Yeah. ROBERT FIELD: That's the resonant stabilization. That's a great thing. OK. So we get a resonant stabilization. And now, the more subtle and wonderful stuff is, we have these eigenvectors. And we have the energies. And we can do stuff with them. And one thing is bond order. And so we have a formula. The bond order between atoms i and j is equal to the sum for mu equals 1, including only the occupied orbitals, of c i mu c j mu. So we have the new molecular orbitals, and we have the adjacent atoms. And this gives you the bond order. And so we can calculate the bond order. And we find that every bond is a pi bond order of 2/3, which is neat, because the non-resonant structure says, three of the bonds have a pi bond order of 0, and 3 have a bond order of 1. And so the average is 1/2. And this is bigger than 1/2. And it's uniform. So you can do this, and you can say, OK. The 1, 2 bond order-- the 2, 3 bond order-- you do the laborious calculation-- always get 2/3. Now, you could also calculate the charge on atom i. And that is, again, mu for the occupied orbitals. And that would be c i mu c i mu. And so for benzene, you would expect this to come out to be 0. And it does. So there is no charge. Let me just make sure that that-- Well, there is an equal charge on each atom, whether it's 0 or 1/6. That I haven't done. And I am a little uncertain about what this is going to come out to be. But they're going to be equal on every atom in benzene. And so now, suppose, instead of benzene, you have amylin. Well, you can do stuff. And you could say, well, the amylin out here is going to affect the alpha value here a lot, and here less and less. And then you do the calculation. And you find when you do this calculation, you get unequal charges. And you get the normal rules for ortho versus meta. And everything is great. And you also, when you do this, you can actually write resonance structures and you could calculate, well, what is the energy of that resonance structure-- in a Huckel-like theory. So there are lots of things you can do in order to say, OK. We do Huckel theory to get this 0 in our picture. And then-- I'm way over time. And then we can add special effects, and we know how to parameterize them. And if we're close to being OK, we'll get results that correspond to experiment. And this is so much better than you deserve, because it's all garbage. But it's empirically-calibrated garbage. And it it's calibrated over an enormous number of molecules and a huge amount of experience. And that's good. That's what we do. OK. So I'll see you on Friday with some perturbation theory. One of my favorite problems, too. OK.
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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: Welcome to the class. We're going to discuss the themes that are going to basically permeate every topic and module we'll talk about here. And one of the central themes of this class is that we're interested in studying the cellular processes of life at a molecular level, right? So as biochemists and chemists, we're interested in this level of understanding. And what we see here is a cartoon depiction of the cell. And we see that there's many types of biomolecules in this environment. So what are our core themes for this year? First, we believe that life must be studied on a molecular level to truly understand it. And so we need to think about the cellular environment, both on a macroscopic scale, and on the molecular level. And this environment is complex, and it always needs to be considered, right? So as experimentalists in biochemistry, often we're doing experiments in aqueous buffer with proteins or some other biomolecule. How does that relate to a context like this one here where the environment is very different and much more complex? Something we'll see, especially in the first half, the first four modules of this course, is that in cells, complex processes are carried out by macromolecular machines and elaborate systems. And these systems are fascinating. You'll see that we know a lot, but as we learn more, there's more and more questions that come up, and more questions we need to address with that. In addition to these macromolecular machines, some additional themes for this course involve homeostasis and signaling. And these will be especially emphasized in the second half of the course when Professor Stubbe takes over there. So how do we think about homeostasis and signaling in these contexts? Something that will come up again and again is how, basically, understanding cellular processes at a molecular level, or the molecular features, can help explain mechanisms of human disease, as well as therapeutics. So an example we'll see in the early part of this lecture involves the ribosome. So many antibiotics target the ribosome. And by understanding ribosome structure and function, we can understand how these small molecule therapeutics work. Another example involves the proteasome which we'll hear about in the second half of the course. So there's therapeutics that target the proteasome, for instance, for cancer. And cholesterol biosynthesis will come up, and how does our understanding of cholesterol biosynthesis lead to ways to treat coronary disease? Something that JoAnne and I really like to think about day-to-day and convey to you in this course is the importance of experimental design, and choice of methods. So as scientists and experimentalists, how do we think about designing an experiment, because that design is really critical to the outcome, and what we can make of the data? And so throughout lectures and recitations, things to keep in mind, and that we'll reiterate, are that all techniques have inherent strengths and limitations. And so it's something we all need to keep in mind when we analyze data and think about how an experiment was done. And these systems we're going to look at in 5.08 are very complex. And what that means is that many different types of experimental method are needed in order to answer complex-- and sometimes not so complex-- questions. So one method alone just often isn't enough. We need insights from many different techniques and types of expertise. And so we look forward to informing you about different types of methods-- whether they be established and quite old or new-- that are important today. And as I alluded to before, something we have to keep in mind when doing biochemistry in the lab is that the test tube is very different from the cell. These environments are vastly different, and so we always need to think about how to relate data back to a cellular or physiological context. If you measure a dissociation constant of one micromolar, what does that mean in a cell versus one picomolar, for instance. Another point to make is that the hypothesis is a moving target. So we have the hypothesis, experiments are designed to test this hypothesis, and there's some outcome. Maybe that supports the hypothesis, maybe not. Or maybe there's some new insight from a related field that really changes how we think about something. So in many cases we're integrating data and insights that are quite new, and Professor Stubbe and I won't have all of the answers. And so that type of uncertainty is something that we aim for you all to gain some level of comfort with. So there's many complexities in primary data, often uncertainties. And that's just an aspect of this course. And scientists, it's something we grapple with every day in our own work. So we're introducing that to you here. And along those lines, just keep in mind, we know so much. And I think it's amazing, and-- if I step back and think about this for some of the systems we'll see-- actually overwhelming. And it's really due to dedicated efforts of many, many people over many, many years. But with that said, there are so many remaining unanswered questions, and we hope that you'll find inspiration in some of these questions as looking forward within this field. There. OK, so what about the cell and macromolecular crowding? Just to emphasize this point a bit more, here we have an E. coli. OK, so E. coli are laboratory workhorses for biochemists. They're fascinating, I love E. coli. But I just show you this simple E. coli cartoon and this depiction here to emphasize how crowded the cellular environment is. So we have an equal E. coli of about two microns long, and maybe half a micron wide, a volume of about a femtolitre. And if we think about the E. coli genome for a minute, it encodes about 4,000 proteins. That's a lot of proteins. And if we think about one E. coli cell of this small size, can just ask a simple question, how many ribosomes are there? So we all know the ribosomes are needed for polypeptide biosynthesis. How many ribosomes are packaged in one E. coli? Any guess? So, 10, 100, a million. AUDIENCE: Order of 1,000? ELIZABETH NOLAN: Pardon? AUDIENCE: Order of like, 1,000? ELIZABETH NOLAN: Yeah, let's say 1,000 times 15 or 20. So there's about 15,000 to 20,000 ribosomes in one E. coli cell. And as we'll see in Friday's lecture, the ribosome is very large. How did they all fit? And there's not only the ribosomes, but there's many, many other players, just as noted here in this cartoon. So you can think about what does that mean in terms of concentrations. We'll bring up concentrations of biomolecules in the cell throughout this course, and what does it mean having them packaged together so much here? So, very different than the test tube. Our goals, some of which I think have been communicated by me so far. But just to emphasize, we're interested in these macromolecular machines and chemical processes responsible for life. We hope by the end of this course, everyone gains an appreciation for the complexity of life, and our current understanding of the topics we present to you this spring. There's close links between basic fundamental research and medicine, and technology development as well. Understanding the experimental basis for understanding, methods and hypotheses. And what we think is something that we hope to achieve, and that you can bring to other places after this course is really to be able to knowledgeably and critically evaluate methods and results, especially primary data. And we also hope that we convince you that biological chemistry is really thought provoking and fun, and hope you all think that right now as well. So what are the actual topics we're going to cover? We organized this course into modules, and these modules are listed here. And different modules will have different numbers of lectures dedicated to them. But where we'll go between now and spring break-- I'll present to you during these weeks-- is that we're going to focus on the lifecycle of a protein for the first three modules. And many of you are familiar with aspects of this. We're going to present these topics, I think, a bit differently than what you've seen before. Again, very much from the standpoint of experimental methods and hypothesis testing. So we'll cover protein synthesis, doing a careful case study of the ribosome. We'll continue with protein folding. So asking the question, after the ribosome synthesizes a polypeptide chain, how does that polypeptide assemble into its native form? What happens when proteins are misfolded? And then we'll move into protein degradation, and we'll look at proteases and machines that are involved in proteolytic degradation. And where we'll close the first half is with module four, which is on synthases, or often called assembly-line enzymology. And this is a different type of template-driven polymerization that's involved in the synthesis of natural products. And then after spring break, Professor Stubbe will take over, and the focus will be on cellular processes that involve homeostasis, metabolism, and signaling. And so these topics will involve cholesterol biosynthesis, and a type of molecule called terpene. And so a third way to make a carbon-carbon bond will be introduced in this section. So you've heard about Claisen and Aldol condensations in prior biochemistry courses, this will be another route. And then, we both love metals and biology, so there's a whole field of bioinorganic chemistry, and it will be introduced to you here with iron homeostasis as a case study. And moving from here, and something quite related, involves reactive oxygen species. So I'm sure you've all heard about these somewhere, maybe in the news, maybe from your lab work. What are these reactive oxygen species? Are they all reactive? What kind of chemistry do they do in a cell? How do we study that here? And then, of course, we'll close with a section, a module on nucleotide and deoxynucleotide metabolism-- excuse me-- as well as regulation. And then an integration of course concepts. So we have a lot of exciting topics and exciting things to tell you about. In terms of level of understanding for this course, as I said, many of these systems are complex. We're going to look at huge macromolecular machines, and multi-step processes. This is a biochemistry course, and we are interested in molecular level, in addition to this big picture. And so things to keep in mind when thinking about structure. You need to think about the amino acids, and please review these if you're a bit rusty. So to know the side chains, PKAs, et cetera, that's all important to have in mind. What are the protein folds? What are the arrangements of these macromolecular assemblies, and how do we study that? In terms of reactivity, we'll see bond-breaking and bond-forming reactions. So again, we need to think about things like PKAs, nucleophiles, and electrophiles. If you need to brush up, organic chemistry textbook or biochemistry textbook is a good place to go. And then something to keep in mind is dynamics. So the macromolecular structures and enzymes and proteins we'll look at are dynamic. Often we only have a static picture or some number of static pictures. But there's conformational change, transient binding occurs, and we always need to think about kinetics. So these are things just to keep in mind when you're reading and questioning to yourself about any given system here. So what about experimental methods? This is just another topic to go over in this course overview. So there's many methods that come up in 5.08. And we don't expect that you have knowledge of any or all of these at the stage of starting the course. The difficulty that comes up is that we can't introduce all of these methods to you at once in a level of detail that's needed for everything we do. OK, so what will happen is that if methods come up in problem sets that haven't yet been addressed, we'll give you enough background information in the problems that material, such that you can think about the questions and answer them. And we'll let you know when a method comes up. You know, you'll hear this in recitation x, or we'll talk about it more in class. So right now, what I'd like to do is just go over a few of the methods that you're going to see multiple times. And the thing to keep in mind is that the context in which these methods are being used may differ, but the underlying principles are the same. And we choose methods that are being used today, and are important. Some of these were developed decades ago, some of these are very, very new, and hot off the press. So if it's an older paper, please don't brush it off as, like, oh, this is old. And so, you know, it's not new. We're all really excited by technology and everything here, but many of these older methods are robust, and used all the time here. So what are some methods and tools that we'll have under our belt? The first to point out are methods involved in macromolecular structure. So we care a lot about structure, because we need structural understanding to be able to comprehend how these systems work. And so one method we'll see a lot-- and you'll discuss in recitation this week-- is x-ray crystallography. And in addition, a method that will come up quite a bit-- and we'll see both of these in the initial discussions of the ribosome-- is electron microscopy. And another method to be aware of-- and if you're curious, talk to your TA, Shiva-- is NMR. OK, so NMR has a lot of applications here within biological chemistry, but we won't discuss that. What can go along with methods is bioinformatics. So how many of you have used BLAST? How many of you know what BLAST stands for? AUDIENCE: Basic Local Alignment Search Tool. ELIZABETH NOLAN: Yeah, Basic Local Alignment Search Tool. So what does this let you do? It lets you find regions of similarity between sequences, whether that's amino acid sequence, a nucleotide sequence. And you can use that information to make hypotheses and design experiments there. So that will come up. I have additional methods and possibilities. What about fluorescence? So how many of you have done an experiment that involves fluorescence, either in lab, or in your research? How many, did that involve a fluorescent protein? What about a small molecule? Yeah. That's fluorescent. So have you thought about why the protein was used, versus maybe why a small molecule, and what are inherent strengths and limitations or one or the other, depending what you want to do? So fluorescence is used in many, many different contexts. We can think about proteins like green fluorescent protein, we can think about using small molecules. And we like fluorescence because it allows us to see. We can get visual information. And so, where fluorescence will first come up in this class is with the ribosome. And in recitation week two, there'll be some discussion about using small molecule fluorophores to label tRNAs, and using fluorescence as a readout of steps in the translation process. And there's a lot of considerations and caveats to that. Do we have a pizza delivery? Thank goodness no. Often in this class, we get pizza deliveries for someone else. I didn't know if that's already starting. Yeah, yeah. We'll also see GFP being used in the proteasome section for degradation. So a folded protein has fluorescence, a degraded protein does not. What about kinetics? So what different types of kinetic studies can be done? So what do we all hear about in introductory biochemistry class? Pardon? AUDIENCE: [INAUDIBLE]. ELIZABETH NOLAN: Yeah, steady state kinetics, right? Turnover. So we have steady state, which I encourage you to review Michaelis-Mentin Kinetics here. And you'll also be introduced in the first weeks of this course, and especially recitation three-- so recitation two is going to build up to this-- pre-steady state kinetics. So here, you've heard about this in 5.07 or another course, introductory course. And we're looking at multiple turnover of an enzyme. And these experiments are set up with an excess of substrate, right, in order to afford conditions that allow multiple turnover. So there's formation of an enzyme substrate complex, and then there's product formation. So review as needed. So what about pre-steady state kinetics? How many of you are familiar with this method? Not so much. So what does the name suggest? Pardon? JOANNE STUBBE: So I'm deaf, you have to speak louder. ELIZABETH NOLAN: Yeah, we're both deaf. JOANNE STUBBE: I'm really deaf. So if you want to say something, so I can hear it. Speak up. AUDIENCE: Yeah, maybe observing single molecule by some spectroscopy. ELIZABETH NOLAN: Yeah, a single turnover, maybe, I think is what. If we're having multiple turnovers here in the steady state, right? If we're before the steady state, what does that mean, right? It means we're in the initial, really initial part of this reaction, where we're looking at a single turnover here. And how would you do this? Basically, you look with subs having limiting substrate rather than excess substrate. And this is just to give a little prelude in terms of thinking about experimental design. So here, look at the first moments of a reaction. So what type of time scale is that? AUDIENCE: Small. ELIZABETH NOLAN: Yeah, small. Maybe a millisecond time scale, compared to a timescale of seconds or minutes. So what does that mean? It means you need some different experimental setup. You can't do pre-steady state kinetics in the way we've done steady state kinetics, say in a lab class for instance. So you need a special apparatus. And what does it let you see? Here you're looking at multiple turnover, products forming. You know, here in the early stages, what can you see? Maybe intermediate formation. And why might that be important for thinking about mechanism? So those will come up in the first weeks of recitation. Another topic that will come up, and is something that you always need to think about, and relates to integrity of materials, is that of purification. So how are proteins purified. For studying the ribosome, how do we get ribosomes that are pure and are correct? Or what if you'd like to use a mutant ribosome? How does that get generated? So here, you can talk about ribosome or protein purification. And so, I'll present to you on ribosomes and mutant ribosomes in week four of recitation. And this topic more generally of proteins will come up in passing again and again. So how many of you have purified a protein? Many. How many of you used an affinity tag? So are they the answer to all problems? No. They can be a huge help, but they can also be problematic in one way or another, right? So with the ribosome we'll look at a case where there was really some elegant work done using an affinity tag approach to allow researchers to obtain new ribosomes. We'll also, though, talk about the limitations of that type of methodology, and the things you need to think about if you're doing protein biochemistry, and how a tag may affect your experiments and data there. In addition, to think about is assay development, and analytical methods. And so there will be many different types of assays that are presented in this course. And something just to think about-- how do you develop the right assay, and what are all the considerations? How do you know your assay is a good one for the question you want to address there? This is actually really complicated. And so there'll be some case studies that come up in the course, but just more broadly to think about. So often in lab classes, you may have an assay, but you might not be aware of all of the considerations that went into actually developing that assay such that it works. And then there's the analytical methods that are used, either for analyzing assay data or other data. And again, these have strengths and limitations. Just some that will come up, to present western blots and immunoprecipitation. So these methods involve antibodies, and so we need to think about the antibodies themselves here. Radioactivity. OK, how does this work? Why do biochemists like to use radioactivity and assay development? And how to think about this productively and correctly. So should you be afraid of iron-55, yes or no? How does that exposure compare to being in an airplane, for instance. Seriously, because there's a lot of fear associated with radioactivity that may or may not be well-founded, depending on what you're doing. And so this gives us a lot of sensitivity. And JoAnne will talk in week two of recitation about radioactivity, and designing experiments that use this as a read out. What else? So affinity measurements. OK, so dissociation constants, or affinity constants, how are these measured? When reading the literature, is the value a good one, or a not-so-good one, and how can you make that distinction? Mass spec and proteomics. So these will be in the later half of the class-- I believe recitations 11 and 12-- and many others. And we're introducing CRISPR this year, in the context of the cholesterol unit as well. So as I said, we can't take care of all of these methods immediately. We'll let you know when they're coming up, when you need to know more details about them as we go through the course here for that. So we can get started. And in the last few minutes, what I'll do is just give you a brief overview of the macromolecular machines we'll look at through modules one through three. And basically, what is the big picture? And then we're going to break that down into looking at individual components. So if we think about the lifecycle of a protein, basically, we'll fast forward to having mRNA from transcription of the genetic code. And then we have the macromolecular machine, the ribosome that allows for translation of this method message to give us a polypeptide chain. So some linear sequence of amino acids. And then what happens? We need to get from a polypeptide chain to some functional unit. And so there's a whole number of interesting players that are involved in protein folding. So we have folding, which is enabled by chaperones, is what we call these proteins that facilitate folding. And that's going to give us some structure that has some function here. And this protein has some lifetime in the cell. So at some time, for some reason, it will be time for this protein to get degraded. In which case, we need machinery that will facilitate the process to break down this folded protein into smaller fragments-- whether that be individual amino acids, or short polypeptide chains of seven to eight amino acids. So from here, we have degradation to give us small fragments. And the players here are proteases and chambers of doom, one of which is the proteasome. And actually, I forgot to mention there will be a second guest lecturer in recitation this year, Reuben Saunders, who is a senior, and does research in the Sauer Lab on one of these chambers of doom, called ClpXP. And so he'll present on single molecule methods, and fluorescence methods to study how this degradation chamber works. So that will be really exciting. He was a student in our course two years ago. So let's just take a look. We have the ribosome here. What are the structural features of this macromolecular machine, and how does it do its job? We'll look at a number of seminal studies that were done. And it is truly fascinating and incredible. What about protein folding? So look at this macromolecular machine here, GroEL, GroES, look at how big this is. So how does this chaperone allow some nascent polypeptide that's unfolded or partially folded to obtain its native structure? And there is many details in this depiction here that probably aren't apparent yet. But by the time we're done with module two, it will be there. Protein degradation. So here is just a cartoon-type depiction of a chamber of doom and its accessory protein from E. coli, ClpZ, ClpP. So look, we have a folded protein here, it's a beta barrel, our friend GFP that emits green light. And somehow, this protein gets threaded through ClpX, enters this chamber-- which has multiple protease active sites-- and that protein gets all degraded. So how does this work? How did ClpX and P work together to allow degradation of this condemned protein? And then finally, where I'll close is on something I think a little bit different for most everyone, and it's a type of template-driven polymerization involved in the synthesis of small molecules like penicillins and erythromycins. So these are antibiotics. So how do we get at molecules like these from simple amino acid precursors, or precursors like those you've seen in fatty acid biosynthesis here? And often, these are described as assembly lines. And something we'll just need to keep in mind in this unit is, are these proteins really acting like an assembly line, or is this just a way to help us think about the templates and what's going on here? So that's where we'll close. OK, so with that I'll finish up, and on Friday we'll begin with looking at the structure of the prokaryotic ribosome.
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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: OK so we're going to do this thing of the hydrogen atom and the algebraic solution. And I think it's not that long stuff so we can take it easy as we go along. I want to remind you of a couple of facts that will play a role. One result that is very general about the addition of angular momentum that you should of course know is that if you have a j1 times j2. What does this mean? You have some states of-- first angular momentum J1 that so you have a whole multiplet with J1 equals little j1. Which means the states in that multiplet have J1 squared, giving you h squared. Little j1 times little j1 plus 1. That's having a j1 multiplet. You have a j2 multiplet. And these are two independent commuting angular momenta acting on different degrees of freedom of the same particle or different particles. And what you've learned is that this can be written as a sum of representations. As a direct sum of angular momenta, which goes from j1 plus j2 plus j1 plus j2 minus 1 all the way up to the representation with j1 minus j2. And these are all representations or multiplets that live in the tense or product, but they are multiplets of J equals j1 plus j2. These states here can be reorganized into these multiplets, and that's our main result for the addition of angular momentum. Mathematically, this formula summarizes it all. These states, when you write them as a basis here-- you take a basis state here times a basis state here-- these are called the coupled bases. And then you reorganize, you form linear combinations that you have been playing with, and then they get reorganized into these states. So these are called the coupled bases in which we're talking about states of the sum of angular momentum. So that's one fact we've learned about. Now as far as hydrogen is concerned, we're going to try today to understand the spectrum. And for that let me remind you what the spectrum was. The way we organized it was with an L versus energy levels. And we would put an L equals 0 state here-- well maybe-- there's color, so why not using color. Let's see if it works. Yeah, it's OK. L equals 0. And this was called n equals 1. There's an n equals 2 that has an L equals 0 state and an L equals 1 state. There's an n equals 3 state, set of states that come within L equals 0 and L equals 1 and an L equals 2. And it just goes on and on. With the energy levels En equals minus e squared over 2a0. That combination is familiar for energy, Bohr radius, charge of electron, with a 1 over n squared. And the fact is that for any level, for each n, L goes from 0, 1, 2, up to n minus 1. And for each n there's a total of n squared states. And you see it here, you have n equals 2, n equals 1, one state. n equals 2, you have L equals 0, one state. L equals 1 is 3 states. So it's 4. Here we'll have 4 plus 5. So 9. And maybe you can do it, it's a famous thing, there's n squared states at every level. So this pattern that of course continues and-- it's a little difficult to do a nice diagram of the hydrogen atom in scale because it's all pushed towards the zero energy with 1 over n squared, but that's how it goes. For n equals 4, you have 1, 2, 3, 4 for example. And this is what we want to understand. So in order to do that, let's return to this Hamiltonian, which is p squared over 2m minus e squared over r. And to the Runge-Lenz vector that we talked about in lecture and you've been playing with. So this Runge-Lenz vector, r, is defined to be 1 over 2me squared p cross L minus L cross p minus r over r. And it has no units. It's a vector that you've learned has interpretation of a constant vector that points in this direction, r. And it just stays fixed wherever the particle is going. Classically this is a constant vector that points in the direction of the major axis of the ellipse. With respect to this vector, this vector is Hermitian. And you may recall that when we did the classical vector, you had just p cross L and no 2 here. There are now two terms here. And they are necessary because we want to have a Hermitian operator, and this is the simplest way to construct the Hermitian operator, r. And the way is that you add to this this term, that if L and p commuted as they do in classical mechanics, that the term is identical to this. And you get back to the conventional thing that you had in classical mechanics. But in quantum mechanics, of course, they don't commute, so it's a little bit different. And moreover this thing, r, is Hermitian. L and p are Hermitian but when you take the Hermitian conjugate, L goes to the other side of p. And since they don't commute, that's not the same thing. So actually the Hermitian conjugate of this term is this. There's an extra minus sign in hermiticity when you have a cross product. So this is the Hermitian conjugate of this, this is the Hermitian conjugate of this second term, here's the first and therefore this is actually a Hermitian operator. And you can work with it. Moreover, in the case of classical mechanics, it was conserved. In the case of quantum mechanics this statement of conservation quantum mechanics is something that in one of the exercises that you were asked to try to do this computation so these computations are challenging. They're not all that trivial and are good exercises. So this is one of them. This is practice. OK this is the vector r. What about it? A few more things about it that are interesting. Because of the hermiticity condition or in-- a way you can check this directly in fact was one of the exercises for you to do, was p cross L-- you did it long time ago, I think-- is equal to minus L cross b plus 2ih bar p. This is an identity. And this identity helps you write this kind of term in a way in which you have just one order of products and a little extra term, rather than having two complicated terms. So the r can be written as 1 over me squared alone, p cross L minus ihp minus r over r. For example. By writing this term as another p cross L minus that thing gives you that expression for r. You have an alternative expression in which you solve for the other one. So it's 1 over me squared minus L cross p plus ih bar p. Now, r-- we need to understand r better. That's really the challenge of this whole derivation. So we have one thing that is conserved. Angular momentum is conserved. It commutes with the Hamiltonian. We have another thing that is conserved, this r. But we have to understand better what it is. So one thing that you can ask is, well, r is conserved, so r squared is conserved as well. So r squared, if I can simplify it-- if I can do the algebra and simplify it-- it should not be that complicated. So again a practice problem was given to do that computation. And I think these forms are useful for that, to work less. And the computation gives a very nice result, where r squared is equal to 1 plus 2 H over me to the fourth L squared plus h bar squared. Kind of a strange result if you think about it. People that want to do this classically first would find that there's no h squared. And here, this h, is that whole h that we have here. It's a complicated thing. So this right hand side is quite substantial. You don't have to worry that h is in this side or whether it's on the other side because h commutes with L. L is conserved. So h appears like that. And this, again, is the result of another computation. So we've learned something. r-- oh, I'm sorry, this is r squared. Apologies. r is conserved. r squared must be conserved because if h commutes with r it commutes with r squared as well. And therefore whatever you see on the right hand side, the whole thing must be conserved. And h is conserved, of course. And L squared is conserved. Now we need one more property of a relation-- you see, you have to do these things. Even if you probably don't have an inspiration at this moment how you're going to try to understand this, there are things that just curiosity should tell that you should do. We have L, we do L squared. It's an important operator. OK. We had r, we did r squared, which is an important operator. But one thing we can do is L dot r. It's a good question what L dot r is. So what is L dot r? So-- or r dot L. What is it? Well a few things that are important to note are that you did show before that you know that r dot L, little r dot L, is 0. And little p dot L is 0. These are obvious classically, because L is perpendicular to both r and p. But quantum mechanically they take a little more work. They're not complicated, but you've shown those two. So if you have r dot L, you would have, for example, here-- r dot L, you would have to do and think of this whole r and put an L on the right. Well this little r dotted with the L on the right would be 0. That p dotted with L on the right would be 0. And we're almost there, but p cross L dot L, well, what is that? Let me talk about it here. P cross L dot L-- so this is part of the computation of this r dot L. We've already seen this term will give nothing, this term will give nothing. But this term could give something. So when you face something like that, maybe you say, well, I don't know any identities I should be using here. So you just do it. Then you say, this is i-th component of this vector times the i-th of that. So it's epsilon ijkpjLkLi. And then you say look this looks a little-- you could say many things that are wrong and get the right answer. So you could say, oh, ki symmetric and ki anti-symmetric. But that's wrong, because these k and i are not symmetric really because these operators don't commute. So the answer will be zero, but for a more complicated reason. So what do you have in here? ki. Let's move the i to the end of the epsilon, so jkipjLkLi. And now you see this part is pj L cross cross L j. Is the cross product of this. But what is L cross L? You probably remember. This is ih bar L L cross L, that's the computation in relation of angular momentum. In case you kind of don't remember it was ih bar L. Like that. So now p dot L is anyway 0. So this is 0. So it's kind of-- it's a little delicate to do these computations. But so since that term is zero, this thing is zero. Now you may as well-- r dot L is 0. Is L dot r also 0? It's not all that obvious you can even do that. Well in a second we'll see that that's true as well. L dot r and r dot L, capital R, are 0. Let's remember-- let's continue-- let's see, I wanted to number some of these equations. We're going to need them. So this will be equation one. This will be equation two, it's an important one. Now what was-- let me remind you of a notation we also had about vectors and their rotations. Vector under rotations. So what was a vector, Vi, under rotations was something that you had LiVj equals ih bar epsilon ijkvk. So there is a way to write this with cross products that is useful in some cases. So I will do it. You probably have seen that in the notes, but let me remind you. Consider this product, L cross V plus V cross L and the i-th component of it. i-th component of this product. So this is epsilon ijk and you have LjVk plus VjLk. Now in this term you can do something nice. If you think of it like expanded out, you have the second term has epsilon ijkVjk. Change j for k. If you change j for k, this will be VkLj. And these would have the opposite order. But this order can be changed up to the cost of a minus sign. So I claim this is ijk-- first term is the same-- minus VkLj. So in the second term, for this term alone, we've done for this term, multiplied with this of course, we've done j and k relabeling. But this is nothing else than the commutator of L with V. So this is epsilon ijkLjVk. That's epsilon ijk, and this is epsilon jkp or LVL. Now, 2 epsilons with 2 common indices is something that it simple. It's a commutator dealt on the other indices. Now it's better if they are sort of all aligned in the same way, but they kind of are because this L, without paying a price, can be put as the first index. So you have jk as the second and third and-- jk as the second and third-- once L has been moved to the first position. So this thing is 2 times delta iL. And there's an h bar, ih bar I forgot here. ih bar. 2 delta ik ih bar iL ih bar VL. So this is 2 ih bar Vi. So this whole thing the i-th component of this thing, using this commutation relation is this. So what we've learned is that L cross V plus V cross L you see go to 2 ih bar V. And that's a statement as a vector relation of the fact that V is a vector in the rotations. So for V to be a vector in the rotations means this. And if you wish, it means this thing as well. It's just another thing of what it means. Now R is a vector in the rotations. This capital R. Why? You've shown that if you have a vector in the rotations and you multiply it by another vector in the rotations under the cross product, it's still a vector in the rotations. So this is a vector in the rotations, this is, and this is a vector in the rotations. R is a vector in the rotations. So this capital R is a vector on the rotations, which means two things. It means it satisfies this kind of equation. So r cross-- or L cross R plus R cross L is equal to ih bar R. So R is a vector in the rotation. It's a fact beyond doubt. And that means that we now know the commutation relations between L and R. So we're starting to put together this picture in which we get familiar with R and the commutators that are possible. So I can summarize it here. L dot R LiRj is ih bar epsilon ijkRk. That's the same statement as this one but in components. And now you see why R dot L is equal to L dot R. Because actually if you put the same two indices here, i and i, you get zero. So when you have R dot L you have R1L1 plus R2L2 plus R3L3. And each of these two commute when the two indices are the same. Because of the epsilon. So R dot L is 0. And now you also appreciate that L dot R is also 0, too. OK. Now comes, in a sense, the most difficult of all calculations. Even if this seemed a little easy. But you can get quite far with it. So what do you do with Ls? You computed L commutators and you got the algebra of angular momentum. Over here. This is the algebra for angular momentum. And this kind of nontrivial calculation, you did it by building results. You knew how R was a vector in the rotation or how p was a vector in the rotation. You multiplied the two of them, and it was not so difficult. But the calculation that you really need to do now is the calculation of the commutator say of Ri with Rj. And that looks like a little bit of a nightmare. You have to commute this whole thing with itself. Lots of p's, L's, R's. 1 over R's, those don't commute with p. You remember that. So this kind of calculation done by brute force. You're talking a day, probably. I think so. And probably it becomes a mess, but. You'll find a little trick helps to organize it better. It's less of a mess, but still you don't get it and-- try several times. So what we're going to do is try to think of what the answer could be by some arguments. And then once we know what the answer can be, there's still one calculation to be done. That I will probably put in the notes, but it's not a difficult one. And the answer just pops out. So the question is what is R cross R. R cross R is really what we have when we have this commutator. So we need to know what R cross R is, just like L cross L. Now R is not likely to be an angular momentum. It's a vector but it's not an angular momentum. Has nothing to do with it. It's more complicated. So what is R cross R quantum-mechanically? Classically, of course, it would be zero. So first thing is you think of what this should be. We have a vector, because the cross product of two vectors. Now I want to emphasize one other thing, that it should be this thing-- R cross R-- is tantamount to this thing. What is this thing? It should be actually proportional to some conserved quantity. And the reason is quite interesting. So this is a small aside here. If some operator is conserved, it commutes with the Hamiltonian. Say if S1 and S2 are symmetries, that means that S1 with h is equal to S2 with h is equal to zero. Then the claim is that the commutator of this S1 and S2 claim S1 commutator with S2 is also a symmetry. So the reason is because commutator of S1 S2 commutator with h is equal actually to zero. And why would it be equal to zero? It's because of the so-called Jacobi identity for commutators. You'll remember when you have three things like that, this term is equal to 1-- this term plus 1, in which you cycle them. And plus another one where you cycle them again is equal to zero. That's a Jacobi identity. And in those cyclings you get an h with S2, for example, that is zero. And then an h with S1, which is zero. So you use these things here and you prove that. So I write here, by Jacobi. So if you have a conserved-- this is the great thing about conserved quantities, if you have one conserved quantity, it's OK. But if you have two, you're in business. Because you can then take the commutator of these two and you get another conserved quantity. And then more commutators and you keep taking commutators and if you're lucky you get all of the conserved quantities. So here R cross R refers to this commutator. So whatever is on the right should be a vector and should be conserved. And what are our conserved vectors? Well our conserved vectors-- candidates here-- are L, R itself, and L cross R. That's pretty much it. L and R are our only conserved things, so it better be that. Still this is far too much. So there could be a term proportional to L, a term proportional to R, a term proportional to L dot R. So this kind of analysis is based by something that Julian Schwinger did. This same guy that actually did quantum electrodynamics along with Feynman and Tomonaga. And he's the one of those who invented the trick of using three-dimensional angular momentum for the two-dimensional oscillator. And had lots of bags of tricks. So actually this whole discussion of the hydrogen atom-- most books just say, well, these calculations are hopeless. Let me give you the answers. Schwinger, on the other hand, in his book on quantum mechanics-- which is kind of interesting but very idiosyncratic-- finds a trick to do every calculation. So you never get into a big mess. He's absolutely elegant and keeps pulling tricks from the bag. So this is one of those tricks. Basically he goes through the following analysis now and says, look, suppose I have the vector R and I do a parity transformation. I change it for minus R. What happens under those circumstances? Well the momentum is the rate of change of R, should also change sign. Quantum mechanically this is consistent, because a commutation between R and p should give you h bar. And if R changes, p should change sign. But now when you do this, L, which is R cross p, just goes to L. And R, however, changes sign because L doesn't change sign but p does and R does. So under these changes-- so this is the originator, the troublemaker and then everybody else follows-- R also changes sign. So this is extremely powerful because if you imagine this being equal to something, well it should be consistent with the symmetries. So as I change R to minus R, capital R changes sign but the left hand side doesn't change sign. Therefore the right hand side should not change sign. And R changes sign and L cross R changes sign. So computation kind of finished because the only thing you can get on the right is L. This is the kind of thing that you do and probably if you were writing a paper on that you would anyway do the calculation. The silly way, the- the right way. But this is quite save of times. So actually what you have learned is that R cross R is equal to some scalar conserved quantity, which is something that is conserved that could be like an h, for example, here. But it's a scalar. And, L. Well once you know that much, it doesn't take much work to do this and to calculate what it is. But I will skip that calculation. This is the sort of thoughtful part of it. And R cross R turns out to be ih bar minus 2 h again. h shows up in several places, like here, so it tends-- it has a tendency to show up. me to the fourth L. So this is our equation for-- and in a sense, all the hard work has been done. Because now you have a complete understanding of these two vectors, L and R. You know what L squared is, what R squared is, what L dot R is. And you know all the commutators, you know the commutation of L with L, L with R, and R with R. You've done all the algebraic work. And the question is, how do we proceed from now to solve the hydrogen atom. So the way we proceed is kind of interesting. We're going to try to build from this L that is an angular momentum. And this R that is not an angular momentum. Two sets of angular momenta. You have two vectors. So somehow we want to try to combine them in such a way that we can invent two angular momenta. Just like the angular momentum in the two-dimensional harmonic oscillator. It was not directly through angular momentum, but was mathematical angular momentum. These two angular momenta we're going to build, one of them is going to be recognizable. The other one is going to be a little unfamiliar. But now I have to do something that-- it may sound a little unusual, but is necessary to simplify our life. I want to say some words that will allow me to think of this h here as a number. And would allow me to think of this h as a number. So here's what we're going to say. It's an assumption-- it's no assumption, but it sounds like an assumption. But there's no assumption whatsoever. We say the following: this hydrogen atom is going to have some states. So let's assume there is one state, and it has some energy. If I have that state with some energy, well, that would be the end of the story. But in fact, the thing that they want to allow the possibility for is that at that the energy there are more states. One state would be OK, maybe sometimes it happens. But in general there are more states at that energy. So I don't-- I'm not making any physical assumption to state that there is a subspace of degenerate states. And in that subspace of degenerate states, there may be just one state, there are two states, there are three states, but there's subspace of degenerate states that have some energy. And I'm going to work in that subspace. And all the operators that I have are going to be acting in that subspace. And I'm going to analyze subspace by subspace of different energies. So we're going to work with one subspace of degenerate energies. And if I have, for example, the operator R squared acting on any state of that subspace, since h commutes with L squared, h can go here, acts on this thing, becomes a number. So you might as well put a number here. You might as well put a number here as well. It has to be stated like that. Carefully. We're going to work on a degenerate subspace of some energy. But then we can treat the h as a number. So let me say it here. We'll work in a degenerate subspace with eigenvalues of h equal to h prime, for h prime. Now I want to write some numbers here to simplify my algebra. So without loss of generality we put what this dimensionless-- this is dimensionless. I'm sorry, this is not dimensionless. This one has units of energy. This is roughly the right energy, with this one would be the right energy for the ground state. Now we don't know the energies and this is going to give us the energies as well. So without solving the differential equation, we're going to get the energies. So if I say, well that's the energies you would say, come on, you're cheating. So I'll put one over nu squared where nu can be anything. Nu is real. And that's just a way to write things in order to simplify the algebra. I don't know what nu is. How you say-- you don't know, but you have this in mind and it's going to be an integer, sure. That's what good notation is all about. You write things and then, you know, it's nu. You don't call it N. Because you don't know it's an integer. You call it nu, and you proceed. So once you have called it nu, you see here that, well, that's what we call h really. h will be-- this h prime is kind of not necessary. This is what-- where h becomes in every formula. So from here you get that minus 2h over me to the fourth is 1 over h squared nu squared. I have a minus here, I'm sorry. 2h minus me to the fourth down is h squared nu squared. So we can substitute that in our nice formulas that hme to the fourth so our formulas four and five have become-- I'm going to use this blackboard. Any blackboard where I don't have a formula boxed can be erased. So I will continue here. And so what do we have? R cross R, from that formula, well this thing is over there minus 2h over me to the fourth, you substitute it in here. So it's i over h bar, one over nu squared L. Doesn't look that bad. And, R squared is equal to 1 minus 1 over h bar nu squared. Like this. L squared plus h squared. 2h, that's minus h squared nu squared. Yeah. So these are nice formulas. These are already quite clean. We'll call them five, equation five. I still want to rewrite them in a way that perhaps is a little more understandable or suggestive. I will put an h bar nu together with each R. So h nu R cross h nu R is equal to ih bar L. Makes it look nice. Then for this one you'll put h squared nu squared R squared is equal to h squared nu squared minus 1 minus L squared. It's sort of trivial algebra. You multiply by h squared nu squared, you get this. You get h squared nu squared minus L squared because it's all multiplied minus h squared. So these two equations, five, have become six. So five and six are really the same equations. Nothing much has been done. And if you wish, in terms of commutators this equation says that the commutator h nu Ri with h nu Rj is equal to ih bar epsilon ijkLk. H cross this thing h nu R, h nu R cross equal iHL in components means this. That is not totally obvious. It requires a small computation, but is the same computation that shows that this thing is really LiLj equal ih bar epsilon ijkLk. In which these L's have now become R's. OK so, we've cleaned up everything. We've made great progress even though at this moment it still looks like we haven't solved the problem at all. But we're very close. So are there any questions about what we've done so far? Have I lost you in the algebra, or any goals here? Yes. AUDIENCE: Why is R cross R not a commutation? Why would we expect that to not be a commutation? PROFESSOR: In general, it's the same thing as here. L cross L is this. The commutator of two Hermitian operators is anti-Hermitian. So there's always an i over there. Other questions? It's good, you have to-- you should worry about those things. Are the units right, or the right number of i's on the right hand side. That's a great way to catch mistakes. OK so we're there. And now it should really almost look reasonable to do what we're going to do. h nu R with h nu R gives you like L. So you have L with L, form angular momentum. L and R are vectors in their angular momentum. Now R cross R is L. And with these units, h nu R and h nu R looks like it has the units of angular momentum. So h nu R can be added to angular momentum to form more angular momentum. So that's exactly what we're going to do. So here it comes. Key step. J1-- I'm going to define two angular momenta. Well, we hope that they are angular momenta. L plus h nu R. And J2, one half L minus h nu R. These are definitions. It's just defining two operators. We hope something good happens with these operators, but at this moment you don't know. It's a good suggestion because of the units match and all that stuff. So this is going to be our definitions, seven. And from these of course follows that L, the quantity we know, is J1 plus J2. And R, or h nu R, is J1 minus J2. You solve in the other way. Now my first claim is that J1 and J2 commute. Commute with each other. So these are nice, commuting angular momenta. Now this computation has to be done-- let me-- yeah, we can do it. J1i with J2J. It's one half and one half gives you one quarter of Li plus h nu Ri with LJ minus h nu RJ. Now the question is where do I-- I think I can erase most of this blackboard. I can leave this formula. It's kind of the only very much needed one. So I'll continue with this computation here. This gives me one quarter-- and we have a big parentheses-- ih bar epsilon iJkLk. For the commutator of these two. And then you have the commutator of the cross terms. So what do they look like? They look like minus h nu Li with RJ, and minus h nu Ri with-- no. So I have minus h nu Li with RJ, and now I have a plus of this term. But I will write this as a minus h nu of LJ with Ri. Those are the two cross products. And then finally we have this thing, the h nu with h nu Rijk. So I have minus h nu squared, and you have then RiRJk. No, I'll do it this way. I'm sorry. You have minus over there, and I have this thing so it's minus ih bar epsilon iJkLk from the last two commutators. So this one you use essentially equation six. Now look. This thing and this thing cancels. And these two terms, they actually cancel as well. Because here you get an epsilon iJR. And here there's an epsilon Ji something. So these two terms actually add up to zero. And this is zero. So indeed, J1i and J2i-- 2J-- is zero. And these are commuting things. I wanted to say commuting angular momentum, but not quite yet. Haven't shown their angular momenta. So how do we show their angular momenta? We have to try it and see if they really do form an algebra of angular momentum. So again, for saving room, I'm going to erase this formula. It will reappear in lecture notes. But now it should go. So the next computation is something that I want to do. J1 cross J1 or the J2 cross J2, to see if they form angular momenta. And I want to do them simultaneously, so I will do one quarter of J1 cross J2 would be L plus minus h nu R cross L plus minus h nu R. OK that doesn't look bad at all, especially because we have all these formulas for products. So look, you have L cross L, which we know. Then you have L cross R plus R cross L that is conveniently here. And finally, you have R cross R which is here. So it's all sort of done in a way that the composition should be easy. So indeed 1 over 4 L cross L gives you an ih bar L. From L cross L. From these ones, you get plus minus with plus minus. It's always plus but you get another ihL. So you get another ihL. And then you get plus minus L cross h nu R plus h nu R cross L. So here you get one quarter of 2 ihL. And look at this formula, just put an h nu here and h nu here and an h nu here. So you get plus minus 2 ih from here and an h nu R. OK so the twos and the fours and the iH's go out and then you get ih times one half times L plus minus h nu R, which is either J1 or J2. So, very nicely, we've shown that J1 cross J1 is ih bar J1 and J2 cross J2 is ih bar J2. And now finally you can say that you've discovered two independent angular momenta in the hydrogen atom. You did have an angular momentum on an R vector, and all of our work has gone into showing now that you have two angular momenta. Pretty much we're at the end of this because, after we do one more little thing, we're there. So let me do it here. I will not need these equations anymore. Except this one I will need. So L dot R is zero. So from L dot R equals zero, this time you get J1 plus J2 is equal to-- no, times-- J1 minus J2 is equal to zero. Now J1 and J2 commute. So the cross terms vanish. J1 and J2 commute. So this implies that J1 squared is equal to J2 squared. Now this is a very surprising thing. These two angular momenta have the same length squared. Let's look a little more at the length squared of it. So let's, for example, square J1. Well, if I square J1, I have one fourth L squared plus h squared nu squared R squared. No L dot R term, because L dot R is 0. And h squared nu squared R squared is here. So this is good news. This is one fourth L squared plus h squared nu squared minus 1 minus L squared. The L squared cancels. And you've got that J1 equals to J2 squared. And it's equal to one fourth of h squared nu squared minus 1. OK. Well the problem has been solved, even if you don't notice at this moment. It's all solved. Why? You've been talking a degenerate subspace with angular momentum with equal energies. And there's two angular momenta there. And their squares equal to the same thing. So these two angular momenta, our squares are the same and the square is precisely what we call h squared J times J plus 1, where j J is quantized. It can be zero, one half, one, all of this. So here comes a quantization. J squared being nu squared, we didn't know what nu squared is, but it's now equal to these things. So at this moment, things have been quantized. And let's look into a little more detail what has happened and confirm that we got everything we wanted. So let me write that equation again here. J1 squared is equal J2 squared is equal to one quarter h squared nu squared minus 1, which is h squared J times J plus 1. So cancel the h squares and solve for nu squared. Nu squared would be 1 plus 4J times J plus 1, which is 4J squared plus 4J plus 1, which is 2J plus 1 squared. That's pretty neat. Why is it so neat? Because as J is equal to zero, all the possible values of angular momentum-- three halves, all these things-- nu, which is 2J plus 1, will be equal to 1, 2, 3, 4-- all the integers. And what was nu? It was the values of the energies. So actually you've proven the spectrum. Nu has come out to be either 1, 2, 3, but you have all representations of angular momentum. You have the singlet, the spin one half-- where are the spins here? Nowhere. There was an electron, a proton, we never put spin for the hydrogen atom. But it all shows up as these representations in which they come along. Even more is true, as we will see right away and confirm that everything really shows up the right way. So what happened now? We have two independent, equal angular momentum. So what is this degenerate subspace we were inventing? Is the space J, which is J1 and m1 tensor product with J, which is J2 but has the same value because the squares are the same, m2. So this is an uncoupled basis. Uncoupled basis of states in the degenerate subspace. And now, you know, it's all a little surreal because these don't look like our states at all. But this is the way algebraically they show up. We choose our value of J, we have then that nu is equal to this and for that value of J there's some values of m's. And therefore, this must be the degenerate subspace. So this is nothing but the tensor product of a J multiplet with a J multiplet. Where J is that integer here. And what is the tensor product of a J multiplet? First, J is for J1. The second J is for J2. So at this moment of course we're calling this N for the quantum number. But what is this thing? This is 2J plus 2J minus 1 plus-- all the way up to the singlet. But what are these representations of? Well here we have J1 and here is J2. These must be the ones of the sum. But who is the sum, L? So these are the L representations that you get. L is your angular momentum. L representations. And if 2J plus 1 is N, you got a representation with L equals N minus 1, because 2J plus 1 is N, L equals N minus 2, all the way up to L equals 0. Therefore, you get precisely this whole structure. So, just in time as we get to 2 o'clock, we've finished the quantization of the hydrogen atom. We've finished 805. I hope you enjoyed. I did a lot. [INAUDIBLE] and Will did, too. Good luck and we'll see you soon.
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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 time for new subject. Let's introduce the subject and pose the questions that we're going to try to answer. And I feel that with identical particles, there's lots to think about, and it makes it into an interesting way to conclude the course. So identical particles. So there is the issue of defining what do you mean by identical particles? And then the issue of treating them. So when do we say that two particles are identical? We say two particles are identical if all their intrinsic properties-- like mass, spin, charge, magnetic moment-- if all these things are the same, these two particles-- we have particle 1 and particle 2-- are said to be identical. For example, all electrons are said to be identical. And if you think about it, well, what does that mean? You can have an electron moving with some velocity and an electron standing here, and they don't look identical. They have different states. Well, they're identical in the sense that-- what we said-- the intrinsic properties are the same. Those two particle have the same mass. They have the same spin-- in principle. They have the same charge. Have the same magnetic moment. Have all these same properties. Now, they can be in different states. One electron can be one momentum state, an electron can be in another momentum state. One electron can be in a spin state-- spin up. This can mean spin down. But what we would mean is that if you would put, by saying these particles are identical, we also mean they're indistinguishable. What that would mean is that if one of you gives me an electron with spin up with some momentum state, and-- yeah, let's say spin up on some momentum state. And another one of you gives me another electron with same spin up, same momentum state, and I have those two electrons and I play with them for a minute and I give them back to you, you have no way of telling you got the same electron or you got your friend's electron. There's no possible experiment that can tell which electron you got. So when we have identical particles like electrons-- elementary particles-- we understand what it means to be identical. Doesn't mean they're in the same state. It means that this is a particle that all the properties intrinsic of them are the same. If you have a more complicated particle, you still can use the concept of identical particles. So, for example, you have a proton. A proton is a more complicated particle. It is made of quarks. And if you have two protons, they are identical in that same sense. All the properties we can give to the proton-- the spin state of the proton, mass of the proton, the dipole moment of a proton, the magnetic moment of a-- all those are the same. If you prepare those protons in identical states, I cannot tell which is proton 1 and which is proton 2. Then you have the neutrons. Neutrons are the same thing. The neutrons can be in several states, but two neutrons are considered to be identical. We can complicate matters more. We can take hydrogen atoms. Are hydrogen atoms identical particles? And in quantum mechanics, we can think of them as identical particles-- or identical atoms, or identical molecules, or if you write the wave function for a hydrogen atom-- a new kind of entity of particle-- we will use the axioms of identical particles even for the hydrogen atom. But you could say, oh, no, but they're not identical. A hydrogen atom can be in the ground state, or it can be in an excited state. But that's the same as saying this electrode is going with little momentum, and this is with high momentum. These are states of the hydrogen atom. Just like an electron has spin up and spin down, hydrogen atom has this state, that state, that state. If you arrange them in the same state, you cannot tell they are different. So however clear these comments can seem-- or confusing, perhaps-- things can be a little subtle. In many ways, for example, physicists used to think of protons and neutrons as the same particle. Said what? Well, that's the way they thought about it. It's a very nice thing. If you're working with energy scales, the proton and neutron mass difference is not that big, first of all. So at some scale for the resolution of some experiments, or physicists that didn't have that many tools, the proton and the neutron were almost identical, and people invented this term called isospin. And you might have heard of it. It's a very famous symmetry of the strong interactions. In fact, for the strong interactions, you have a nucleus. Whether you're a proton or a neutron doesn't make that much difference. So people used to think of this thing as an isospin state. Just the spin one half-- you have spin up, spin down. Isospin means spin in some new direction that is unimaginable. But the isospin up would be the proton, the isospin down would be the neutron. And you will have a doublet. So people used to think of these two particles as different states of a nucleon, and then they would say, all nucleons are identical. Why do you complain? A proton is the same as a neutron. It's just a different state of the isospin, just like the spin electron up or spin down is the same. So the power of the [? formalism ?] in quantum mechanics is that it allows you to treat these things as identical particles, and this makes sense. For you-- for your experiments-- these are identical things that you can think of different states of that. You may as well treat them that way. So this is basically what happens. Now, so we define the identical-- I didn't write anything here. I'm going to get notes out today on scattering, and some of these things as well. So in classical mechanics-- classical mechanics-- identical particles are distinguishable. And that's the main thing. How are they distinguishable? Well, you have two particles, and I can follow-- whenever they're moving, I can say, OK, this is particle 1. This is particle 2. With quantum mechanics, you can do the same thing when they are really far away-- those particles-- and they don't come close together. There's some sense in which classical mechanics sometimes applies, and that's when they're far away. Now, when the particles in quantum mechanics get close to each other, then you lose track which one is which. They occupy the same position. But in classical mechanics, they are distinguishable, because you can follow their trajectories, which is very nice. You have an experiment. You follow the particle, say, oh, this is particle 1. This is particle 2. They're here together. They're going around. And they split, you follow the trajectories from the beginning to the end. In quantum mechanics, there's no such thing as the trajectories. There's these waves. The waves mix together. They do things, then they separate out, and you just can't tell what they do. There's another technique that we use in classical physics that it's probably also relevant. We can tag the particles. That means, if you're doing an experiment with billiard balls colliding, you could take a little marker and put a red dot on one of them and a black dot on the other one, and that tagging doesn't affect the collisions. And you can tell, at the end, where is the red ball, and which is the black ball. We do the same with quantum mechanics. There's no way anyone has figured out the tagging a particle without changing drastically the way interactions happen. So it's a nice option in classical physics, but doesn't work in quantum mechanics. Even in classical physics, we have something that survives. If you have a Hamiltonian for identical particles-- R2, P2-- that Hamiltonian is symmetric under the exchange. Whatever the formula is, it's not changed if you put r2, p2 and r1, p1. It's a symmetric thing. The Hamiltonians have that symmetry, and there's no way to do this. So let's get to the bottom-- the real problem with identical particles with quantum mechanics. We cannot tag them. Once these particles get together, you don't know what they did. You do an experiment of scattering in the classical mechanics, and you put two particles coming in, two detectors, and you tag the particles and you see what they do. You do it in quantum mechanics, and you don't know if particle 1 did that and particle 2 did this, or if particle 1 did that and particle 2 did that. It's just not possible to tell. They're very different. So how do we deal with this? Well, that's the subject of what we're going to do, but let's just conclude today by stating the problem. So the problem is that when we had distinguishable particles in quantum mechanics, we used that tensor product to describe a state. So for distinguishable particles, this thing which all particles, say, 1 up to n, we would use the tensor product and write PSI i1 for the particle one, PSI i2 for the particle two, PSI in for the particle n. And this says particle one is in the state PSI i1, particle two is in the state PSI i2, PSI in. And these states are one of many states, for example. That's all good. And this is for distinguishable particles. And that's all correct. Now, suppose you have two electrons, one up and one down. If they are indistinguishable, how are we supposed to write the state of the two electrons or to spin one-half particles, or maybe, in some cases, maybe some other particles. Am I supposed to write that the first particle is in state up and the second is in state down? Or are I supposed to write that the first particle is in state down and the second particle is in state up? How do I describe the state with this one or with this one? They look equally plausible. So if you were in charge of inventing quantum mechanics, one possibility that it may occur to you is that if the particles are defined to be identical, then those two states should be identical. They should be indistinguishable, identical, physically equivalent. And this might be a good hypothesis to consider. Unfortunately, that does not work. You say, why not? It seems so logical. I cannot tell the difference between these two. Can I say that they're equivalent? Well, no. If they would be equivalent, you could form a state PSI alpha beta, which is alpha times the first plus minus plus beta times the second minus plus. And I'm not going to write all the subscripts nor the tensors sometimes. With alpha and beta having this for normalization. That's a normalized state. Now, if all those are equivalent, if those two are equivalent, all these are equivalent for all values in alpha and beta because the superposition of equivalence state is an equivalent state. But then let's ask for what is the probability that we find the two particles in the state PSI 0, which is plus along the x direction times plus along the x direction. You know, whatever I do whether this hypothesis there that those two states are equivalent is correct, a state that is plus and plus can only be described one way-- plus plus. So I ask, what is the probability that this state PSI alpha beta in the plus along x and in the plus along x. Now, you'll remember those pluses are states like plus, plus minus, 10 surplus plus minus with a 1 over square root of 2. And that becomes this. So this is 1/2 of plus plus plus plus minus, plus minus plus plus minus minus. That's the state. So what is the probability that PSI alpha beta is found in the state PSI 0? It's this number. Now, if you do the inner product of these two vectors, only the mixes ones go with each other. And this gives you 1/2 of alpha plus beta squared. And 1/2 of alpha plus beta squared is the inner product of these two states. And now you see that it depends on the values of alpha and beta because this is in fact 1/2 of alpha squared plus beta squared plus 2 real of alpha beta star. And since alpha and beta are normalized, this is 1/2 plus real of alpha beta star. So this hypothesis that these two states are equivalent would mean that these states are equivalent for all alpha beta that are normalized. And then you would have that the probability to be found in PSI 0 would depend on what the values of alpha and beta you choose. So it's a contradiction. So we cannot solve the problem of the degeneracy of identical particles by declaring that all the states are the same. So we have to find a different way to do it. And that's what we will do next time.
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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: Thermodynamics, all right, let's start. Thermodynamics is the science of the flow of heat. So, thermo is heat, and dynamics is the motion of heat. Thermodynamics was developed largely beginning in the 1800's, at the time of the Industrial Revolution. So, taming of steel. The beginning of generating power by burning fossil fuels. The beginning of the problems with CO2 and [NOISE OBSCURES] global warming. In fact, it's interesting to note that the first calculation on the impact of CO2 on climate was done in the late 1800's by Arrhenius. Beginning of a generation of power moving heat from fossil fuels to generating energy, locomotives, etcetera. So, he calculated what would happen to this burning of fossil fuels, and he decided in his calculation, he basically got the calculation right, by the way, but he came out that in 2,000 years from the time that he did the calculations, humans would be in trouble. Well, since his calculation, we've had an exponential growth in the amount of CO2, and if you go through the calculations of -- people have done these calculations throughout times since Arrhenius, the time that we're in trouble, 2,000 years and the calculation, has gone like this, and so now we're really in trouble. That's for a different lecture. So, anyway, thermodynamics dates from the same period as getting fossil fuels out of the ground. It's universal. It turns out everything around us moves energy around in one way or the other. If you're a biological system, you're burning calories, burning ATP. You're creating heat. If you're a warm-blooded animal. You need energy to move your arms around and move around -- mechanical systems, obviously, cars, boats, etcetera. And even in astrophysics, when you talk about stars, black holes, etcetera, you're moving energy around. You're moving heat around when you're changing matter through thermodynamics. And the cause of some thermodynamics have even been applied to economics, systems out of equilibrium, like big companies like Enron, you know, completely out of equilibrium, crash and burn. You can apply non-equilibrium thermodynamics to economics. It was developed before people knew about atoms and molecules. So it's a science that's based on macroscopic properties of matter. Since then, since we know about atoms and molecules now, we can rationalize the concepts of thermodynmamics using microscopic properties, and if you are going to take 5.62, that's what you'd learn about. You'd learn about statistical mechanics, and how the atomistic concepts rationalize thermodynamics. It doesn't prove it, but it helps to getting more intuition about the consequences of thermodynamics. So it applies to macroscopic systems that are in equilibrium, and how to go from one equilibrium state to another equilibrium state, and it's entirely empirical in its foundation. People have done experiments through the ages, and they've accumulated the knowledge from these experiments, and they've synthesized these experiments into a few basic empirical rules, empirical laws, which are the laws of thermodynamics. And then they've taken these laws and added a structure of math upon it, to build this edifice, which is a very solid edifice of thermodynamics as a science of equilibrium systems. So these empirical observations then are summarized into four laws. So, these laws are, they're really depillars. They're not proven, but they're not wrong. They're very unlikely to be wrong. Let's just go through these laws, OK, very quickly. There's a zeroth law The zeroth law every one of these laws basically defines the quantity in thermodynamics and then defines the concept. The zeroth law defines temperature. That's a fairly common-sense idea, but it's important to define it, and I call that the common-sense law. So this is the common-sense law. The first law ends up defining energy, which we're going to call u, and the concept of energy conservation, energy can't be lost or gained. And I'm going to call this the you can break even law; you can break even law. You don't lose energy, you can't gain energy. You break even. The second law is going to define entropy, and is going to tell us about the direction of time, something that conceptually we, clearly, understand, but is going to put a mathematical foundation on which way does time go. Clearly, if I take a chalk like this one here, and I throw it on the ground, and it breaks in little pieces, if I run the movie backwards, that doesn't make sense, right? We have a concept of time going forward in a particular way. How does entropy play into that concept of time? And I'm going to call this the you can break even at zero degrees Kelvin law. You can only do it at zero degrees Kelvin. The third law is going to give a numerical value to the entropy, and the third law is going to be the depressing one, and it's going to say, you can't get to zero degrees. These laws are universally valid. They cannot be circumvented. Certainly people have tried to do that, and every year there's a newspaper story, Wall Street Journal, or New York Times about somebody that has invented the device that somehow goes around the second law and makes more energy than it creates, and this is going to be -- well, first of all, for the investors this is going to make them very, very rich, and for the rest of us, it's going to be wonderful. And they go through these arguments, and they find venture money to fund the company, and they get very famous people to endorse them, etcetera. But you guys know, because you have MIT degrees, and you've, later, and you've taken 5.60, that can't be the case, and you're not going to get fooled into investing money into these companies. But it's amazing, that every year you find somebody coming up with a way of going around the second law and somehow convincing people who are very smart that this will work. So, thermo is also a big tease, as you can see from my descriptions of these laws here. It makes you believe, initially, in the feasibility of perfect efficiency. The first law is very upbeat. It talks about the conservation of energy. Energy is conserved in all of its forms. You can take heat energy and convert it to work energy and vice versa, and it doesn't say anything about that you have to waste heat if you're going to transform heat into work. It just says it's energy. It's all the same thing, right? So, you could break even if you were very clever about it, and that's pretty neat. So, in a sense, it says, you know, if you wanted to build a boat that took energy out of the warmth of the air, to sail around the world, you can do that. And then the second law comes in and says well, that's not quite right. The second law says, yes, energy is pretty much the same in all this form, but if you want to convert one form of energy into another, if you want to convert work, heat into work, with 100% efficiency, you've got to go down to zero degrees Kelvin, to absolute zero if you want to do that. Otherwise you're going to waste some of that heat somewhere along the way, some of that energy. All right, so you can't get perfect efficiency, but at least if you were able to go to zero degrees Kelvin, then you'd be all set. You just got to find a good refrigerator on your boat, and then you can still go around the world. And then the third law comes in, and that's the depressing part here. It says, well, it's true. If you could get to zero degrees Kelvin, you'd get perfect efficiency, but you can't get to zero degrees Kelvin, you can't. Even if you have an infinite amount of resources, you can't get there. Any questions so far? So thermodynamics, based on these four laws now, requires an edifice, and it's a very mature science, and it requires that we define things carefully. So we're going to spend a little bit of time making sure we define our concepts and our words, and what you'll find that when you do problem sets, especially at the beginning, understanding the words and the conditions of the problem sets is most of the way into solving the problem. So we're going to talk about things like systems. The system, it's that part of the universe that we're studying. These are going to be fairly common-sense definitions, but they're important, and when you get to a problem set, really nailing down what the system is, not more, nor less, in terms of the amount of stuff, that's part of the system, it's going to be often very crucial. So you've got the system. For instance, it could be a person. I am the system. I could be a system. It could be a hot coffee in a thermos. So the coffee and the milk and whatever else you like in your coffee would be the system. It could be a glass of water with ice in it. That's a fine system. Volume of air in a part of a room. Take four liters on this corner of the room. That's my system. Then, after you define what your system is, whatever is left over of the universe is the surroundings. So, if I'm the system, then everything else is the surroundings. You are my surroundings. Saturn is my surroundings. As far as you can go in the universe, that's part of the surroundings. And then between the system and the surroundings is the boundary. And the boundary is a surface that's real, like the outsides of my skin, or the inner wall of the thermos that has the coffee in it, or it could be an imaginary boundary. For instance, I can imagine that there is a boundary that surrounds the four liters of air that's sitting in the corner there. It doesn't have to be a real container to contain it. It's just an imaginary boundary there. And where you place that boundary becomes important. So, for instance, for the thermos with the coffee in it, if you place the boundary in the inside wall of the glass or the outside wall of the glass and the inside of the thermos, that makes a difference; different heat capacity, etcetera. So this becomes where defining the system and the boundaries, and everything becomes important. You've got to place the boundary at exactly the right place, otherwise you've got a bit too much in your system or a bit too little. More definitions. The system can be an open system, or it can be a closed system, or it can be isolated. The definitions are also important here. An open system, as the name describes, allows mass and energy to freely flow through the boundary. Mass and energy flow through boundary. Mass and energy -- I'm an open system, right? Water vapor goes through my skin. I'm hot, compared to the air of the room, or cold if I'm somewhere that's warm. So energy can go back and forth. The thermos, with the lid on top, is not an open system. Hopefully, your coffee is going to stay warm or hot in the thermos. It's not going to get out. So the thermos is not an open system. In fact, the thermos is an isolated system. The isolated system is the opposite of the open system, no mass and no energy can flow through the boundary. The closed system allows energy to transfer through the boundary but not mass. So a closed system would be, for instance, a glass of ice water with an ice cube in it, with the lid on top. The glass is not very insulating. Energy can flow across the glass, but I put a lid on top, and so the water can't get out. And that's the closed system. Energy goes through the boundaries but nothing else. Important definitions, even though they may sound really kind of dumb, but they are really important, because when you get the problem, figuring out whether you have an open, closed, or isolated system, what are the surroundings? What's the boundary? What is the system? That's the first thing to make sure that is clear. If it's not clear, the problem is going to be impossible to solve. And that's also how people find ways to break the second law, because somehow they've messed up on what their system is. And they've included too much or too little in the system, and it looks to them that the second law is broken and they've created more energy than is being brought in. That's usually the case. Questions? Let's keep going. So, now that we've got a system, we've got to describe it. So, let's describe the system now. It turns out that when you're talking about macroscopic properties of matter, you don't need very many variables to describe the system completely thermodynamically. You just need a few macroscopic variables that are very familiar to you, like the pressure, the temperature, the volume, the number of moles of each component, the mass of the system. You've got a magnetic field, maybe even magnetic susceptibility, the electric field. We're not going to worry about these magnetic fields or electric fields in this class. So, pretty much we're going to focus on this set of variables here. You're going to have to know when you describe the system, if your system is homogeneous, like your coffee with milk in it, or heterogeneous, like water with an ice cube in it. So heterogeneous means that you've got different phases in your system. I'm the heterogeneous system, soft stuff, hard stuff, liquid stuff. Coffee is homogeneous, even though it's made up of many components. Many different kinds of molecules make up your coffee. There are the water molecules, the flavor molecules, the milk proteins, etcetera. But it's all mixed up together in a homogeneous, macroscopic fashion. If you drill down at the level of molecules you see that it's not homogeneous. But thermodynamics takes a bird's eye view. It looks pretty, beautiful. So, that's a homogeneous system, one phase. You have to know if your system is an equilibrium system or not. If it's an equilibrium system, then thermodynamics can describe it. If it's not, then you're going to have trouble describing it using thermodynamic properties. Thermodynamics talks about equilibrium systems and how to go from one state of equilibrium to another state of equilibrium. What does equilibrium mean? It means that the properties of the system, the properties that describe the system, don't change in time or in space. If I've got a gas in a container, the pressure of the gas has to be the same everywhere in the container, otherwise it's not equilibrium. If I place my container of gas on the table here, and I come back an hour later, the pressure needs to be the same when I come back. Otherwise it's not equilibrium. So it only talks about equilibrium systems. What else do you need to know? So, you need to know the variables. You need to know it's heterogeneous or homogeneous. You need to know if it's an equilibrium, and you also need to know how many components you have in your system. So, a glass of ice water with an ice cube in it, which is a heterogeneous system, has only one component, which is water, H2O. Two phases, but one component. Latte, which is a homogeneous system, has a very, very large number of components to it. All the components that make up the milk. All the components that make up the coffee, and all the impurities, etcetera. cadmium, heavy metals, arsenic, whatever is in your coffee. OK, any questions? All right, so we've described the system with these properties. Now these properties come in two flavors. You have extensive properties and intensive properties. The extensive properties are the ones that scale with the size of the system. If you double the system, they double in there numerical number. For instance, the volume. If you double the volume, the v doubles. I mean that's obvious. The mass, if you double the amount of stuff the mass will double. Intensive properties don't care about the scale of your system. If you double everything in the system, the temperature is not going to change, it's not going to double. The temperature stays the same. So the temperature is intensive, and you can make intensive properties out of the extensive properties by dividing by the number of moles in the system. So I can make a quantity that I'll call V bar, which is the molar volume, the volume of one mole of a component in my system, and that becomes an intensive quantity. A volume which is an intensive volume. The volumes per mole of that stuff. So, as I mentioned, thermodynamics is the science of equilibrium systems, and it also describes the evolution of one equilibrium to another equilibrium. How do you go from one to the other? And so the set of properties that describes the system -- the equilibrium doesn't change. So, these on-changing properties that describe the state of the equilibrium state of the system are called state variables. So the state variables describe the equilibrium's state, and they don't care about how this state got to where it is. They don't care about the history of the state. They just know that's if you have water at zero degrees Celsius with it ice in, that you can define it as a heterogeneous system with a certain density for the water or certain density for the ice, etcetera, etcetera. It doesn't care how you got there. We're going to find other properties that do care about the history of the system, like work, that you put in the system, or heat that you put in the system, or some other variables. But you can't use those to define the equilibrium state. You can only use the state variables, independent of history. And it turns out that for a one component system, one component meaning one kind of molecule in the system, all that you need to know to describe the system is the number of moles for a one component system, and to describe one phase in that system, one component, homogeneous system, you need n and two variables. For instance, the pressure and the temperature, or the volume and the pressure. If you have the number of moles and two intensive variables, then you know everything there is to know about the system. About the equilibrium state of that system. There are hundreds of quantities that you can calculate and measure that are interesting and important properties, and all you need is just a few variables to get everything out, and that's really the power of thermodynamics, is that it takes so little information to get so much information out. So little data to get a lot of predictive information out. As we're going on with our definitions, we can summarize a lot of these definitions into a notation, a chemical notation that that will be very important. So, for instance, if I'm talking about three moles of hydrogen, at one bar 100 degrees Celsius. I'm not going to write, given three moles of hydrogen at one bar and three degrees, blah, blah, blah. I'm going to write it in a compact notation. I'm going to write it like this: three moles of hydrogen which is a gas, one bar 100 degrees Celsius. This notation gives you everything you need to know about the system. It tells you the number of moles. It tells you the phase. It tells you what kind of molecule it is, and gives you two variables that are state variables. You could have the volume and the temperature. You could have the volume and the pressure. But this tells you everything. I don't need to write it down in words. And then if I want to tell you about a change of state, or let's first start with a mixture. Suppose that I give to a mixture like, this is a homogeneous system with two components, like five moles of H2O, which is a liquid, at one bar 25 degrees Celsius, plus five moles of CH3, CH2, OH, which is a liquid, and one bar at 25 degrees Celsius. This describes roughly something that is fairly commonplace, it's 100-proof vodka 1/2 water, 1/2 ethanol -- that describes that macroscopic system. You're missing all the impurities, all the little the flavor molecules that go into it, but basically, that's the homogeneous system we were describing, two component homogeneous systems. Then you can do all sorts of predictive stuff with that system. All right, that's the equilibrium system. Now we want to show a notation, how do we go from one equilibrium state like this describes to another equilibrium state? So, we take our two equilibrium states, and you just put an equal sign between them, and the equal sign means go from one to the other. So, if we took our three moles of hydrogen, which is a gas at five bar and 100 degrees Celsius, and, which is a nice equilibrium state here, and we say now we're going to change the equilibrium state to something new, we're going to do an expansion, let's say. We're going to drop the pressure, the volume is going to go up. I don't need to tell you the volume here, because you've got enough information to calculate the volume. The number of moles stays the same, a closed systems, gas doesn't come out. Stays a gas, but now the pressure is less, the temperature is less. I've done some sort of expansion on this. I've gone from 1 equilibrium state to another equilibrium state, and the equal sign means you go from this state to that state. It's not a chemical reaction. That's why we don't have an arrow here, because we could go back, this way too. We can go back and forth between these two equilibrium states. They're connected. This means they're connected. And when I put this, I have to tell you how they are connected. I have to tell you the path, if you're going to solve a problem. For instance, you want to know how much energy you're going to get out from doing this expansion. How much energy are you going to get out, and how far are you going to be able to drive a car with this expansion, let's say, so that's the problem. So, I need to tell you how you're doing the expansion, because that's going to tell you how much energy you're wasting during that expansion. It goes back to the second law. Nothing is efficient. You're always wasting energy into heat somewhere when you do a change that involves a mechanical change. All right, so I need to tell you the path, when I go from one state to the other. And the path is going to be the sequence, intermediate states going from the initial state the final state. So, for instance, if I draw a graph of pressure on one axis and temperature on the other axis, my initial state is at a temperature of 100 degrees Celsius and five bar. My final stage is 50 degrees Celsius and one bar. So, I could have two steps in my path. I could decide first of all to keep the pressure constant and lower the pressure. When I get to 50 degrees Celsius, I could choose to keep the temperature constant and lower the pressure. I'm sorry, my first step would be to keep the pressure constant and lower the temperature, then I lower the pressure, keeping the temperature constant. So there's my intermediate state there. This is one of many paths. There's an infinite number of paths you could take. You could take a continuous path, where you have an infinite number of equilibrium points in between the two, a smooth path, where you drop the pressure and the temperature simultaneously in little increments. All right, so when you do a problem, the path is going to turn out to be extremely important. How do you get from the initial state to the final state? Define the initial state. Define the final state. Define the path. Get all of these really clear, and you've basically solved the problem. You've got to spend the time to make sure that everything is well defined before you start trying to work out these problem. More about the path. There are a couple ways you could go through that path. If I look at this smooth path here. I could have that path be very slow and steady, so that at every point along the way, my gas is an equilibrium. So I've got, this piston here is compressed, and I slowly, slowly increase the volume, drop the temperature. Then I can go back, the gas is included at every point of the way. That's a reversible path. That can reverse the process. I expand it, and reverse it, no problem. So, I could have a reversible path, or I take my gas, and instead of slowly, slowly raising it, dropping the pressure, I go from five bar to one bar extremely fast. What happens to my gas inside? Well, my gas inside is going to be very unhappy. It's not going stay in equilibrium. Parts of the system are going to be at five bar. Parts of it at one bar. Parts of it may be even at zero bar, if I go really fast. I'm going to create a vacuum. So the system will not be described by a single state variable during the path. If I look at different points in my container during that path, I'm going to have to use a different value of pressure or different value of temperature at different points of the container. That's not an equilibrium state, and that process turns out then to be in irreversible process. Do it very quickly. Now to reverse it and get back to the initial point is going to require some input from outside, like heat or extra work or extra heat or something, because you've done an irreversible process. You've wasted a lot of energy in doing that process. I have to tell you whether the path is reversible or irreversible, and the irreversible path also defines the direction of time. You can only have an irreversible path go one way in time, not the other way. Chalk breaks irreversibly and you can't put it back together so easily. You've got to pretty much take that chalk, and make a slurry out of it, put water, and dry it back up, put in a mold, and then you can have the chalk again, but you can't just glue it back together. That would not be the same state as what you started out with. And then there are a bunch of words that describe these paths. Words like adiabatic, which we'll be very familiar with. Adiabatic means that there's no heat transferred between the system and the surrounding. The boundary is impervious to transfer of heat, like a thermos. Anything that happens inside of the thermos is an adiabatic change because the thermos has no connection in terms of energy to the outside world. There's no heat that can go through the walls of the thermos. Whereas, like isobaric means constant pressure. So, this path right here from this top red path is an isobaric process. Constant temperature means isothermal, so this part means an isothermal process. So then, going from the initial to final states with a red path, you start with an isobaric process and then you end with an isothermal process. And these are words that are very meaningful when you read the text of a problem or of a process. Any questions before we got to the zeroth law? We're pretty much done with our definitions here. Yes. STUDENT: Was adiabatic reversible? PROFESSOR: Adiabatic can be either reversible or not, and we're going to do that probably next time or two times. Any other questions? Yes. STUDENT: Is there a boundary between reversible and irreversible? PROFESSOR: A boundary between reversible and irreversible? Like something is almost reversible and almost irreversible. No, pretty much things are either reversible or irreversible. Now, in practice, it depends on how good your measurement is. And probably also in practice, nothing is truly reversible. So, it depends on your error bar in a sense. It depends on what what you define, exactly what you define in your system. It becomes a gray area, but it should be pretty clear if you can treat something is reversible are irreversible. Other questions, It's a good question. So the zeroth law we're going to go through the laws now. The zeroth law talks about defining temperature and it's the common-sense law. You all know how. When something hot, it's got a higher temperature than when something is cold. But it's important to define that, and define something that's a thermometer. So what do you know? What's the empirical information that everybody knows? Everybody knows that if you take something which is hot and something which is cold, and you bring them together, make them touch, that heat is going to flow from the hot to the cold, and make them touch, and heat flows from hot to cold. That's common sense. This is part of your DNA, And then their final product is an object, a b which ends up at a temperature or a warmness which is in between the hot and the cold. So, this turns out to be warm. You get your new equilibrium state, which is in between what this was, and what a and b were. Then how do you know that it's changed temperature, or that heat has flowed from a to b? Practically speaking, you need some sort of property that's changing as heat is flowing. For instance, if a were metallic, you could measure the connectivity of a or resistivity, and as heat flows out of a into b, the resistivity of a would change. Or you could have something that's color metric that changes color when it's colder, so you could see the heat flowing as a changes color or b changes color as heat flows into b. So, you need some sort of property, something you can see, something you can measure, that tells you that heat has flowed. Now, if you have three objects, if you have a, b, and c, and you bring them together, and a is the hottest, b is the medium one, and c is the coldest, so from hottest to coldest a, b, c, -- if you bring them together and make them touch, you know, intuitively, that heat will not flow like this. You know that's not going to happen. You know that what will happen is that heat will flow from a to b from b to c and from a to c. That's common-sense. You know that. And the other way in the circle will never happen. That would that would give rise to a perpetual motion machine, breaking of the second law. It can't happen. But that's an empirical observation, that heat flows in this direction. And that's the zeroth law thermodynamic. It's pretty simple. The zeroth law says that if a and b -- it doesn't exactly say that, but it implies this. It says that if a and b are in thermal equilibrium, if these two are in thermal equilibrium, meaning that there's no heat flows between them, so that's the definition of thermal equilibrium, that no heat flows between them, and these two are in thermal equilibrium, and these two are in thermal equilibrium, then a and c will be also be in thermal equilibrium. But if there's no heat flowing between these two, and no heat flowing between these two, then you can't have heat flowing between these two. So if I get rid of these arrows, there's no heat flowing because they're in thermal equilibrium, then I can't have an arrow here. That's what the zeroth law says. They're all the same temperature. That's what it says. If two object are in the same temperature, and two other object are in the same temperature, then all three must have the same temperature. It sounds pretty silly, but it's really important because it allows you to define a thermometer and temperature. Because now you can say, all right, well, now b can be my thermometer. I have two objects, I have an object which is in Madagascar and an object which is in Boston, and I want to know, are they the same temperature? So I come out with a third object, b, I go to Madagascar, and put b in contact with a. Then I insulate everything, you know, take it away and see if there's any heat flow. Let's say there's no heat flow. Then I insulate it, get back on the plane to Boston, and go back and touch b with c. If there's no heat flow between the b and c, then I can say all right, a and c were the same temperature. B is my thermometer that tells me that a and c are in the same temperature. And there's a certain property associated with heat flow with b, and it didn't change. And that property could be color. It could be resistivity. It could be a lot of different things. It could be volume. And the temperature then is associated with that property. And if it had changed, then the temperature between those two would have changed in a very particular way. So, zeroth law, then, allows you to define the concept of temperature and the measurement of temperature through a thermometer. Let's very briefly go through stuff that you've learned before. So, now you have this object which is going to tell you whether other things are in thermal equilibrium now. What do you need for that object? You need that object to be a substance, to be something. So, the active part of the thermometer could be water. It could be alcohol, mercury, it could be a piece of metal. You need a substance, and then that substance has to have a property that changes depending on the heat flow, i.e., depending on whether it's sensing that it's the same temperature or different temperature than something else. And that property could be the volume, like if you have a mercury thermometer, the volume of the mercury. It could be temperature. It could be resistivity, if you have a thermocouple. It could be the pressure. All right, so now you have an object. You've got a property that changes, depending on the heat flow. It's going to tell you about the temperature. Now you need to define the temperature scales. So, you need some reference points to be able to tell you, OK, this temperature is 550 degrees Smith, whatever. So, you assign values to very specific states of matter and call those the reference points for your temperature. For instance, freezing of water or boiling of water, the standard ones. And then an interpolation scheme. You need a functional form that connects the value at one state of matter, the freezing point of water, to another phase change, the boiling point of water. You can choose a linear interpolation or quadratic, but you've got to choose it. And it turns out not to be so easy. And if you go back into the 1800's when thermodynamics was starting, there were a zillion different temperatures scales. Everybody had their own favorite temperature scales. The one that we're most familiar with is the centigrade or Celsius scale where mercury was the substance, and the volume of mercury is the property. The reference points are water, freezing or boiling, and the interpolation is linear, and then that morphed into the Kelvin scale, as we're going to see later. The Fahrenheit scale is an interesting scale. It turns out the U.S. and Jamaica are the only two places on Earth now that use the Fahrenheit scale. Mr. Fahrenheit, Daniel Gabriel Fahrenheit was a German instrument maker. The way he came up with his scale was actually he borrowed the Romer scale, which came beforehand. The Romer scale was, Romer was a Dane, and he defined freezing of water at 7.5 degrees Roemer, and 22.5 degrees Romer as blood-warm. That was his definition. Two substances, blood and water. Two reference points, freezing and blood-warm, you know, the human body. A linear interpolation between the two, and then some numbers associated with them, 7-1/2 and 22-1/2. Why does he choose 7-1/2 as the freezing point of water? Because he thought that would be big enough that in Denmark, the temperature wouldn't go below zero. That's how he picked 7-1/2. Why not? He didn't want to use negative numbers to measure temperature in Denmark outside. Well, Fahrenheit came along and thought, well, you know, 7-1/2, that's kind of silly; 22-1/2 that's, kind of silly. So let's multiply everything by four. I think it becomes 30 degrees for the freezing of water and 22.5 x 4, which I don't know what it is, 100 or something -- no, it's 90 I think. And then for some reason, that nobody understands, he decided to multiply again by 16/15, and that's how we get 32 for freezing of water and 96 in his words for the temperature in the mouth or underneath the armpit of a living man in good health. What a great temperature scale. It turns out that 96 wasn't quite right. Then he interpolated and found out water boils at 212. But, you know, his experiment wasn't so great, and, you know, maybe had a fever when he did the reference point with 96, whatever. It turns out that it's not 96 to be in good health, it's 98.6 -- whatever. That's how we got to the Fahrenheit scale. All right, next time we're going to talk about a much better scale, which is the ideal gas thermometer and how we get to the Kelvin scale.
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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 today, we're going to talk about immunity, which is important, especially at this time of the year. So immunity is the resistance to disease based on a prior exposure. Based on prior exposure. And of course, this is the principle behind vaccination. So humans have been sort of using the properties of the immune system to prevent themselves from getting disease for centuries. One of the first very clear examples of this is back in the 18th century with the English physician, Edward Jenner, and Edward Jenner found out or came to the realization that farm hands on farms, specifically milk maids, that were exposed to a variant of smallpox from cows, which is known as cow pox, could become immune to smallpox. So cow pox is a less severe form of the disease. And what Jenner did was to take pustules from individuals who had the cow pox disease and inject them into an eight-year-old boy, and then infect that boy with smallpox to show that the boy was immune to smallpox after having received the cow pox material. And so this is the first example of the vaccine. And because the vaccine was derived from basically someone with cow pox, the word vaccine is from the Latin root of vaca, which is cow, so that's where the word vaccine comes from, OK? So today, we're going to talk about the systems that our bodies have to fight disease and there are several different all levels of the immune system. So I'm going to talk about two of them. So I'll talk about two levels of immunity. The first I want to mention is the one that we're all just like born with, which is known as innate immunity. [SNEEZING] [SNEEZING] Bless you. So innate immunity, as the name implies, is something that we are born with, so this is inborn. It also doesn't have a delay in when it's activated, right? So if you have an infection, this is sort of the first line of defense, right? There is an immediate response, and that's the innate immune system, so this is immediate. Here, I'll put this down here. It's immediate. So one example is of an innate immune response. This is not in the body but ex vivo, but here you see a human neutrophil, and neutrophils are part of our innate immune system. And neutrophils hunt and kill bacteria, right? You see that neutrophil chasing after that bacterium, and it's going for it. It's really trying to get it, but that bacteria really wants to get away, but got it! OK, great. So these neutrophils are part of the innate immune system. It's inborn, it's immediate. And in addition, the response of the innate immune system doesn't really change if you've been exposed to an infectious agent prior to the-- OK, so this does not change. I'm using the Greek delta for change. Does not change with prior exposure. OK, so it's sort of like a constant surveillance mechanism in your body that will go after foreign agents, OK? Now, this is very different from the next level of immunity, which is known as adaptive immunity. And as the name adaptive immunity implies, this is a type of immunity that does change. It adapts, OK? And this type of immunity is acquired, so it's also known as acquired immunity, but it's acquired with exposure to a foreign agent, OK? So this involves a change in immunity, this one does not, but the innate immune response is immediate, whereas adaptive immunity takes time. There's a delay, so this is also delayed. It's also highly specific, OK? So it's highly specific to the foreign agents that you are infected with. The innate immune system is less specific. It'll recognize, like, things like bacteria, but it won't be able to necessarily distinguish between different types of bacteria. So this is more specific than the innate immune system. This is why every year, you have to get a flu shot, because the flu virus is constantly changing. And our immune system is so specific that unless we get a new vaccination, our bodies will not be able to recognize it, OK? So this is-- so now I'm going to break down adaptive immunity into two branches. One is known as humoral immunity, and humoral immunity is basically protein-mediated, and there are proteins that mediate this are called antibodies. These antibodies are proteins, and it's called humoral because the antibodies can be secreted into the fluids or humors of our body, which is basically the blood, OK? So there is humoral immunity. The other type of adaptive immunity is cell mediated, and one thing I want to point out that the types of cells that make an antibody are known as B cells. What the B stands for isn't really important, but one thing that's helpful is that these cells mature in the bone marrow, and B stands for bone marrow, OK? So you can always remember where they mature. Now cell-mediated immunity, in contrast, involves a different type of cell called a T cell, and the T of T cell stands for thymus because these cells mature in the thymus. And I just want to point out where these cells come from. So we talked about adult stem cells earlier, and in this case, these T and B lymphocytes over here are derived from a multipotent hematopoietic stem cell, which generates a whole bunch of different types of cells. Many of them are involved in the innate immune response, but this common lymphoid progenitor over here gives rise to D-- T lymphocytes and B lymphocytes, which are involved in adaptive immunity. OK, so it's not important that you remember where-- what all these cells come from or what they, like, what the tree is, but that these cells arise from a common progenitor cell. OK, so both of these branches of the adaptive immune system have what are known as antigen receptors. I'll abbreviate antigen, AG. So they have antigen receptors, meaning that they have things on them that recognize specific antigens, and antigens are basically things that result in an immune response. They could be proteins. Antigens are substances that activate the immune system. That's just immune system, OK? Another abbreviation that I'll use is when I refer to an antibody, I'm going to abbreviate it, AB. All right, so we have these two branches of the immune system, and they each have the type of antigen receptor, so now I want to go through what these different types of antigen receptors look like, OK? And I'm going to start with the B-cell antigen receptor, also known as the antibody, also known as an immunoglobulin. OK, so another-- these are all synonymous, but you will see them in different contexts. Immunoglobulin is abbreviated IG. And what the antibody looks like structurally, it looks like this, and I'll just draw it out for you down here. So I'm drawing a lipid bilayer that represents the plasma membrane. The outside of the cell is going to be up, so that's the exoplasm up here. The inside down here is the cytoplasm. And this would be a B cell, then, we're talking about here. I'm going to draw just a segment of the B cell plasma membrane. And the B-- the antibody can have a transmembrane domain that spans the plasma membrane, and then there are domains-- and what I'm drawing here is a circle, is an IG domain, so this is going to equal an IG domain. It's just a type of protein fold that is modular, OK? So you can see up on my diagram here, right? You see these like here there are these two green segments labeled V and C. Each of those is a single IG domain. OK, it's just a modular fold that is separate from the other part of the protein. OK, so here we have along-- this is one polypeptide chain that has a transmembrane domain, and it is inserted into the plasma membrane. The N-terminus is here, the C-terminus is down here, and each antibody protein has two of these long peptides. And because they're the longest part of the molecule, they're known as heavy chains, so these are the heavy chains. And each antibody protein is composed of two identical heavy chains, OK? So these are identical. And then also there's another component, which is present up here, and this is a shorter polypeptide. And because it's shorter and smaller, it's known as the light chain. OK, that's the light chain. OK, so that's more or less what an antibody looks like. The part of this antigen receptor that recognizes the antigen are the tips right here, so this is where the antigen binds, and it can bind on either this side or this side. This molecule is laterally symmetric. One side is identical to the other, OK? Now, the T-cell receptor looks different, and the T cell receptor has fewer names. It's just called the T-cell receptor, or the TCR, for short. And the T-cell receptor is structurally very different, so now I'm drawing here a T-cell plasma membrane. Here's the plasma membrane. The exoplasm, again, is up. The cytoplasm is down below this plasma membrane. And the T-cell receptor has two chains. One is called alpha and the other is beta, and it has fewer immunoglobulin repeats, so that you can see you just have this sort of smaller system here, where you have an alpha and a beta chain. And in this case, this region here recognizes the antigen, OK? So basically the T-cell receptor, or the tip of it, interacts with the antigen. Now, the B-cell receptor, or the antibody, has different forms, so let's talk about the different forms. And these are shown up on my slide above, right? So you see over here, here is an antibody that has a transmembrane domain and is anchored in the plasma membrane, but there's another form that lacks that transmembrane domain, and instead of being an integral membrane protein, is instead secreted into the blood, OK? So the forms of the B cell receptor are both a membrane-bound form, which is initially how this antibody is presented, but later on, it can be secreted, and this often changes when there is an infection, OK? So once you have a virus or bacteria in your system, then you get the B cells sort of pumping out the secreted form of the antibody in order to fight the infection, OK? In contrast, for the T-cell receptor. For the T-cell receptor, there's only one form, which is the membrane-bound form, OK? So for T-cell receptors, it's membrane only. OK, another thing that differs between these antigen centers receptors is the types of antigens that are recognized. So antibodies can recognize all sorts of different molecules, OK? They're very promiscuous, but they-- and a given antibody is not promiscuous. A given antibody will recognize a very specific structure, but the possibility for antibodies is that they can recognize small molecules. They can recognize proteins, they can recognize DNA, they can recognize carbohydrates, you get the idea, right? They really can recognize a whole range of different types of molecules. In contrast, the T-cell receptor is more restricted in that T-cell receptors will recognize peptides or short sequences of amino acids. So it recognizes peptides, and these peptides are presented to the T cell on a type of molecule presented by the MHC complex. There are two classes, 1 and 2, and we're going to talk about this in detail in Friday's lecture. So I just want to point out the difference in the types of antigens that can be recognized here, and we'll talk about exactly what that means on Friday. OK, so now we have to talk about the amazing properties of the immune system. The first is how specific it is, its specificity, and I think this is a really amazing property, the ability to really discriminate between very closely related molecules, right? And this is essential for immunity to work well. You want to recognize things that our foreign agents that have like invaded your system. You don't want to be recognizing proteins and structures that are natively present in your body, because if your immune system did that, you'd have an autoimmune disease, so this specificity is really crucial for the function of the immune system. So now I want to talk how is it that the immune system achieves such high levels of specificity, and the way I want to illustrate this is I want to bring this down quickly. So if we consider the structure of the antibody, these different domains are different in that-- in how variable they are, so some are variable. So this domain here for the heavy chain is the variable domain of the heavy chain, which I'll just abbreviate VH, and then these other immunoglobulin domains are constant, meaning they don't have a lot of variation in sequence. Like the heavy chain, there is a variable domain for the light chain, which I'll abbreviate VL, and then there is a constant domain for the light chain, OK? And so what I want to do now is consider what the sequence variation is here on this antibody is the same over here. This is the same thing over here. You have a variable domain for the heavy chain and a variable domain for the light chain. So let's consider the amino acid sequence of the antibody molecule specifically at that variable part of the protein. So let's say we could take individual antibodies and define their sequence from end to C-terminus. That would be from tip towards the end here. So if we take a number of different antibodies and align their amino acid sequence-- so what I am-- I'm not writing out an amino acid sequence, but I'm just illustrating like a particular type of computational experiment you could do. So these would be aligned amino acid sequences where each of these represents a different antibody, let's say, heavy chain polypeptide that's produced from a different B cell, OK? So each of these is a different antibody from a unique B cell. And then we just consider the residue number and how much each amino acid residue varies along this sequence. So if we were to align antibody gene stretches like this and look at how much variation there is, you'd get a graph that looks like this, OK? So the y-axis is the amount of variation and the x-axis here is the residue number along this polypeptide sequence. And what you see, probably even without the color here, is that there are these three regions where there's a lot of variation in the sequence of different antibodies, OK? So here you see the blue segment here has a lot of variation, the yellow segment has a lot of variation, and the reds segment here has possibly the most variation. And what these are known as are hypervariable regions, meaning that they exhibit a lot of variation. Another name for them is that they are complementarity-determining regions. Complementarity. Complementarity-determining. Determining regions, or CDRs, and there are three of them, 1, 2, and 3, OK? So there are regions in this antibody molecule which are much more variable than others, OK? So what are these regions? Well, this is a sort of crystal structure of the-- of an antibody, and you can see how the antigen is bound at the end. That would be this end of the molecule or this end of the molecule. And here you see a ribbon diagram of the structure of the antibody, and the complementary complementarity determining regions are the regions here that contact the antigen. And what they are are basically here's an IG fold, this whole thing, and there are these three loops that extend out of the end of this molecule, and you can think of them as three fingers, OK? Then these fingers are able to reach out and sort of grab on to like a foreign particle and/or any particle and stick to it, OK? So these are the variable regions, and they have differences in amino acids-- in amino acid sequence, and even very small differences in the amino acid sequence at this particular part of the antibody can have a huge effect on whether or not they're able to stick to something, right? You can imagine if I lost my thumb, then right now, I'm not able to sort of stick to that anymore, OK? So small differences in amino acid sequence result in large changes in the affinity of this antibody for an antigen. And antibodies have different sequences, meaning that they're able to bind to specific substances differently. So if an antibody has one set of sequence, it might recognize one structure. If it has another sequence, it might recognize another structure. So just by changing the sequence at these complementarity-determining regions has a huge influence on what these proteins will bind to, OK? Now, each B cell expresses a unique antibody and just one unique antibody. So each B cell in our body expresses one and only one antibody protein, and that antibody protein has a unique sequence at the CDR region, and this one antibody has unique specificity for an antigen, OK? So here you can see in my diagram, I have a whole bunch of B cells here. They all express a different antibody, and you can see that the way you could get more of a given antibody is to clonally expand one of these cells, and all of the cells that result from that colonial expansion will express the exact same antibody. And when you have a clonal population of a cell that all has the same antibody, that's known as monoclonal, OK? So each B cell will have a B-cell receptor or an antibody with unique specificity. So now the question becomes, OK, so I told you how you get specificity, but in order to have a functioning immune system, you need to have lots of different cells that each express a different cell receptor, so there needs to be a way to generate diversity. And the answer to how we generate diversity has an MIT connection. The research wasn't done at MIT, but the person who discovered the mechanism is now at MIT. This research was performed by Susumu Tonegawa, and Professor Tonegawa, for his work on how this diversity is generated, was awarded the Nobel Prize in medicine in 1987. OK, so Professor Tonegawa did this research elsewhere, but now he is a faculty member here at MIT. All right, so diversity. The problem of diversity, right? We have millions of B cells that have a unique antibody. OK, so one solution to this problem would be we have a million different antibody genes, and each B cell clone sort of expresses one of them OK how many genes do we have? Anyone know, roughly, on the order of magnitude? Do we have a million? What's that? AUDIENCE: 30,000? ADAM MARTIN: Exactly. Yeah, so Mr. George has suggested-- Miles, I believe. Yeah, OK, good. Miles suggested 30,000, which is the good upper limit, right? So having a million antibody genes sounds a little bit unfeasible, OK? And so it's basically unfeasible for us to express as many antibody genes or have as many antibody genes as we have antibodies. We just don't have enough real estate in our genome, OK? But there's another solution to generate the diversity, which is essentially a form of shuffling. So we have a single heavy chain gene for antibodies, and we have two genes for the light chain, but these genes are composed of multiple gene segments. There are multiple gene segments. Specifically, the segments that make up-- that generate this variable domain is composed of multiple gene segments, and these gene-shaped segments are shuffled during the development of the B cell to give rise to different proteins. OK, so these gene segments are shuffled to generate this diversity. OK, so now I'm showing you on the top here, this is the human immunoglobulin heavy chain locus here. You can see it's pretty big. There are lots of components. I want you to focus on this. So there is-- you see in orange, there's this variable gene segment, and there are 45 variable gene segments here. There's this diversity, or D segment here, which there are 23 of, and then there are six of these joining or J segments. OK, so these are all distinct parts of the gene. They're all distinct parts of the exon that encodes this variable region of the antibody, OK? So you have multiple V, D, and J gene segments. And in order to generate a functional antibody, one V has to be brought together with one D, which has to be brought together with one J for that heavy chain, OK? So you have multiple V-D gene segments, and they have to be brought together to form a functional antibody. OK, that's illustrated right here. So here you see this is the light chain. For the light chain, there are only V and J gene segments. V For the heavy chain, there there's V, D, and J. And so most of the cells in our body and the cells of our germline, at the very earliest stages of development, all have this arrangement, where you have everything still intact. But during lymphocytes development, specifically in lymphocytes, there is a recombination event that brings together V and J segments or V, D, and J segments, OK? So this is mediated by recombination at the heavy and light chain genes for that antibody, OK? And so this is very different from the recombination we talked about earlier in the semester, where recombination is happening during meiosis and the formation of the gametes, right? In that case, recombination is happening between homologous chromosomes. Here we're not talking about recombination between homologous chromosomes. We're talking about recombination that brings together and deletes segments along a single chromosome to bring these V and J segments together, OK? So this is sort of a intra-chromosomal recombination, which deletes the intervening sequences and brings these gene segments together to form a functional antibody protein. So this process is known as V(D)J recombination, and this is lymphocyte specific. OK, and that's because during the development of B and T cells, there is an induction of recombinases that mediate this recombination. So in this case, there is recombination, which is mediated by recombination-activating genes 1 and 2, called RAG1 and 2, OK? So there are these are lymphocyte-specific recombinases which mediate this rearrangement, which bring together a unique V, D, and J segments together, OK? So the diversity comes from the fact that each of these V, D and J segments, each V segment-- you could-- this also applies to D segments and also J segments-- has a unique sequence. So it encodes for a unique amino acid sequence, meaning that if you bring together different combinations of Vs, Ds, and Js, you get a distinct protein, OK? Now even if you had all of the combinations of V, Ds, and Js, you still don't have the diversity that we see in the human body. So there is another process that further generates diversity, which is the fact that when these segments are getting shuffled, it's imprecise in that nucleotides can be inserted or deleted as these segments are joined, which generates greater amino acid diversity, and this is called-- it's called junctional imprecision. So this recombination is not precise, but it leads to the insertion or deletion of nucleotides of nucleotides. And if there's a multiple of 3 nucleotides either inserted or deleted, then you get a functional antibody. Why is it that it has to be a multiple of 3? Jeremy? AUDIENCE: Otherwise, you end up with a frameshift mutation. ADAM MARTIN: Exactly. Right? This is and the-- this is on the more sort of like on the N-terminus side of the gene, right? So if you inserted one nucleotide between V and J, then the downstream portion of the gene, the downstream part of the open reading frame would be out of frame and wouldn't generate a functional protein. OK, so it has to be a multiple of 3. Yeah, Georgia? AUDIENCE: How is functional precision lymphocyte-specific? Or is it not? ADAM MARTIN: It's just the RAG1 and RAG2 are turned on specifically in the lymphocytes as they mature. AUDIENCE: And that also affects the insertion, deletion? ADAM MARTIN: Well, if you don't have recombination, you can't get junctional precision, right? So the junctional imprecision-- or junctional imprecision. The junctional imprecision is a consequence of the recombination process itself, right? So if you're not having recombination, you're not having any junctional imprecision because you're not generating a junction. OK, now there's one more thing that's important here, which is something that happens not as a consequence of this recombination process but as a consequence of activating the T cell response, which is that in addition to these variations, there's also something known as somatic mutation. So there's an elevated mutation rate at the IG locus that further increases the diversity of the amino acid sequence at these variable regions of the antibody, OK? Another way this is referred to is because it can increase the affinity of the antibody for a antigen, it's also known as affinity maturation, so these are synonymous. Maturation. Maturation. OK, so-- and this depends on the T cell, the cell-mediated branch of adapted immunity, so this is T-cell mediated. So one other aspect of this process that I want to talk about is until this recombination happens, the immunoglobulin gene is not expressed, so it's this recombination that leads to the expression of the-- either the heavy chain or the light chain gene, OK? And that's because the enhancer is sort of downstream in the gene, and by deleting the intervening sequence here, you bring the promoter in range of the enhancer, and now this gene is expressed, OK? But remember you have two copies of each of these genes. You have a parental copy and a maternal copy, and another feature of this system is that there is what is known as allelic exclusion. So the system is such that a B cell expresses only one antibody, and so if you had both alleles expressing, that wouldn't be the case, OK? So allelic exclusion makes it that if you get a recombination event that leads to a functional antibody for one of your sort of inherited copies of the gene, one of your alleles, it suppresses recombination on the other one, OK? So you will only get one of these genes, one heavy chain and one light chain, expressed per B cell. OK, so only one gene expressed so that each B cell only has one antibody. OK, I just wanted to point out, finally, that these junctions between V-D and J segments fall right in this CDR-3 region, so they're responsible for the high level of variability at the CDR or hypervariable 3 region. OK, and because of the allelic exclusion, each B cell expresses only one antibody, OK? So all of the antibody proteins expressed by that cell will be exactly the same. OK, so now the last property of the immune system we need to talk about is memory. And so the immune system needs to be able to recall past infectious agents that it's experienced, and so it needs-- I guess we're kind of personifying here, but it needs some sort of memory, right? It needs the ability to recall this, OK? And this is the principle behind vaccination, right? The way vaccines work is to put in one of these attenuated or inactivated foreign agents, such that your body is able to remember that later on when you get the real deal, and it's able to fight it off, OK? So the body has to be able to remember. And several ways in which this manifests itself, if we compare a primary infection, the first time you've seen an infectious agent, versus a secondary infection, they have very different responses from the standpoint of the adaptive immune system, OK? So if we consider the lag before your adaptive immune system really takes off, the primary response takes about five to 10 days, so it's a bit delayed, whereas the secondary response can be one to three days, OK? So it's faster. It's able to react faster when you see an infectious agent the second time. If we also just consider the magnitude of the response by considering how much antibody, the antibody concentration that's like put into your system, then the primary response is smaller and the magnitude of the secondary response is larger. So you basically-- your body's able to produce more antibody against an infectious agent the second time it sees it. Not only is the antibody amount better the second time, but actually the antibodies themselves are better antibodies, OK? And we can show that by thinking about antibody affinity, which is how tightly the antibody recognizes the antigen, and I'll give you numbers that represent the dissociation constant for an antibody to a given antigen. So the lower that number is, the tighter the binding. So for the primary infection, the antibody affinity is weaker on the order of 10 to the negative 7th molar in terms of KD, and this secondary infection generates antibodies that are functionally quite better. They bind much tighter. It can be less than 10 to the negative 11th molar, which is sub-nanomolar. Right? That's a really tight interaction between two molecules. So the antibodies, you get more of them, and they're better antibodies, OK? So what makes this memorable is that when-- what lasts in your body from the first time you see the agent to the next is there's a type of B cell known as a memory B cell, and this memory B cell will express a given antibody, and that antibody will be specific to the substance you saw previously. And because recombination is-- this recombination is irreversible, then that B cell is going to remember that antibody because it's still encoded in the genome. So the memory results from V(D)J recombination being irreversible and the fact that these memory B cells stay in your body, even if the antigen is not present, so these also stay in the body. OK, so effective vaccines generate these types of cells, these memory B cells. OK, that's important if you want an effective vaccine, that you have these B cells that retain information about the past infection. All right, so what exactly is it that the antibodies do? So I'll talk about effector functions of antibodies. So antibodies can bind to a foreign substance and interfere with the normal function, right? If you have a bacteria and maybe the antibody binds to some part of the bacteria to interfere with that bacteria getting into the cell, and this type of effect is known as neutralization. If you had an antibody that bound to something like a bacteria, you could also have it recruit phagocytic cells to internalize that bacteria, and so you could also induce phagocytosis. In addition, antibodies, when bound to a foreign substance, if that foreign substance is a cell, then it could recruit a killing cell to kill that cell, so there's also a killing aspect to this, OK? So what's in this diagram here is a type of cell known as a natural killer cell that is killing its target cell, and so you can kind of think of this cell as the Terminator, OK? So right, if the natural killer cell recognizes this target here, then it's hasta la vista, baby, and that cell is dead, OK? I just want to point out one thing that I mentioned before, which is that antibodies can be leveraged to generate treatments for certain types of diseases. And we talked about a drug called Herceptin-- or not a drug but a-- it's an antibody, but it's a treatment for HER2 positive breast cancer, so this is used to treat HER2-positive breast cancer. And it's really been a nice success story in the cancer field because what this-- what Herceptin is-- it was derived from a mouse antibody, so this is a mouse monoclonal antibody that recognizes this HER2 growth factor receptor, which is over expressed on 30% of human breast cancers. And what Herceptin is is that researchers took this mouse antibody and engineered a human antibody to have the mouse sequence at its complementarity-determining regions, such that you have a human antibody that won't be sort of removed by the human immune system but will recognize HER2 and recruit human immune cells to HER2 positive cells, possibly killing those cells or binding to HER2 and somehow neutralizing the activity of HER2 on these cancer cells. So antibodies can be very useful for therapeutics, as well as being useful in our own bodies to mediate immunity. OK, we'll talk about T cells on Friday. Remember to bring your projects.
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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: So in this final lecture, I want to think a bit sort of with an eye towards thinking about how one might actually make measurements that prove the nature of the black hole spacetime that was discussed the previous lecture. I'm going to discuss motion in a black hole spacetime. We touched on this a little bit in the previous lecture, where we discussed the motion of radial light rays, OK? We, in fact, used radial light rays as a critical tool for describing the properties of the spacetime. We use that to help us understand the location of events horizons. But I want to think a little bit more generally. What might it look like if I have material orbiting in the vicinity of one of these black holes? What if it's not light? What if it's made out of matter? And so what this is going to boil down to is understanding the behavior of geodesics in a black hole spacetime. And the naive approach to doing this is not wrong, but naive. What you do is you would just take the spacetime-- take your Schwarzschild or take your Kerr spacetime-- and turn a very large crank, grind out all of the connection coefficients, evaluate the geodesic equation, integrate it up. Solve for the geodesics. Boom, you got yourself your motion, OK? And that is absolutely correct. You can do that using your Kerr space time or your Schwarzschild spacetime. In fact, if you do that for-- you can get the connection coefficients describing this. Those are relatively easy to work out. I think they are listed in Carroll, in equation 5.53, according to my notes. That may just be for Schwarzschild. But at any rate, they're all listed there, and have a blast. This approach is not wrong, but-- my notes say, but it is not useful. That's not really true. I'll just say that there is a more useful approach to this. A more fruitful approach is to exploit the fact that these are highly symmetric spacetimes. Exploit the symmetries and the Killing vectors, and see how they can be used to reduce the number of degrees of freedom that you need to describe. In this lecture, I'm going to go through this in quite a bit of detail for Schwarzschild. The concepts that I'm going to apply work for Kerr and for Kerr-Newman as well, OK? Schwarzschild is just a little easier to work with. It's something that I can fit into a single lecture. In particular, one of the nice things about Schwarzschild-- so let's go ahead and write down that spacetime. So one of the nice things about Schwarzschild is it is spherically symmetric. This means I can always rotate coordinates such that-- well, when me think about it-- let's just back up for a second. Imagine that I have some kind of a body orbiting a Schwarzschild black hole. Spherical symmetry tells me that there must be some notion of a conserved angular momentum such that that orbit always lies within a given plane. Put it another way. Because it is fairly symmetric, there cannot exist a torque. The black hole cannot exert a torque that changes the orientation of the orbital plane. In particular, because it is fairly symmetric, there is no unique notion of an equator to this object. And so you might as well define any orbits to live in the theta equals pi over 2 plane. You can always rotate your coordinates to put any orbit in the theta equals pi over 2 plane. It's actually a pretty simple exercise. I won't do this, but if you start with an orbit that is in the theta equals pi over 2 plane, and it is moving such that its initial velocity would keep it in a theta equals pi over 2 plane, it's a very simple exercise using the geodesic equation to show that it will always be in the theta equals pi over 2 plane. So spherical symmetry says, you know what, let's just forget about the theta degree of freedom. I can always define my coordinates in such a way that it lives in the theta equals pi over 2 plane. Boom, I have reduced my motion from, in general, being three spatial dimensions to two spatial dimensions. That's another one of the reasons why, for pedagogical purposes, it's nice to start with Schwarzschild. For Kerr, this is not the case, OK? Kerr is a little bit more complicated. You have to treat the theta motion separately. It's not strictly symmetric, and so you can't do that. I have actually spent a tremendous amount of my career studying the orbits of objects around Kerr black holes. And I do have to say that the additional complications that arise from this lack of sphericity, they're really beautiful, OK? There's an amazing amount of fun stuff you can do with it. You know, there's a reason why I just keep coming back to this research problem, and part of it is it's just bloody fun. But if you're teaching this stuff for the first time, it's not where you want to begin. All right, so Schwarzschild allows us to reduce it from a three-dimensional problem to a two-dimensional problem. And Schwarzschild also has two Killing vectors. The time derivative of every metric component is equal to 0. That means p downstairs t is constant. So there is a timelike Killing vector. So p downstairs t is constant every level along the orbit. It means there exists a timelike Killing vector. And so what we do is we associate this with the energy of the orbit. We call this up to a minus sign. And we choose that minus sign because if we imagine orbits have very, very large radius, the spacetime is nearly flat, and we want to sort of clear out the minus sign associated with lowering our index here. We're going to call that constant negative of the energy of the orbit. The spacetime is also independent of the angle phi. And so p sub phi is a constant. This means that the spacetime has an axial Killing vector, something associated with motions around a symmetry axis. So I'm going to call this L-- well, actually, I will tend to call it L sub z. You can kind of think of this as, after I have put everything in the theta equals pi over 2 plane, this is like an angular momentum on the-- angular momentum parallel to the axis normal to that plane, and I call that the z-axis. It's worth noting that all three of these things are also true for Reissner-Nordstrom black holes. This one is not true for Kerr, but these two are, OK? So although the details change, many of the concepts will carry over when you look at different, more complicated classes of black holes. Let's now think about the forward momentum of a body moving in a Schwarzschild spacetime. So let's say it's got a rest mass m, then it will have three components. Remember, I have put this thing in a plane where there is no theta motion. It will have three theta motions describing its motion with respect to the time coordinate, the radial coordinate, and the actual coordinate phi, OK? So before I do anything with this, let's take advantage of the fact that we have these quantities that are constants. So using the fact that p downstairs mu is what I get when I hit this guy-- that be a mu, pardon me. This is what I get when I hit this guy with the metric. I can write out a p sub t and p sub phi. So let's do that on another board. Well, let's do a p sub t first. So p sub t is going to be gtt, p sub-- whoops. OK, this is minus. There's an m from that-- minus sign from my metric. And this whole thing I define as the negative energy, e. OK, so that combination-- we'll do something at the rest mass in just a moment. But you should basically look at this and saying that dt d tau, which tells me about how the body moves with respect to the Schwarzschild time coordinate-- that complement of the four velocity times 1 minus 2gm over r is a constant. Let's look at p downstairs phi. So this is m r squared sine squared theta. Ah, we've chosen our orbital plane, 1 times d phi d tau. This is equal to [INAUDIBLE] momentum L sub z. In my notes, I might flip around a little bit between L sub z and L. So let's massage these a little bit more. So I can take these two expressions and I can use them to write dt d tau and d phi d tau in terms of these conserved quantities. So dt d tau is equal to-- I'm going to call it e hat divided by 1 minus 2gm over r. d phi d tau is going to be Lz hat over our squared. And so these quantities with a hat are just the conserved values normalized to the rest mass, OK? They all are proportional to the rest mass, but it's actually if you want to know what the-- this is essentially telling me about how clocks on the orbit tick relative to clocks that are infinitely far away. And that can't depend on the mass of the orbiting object. This is telling me about how this small body is moving according to the clock of the orbiting observer. And again, that can't depend on the mass of the object. OK, so that's kind of cool. So I've now managed to relate three-- excuse me, two of the three components of the forward momentum to functions of r and quantities that are known to be constant. That's good. We have one more overall constraint. We know that if I take p dot p, I get negative of the rest mass squared. So this, when I write it out-- OK, notice every single term is proportional to m squared. So I can divide that out. I can insert dt d tau. I can replace for this e divided by this guy. I can replace d phi d tau by my Lz hat divided by r squared. Doing so, manipulating a little bit, I get-- we get something like this. And let me rearrange this a tiny bit. All right, I have managed to reduce this to a one-dimensional problem, OK? So going from that line over to there, basically all I did was insert the relationship between dt d tau and E, d phi d tau and L, cleared out some overall factors of things like 1 minus 2GM/r, manipulate, manipulate, manipulate, and what you finally get is this lovely equation here that tells you how the radial velocity, the radial velocity with respect to proper time, how it depends as a function of r given the energy and the angular momentum. I have written it in this form because the problem is very strikingly reminiscent to the Newtonian problem of understanding the motion of a particle in a 1/r potential, which we often describe as having an effective potential that has a gravitational term-- Newton's gravity, or if you're doing things like quantum mechanics, the Coulomb potential, and a Coulomb barrier associated with the angular momentum of an orbit. So one way to approach what we've got now, one thing that we could do, is essentially just pick your energy and your Lz-- pick your energy and your angular momentum-- pick an initial position-- let's imagine you synchronize the clocks at t equals 0-- and then just integrate. You've got your dr d tau given here. Don't forget, you also have d phi d tau and dt d tau related to the energy and your angular momentum, like so. Boom. It's a closed system. You can always do just sort of a little numerical integration of this. In a certain sense, this completely specifies the problem. But there's so much more we can do. In particular, what we see is that all of the interesting behavior associated with this orbit is bound up in this function V effective. So something that's really useful for us to do is to take a look at what this V effective looks like. So suppose you are given a particular value for E hat and Lz hat, and you plot V effective versus r. Well, what you typically find is that it's got a behavior that looks kind of like this, where this value right here is V effective equal to 1. Notice as r goes to infinity, you get 1 times 1. So this asymptotes to 1. Notice E hat squared has the same dimensions as V effective. In fact, in the unit choices I've used here, they are both dimensionless. So what we can do is plot-- let's imagine that we have-- you know what? I'm going to want to sketch this on a different board. Let me go over here. So I'm going to want to look at a couple of different values of V effective-- excuse me, a couple of different values of E hat. OK. Here's an example. This guy is asymptoting at 1. So since E hat, as I said, has the same units as V effective, let's plot them together. So example one-- imagine if E hat lies right here. OK. So let's call this E hat 1. So this is some value that is greater than 1. What is the point of doing this? Well, notice-- I'm going to flip back and forth between these two middle boards here. Let's look at the equation that governs the radial motion of this body. dr d tau has to be a real number. This has to be a real number. So we have to have E hat squared greater than or equal to V effective in order for dr d tau to have a meaningful solution. So let's look at my example here, E hat 1. E hat 1 is greater than my effective potential everywhere at all radii until I get down to here. Let's call that r1. So in this case, dr d tau, if I think about this thing-- so note that defines dr d tau squared. Let's suppose we take the negative square root. dr d tau is positive and inward-- or, well, it's negative-- negative and real, negative and real, negative and real, negative and real, negative and real-- boom. It's 0. Can it go into here? No. It cannot go into there, because there E hat squared is less-- sorry. That should have been squared. Inside here, E hat squared is less than the effective potential. dr d tau is imaginary. That doesn't make any sense. The only option is for this guy to change sign and trundle right back out. So E hat greater than 1 corresponds to a body that comes in from large radius, turns around at particular radius where E hat is the square root of the effective potential, and then goes back out to infinity or back out to-- let's not say infinity-- goes back out to large radius. OK? In Newtonian gravity, we would call this a hyperbolic orbit. This corresponds to-- so remember, when I did this I have not-- I'm not actually-- I'm only computing the radial motion. I'm not looking at the phi motion. So this actually, when you look at both the radial motion and the phi motion, what you see is that this is a body that comes in and then sort of whips around that small radius and goes right back out. Let's look at another example. Let's call this E2-- some value that is less than 1. OK. Well, for E hat 2, it's a potential. The potential is underneath E hat 2 squared only between these two radii, which I will call r sub p and r sub a. What we expect in this case is motion of this body essentially going back and forth and turning around at periastron and apoastron. This is a relativistic generalization of an elliptical orbit. In general relativity, they turn out generally not to be ellipses. So we call this an eccentric orbit. In the weak field limit, if you imagine r being very, very large, it's not hard to show that the motion is nearly an ellipse, but it's an ellipse whose long axis is slowly precessing. This actually leads to the famous perihelion precession of Mercury that Einstein first calculated. And this is an exercise that I am asking you to do on one of the final P-sets of this course. Using what I have set up here, it's really not that difficult to do. Let me go to another board. And note that one could imagine an energy such that dr d tau is exactly 0. So if you choose your energy so that you sit right here at the minimum of the potential-- I will label this as point s-- the energy that corresponds to exactly that point is what would be a-- that is, there is a single point at which dr d tau equals 0. Anywhere away from that, dr d tau would be imaginary, so that's not going to work. But right at that point, dr d tau equals 0, and you get a circular orbit. Notice there's a second point at which that can happen. Let's call this point u. Perhaps you can guess why I called these points s and u. If you imagine that you add a tiny amount of energy right here, well, what it will do is it will execute small oscillations around the point s. It will sort of move it up to something that's similar to what I drew up there as E2, but with a very small eccentricity. So if I slightly disturb a circular orbit down here at s, I essentially just oscillate in the vicinity of that circular orbit. S stands for stable. If I have an orbit up here at u and I very slightly perturb it, well, it'll either go in and eventually reach r equals 0, or it'll go out, and then it's completely unbound, and it will just keep trundling all the way out essentially forever. This guy is unstable. Stable orbits are particularly interesting and important. So let's look at these orbits with a little bit more care. So the very definition of a circular orbit is that dr d tau equals 0. Its radius does not change. If dr d tau equals 0, then it must have E hat equal the square root of the effective. Both of these orbits happen to live at either a minimum or a maximum of this potential curve. So I'm going to require that the partial derivative of that potential with respect r be equal to 0. So let's take a look at this condition. My effective potential is given up here. Do a little bit of algebra with this, set this guy equal to 0. What you'll find after your algebraic smoke clears is that you get this condition on the angular momentum. Notice as r gets really large-- oh, shoot. Try it again. Notice as r gets really large that this asymptotes to plus or minus the square root of GM r. That is indeed exactly what you get for the angular momentum of a circular Newtonian orbit. OK. So it's a nice sanity check. It appears to be somewhat pathological as r approaches 3GM, though. Hold that thought. OK. So now let's take that value of L, plug it back into the potential, and set E equal to the square root of that. A little bit of algebra ensues. And what you find is that this equals 1 minus 2GM/r over, again, that factor under a square root of 1 minus 3GM/r. Again, we sort of see something a little bit pathological happening as r goes to 3GM. Let me make two comments about this. So first of all, notice that this energy is smaller than 1. I can intuitively-- you can sort of imagine-- remember, E hat is the energy per unit rest mass. You can think of this as something like total energy over M-- so the energy associated with the orbiting body. It's got a rest energy, a kinetic energy, and a potential energy divided by m. For an orbit to be bound, the potential energy, which is negative, must have larger magnitude than the kinetic energy. So for a bound orbit, E kinetic plus E potential will be a negative quantity. So the numerator is going to be something that, when normalized to m, is less than 1. So this is exactly what we expect to describe a bound orbit. Notice, also-- so if you take this formula and look at it in the large r limit, it goes to 1 minus GM over 2r. This is in fact exactly what you get when you look at the energy per unit mass throwing in-- sort of by hand-- a rest mass. The minus GM/2r exactly corresponds to kinetic plus potential for a Newtonian circular orbit. So lots of stuff is hanging together nicely. So we've just learned that we can characterize the energy and angular momentum of circular orbits around my black hole. Let's look at a couple other things associated with this. So these plots where I look at the radial motion, this effective potential, as I mentioned a few moments ago, there's additional sort of degrees of freedom in the motion that are being suppressed here. So this thing is also moving in that plane. It's whirling around with respect to phi. We've lost that information in the way we've drawn this here. Let's define omega to be the angular velocity of this orbit as seen by a distant observer. OK. Why am I doing it as seen by a distant observer? Well, when things orbit, there tend to be periodicities that imprint themselves on observables. It could be the period associated with the gravitational wave that arises out of this. It could be the period associated with peaks and a light curve if this is a star orbiting around a black hole. It could be oscillations in the X-ray flux if this is some kind of a lump in an accretion disk of material orbiting a black hole. So if this is the angular velocity seen by distant observers-- remember, the time that distant observers use to-- the time in which the distant observers' clocks run is the Schwarzschild time t. So this will be d phi dt, which I can write as d phi d tau-- this is the angular velocity according to the orbit itself-- normalized to dt d tau. Now, these are both quantities that are simply related to constants of motion. What I've got in the numerator here is L hat over r squared. And what I've got in the denominator here is E hat over 1 minus-- I dropped my t. It would happen at some point-- 1 minus 2GM/r. So let's go ahead and take our solution here. My E hat is 1 minus 2GM/r divided by square root of blah, blah, blah. The 1 minus 2GM/r cancels. My L hat is square root GM r divided by, again, that square root 1 minus 3GM/r. Notice, the square root 1 minus 3GM/r factors all cancel. So this becomes plus or minus 1 over r squared square root GM r. Looks like this. Plus and minus basically just correspond to whether this motion sort of is going in the same sense as your phi coordinate or in the opposite sense. There's really no physics in that. It just comes along for the ride that both behave the same. If you guys get interested in this, and you do a similar calculation around a Kerr black hole, you'll find that your prograde solution gives you a different frequency than your retrograde solution because of the fact that the dragging of inertial frames due to the spin of the black hole kind of breaks that symmetry. Something which is interesting and-- well, I'll make a comment about this in just a second, which is kind of interesting. And here's what I'll say-- sometimes people think this is more profound than it should be-- is this is, in fact, exactly the same frequency law that you get using Newtonian gravity. This is actually exactly the same as Kepler's law. That seems really, really cool. And it is. It's actually really useful. It makes it very easy to remember this. But don't read too much into it. More than anything, it is a statement about a particular quality of this radial coordinate. So remember, in Newtonian gravity r tells me the distance between-- if I have an orbit at r1 and an orbit at r2, then I know that the distance between them is r2 minus r1. In the Schwarzschild spacetime, the distance between these two orbits is not r2 minus r1. However, r2 labels a sphere of surface area 4 pi r2 squared. And r1 labels a sphere of surface area 4 pi r1 squared. It's easy to also show that the circumference of the orbit at r2 is 2 pi r2, the circumference of the orbit at r1 is 2 pi r1. That, more than anything, is why we end up reproducing Kepler's law here, is that this areal coordinate is nicely amenable to this interpretation. So is this orbit-- so I described over here an orbit that is unstable and an orbit that is stable. I have described how to compute-- if I wanted to find a circular orbit at a given radius, those formulas that I derived over there on the right-most blackboards, they tell me what the energy and the angular momentum need to be as a function of r. Is that orbit stable? Well, if it is, I can check that by computing the second derivative of my effective potential. So my orbits are stable if the second derivative with respect to r is greater than 0, unstable if this turns out to be negative. Let's look at the crossover point from one to the other. What if there is a radius where, in fact, the stable and the unstable orbits coincide? In fact, what one finds, if you look at the effective potential-- you imagine just sort of playing with L sub z. So let's say we take that L sub z, and we just explore it for lots of different radii of the orbits. You find that as the orbits radius gets smaller and smaller, your stable orbit tends to go up, and this minimum sort of becomes flatter, and your unstable orbit just kind of comes down. They sort of approach one another. There is a point just when they coincide-- this should have an "effective" on it. My apologies. Right when they coincide, this defines what we call the marginally stable orbit. I may have put this one on a problem set. But it might be one of the ones I decided to drop. So I'm just going to go ahead and do the analysis. When you compute this, bearing in mind that your angular momentum is a constant-- so take this, substitute in now your solution for L sub z, which I've written down over there-- what you find is that the marginally stable orbit-- let's call it r sub ms-- it is located at a radius of 6GM. This is a profoundly new behavior that doesn't even come close to existing in Newtonian spacetime-- spacetime-- doesn't come close to existing in Newtonian gravity, excuse me. [SIGHS] I'm getting tired. The message I want you to understand is that, what this tells us is that no stable circular orbits exist inside r equals 6GM. So this is very, very different behavior. If I have-- let's just say I have a very compact body but Newtonian gravity rules. I can make circular orbits around it, basically go all the way down until they essentially touch the surface of that body. And you might think based on this that you would want to make orbits that basically go all the way down, that sort of kiss the edge of r equals 2GM. Well, this is telling you you can't do that. When you start trying to make orbits that go inside-- at least, circular orbits-- that go inside 6GM, they're not stable. If someone sneezes on them, they either are sort of blown out to infinity, or they fall into the event horizon. And in fact, one of the consequences of this is that, in astrophysical systems, we generically expect there to be kind of-- if you imagine material falling into a black hole, imagine that there's like a star or something that's just dumping gas into orbit around a black hole, well, it will tend to form a disk that orbits around this thing. And the elements of the disk are always rubbing against each other and radiating. That makes them get hot. They lose energy because of this radiation. And so they will very slowly sort of fall in. But they make this kind of-- it's thought that in most cases they'll make this kind of thick disk that fills much of the spacetime surrounding the black hole. But there will be a hole in the center. Not just because the thing is a black hole. I don't mean that kind of a hole. There'll actually be something surrounding the black hole's event horizon, because there are no stable circular orbits. Once the material comes in and hits this particular radius, 6GM in the Schwarzschild case, it very rapidly falls in, reduces the density of that material tremendously. And you get this much thinner region of the disk where essentially things fall in practically instantly. It should be noted that this 6GM is, of course, only for Schwarzschild. If you talk about Kerr, you actually have two different radii corresponding to material that goes around parallel to the black hole's spin and material that goes anti-parallel to the black hole's spin. And it complicates things somewhat. There's two different radii there. The one that goes parallel tends to get a little bit closer. The one that's anti-parallel tends to go out a little bit farther. But the general prediction that there's an innermost orbit beyond which stable orbits do not exist, that's robust and holds across the domain of black holes. So let me conclude by talking about one final category of orbits-- photon orbits. So let's recall that when we were talking about null geodesics, we parametrized them in such a way that d-- sort of the tangent to the world line has an affine parameter attached to it, such that we can write p equals dx d lambda. These guys are null. So p dot p equals 0. For the case of orbiting bodies, that was equal to minus mass squared. Mass is 0 here. So we're going to follow a very similar procedure. The spacetime is still time independent, so there is still a notion of a conserved energy. It is still actually symmetric, so there is still a notion of axial angular momentum. But because p dot p is 0 now, rather than minus m squared, when we go through the exercise of-- that sort of parallels what we did for our massive particle, we're going to derive a different potential. So let's go ahead and evaluate this guy again. And I get 0 equals minus 1 minus 2GM/r dt d lambda squared plus 1 minus 2GM/r dr d lambda squared. I'm still going to require the thing to be in the theta equals pi/2 plane so that we know theta term. And I have set theta to pi/2. So my sine squared theta is just one. So I get this. I'll remind you that I can relate-- so the relationship between the time-light component of momentum and energy, the axial component of momentum, and the angular momentum, it's exactly the same as before. There's no rest mass appearing. And so now I find-- so my energy, I don't put a hat on it. It's not energy bringing it mass, because there is no mass. That looks like so. And my L looks like so. Using them, I can now derive an equation governing the radial motion of my light ray. And sparing you the line or two of algebra, it looks like this. Pardon me just one moment. OK. OK. So kind of similar to what we had before, if you look at it you'll see the term involving the angular momentum is a little bit different. As you stare at this for a moment, something should be disturbing you. Notice that the equation of motion appears to depend on the energy. OK. That should really bug you. Why should gamma rays and infrared radiation follow different trajectories? As long as I am truly in a geometric optics limit, I shouldn't. Now, it is true that if you consider very long wavelength radiation, you might need to worry about things. You might need to solve the wave equation in the spacetime. But as long as the wave nature of this radiation is-- if the wavelength is small enough that it's negligible compared to 2GM-- I shouldn't care what the energy is. Energy should not be influencing this. So what's going on is I need to reparametrize this a little bit. What I'm going to do to wash away my dependence on-- wash away my apparent dependence on the energy here, is I'm going to redefine my affine parameter. So let's take lambda-- so L divided by lambda. And I am going to define b to be L divided by E. This is the quantity that I will call the impact parameter. And I'll describe why I am calling it that in just a few minutes. So I'm going to take this entire equation, divide both sides by L squared. And I get something that looks like this. What I'm going to do is say that the impact parameter, b, is what I can control. It's the parameter that-- all I can control that defines the photon that I am studying here. And everything after this, this is my photon potential. Notice that the photon potential has no free parameters in it. If I plot this guy as a function of r, it kind it just looks like this. Two aspects of it are worth calling out. This peak occurs at r equals 3GM. Remember the way in which our energy-- oh, still have it on the board here-- things like the energy per unit mass and the angular momentum per unit mass all blew up when r equal 3GM. 3GM is showing up now when I look at the motion of radiation, look at massless-- radiation corresponding to a massless particle, so to speak. If I look at the energy per unit mass, and the mass goes to 0, I get infinity. So the fact that I was actually seeing sort of things blowing up as r goes to 3GM was kind of like the equations hinting to me in advance that there was a hidden solution corresponding to radiation that could be regarded as a particular limit that they were sort of struggling to communicate to me. The other thing which I want to note is that the potential, the height of the potential right here, it has a peak value of 1 over 27 G squared M squared. Hold that thought for just a moment. Actually, let me write it in a slightly different way. This equals 1 over 3 root 3 GM squared. So now, to wrap this up I need to tell you what is really meant by this impact parameter. So go back to freshman mechanics. And impact parameter there is-- like, let's say I have a problem where I am looking at an object that is on an infall trajectory. There is a momentum p that is describing this. And it's moving in such a way-- it's not moving on a real trajectory, right? It's actually moving in a sense that is a little bit off of true to the radial direction. This guy actually has an angular momentum that is given by-- let's say that this is equal to px x hat. Let's say this is b y hat. This guy's got an angular momentum of b cross p. That's the impact parameter associated with this thing. It sort of tells me about the offset of this momentum from being directly radial towards the center of this object. Well, I'm going to use a similar notion to give me a geometric sense of what this impact parameter here means. Suppose here is my black hole. It's a little circle of radius 2GM. And I'm sitting way the heck out here, safely far away from this thing. And what I'm going to do is shoot light-- not quite radial. What I'm going to do is I'm going to have sort of an array of laser beams that kind of come up here-- an array of laser beams. And I'm going to shoot them towards this black hole. And what I'm going to do is I'm going to offset the laser beams by a distance b. And I'm going to fire it straight towards this thing. Go through sort of a careful definition of what angular momentum means, and you'll see that that definition of impact parameter gives you a very nice sense of the energy associated with the trajectory of this beam and a notion of angular momentum associated with this. So let's flip back and forth between a couple of different boards here. There are three cases that are interesting. Suppose b is small. In particular, suppose I have b less than 3 root 3 GM. So I start over here. Let's take the limit. What if b equals 0? Well, if b equals 0, this guy just goes, zoom, straight into the black hole. As long as it is anything less than 3 root 3 GM, what will happen is, when I shoot this thing in, it goes in, and it bends, perhaps, but it ends up going into the black hole. Let's look at this in the context of the equation of motion, the potential, and the impact parameter. If b is less than 1 over 3 root 3 GM, then 1 over b squared will be higher than this peak. And this thing is just going to be, whoo [fast motion sound effect], it's going to go right over the top of the peak. And it shoots into small radius. OK. Remember, L and E are constants of the motion. So that b is a parameter that defines this light for the this entire trajectory. 1 over b squared is less than V phot everywhere-- excuse me, greater than V phot everywhere. A rather crucial typo. And as such, it shoots the light ray, and it goes right over the peak into the black hole, eventually winds up at r equals 0. If b is greater than 3 root 3 GM, then 1 over b squared is less than V phot at the peak. This corresponds, in this drawing, to a beam of light that follows a trajectory that kind of comes in here at this point. dr d tau, if we were to continue to go into smaller radii, it would become imaginary. That's not allowed. So it switches direction and trundles right back out to infinity. On this diagram, that corresponds to a light ray that's perhaps out here. This guy comes in, and he gets bent by the gravity a little bit, and then just, shoo [light ray sound effect], shoots back off to infinity. The critical point, b equals 3 GM, is right where the impact parameter hits the peak. And so what happens in this plot is this guy comes in, hits the peak, and just sits there forever. You get a little bit more of a physical sense as to what's going on by thinking about it here. This guy comes in, [INAUDIBLE] this, and then just whirls around, and lies on what we call the photon orbit. What's the radius of that photon orbit? r equals 3 GM. Now, astrophysically, a more interesting situation is, imagine you had some source of light that dumps a lot of photons in the vicinity of a black hole. Some of those photons are going to tend to-- some of them are going to go into the black hole, some are going to scatter a little bit and shoot off to infinity. But if you imagine that there will be some population of them that get stuck right on the r equals 3GM orbit-- now, that is an unstable orbit. And in fact, if you look at it, what you find is that your typical photon is likely to whirl around a bunch of times, and then, you know, if it's not precisely at 3GM but it's 3.00000000000000001GM, it will whirl around maybe 10 or 15 times, and then it'll go off to infinity. So what we actually expect is if we have an object that is illuminated like this, then we will actually see these things come out here, and we will see-- so bear in mind, this is circularly-- this is symmetric. So take this and rotate around the symmetry axis. We expect to see a ring whose radius is twice the critical impact parameter. So it would have a diameter of 6 root 3 GM. It would be essentially a ring or a circle of radius 3 root 3 GM. This is, in fact, what the Event Horizon Telescope measured. So last year when I was lecturing this class, this was announced almost-- I mean, they timed it well. They basically timed it to about two or three lectures before I discussed this aspect of black holes. So that was-- thank you, Event Horizon Telescope. That was very nice of them. And of course, we don't expect-- so Schwarzschild black holes are probably a mathematical curiosity. Objects in the real universe all rotate. We expect Kerr to be the generic solution. And so there's a fair amount of work that goes into how you correct this to do-- so doing this for Schwarzschild is, because of spherical symmetry, it's beautiful and it's simple. Kerr is a little bit more complicated. But, you know, it's a problem that can be solved. And a lot of very smart people spend a lot of time doing this to sort of map out what the shadow of the black hole looks like, what this ring would look like in the case of light coming off of a spinning black hole. So that's it. This is all that I'm going to present for 8.962 in the spring of 2020 semester. So to everyone, as you are scattered around the world attempting to sort of stay connected to physics and your friends and your classwork, I wish you good health. And I hope to see you again at a time when the world is a little less crazy. In the meantime, enjoy our little beautiful black holes.
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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Let's examine the velocities of objects in the center of mass reference frame. So we have object 1 and object 2. And we have CM. And we have r1 prime. And we have r2 prime. And recall that we found that the position vector, r1 prime, was equal to the reduced mass over mu m1 times the relative position to r1, 2. I'll draw that up here, r1, 2 from 2 to 1. And just to remind you, r1 minus r2. And likewise the position vector r2 prime is minus mu m2 over r1, 2. Well, now we can differentiate the positions in the center of mass frame. And we find that the V1, which is, of course, the derivative of the position, is just mu over m1 V1, 2. And V2 prime is equal to mu over m2 with a minus sign V1, 2. Now the significance of this result is that the velocity of the object in the center of mass frame is equal to some constant times the relative velocity. Likewise, velocity V2 prime in the center of mass frame is also proportional to a constant. And it's in minus the relative velocities. So the significance of this result is that when we looked at an elastic one-dimensional collision, we already have the result that the relative velocities, V1, 2 initial, is minus the final relative velocity. And we also know from what we saw that the relative velocity is independent of the reference frame. So this condition about relative velocity is true in any reference frame. And in particular, it's true in the center of mass frame. So what that shows us is that these conditions, this statement, is that the magnitude of the relative velocity is constant. And the conclusion is because in the center of mass frame, V1 prime is proportional to the relative velocity. That tells us that in a center of mass reference frame that the initial velocity of object 1 in the center of mass frame is equal to the final velocity of object 1. So the quantity, the speed of objects 1 and objects 2 in an elastic collision in the center of mass reference frame do not change magnitude. We can call this quantity V2 prime and likewise V prime 2 initial is equal to V2 final prime V2 prime. And this will make our analysis of collisions in the center of mass reference frame very easy. Two objects, when they're colliding in the center of mass reference frame, their speeds do not change. Only the direction of their velocities change.
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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: All right, now for something completely different. Before beginning though, I would like to raise a procedural question. It was my intent that quiz number two, which is like a little more than a week away, was going to be part symmetry and part tensors. I don't want to have your fortunes based by weight of 2:1 on symmetry theory as opposed to properties. What I would suggest, and you can express your opinion either way, is that we postpone quiz number two by maybe as much as 10 days, so that it can be half symmetry and half the introductory discussion of tensors. And if that does not create a conflict with some of your other classes, I would propose that as a suggestion. If not, we can just have on the quiz what we're going to cover of tensors in the next couple days and have the rest symmetry. So what I am suggesting is the quiz was scheduled for a week from November 8, which is a Tuesday. What I would suggest is putting that off to the 18th, and that will give us four extra lectures-- six extra lectures-- on tensors and we can make the quiz 50-50. AUDIENCE: What does that do to quiz number three? PROFESSOR: Quiz number three is going to stay in place because it's right at the end of the term. AUDIENCE: What about the subject in three? PROFESSOR: It will be all tensors. It will be all tensors and properties. And the third quiz is set up for longer than you think. It's December 8, so it'll be about three weeks. AUDIENCE: So everything from here on? PROFESSOR: Yeah. Does that make sense? Or is somebody going to have a big, traumatic responsibility at that point? AUDIENCE: I don't know, but I think it would be probably be easiest for us not to have it three days before close of exams. PROFESSOR: OK, OK, that's what I wanted to hear. How about if we went just a little bit further then and did it not on the-- when was the suggestion? The 18th. Let's say we did it on Tuesday the 15th? Does that conflict with anything? AUDIENCE: That's a little bit. AUDIENCE: We were doing it on the 18th. AUDIENCE: Is that Friday? [INTERPOSING VOICES] PROFESSOR: I'm sorry, 18th. I said it would be 17th. Excuse me. AUDIENCE: That definitely would be a little better. PROFESSOR: OK, let me allow you to think about that, and that would be pushing it back one week and maybe it'd be-- we'll see how much material on tensors we cover. So let's table it for now but let's say, tentatively, let's consider seriously moving it to the 15th. OK? But you guys have final say if there's a preference that you have. OK let's talk about properties in first a general and rather philosophic way. When we discuss properties we very commonly lump together a set of behaviors which have some common phenomenology. So for example, we will talk about mechanical properties. And that's a basket in which we place many different sorts of behavior. We place things like fracture toughness, yield strength, elastic constants, a whole variety of things involving strength deformation and so on. Or another thing we very often lump together in a basket is something called electrical properties. And we talk about electrical conductivity, ionic conductivity, electronic conductivity. We talk about dielectric constants, we talk about permittivity and permeability, magnetic properties-- a whole bunch of other classes of behavior, which have some sort of root or description in terms of electromagnetism. There are though a lot of strange aspects to properties. There are some properties that are determined for material which no longer exists when you're through determining the property. You have a sample you want to know what the yield strength is. So you put the gradually increasing load on it until finally, POP, it breaks. And now you know what the yield strength is for a material that no longer exists. Another example is an old one. Back before the days of X-ray diffraction when people would try to characterize material, anything that is very intensely colored usually looks black, and that's not a very definitive property. The color of something is a very prominent characteristic. So for mineralogists, in particular, anybody going out into the hills and wanting to be prepared to determine what a particular rock that they tripped over was had something on a rope around their neck called a streak plate. And what the streak plate was was a little rectangle of porcelain. And they would take this black mineral and rub it on this piece of porcelain and it would leave a mark-- that was called the streak. And what this was was actually little bunch of fragments that rubbed off on to the porcelain because the porcelain was white and the fragments were very small. A black piece of rock could give you a streak that was brown or green or deep orange. And so you really were determining the color of the material when it was in a form finely divided enough that light could pass through it. So the streak test was a very important diagnostic for determining minerals. There are some properties that we refer to as structure sensitive. In the sense that their value-- conductivity is a good example-- the value of ionic conductivity depends on the impurities and point defects that are present in the material. So to say that the ionic conductivity of a certain material is such and such, you have to specify the purity and perfection of the material or the property's meaningless. There are some properties that are not even single-valued. And the best example there are dielectric or magnetic properties. If you plot the magnetization of a material, the magnetic moment per unit volume as a function of the applied magnetic field B, the material is initially non-magnetized. When you apply B, you get a magnetic moment per unit volume that eventually saturates. And then if you remove the magnetic field, the material keeps some of its magnetization. It doesn't go to zero when you reduce the field to zero. You have to reverse the field to get the property to go to zero and then magnetize it in another direction. And if you continue to cycle the magnetic field, you get a behavior like this. This is hysteresis, and this type of behavior is called generally hysteretic. Same thing would be a relation between the displacement vector and an applied electric field. But clearly you can have any value of the magnetization in this range. And if you relate the magnetization as a function of the applied field, you can get any value of the proportionality constant you like between a negative maximum and a positive maximum including zero. So there's a property that's not even single-valued. Depends on the past history of the sample. And then there are even more peculiar properties. There are properties that are-- we call them composite properties and very often qualitative. What do they mean by a composite property. Let me give you one example-- The property fuzzy. If I say fuzzy, you know exactly what I mean. It means something that has a diffuse reflectivity, something that has a surface texture that's yielding. It's not like rubbing a wire brush. It's soft and giving, a whole collection of different properties. But yet when I say fuzzy, you know exactly what I mean. Let me not be so facetious as talking about a fuzzy property. People such as ceramists or powder metallurgists very often will take a powder of a material and consolidate it by compacting it and heating it, very often subject to a compacting stress. And when you do that little necks grow between the particles and they hold together and the material is centric. So you refer to a property of a powder as being centerable or the centerability of a powder. You know exactly what somebody means by that. But what does it depend on? It depends on the surface energy, it depends on vapor pressure, depends on bulk diffusion coefficients. It depends on surface diffusion coefficients. And all of these things have to be just right to make the powder something that easily densifies upon heating in the application or not of pressure. So you know exactly what I mean by centerability, but it is a very complex property. And was one that really was not understood until probably the late 1950s. Until then, it was an art that was entirely empirical. And it was somebody here at MIT, a fellow named Robert Coble, who developed a theory of centering that was the first really workable theory that described densification by heating and compaction. OK, composite property that depends on many different individual properties of the material. OK, what we are going to examine here are equilibrium properties that can be rigorously defined and measurable. So we're going to leave out of the picture things like fuzzy and centerability. And let me give a very nice, obscure definition of what I mean by a property. And what I mean by a property, in terms of a formal statement, is the response of a material to a specific change in a given set of conditions. So the response of the material to a specific change in a set of conditions that relates independent properties and dependent properties for a particular process. So I'll state that again because I think it's terribly elegant. So a property is the response of the material to a specific change in a given set of conditions that relates independent and dependent quantities in a particular process. Now there are a lot of properties that have their roots solidly embedded in thermodynamics. So let me give you a few of those. What we'll talk about when we specify a property is something that we will refer to as a displacement. We'll talk about a generalized displacement in response to a generalized force. So some of the thermodynamic quantities that are related in this fashion-- if we list some forces and some displacements, the thing that happens as a result of that stimulus that's applied to the material. Temperature can be regarded as a force that results in all sorts of processes as a result-- thermal energy flow, thermal expansion, all sorts of things. But one of the things that will happen in response to a temperature change thermodynamically is an entropy. So you can view entropy as a generalized displacement resulting from the application of temperature as a generalized force. Another example is electric field and the thing that happens there, and electromagnetism, is the quantity, D, which is displacement. Still another example, stress, and the result of applying a stress, among other things, is a strain. What is special about these forces in this displacement is that their product is, in each of these cases, energy, the change in internal energy of the particular body. And when that is the case, these are said to be conjugate-- a conjugate force and displacement. And there are other examples that one can come up .with. Now to talk about a specific set of properties. Very often, and this crept into the earlier discussion, very often the thing that we do to a material, as a generalized force, is a vector. So very often the thing that we do to the material, apply an electric field, apply a magnetic field, apply a temperature gradient, apply a tensile stress, it has the characteristics of a vector, magnitude and direction. And in many cases, the thing that happens as a generalized displacement, we may call this in general, q, is also a vector. An example is if we apply an electric field as a generalized force, one thing that might happen is a current flow, which is also a vector. And we are accustomed to writing, q, the generalized displacement, as a proportionality constant, sigma, which is in the case of electric field and current flow, the electrical conductivity. Let me call it, in general terms, proportionality constant, a, times the applied vector, p. Is this something we would like to stick with as a general relation? Well let me submit that writing an expression of this form makes an inherent assumption. Namely that this will be true for small, and we'll have to define for each property what we mean by a small, applied vector. Let me give you an example. If p were-- the applied vector was electric field, and the resulting vector was current flow, and the relation between those is the conductivity-- a relation of this sort says that if you double the field, you double the conductivity. You triple the field, you triple the conductivity. Obviously this can't go on indefinitely because all a sudden, POW, dielectric breakdown. The sample evaporates at the smoke. And again, you have the property of a material which no longer exists. So a lot of properties, we inherently assume that the applied vector is small in order to write something in an expression of this form. So for conductivity, dielectric breakdown is going to destroy the nice linear relation between current flow and electric field. That's not always the case. Let me give you an example of another property. And this is a property, magnetic susceptibility, which relates the magnetization, which is magnetic moment per unit volume, and relates that to an applied electric field-- an applied magnetic field, H. And the proportionality constant, represented by a Greek chi, is the magnetic susceptibility. So let me give you a specific example. Suppose we have a chunk of glass and the glass contains a dilute concentration of iron. And iron carries a permanent magnetic moment. And the reason I want to make it a dilute concentration of iron is I don't want these magnetic moments close enough that they can interact with one another. I want this to, therefore, be a dilute system. So we have different iron atoms in this. Each iron atom has a magnetic moment. And then we put this in a magnetic field, H. And the magnetic field acts on each of these little magnetic moments just as though it were a compass needle. And so it will try to take each of these moments and drag it into coincidence with the magnetic field. But at a finite temperature, temperatures making these magnetic moments jiggle around, so at a finite temperature, the magnetic moments will just not simply zing into coincidence with the magnetic field. You'll have to increase the magnetic field to make it larger. When that happens, more and more of the magnetic moments will come into alignment. And if you put on a really strong magnetic field, then every single magnetic moment will be dragged into exact alignment with the magnetic field and the system has saturated. There's no way you could squeeze further magnetic moment per unit volume out of it. So again, you would expect, for this particular property, magnetic susceptibility of a material, if you plot it as a function of the applied magnetic field-- linear maybe be at low applied fields-- but eventually if you make the field strong enough, the system is going to saturate. And then again, no longer will the direction of the magnetization, and that magnetic moment, be parallel to H. But the property becomes nonlinear if the applied vector is strong enough. This is an example of a property where you would not go wrong at all by stating direct proportionality. For ordering to occur, temperature and applied field go hand in hand. Increased temperature tends to create more disorder. Increased magnetic field tends to align the moments. In the days when MIT had a national magnet laboratory, the magnet laboratory held the record for the strongest magnetic field ever produced artificially by man. And if you took that magnetic field and applied it to this system of dilute iron in a glass, you would have to lower the temperature of the sample to about 3 Kelvin before you would begin to see saturation. So even the strongest field that you could produce in a laboratory environment would not succeed in producing non-linearity until you cooled the sample down almost to absolute zero. So here would be a case where under any practical consideration whatsoever, assumption of strict proportionality would be right on the money. You'd be absolutely correct. But there's another assumption built into this statement. P, the generalized force, is a vector in the cases we're discussing now, and q is also a vector. So when we write an expression of this form you're making another assumption. And that is that the vector displacement that results is always exactly parallel to the vector that you apply. Do I make a big deal out of this? Isn't that always going to be the case? I mean whoever heard of taking a piece of metal and putting an electric field on it in this direction and having the current run off in this direction. It's absurd. Or maybe it isn't so absurd. So let's think of some of the atomistics of this process. Now, since I know I'm among friends, I will not hesitate to display my ignorance, total ignorance, of polymer chemistry. So suppose we had a piece of polymer. That's what a polymer molecule looks like. It's a more or less linear molecule, and so these might, in a very highly ordered polymer, be chains that lined up like this. And suppose we now put an electric field on this polymer and asked how the current will flow. Again, it would not be absurd to say that an electron sitting on this polymer chain in response to this field would be constrained only to flow in the direction of the chain and would find it rather difficult to hop from one chain to another. So maybe, just maybe, we could apply an electric field to a polymer and it would have a flow in this direction, that would be pointing in this direction, and the current flow, J, would be in that direction. So should we maybe rethink this idea of the direction of the generalized displacement being not parallel to the direction of the generalized force. OK, these are hypothetical examples. Let me now give you an example of a real property for a real material, for which the thing that happens, the vector that happens, is decidedly not parallel to the direction of the applied vector. OK, thermal conductivity is something that relates a heat flux, usually represented by the symbol K, and relates that to a temperature gradient, dt dx, which is a vector. The thing that gives rise to thermal conductivity can be either propagation of radiation as in propagation of light traveling through a transparent material. But the other mechanism for thermal conductivity is modes of lattice vibration. When a material is hot, the atoms are jiggling around and you can make the displacement of an individual atom be represented by a sum of waves moving in all different directions with a variety of wavelengths in a variety of amplitudes. Take all those waves, add them together at a particular time, and you get the displacement of the atom. Propagation of heat by this mechanism, by modes of thermal vibration, can be very, very anisotropic and probably the best example of this is-- get a single crystal of graphite. And have the layers, the graphite sheets, which are hexagonal rings in which each carbon has three neighbors, and have the sheets be parallel to the surface of an extended two-dimensional slab. Now we can't do that. Single crystals of graphite don't occur in sizes like that. But what you can do is make a material called pyrolytic graphite that you make by having a reaction in the vapor phase and having the soot that's formed settle down onto a substrate. And what happens is you nucleate a graphite crystal in one orientation, and that grows preferentially with its layers parallel to the surface on which it's nucleated. And these nuclei occur at random. So you get a bunch of single crystals of graphite, which all have their layers, their three coordinated layers at parallel, but they are oriented at random about their c-axis. And that's a material that's very easy to prepare. And it's called pyrolytic graphite, and it has interesting applications and properties. OK, now I'm going to suggest an experiment to you and I'll caution you, please do not try this at home. Get yourself a piece of pyrolytic graphite. Put a Bunsen burner underneath it. Play the Bunsen burner on the bottom of the sheet of pyrolytic graphite. And to determine temperature, we'll take a ball of cotton and put it directly above the flame. And then take another ball of cotton and put it down at the end of the sheet. If you do that and bring up the Bunsen burner, this piece of cotton will sit there and sit there and sit there and sit there. And this piece of cotton will instantly burst into flame. So here's a case where dt dx, the thermal gradient, points in this direction. And the heat flow goes off in a direction at right angles to the temperature gradient, almost exactly at right angles. It turns out that the thermal conductivity in this direction, the value of K in the-- I want to put crystallographic directions on this. But K parallel to the layers is equal to 10 to the third times K, perpendicular to the layers. So there's an anisotropy of the property that amounts to a factor of 1,000. Very dramatic. So what I'm leading up is that in a general relation between a generalized displacement, p, and a generalized force-- I'm sorry, I'm using q-- p and q-- we just can't simply put some constant in front. Because that assumes inherently that the direction of what happens, the vector that happens, is exactly parallel to the direction of the applied vector and that generally is not true. The difference may be small but it has to be taken into consideration. All right, let me now suggest a way of patching this up. What I'm going to do is assume-- and it is an assumption-- we're going to assume that each component of the vector that happens is given by-- in fancy terms, the vector that results as the generalized displacement is given by a linear combination of every component of the vector that's applied. And that's what we've defined as the generalized force. So how would we express this analytically? I am going to use as my coordinate system, not x, y, and z as we usually do, I'm going to call it x1, x2, and x3. The reason is we'll see some unique properties of these indices that are very useful algebraically. So I'm going to now take my applied vector, p, and I'm going to assume that it has three components, p1, p2, and p3 along x1, x2, x3, respectively. My coordinate system, although I did not state it explicitly, is going to be a Cartesian coordinate system, not a crystallographic coordinate system. So what I'm proposing here is that we take each component of the resulting vector q-- let's say q1-- and we'll assume that that's given by a linear combination of all three components of the vector p. So it'll be some number times p1 plus some number times p2 plus some number times p3. And those numbers in general will be different. So let's say I call the coefficient a, and now I'm going to define a convention that will stay with us. I'm going to define each of these coefficients, a, in terms of the index of the component of the generalized displacement which is being computed, and the coefficient modifies the component of the generalized force for that particular term. So I'm going to call this a11, where the 1 goes with this and the 1 goes with this. I'm going to call this a12, so that the 1 again says it's a contribution to q1. The 2 says that this term modifies p2. . And this similarly would be a13. For the term component of the generalized displacement, q2, I'll use the coefficients a21 times p1 plus a22 times p2, plus 23 times p3. And q3 similarly will be a31 times p1, plus a32 times p2 plus a33 times p3. So I've got three simultaneous equations then, one for each component of the vector that results, the generalized displacement. So I can sum up this set of three equations by saying that the i-th component of q, where this is some particular component, q1, q2, or q3, is given by the sum over j from 1 to 3 of aij times p sub j. So I've got ai1 times p1, ai2 times p2 plus ai3 times p3. So this is in a nice, compact little nugget the expression that we are assuming will apply for all sorts of physical properties in which this is a vector and this is a vector-- electrical conductivity, magnetic susceptibility, thermal conductivity, and so on. OK, this is a bad point to introduce something as hairy as what comes next. But I've got one minute left, and I think I can do it. I'm going to introduce something called the Einstein convention after old Albert Einstein himself. I'm sorry, but nobody said everything this term had to be easy. So let me introduce, if you're ready for it, the Einstein convention. Old Albert, I think, was just as lazy as anybody else. And he said it is going to be a bloody pain in the butt-- I don't know if he put it exactly this way. It's going to be a pain in the butt to write this summation every time we want to combine three terms in a linear combination. So the Einstein convention is let us throw out the sigma and write this expression just as qi times aij p sub j. And whenever we see a subscript repeated, summation over repeated subscript is implied. OK so that's the Einstein convention. It will save us the trouble of writing in a lot of summation signs. But I would caution you that this convention, compact and convenient as it is, can define some polynomials that are nightmares. So let me give you one example of this, and this is actually a physical property. Let me say that Cijkl is given by ai capital I aj capital J ak capital K, al capital L times let's say D, capital I, capital J, capital K, capital L. That actually, believe it or not, means something physically and we're going to get to that in due course. But what this is, taking the Einstein convention into account, this is a quadruple summation over capital I, capital J, capital K, and capital L. These are very often referred to in this business as dummy indices, meaning that they don't really mean anything physically. They're just indices of summation. I would caution you that this is a term that does not permute. If I say those are dummy indices, it means one thing. If I say those are indices, dummy, it means something completely different, and you're apt to get a poke in the nose. OK so these are dummy indices. Now, what does this represent? I won't say what it represents physically, but this consists of terms in four variables times a coefficient. So there are five quantities in each term. If I sum over capital I, J, and K from 1 to 3, there are going to be 81 such terms, each with five elements in each term. So it's going to be on the order of 405 terms in this summation, and we've collapsed the whole thing down, 405 terms in this nice summation. So it is a great facility for writing down expressions explicitly, but these terms can hide a terribly, terribly complex polynomial. All right, I think that's a good place to quit. It continues to amaze me, though, how some absolutely trivial convention, when first proposed by a great man, will carry that man's name no matter how stupid it is. So this is called the Einstein convention by everybody who works in this field. Another one, just briefly to finish up, when you discuss dislocations, you talk about a circuit of steps around the dislocation. That's called a Burgers circuit. And if you do the same circuit in a corresponding piece of material that doesn't have a dislocation, if this circuit closes, this circuit fails to close by something that's called the Burgers vector. Every book on dislocation theory says, Shockley called this good material. I'm sorry, Shockley called this good material, and Shockley called this bad material. Big deal-- good material, bad material because it's got a dislocation in it. But that is identified with Shockley's name in every discussion of dislocation. I'm just jealous because nobody has ever said according to me, such and such is-- all right, see you later in the week for more great things.
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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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Welcome back to 8.701. In this lecture, we'll talk about CP symmetry or CP violation. In previous lectures, we discussed that the weak interaction is not invariant under parity and charge conjugation transformation. But now we can ask the question, how about CP-- so transformation which does change conjugation and parity transformation. The classical example to show parity violation is the decay of a pion. So we have here this charge pion with spin 0. And it decays into a muon and a neutrino, an anti-muon and a neutrino. And so since the neutrino is left-handed, the-- [INAUDIBLE] decay coming onto muon needs to be left-handed as well. So if you do parity transformation of this decay, you see that the outcoming muon would be right-handed. On the other hand, there is no right-handed neutrino. And therefore, this decay is not possible. So this is not-- so this mirror symmetry is not realized in nature, as a consequence of the weak interaction. So similarly, we could do a charge transformation, a charge conjugation, of this decay. So you turn particles into the antiparticles. And you find here this antineutrino, which is left-handed. And also, those don't exist in nature. So parity or charge conjugation doesn't really work on those decays, those [INAUDIBLE] decays. But what does work, if you apply the parity and a charge conjugation, we turn the positively-charged pion into a negatively-charge pion, the antimuon into a muon, and the neutrino into an antineutrino. And you see here that the antineutrino is right-handed. So is the muon. And so that decay is actually observed in nature. Good. So we saved the day. It seems like that CP, that the weak interaction is invariant in the CP transformation. However, that's not quite true. Gell-Mann and Pais noted that in systems of neutral kaons, there's an artifact. And the fact is that a particle, a K0, can turn into an antiparticle by changing the strangeness. And that's possible in this kind of box diagrams, which include a box with a couple of W's. And it's easy to see that if you could prepare a kaon, it will oscillate, because those diagrams are possible, into an antiparticle. So now what is happening now to CP here, if I apply CP on a kaon, I find a minus sign and an antikaon. So if you want to analyze this further, you might want to find the eigenstates to this. And so the eigenstates can be found, as well as K1 and K2, which are admixtures of the K0 and the anti-K0. And you find this symmetric, the symmetric and the anti-symmetric states. Good. So if you apply CP on the eigenstates, you find eigenvalues of 1 and minus 1. It turns out that the lifetime of decay 1 decay 2, those eigenstates, is very different. One is 10 to the minus 10, and one is 5 times 10 to the minus 8. So decay 1 decays much, much quicker than decay 2. So this, then, sets the stage to a test of CP violation. So what you're going to do is prepare a beam of K0's and let them decay. And only after some time, you study the beam again, which then should be made up solely of K2's. So if you in that beam observe decays of the K2 into two pions, you noted that there is an admixture again, which violates CP. So you have an admixture of K1's in a beam which should just be of K2's. So that mixture, then, will violate CP invariance. And exactly that was done. So Croning and Fitch picked up this idea. They set up an experiment in which they produced kaons. They had them decay. And then they studied later in the beam whether or not they could find two pion decays. And they did, indeed, observe 42-- 45 pion decays, two-pion decays, in a total of 22,700 decays. So that means that this K long beam, the long-lived kaon beam, is actually an admixture of K2's with a small additional component of K1's. So here they observed CP violation through the mixture of those states. And this epsilon gives you, you know, size of the strength of the CP violation. So here is a note of the paper. We'll have another discussion of this in class by a student presentation, Croning and Fitch. Here this is Croning, and this is Fitch. It turns out that Croning is actually a student, or was a student, of Enrico Fermi and also worked in Chicago. So that's quite an interesting family tree here, to which also Jerry Friedman belongs. Jerry Friedman is a retired faculty at MIT and discovered that protons are made out of quarks. So this is a very interesting family tree. If you have some time, you might want to look into this. But here's the experiment. So you take protons, you dump them into a beam. You try to, with this magnet, filter out a neutral component, get rid of all photons, and then let this beam decay, and look in the spectrometer for decays of two photons. Here's a bigger picture of the same spectrum. So this is actually a blow-up view of this. So you have your kaons, neutral kaons coming in, the K2's. And then you look for pion decays. The instrumentation and how we actually would do this is part of later discussions where we actually talk about detectors in more detail. All right. So we just saw that Croning and Fitch observed CP violation in mixture of states. But we can also observe CP violation in direct decays. And the classical example here is the case of the K long and semileptonic decays. So semileptonic here means we have a decay of the K long, a neutral particle and a charged pion, an electron and antineutrino-- or it might very well also decay into a pi minus, a positron, and a neutrino. And it turns out, when you really count those events and perform a precise experiment, that the K longs prefer decays to positrons over decays to electrons. And so the fractional amount of this imbalance is 3 times 10 to the minus 3. So this is a rather small effect, again, of CP violation here in direct decays. Since then, CP violation has also been shown in the decay of B mesons. And the program of studying B mesons is a big part of the LHC experiment at the LHC. There's also experiments in Japan going on right now which study B mesons in order to learn further about B systems. Tests are also underway, for those who listened to the colloquium on Monday, in the neutrino sector. So here we have a completely different part, so not quarks are involved in weak interaction but neutrinos. And so the question is whether or not in that sector of physics, that sector of the standard model, there is CP violation. Those are aspects we'll discuss later on when we talk about neutrinos specifically. Before I close, a few more remarks on the matter-antimatter symmetry. So one of the biggest mysteries in physics, I would claim, is the fact that we're even here to ask this question. So there is apparently more matter in the universe than antimatter. You start from a big bang, there was this symmetry, and now we live in a universe which is dominated by matter. So how is this possible? So in 1967, Zakharov proposed that this is possible in a system where baryon number is violated. So this is almost a trivial statement. If you start from an equal number of baryons and antibaryons, the sum is then 0. The baryon number is 0 of this system. And you end up in a system which is dominated by baryons, then baryon number needs to be violated. But there's also the need of CP violation in this. So we just saw that this is realized in nature. But the amount of CP violation we observe in the system I just discussed is not sufficient to explain the matter-antimatter symmetry we observe in nature. So there is more to be found. There's new physics to be looked for in CP violation on this overall question. And there's also a need for the actions to be out of equilibrium, meaning that you don't revert the processes as you go forward. Yet another point of discussion, which I will not go into much detail in this lecture, is that our quantum field theory, which describes quasi-standard model and describes the interaction of particles, is invariant, and the CPT transformations. That means that if CP is violated, time reversal cannot be a symmetry. So meaning, going backwards and forwards in time is not symmetric. And you can test this. You can design experiments which test [INAUDIBLE] the fact. You can also design experiments which test CPT directly. But this is-- those are all interesting questions, but we will not go into any of those in this lecture. We will, however, come back to understanding the origin of CP violation in the standard model when we talk in more detail about the weak interaction.
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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: All right, great. I looked over some of the Mud cards we had from last time. And there were some really good questions, some really good points, and one lingering one from the previous time that I forgot to address. Someone said, could we please just move the desks into an arrangement where we could see each other. It was a fabulous suggestion. Today we can't do it. But I'm going to try to do it in the future whenever the logistics allow, because I think you're absolutely right. A little bit more of a U, or a circle, would be great for this class. So thank you for that suggestion. Another one was that we didn't get to motivation theory. And I think I mentioned this last time, the idea that there's a huge amount of stuff that's been written and published on motivation and theories of motivation. And I'd be more than happy to discuss this as the semester goes on. And it might be a great topic for one of those lunches that somebody was going to organize, one of those informal lunches that someone was going to organize. Or maybe we can carve out some space at the end of the semester and pop it in, take something out and pop it in, because it's really fascinating. It's really interesting, but it doesn't quite fit anywhere in particular. So thank you for bringing that up. And keep bugging me about it. The other thing is someone asked if all those learning theories that we discussed were equally valid. And yes, they are equally valid. I mean, there are situations where being behaviorist could help. There are situations where being constructivist could help. It depends on what you're trying to have students learn. So you want to make sure that you tailor the way that you're teaching to support what it is you want students to learn. And we'll get into that in a lot more detail today when we talk about developing learning outcomes, and then also when we talk about tailoring your instructional activities to support those learning outcomes. So this class, this particular class today is a balance between developing course design, developing intended learning outcomes, which for me is at the heart of it. If you can come up with some good intended learning outcomes, you basically have designed your course. And I'll show you how. I mean, you can't just ignore it, but that's the hard work, and then the act of writing constructing a syllabus. So once you've got those learning outcomes, you have an incredible anchor for the rest of your course. And there was just a very timely haiku. I put it on the wiki. But there's this Twitter feed, academic haikus, which if you want to waste any time and read them, it's a way to waste time. But there was one here, "that question you have? I know where the answer is-- on the syllabus. So it was, like I said, it came out yesterday. I thought it was pretty timely. And it's true. If you write a good syllabus, it should really clear up a lot of, not the content, not the understanding level issues students have, but the issues with respect to when's the test, do we have to know this, how will this be graded, what's important? All those things should be in your syllabus. So that's why that's there. So here's a little slide, a graph that I saw a few years ago, and I thought it's a really good way to start this discussion. So on the y-axis we have sort of an assessment, a proclamation of students understanding. They got it, they didn't get it. And I think the word in French is [FRENCH]. So for some reason I find it very cool to say [FRENCH]. Got it. I don't know. So whether they get it, like whether after the tests, you've given a test, and you go, man, that student really got it. That student really gets this material, and whether they passed the exam or failed the exam. So the size of the circle is more or less the number of people that fall into those quadrants. So anybody surprised by this plot? Yes, Rachel. AUDIENCE: [INAUDIBLE] JANET RANKIN: OK. All right. Well, there's the segue. So we want, we don't want the green, but we more or less expect the green, right? If you get it, you pass the exam. If you pass the exam, you get it. That's that, and we like to have lots of people have that happen too. And then we also acknowledge that somebody isn't going to get it, and they're not and they're not going to pass the exam. And that kind of maybe makes us sad, but it's not surprising. And then anybody can kind of have a bad day or kind of a fluke. So sometimes, sure, somebody will fail the exam, and actually really understand the material. Generally speaking, that doesn't happen, and then there's often a reason for it. But it's that red circle that Rachel mentioned that is troublesome, at the least, troublesome. So why might that happen? Why so many red circles? Yeah, Dave. AUDIENCE: Because the test was set up for basic memorization, or you could figure out with a lucky guess with multiple choice. So you could easily pass just with good test-taking skills or a good memory. JANET RANKIN: Excellent. So you pass with sort of good test taking skills. Now I want us to hang on to that. There's an underlying assumption in Dave's comment. So I'd like to hang on to that thought, but let's hear a few other conjectures and see. Rachel, why do you think you might have passed it? AUDIENCE: So often I feel like TAs are either really good or really terrible when it comes to the exam reviews, and they're mainly good. They often told us too much of what was going to be on the exam. And so then you kind of just knew what the test was going to be if you could well on it if you [INAUDIBLE]. JANET RANKIN: OK. That's interesting. Yes, Gordon. AUDIENCE: Yeah, it's possible that [INAUDIBLE] tests [INAUDIBLE]. When it comes to true life situations and something that [INAUDIBLE] knowledge that was gained from that material, then [INAUDIBLE]. JANET RANKIN: OK. So Gordon is sort of kind of flirting with this idea of you as the instructor, this real life knowledge, this is what it means. They have to have this practical working knowledge for us as the instructors to say they got it, right? But the exam isn't measuring it. And that's exactly what, that's what Dave was saying as well. This idea that the exam is measuring one thing, but it's not really measuring whether they got it. When we go home and think what does it mean to be a good material scientist, or what does it mean to be a good chemist, we don't think, oh, it means that you can define what a mitochondria is, or you can balance a simple equation. And yet, we put that on the exam. It's also a bit of the idea that students are good at taking tests. They're good at gaming systems. They're good at listening to TAs or TAs may have not given out appropriate information. So it comes from both sides. It comes from the students and it comes from the professors. But you have control over this. You shouldn't be writing an exam that people can get a good grade on, but then when you go home and say, oh, they got a good grade on this test, but they didn't get it. You need to define what it means to get it in your field, to get this subject, whatever it is that the test is about. And then you need to write a test that measures that. And too many times we don't do it. We're kind of rushed, or we don't really know. We haven't really thought about how to measure whether somebody gets it. So the first step in all this is to define what it is that means that they got it. And that's the first step, and that's the hardest step. But once that's done, the rest is easier, and it will save you from this. OK? OK. So that's the topic of today's class. And if you needed a little more motivation, I'd like you to read this paragraph, and I'd like you to tell me what it's talking about. But I'm just going to give you about 2 and 1/2 minutes to read it. And then we'll discuss what it means. What do you think this is talking about? Yes, David. AUDIENCE: Yeah, I think it stresses about how we can arrange by say, maybe schedules or activities that put them into groups. Then maybe if we form a group, how we plan [INAUDIBLE]. JANET RANKIN: OK. So arranging schedules or activities. OK. OK. Something else? AUDIENCE: When you're planning, how to go about teaching a module, [INAUDIBLE]. JANET RANKIN: Planning teaching. What else? Anything? AUDIENCE: It's like a mailroom operation or something. JANET RANKIN: Mailroom. Something from the back, maybe? You know you're smiling. AUDIENCE: I couldn't think of anything. JANET RANKIN: All right. Well, then don't say anything. AUDIENCE: The first is the course content, the course outline, and what you are going to teach the students, [INAUDIBLE]. JANET RANKIN: OK. So students, grouping students-- Adam, right? Adam. OK. So content, arranging content. Anybody in the back? Flyer? Yes, Gordon. AUDIENCE: I think you talked about sorting materials. JANET RANKIN: Sorting materials. So we'll go with maybe sorting. So if I tell you-- so these are all great guesses. They're as good as what I'm going to tell you the answer is. So this was written to describe the process of doing your laundry. Like doing the wash. You take the clothes and you put, maybe you put the dark clothes in one bin and the white clothes in another bin. And then you fold them and you put them away. It's impossible. This happens almost every time I show this little piece of writing. People don't generally get it. They get the fact that it has to do with sorting, and some planning, some arranging things, but they don't get that it's about the laundry. If you skim it-- you don't have to totally reread it-- but if you skim it with the idea that it's about doing the laundry, you probably can understand it better, maybe. So now I'm going to ask why did I show you this now, today, before the third, the beginning of the third class? Yes. AUDIENCE: I think it's sort of about arrangement of course content [INAUDIBLE]. JANET RANKIN: So it could be about how you arrange the content in the syllabus. [? Hina. ?] AUDIENCE: I think it's easy to describe methodology as a person more in a different field. But if there's no context for the audience about what this is, it doesn't matter that you've laid out everything. No one knows [INAUDIBLE]. JANET RANKIN: Right. It's very detailed at a certain level, but it's still really hard to understand, right? Any other ideas? Yes. AUDIENCE: I think it's interesting. There's a lot of value judgements in it. Like, it's really easy. And one can never tell with comments like that. Because I'm trained to figure out what they're saying, it's actually just frustrating me more, because I don't know what they're talking about. And it's really easy-- JANET RANKIN: All right. So that's a great point. So we're talking about writing learning outcomes, about organizing the course around writing learning outcomes. If you're using phrases, or you're using language that's just, there's just like throwaway words in there or value judgments, it's not doing your students any good, right? And also, that if you have this beautiful course that's beautifully organized even, wonderfully organized, but if you don't tell them what it is you're trying to do with it all-- what it is they're going to be able to do with it all, whatever it is you expect them to do-- they're not going to be able to follow you. So imagine a procedure as fairly straightforward as doing the laundry. Now imagine, OK, you're going to be able to select the right, the appropriate method to integrate equations, so methods of integration, by separation, by card or substitution or whatever. So if you don't tell them what it is you're expecting them to do, even if you're very good at explaining what it is they're going to do, they won't get it. So this is my sell for why when you write a learning outcome, you have to write a learning outcome that's clear and that's understandable to the students, that doesn't have a lot of extra words and that tells them exactly what they're supposed to do. So I have some learning outcomes for today and I hope that they're clear. You'll be able to describe the components of the constructive alignment and backward design processes. And those, constructive alignment anyway, was in the pre-class reading. You'll be able to classify the content of a course that you might teach. So what do I expect students to do with this content, to be able to do with this content? You'll be able to create measurable, specific, and realistic learning outcomes for a class you will teach, and you'll be able to state the components of a syllabus. So I feel that those are pretty understandable. You may disagree with me, or maybe you want to wait. So constructive alignment, it was in the pre-class reading, right? Anybody want to just give a brief summary of it? I mean it's kind of up here, but in your own-- AUDIENCE: I think constructive alignment is about aligning assessment, method of assessment, with course content. [INAUDIBLE] assess [INAUDIBLE] type of assessment so that what the teacher has in mind, the course content or the content I want them to get, to test it with the assessment process. Then we would also do the assessment, and then come up with the intended outcome. JANET RANKIN: Right. They'll achieve the learning outcomes. And if you do this right, you get rid of that red circle on that graph. You get rid of students who did well on the assessment but don't really know what's going on, because you have aligned the assessment with what it was you wanted them to know or be able to do. So that's important. I just think this is a very, very compelling image to keep in your head, that these three things have to be linked. And as I said before, this, you will see that the top part, writing the learning outcomes is the tricky part, it's the hard part. It's the part that requires a lot of thinking and work. When you get a good learning outcome, the other two things almost take care of themselves. OK? Did anybody have any, does anybody have any questions about something from the reading associated with constructive alignment or some other idea from the reading? Yes, Gordon. AUDIENCE: I just wanted to make a contribution. I think that this constructive alignment is one of the things that the intended outcome should measure. So I think maybe that's confusing [INAUDIBLE] and also how we can test and be sure and confident. JANET RANKIN: Right. Right. I think the measurability part for scientists and engineers is kind of intuitive. Like why would you say something if you couldn't tell whether or not it was true, or not from a scientific point of view. What's the point of saying, making a proclamation about what somebody should be able to do if you can't measure it in the end? So it's comforting to say, OK, I'm going to make learning outcomes that are measurable. But it also can be a bit challenging. But we'll do that. We're going to do that today. Some other comment? Nina. AUDIENCE: I guess as far as [INAUDIBLE] Do people write learning outcomes that have no constructive alignment process, like choosing to write learning outcomes, are you going through the prcoess? It seems decoupled a little bit in the reading, but it seems like you can't have one without the other. JANET RANKIN: You owe it to yourself-- and I don't know what people do, and I guarantee you people write learning outcomes without thinking about whether they can assess them or measure them, I guarantee that happens in real life-- you owe it to yourself to try very hard not to do that. OK? And I think the important thing to do is, often we say, oh, my god, this is the best assignment. This is the greatest question. It's this project, and they're going to do this, and they're going to whatever. They're going to build a race car. And if you sit down and say, well, what do I want them to get out of that? If you can't articulate it in a way that's consistent with your class, consistent with the course as it sits in this series of their courses, or as it sits in the institution, then you either need to rethink the assignment, rethink the project, or think a lot harder about what learning outcomes that activity advances. That's a very good point. It's really kind of iterative. And I think it might be easier to see with this diagram. This is another-- so John Biggs is from the UK. And so there's sort of the UK camp of people and they like John Biggs. And then there's Wiggins and McTighe. Grant Wiggins and Jay McTighe are from the US. They wrote primarily for the K through 12 audience, but it's completely applicable. And they're from the US, so they have kind of a US following. It's just how it works. It's essentially the same thing. It's identify the results, so the intended learning outcomes. Figure out how you're going to get them there. So the black is Wiggins and McTighe and the red is Biggs. So plan the learning experiences that are going to support them, achieving the learning outcomes, and really help you measure whether they got them or not. And then determine what evidence will be that acceptable evidence that they achieved the learning outcomes and what are the assessments. So real often we do have this project that we want. Oh, this is a great problem. Often in thermo-- I taught a lot of thermo-- there's like, oh, they're going to do this double loop, double reactor cycle with the refrigerant and a heating cycle. And it's going to be coupled and they're going to use the output from one for the input in the other and all this crazy stuff. Like, this is a great assignment. And you kind of, I know this is a great assignment. If I can't back up from this assignment-- whether or not I put it here or here-- if I can't back up and say after a successful completion of this assignment, you will be able to describe the components of the Brayton cycle, you will be able to define the increase in entropy for a system with x boundaries, If I can't articulate those things, then I need to either change the assignment, or, well, then I need to change the assignment. OK? But you can start with the activity, and then back up and say, well, what learning outcomes does that support, and then, am I OK with that? Is that relevant to this class? So it's often a very iterative, the arrows kind of go this way, but it's often iterative. So I had asked that you come to class with some topics from a class you're likely to teach. That was on the assignment page. If not, start thinking now. And so you want to have, for most classes, most undergraduate classes, there's a bunch of topics. You might have 10 topics, 20 topics. And I don't care right now about the scale of the topics, whether they're really big. You probably don't want to go with like first law, second law. That's probably a little bit too big, but on the order of 5 to 15 topics. Jot down the topics. And then I have a chart here. So the chart has two sides, and the columns are Bloom's Taxonomy, which you don't have to worry about that specifically, but you can look at the words that are associated with it. And then the second column is what students should be able to do with each topic. So let's say my topic is first law for closed systems. So I might think about, OK, do I want them to be able to define, to just state the first law for closed systems? Do I want them to be able to use the first law for closed systems in order to solve certain kinds of equations? Do I want them to be able to derive the first law? Do I want them to be able, whatever it is. So you're going to look at the columns, the categories on the left side, and you're going to think about your topic. And you're just going to decide which box in the middle column to put the topics into. You don't have to worry about writing anything. You're just going to distribute the topics to the level of, it's really cognitive processes that are associated with the things on the left. Remember, understand, apply, analyze, evaluate, and create. So we'll take about five minutes. And you just do that by yourself. And just fill in that middle column of the table with the topics from your course. Do you want more explanation, Rachel? It's starting to make you think about this part of the beginning circle. But you don't want to write any sentences yet. But it's starting to help you think about what are they going to do with these topics. What do I want them to do? Is it remember, or is it create something using these things? Where is it? Those are two ends of the spectrum. And in your chart, the things that at the top are the more remember, right? It's more about kind of a recall information. And then the things in the back of the chart tend to be more higher order cognitive processes, like synthesize, apply, create. Yes. AUDIENCE: [INAUDIBLE] JANET RANKIN: So don't worry about the third column. Don't worry about the last column yet. We're going to do that later. So all you have to do is just say where each topic falls loosely. OK. Does everybody have a chart? OK. AUDIENCE: Do you want us to write as many topics as possible? JANET RANKIN: Write as many topics as possible, but distribute them through the chart. Right? Not everything is going to be remember, right? So some things will be remember, some things will be apply, some things will be analyze, some things will be create. OK? If you have any questions, just raise your hand. All right. So what we're going to do is I'm going to ask you to just keep those with you. We're going to do another activity later on using those as a starting point, but I want you to just keep that sorting with you. Any comments or questions about the experience, about what you just did? Was it easy? No. It can sometimes be a challenge. Well, what the heck you were supposed to do with this anyway? What do I really want them to do with it? This is the beginning of that process whereby you define exactly what it is you want students to do, know or be able to do. And then you can start to think about how you're going to assess whether or not they can do it or not. But it's very different if I were to say I want everybody to be able to state the first law of thermodynamics versus I want you to be able to use the first law of thermodynamics to define chemical, mechanical, and thermal equilibrium in a simple composite system. That's different. That's very different. So that's what we're starting to do. So now we have kind of the seeds. We thought about which topics kind of we want to have, what we want to do with each of the topics. But now we really have to turn them into learning outcomes, into real statements. By the end of this class, you will be able to, by the end of this course, you will be able to. And just as an aside, remember that sometimes you'll notice that for every class, every session we have, I have learning outcomes. But there's also overarching learning outcomes for the course. So I will use them interchangeably, but they'll be at a slightly different scale. The ones for the class are a little bit more fine grained. As was mentioned in the reading, and as Gordon mentioned, they have to be specific, measurable, and realistic. If they don't have those properties they're not going to be particularly useful, or as useful. All right. And we'll see some examples of that in just a sec. So here's what they're not. They're not topics. So you came with topics. You will leave today with learning outcomes. But topics are not learning outcomes. The first law of thermodynamics is not a learning outcome. You need a verb with it. What do you want me to do, what do you want your students to do with the first law of thermodynamics? What do you want them to know about the first law of thermodynamics? So they're not topics. They're not things that you will do. It's easy to say, I could have a list a mile long. I will talk about the first law of thermodynamics. I can check that off pretty easily. I mean I just did. I just talked about the first law of thermodynamics. It says nothing about what you learned about it, what your students learned about it. So it's not about you. Get over it. That was a joke. It's not value statements. As Adrian pointed out, some of these think this is easy, this is good, this is useful. Any of those things don't belong in a learning outcome. And it's not your hopes and dreams for your students. I mean, ultimately, it is. You hope that they will be able to use the first law of thermodynamics, da, da, da. But it's really about what they're doing, about what they're going to do. So, no. Here's a little example. This is just from like an earth and planetary course. Students will understand plate tectonics. How's that? Is that OK for a learning outcome? Why not? AUDIENCE: It's too broad. JANET RANKIN: It's not very specific. What the heck about plate tectonics? What else? AUDIENCE: Measurable. JANET RANKIN: Thank you. OK, the measurable part. This word understand, strike it from your brains. It's a property that we love. We love understand. But you can't flippin' measure it. You can't go in there and measure it. You can measure lots of things that tell you whether students understand, but you can't measure understanding. And I think we'll see that with some more examples. I don't think anybody is earth and planetary here. So how about this one? Students will be able to interpret unfamiliar tectonic settings based on information on volcanic activity and seismicity. Again, I don't think any of us are geologists. But the idea that I've told, or the person has told us what we're going to use to interpret these situations, and then we're going to make predictions based on those interpretations, I mean, that's an assumption. So it's specific and now it's measurable. Even though we're not geologists, I suspect we can think of how we could write a test question that would ask students to do this. And then we could see whether they did it or not. So that's measurable. And probably, by doing this, by having students do this, we get a pretty good measurement about whether they understand plate tectonics. But it's much more specific and it's much more measurable. From here, this is from the University of Minnesota, which has a nice website on learning outcomes. They're not from your point of view, as we mentioned. They're not what you're going to do. I'm going to introduce students to the major turning points and processes in North American history. You can say anything you want, but it doesn't mean the students learned anything. So it's not about you. It's what they're going to do. They're going to list and describe the turning points and processes in North American history. I'm going to create an understanding of the formal constructs of physical design. That one is like, this one is, it's like, I don't even know what it means, right? Create an understanding of the formal constructs of physical design. I don't even know what it means. And so it's like that laundry example. Your students aren't going to know what it means, and it's not going to help them. It's not going to help them access the material. So the ones on the left are all teacher-centered and they're not very useful. The ones on the right are student-centered and they are generally useful. But there's some things wrong with the things on the right. AUDIENCE: Understand. JANET RANKIN: Understand, yes. So what would you write instead of understand the formal constructs of physical design, whatever the heck that means? What might you use? AUDIENCE: Identify. JANET RANKIN: Identify. OK. That might be like a list, right, or identify or list, pick it out of the lineup kind of thing. This is a nice one. Explain to an intelligent non-expert, an INE. And so explain this to somebody who's not in the field. If you could do that, it probably means that you get it. Somebody I know used to say, explain this to my four-year-old brother. Explain-- what was it-- angular momentum, to my four-year-old brother. I would argue if you can explain it to a four-year-old kid, you probably understand it. So you can use that. Use formal constructs to design something. I mean, that's kind of an inversion. How about the one on the bottom. Understand gender, race, ethnicity, socioeconomic class, understand how they have shaped Americans' lives. Completely unmeasurable, right? So what would you write instead? AUDIENCE: Describe. JANET RANKIN: Describe how, sure. Anything else? AUDIENCE: [INAUDIBLE] it seems like you're trying to get [INAUDIBLE]. JANET RANKIN: Right. So that's a great point, is that we don't know where this faculty member put this on their table. We don't know whether it really just states some ways that, list some ways that these things have affected Americans, North Americans' lives, or Americans' lives-- which is kind of a remember thing, just make a list-- or whether it's do some analysis on some situation. We don't know that, but you as the instructors get to decide that for your classes. So that's an important thing. You get to call all the shots. You get to make the decision. We can only guess about what this person meant, and therefore, his or her students can only guess about what he or she means. So you have on your table this hierarchy of Bloom's Taxonomy. It's often shown in this triangle, this pyramid. There's a handout which has exactly the same strata on it. And it just has some handy dandy verbs associated with it. But Benjamin Bloom came up with this classification of cognitive processes way back when, in the 1950s. And it is this idea that it is this pyramid, that at the base of the pyramid are these ideas like remember. And those are basic cognitive processes, remember, list. And then it moves up the pyramid to things that require a higher level of cognitive processing. Rather than just pulling facts out of storage and stating them, you're going to do something with those facts. So as you go higher and higher up the pyramid, note that he chose-- really unfortunately-- to call the second level understand. However, there are words that can help you. If it really is understand that you want to measure, there's words that describe that a little bit better. They're in your table, and they're also in that second table that I handed out. So arrange, list, label. For understand-- describe, relate, recognize, explain, those things. Gordon. AUDIENCE: Just a little [INAUDIBLE]. When it comes to understand, maybe if you put it like [INAUDIBLE]. JANET RANKIN: Understand to what? AUDIENCE: Understand to an [INAUDIBLE]. Maybe that's what we can measure. Understand to this level or to this [INAUDIBLE]. JANET RANKIN: But how, can you give me a specific example? AUDIENCE: I wish I could go back to the last slide [INAUDIBLE] can describe [INAUDIBLE]. JANET RANKIN: OK. So I see where you're going. Does someone have a suggestion? So think about that laundry. Think about that description of doing a laundry. Can somebody take what Gordon said and get where he wants to go, but get there in perhaps a more direct way. Katherine. AUDIENCE: [INAUDIBLE] JANET RANKIN: Exactly. So if you're going to do that, and I see where you're coming from, understand so that you can describe, in order to describe, or to the level of describe. Just forget the understand part, and just say describe, or define, or whatever it is you want. But yes, I mean, that first step is saying, what do I mean by understand? And when you say that, then it generally rewrites itself. And so the more direct you can be, generally, the better. And keep that laundry example in your head. Yes. AUDIENCE: You're asking a student [INAUDIBLE]? JANET RANKIN: So I would say, well, let's just open that up. What do you think about that, of asking students? So do you understand this stuff about learning outcomes? AUDIENCE: I think as teachers, when you, in the front of the class, you're teaching students and you say do you understand, they say yes. But you see from psychology you know who doesn't understand. JANET RANKIN: So David says as I ask you that, do you understand? Most people are going to say, hey, I understand, sure. Don't think I'm an idiot. I understand. Right? So I completely agree. Students are not likely to say they don't understand. What else? AUDIENCE: That's difficult for the teacher to [INAUDIBLE] JANET RANKIN: Well, so do you want to say something about that? AUDIENCE: Yeah, I agree. I think the part that we do understand as nonspecific is the learning understanding [INAUDIBLE]. As an instructor you have to know, understand what you're trying to get at, which I think is where the clicker question is beautiful, because you know already what you're trying to evaluate that they're catching in class in real time, if you can see who's getting at the nuance of what you're teaching. JANET RANKIN: Yes. I think that's an excellent, excellent point, that many of these activities that we do, these small group activities, or the clicker questions or other activities that students are doing in class are helping them learn. But they're also formative assessments of whether or not they get it or not, whether or not they understand. And you as the instructor can see it. If I walk around, if I listen, if I look at who answered what question on the clickers, or on the raising your hand, multiple choice questions, whatever methods I used here, I get a better, a much better sense of who understands or doesn't understand. If all I'm doing is lecturing I can't tell. And I can ask do you understand, but they're not likely to tell me. Which is kind of the other point is that sometimes-- and I think you were alluding to this, Gordon-- that sometimes students don't know whether they understand or not. So we need to give them the opportunities to find out whether they understand or not. And arguably, if you're just lecturing and you're a great lecturer-- you're eloquent, you don't stumble, the writing on the board is awesome, all this stuff-- students will think they understand maybe when they don't, because they're not having to confront any misunderstanding, and it all seems very nice. So these activities where you actually have students do something, say something, where you ask them a specific question, that's when you can determine whether or not they understand or not. And because they are not likely to say they don't understand, either because they just don't want to admit it, or because they really don't know whether they don't understand. So you need to give them measurements to find out whether they get it or not. Yes. AUDIENCE: Sometimes, for example lecture and the activity has [INAUDIBLE]. For example, we have lectures and the teacher just [INAUDIBLE] and then we have expectations as we do exercises. So do you recommend that we also do some activities during the lecture or just [INAUDIBLE]? JANET RANKIN: No, do them during lecture. I mean, the model that we're trying to do here is the model that you should do even if you have a hundred or 200 people in your class. You can use clickers. You can use pair share discussions. You can do all sorts of things to break it up, and it's effective. There's a paper coming up for next week, a meta-analysis, active learning strategies, how they actually, students learn more in a lecture if you actually ask them to talk to each other, break up, share, et cetera. And I think when we talked about misconceptions, remember that example with the coin? How many students got it, could draw the free body diagram before and after? I mean, before and after in a straight lecture, there's hardly any learning gain. But if you incorporate active learning, the learning gains go way up. So that's Bloom's Taxonomy. And an interesting take on Bloom's Taxonomy, there's a woman, Kathy Schrock, who-- it's a nice website and the slides we posted so you have the link-- but she decided, she has this theory that it's not really so pyramidal. That everything really is dependent, these cognitive processes are dependent on each other at a certain level. And so she's drawn it as a gear, where everything sort of supports the act of creating, which is up at the top. But the other ones are sort of around the side, but they all contribute to the ultimate cognitive process of creating. So that might be a nice way to think about it. And then she's taken the specific words, like analyzing, and said, OK, what's an instantiation of analyzing, outlining, deconstructing, organizing, structuring, surveying, whatever? So it's the same idea, it's just a different graphical representation. So what you've done is you've taken your topics, put them in those cells in the table. And I just want to point out kind of how as the instructor you have a lot of power about where you're putting those, where you've decided to put those. And it may be different for a first year undergraduate class, or a second or third year graduate course. It could be the same topic, but you're going to put them in a different category. So that's why when I asked you to bring a topic, it's for a specific class. It's for a specific group of students. So let's say we take the idea of interstitial sites. So I have a crystalline structure, it's got an order of periodicity of atoms. And then I have spaces between the atoms, those are the interstitial sites. And so I could say you'll be able to identify the interstitial sites. So I could say where's the tetrahedral site in a face-centered cubic or something, whatever? And students just really have to find it. I could say calculate the maximum size of an interstitial atom that could occupy that space, an ion that could occupy that space in a particular crystal structure. So those are two very different levels, but it's the same topic. X-ray diffraction, I could say apply Bragg's Law to calculate d-spacing. So that's very much a turn the crank kind of thing. I know the equation for Bragg's Law, I stick things in, I solve for the unknown. I could then also ask you to index unknown diffraction patterns, meaning identify an unknown crystal structure based on the diffraction pattern. That's a completely different activity, completely different set of cognitive processes than just using Bragg's Law. But you have to decide what it is. And arguably, if all you've had students do during the class is sort of problems that they applied Bragg's Law and they solved for d-spacing-- so they solved for the incident angle or they solved for the wavelength or whatever it is-- you really can't throw this at them on the exam, right? Or conversely, if you spent the whole time, they've indexed, they've been in the lab indexing unknown diffraction patterns, and then this is the question you ask them on the exam. Well, then you have people that perhaps get it without really, perhaps pass the exam without really getting it. So that's just something to keep in mind. The other thing to note is let's say this is my intended learning outcome-- calculate the maximum size of the interstitial atoms in a variety of crystal structures. How do I know whether students get that or not? What's that? AUDIENCE: It's easy. JANET RANKIN: It's very easy. I give them a crystal structure and I say calculate the maximum size of the ion that can fit in the space. So I have figured out how, the measurement is easy, it's cake. It's totally straightforward. Now I haven't told them, OK, I'm going to give you this crystal structure and I'm going to ask you this spacing, this ion size. I haven't given it away, but I have told them what's expected of them. And my colleague wrote these for VSEPR theory. It's the same topic, and it's different. Identify the common geometric shapes found in simple molecules. Explain the assumptions of this theory. Apply the theory to predict 3D structures. Compare and contrast the geometry of a certain molecule as predicted by two different theories. Evaluate the accuracy of each theory for a particular set of compounds, and then create some recommendations. So it's the same topic, but you decide what it is that students should be able to do. OK? So I have some, just a little exercise for us. So they should be specific, measurable, realistic. I'm going to put them up here and we're going to say, you're going to tell me what's wrong with them. So number one, t-tests are like a statistical analysis. So is anybody a t-test kind of person? We can skip that one. How about gaining appreciation for the use of linearization techniques? What's wrong with it? AUDIENCE: It's not specific. JANET RANKIN: It's not specific. AUDIENCE: It's not measurable. JANET RANKIN: It's totally not measurable, exactly. I mean, how do I know that you appreciate it? Don't do it. All right. Great. Have an intuition for the most effective method of integration for a given problem. AUDIENCE: It's vague. It's not measurable. JANET RANKIN: So I've heard it's not specific, it's not measurable, and it's not vague, I mean, and it's vague. So right, so have an intuition, I cannot measure that. I have not figured out a way to measure intuition. I can tell whether you can do it, right, but I can't tell you whether you have an intuition for it. Well, how would you rewrite it? Actually, how would you rewrite 2? AUDIENCE: Use [INAUDIBLE] JANET RANKIN: So it could be use the techniques. Another suggestion? AUDIENCE: Make something using the evaluation techniques. JANET RANKIN: Calculate something. Make a first pass estimate using a particular linearization technique. AUDIENCE: Describe when you would use a linearization technique. JANET RANKIN: Describe when you would use it, describe why you would use it, all of those things. AUDIENCE: List those areas where [INAUDIBLE] JANET RANKIN: List the areas where it's useful, yes. Great. All right. For number 3, the gain an intuition, we decided it's vague, it's not measurable. So how would you rewrite it? AUDIENCE: Compare. Compare the areas integration technique [INAUDIBLE] Solve a particular problem using different kind of [INAUDIBLE]. JANET RANKIN: OK. So solve a problem, and then we need a little more. Did you have something? AUDIENCE: Identify the most effective method of-- JANET RANKIN: Let's hear what Rachel had to say. Rachel. AUDIENCE: Yeah. Learn how to identify the simplest [INAUDIBLE]. JANET RANKIN: Right. So you probably don't want to say learn how. You would say be able to identify. Right? Gordon. AUDIENCE: [INAUDIBLE] JANET RANKIN: Right. I mean, you could just say integrate-- AUDIENCE: [INAUDIBLE] JANET RANKIN: Integrate a problem using whatever. Select a technique to integrate a particular equation. And again, those things end up being more measurable. So 4, provide problem solving tools and strategies. What's wrong with that one? AUDIENCE: It's too wide. JANET RANKIN: It's too wide. AUDIENCE: It's instructor centered. JANET RANKIN: It's instructor centered. What else? AUDIENCE: But it's measurable. JANET RANKIN: Well, I can measure that I provided it. Here, I provided these handouts, right? OK. Check. It happened. Whether or not you could do anything with them, I don't know at this point, right? That's the problem. AUDIENCE: It doesn't tell us what kind of problem and what kind of tools we're going to use. Just who gets to choose the strategies. JANET RANKIN: Right. Right. It's very, very general, totally not specific. So what could we do? I mean, we don't know what this instructor had in mind, but what could you do if this was your class? What would you say? AUDIENCE: What about develop instead of provide? JANET RANKIN: Develop. And again, it would have to be student centered, so the student will be able to develop problem solving tools and strategies. That's for x, exactly. We want to make it a little more specific and a little more focused, because that's pretty broad. I mean, I have developed a problem solving strategy that lets me get out the door without tripping. Does that mean I satisfied your learning outcomes? Probably not. So in order to do something-- Use thermodynamics to solve engineering problems. AUDIENCE: That's not specific as well. JANET RANKIN: It's not specific at all. AUDIENCE: Engineering is [INAUDIBLE]. JANET RANKIN: Yeah. I mean, it's crazy broad on a number of fronts, right? I mean, it's like engineering is huge. Thermodynamics is pretty darn big, too. So which part of thermodynamics, which part of engineering? Build an SAE race car. AUDIENCE: [INAUDIBLE] JANET RANKIN: What? AUDIENCE: It's not realistic. JANET RANKIN: Right. It's pretty specific. Build a functioning SAE race car. It's very specific, and I can totally think of the metric of how I would measure whether students could do it. I would just ask them to build the race car. But they're never going to do it in a class, in one class period, in one course. It's unrealistic, exactly. Learn to use Laplace transforms to solve differential equations. AUDIENCE: [INAUDIBLE] JANET RANKIN: It's funny. I mean, you can imagine the instructor saying you will learn to use that. But let's go, Katherine, work your magic with this one. AUDIENCE: [INAUDIBLE] JANET RANKIN: Right. You're my go-to person for when you want to get rid of all the extra words. I'm going to call on you. Right. Use Laplace transforms. That you can measure. You can give them an equation and see if they can use Laplace transforms to solve it. And yes, you might want to be a little more specific about the kinds of differential equations they use, and know how to upper diagonalize a matrix. Katherine. AUDIENCE: . Upper diagonalize a matrix. JANET RANKIN: Upper diagonalize a matrix. That's all you have to say, right? Forget the extra words. And that is really, I have just written the assessment question for that. I know it's totally measurable. I give you a matrix and I ask you to upper diagonalize it. I haven't given it away. I haven't told you you will upper diagonalize this matrix, right? Memorize these equations that you're going to need to write down in order to upper diagonalize it. But I have told you that that's what's expected of you. AUDIENCE: I have a comment on it. It's pretty useful when a student is trying to evaluate the use of the class. For example, the career fair tomorrow [INAUDIBLE]. There's a million [INAUDIBLE] to get jobs. OK, how are they going to evaluate me? And if you could come in and say, this happened and this is what we did, as opposed to having, or have them evaluate me on some random knowledge. And then I don't know that I could respond, but I do know stuff. So this is really helpful in kind of a broader sense. JANET RANKIN: Right. Great. No, I'm glad. And I think it's so true. And people often think, well, you're giving it away. But you're not giving anything away. You're just telling them the expectations. You're not giving them the problems, you're not teaching to the test, but you're just telling them what they're going to be able to do at the kind of high level, or medium level. You're telling what they're going to be able to do. I will say-- so ABET, the Accreditation Board for Engineering Teaching-- every accredited engineering program has to list, they have A through K outcomes. And if you look it up, ABET-- and I'm sure you see us as accredited-- if you look up, they have to, they used to do a lot more bean counting. But what they would do is the outcomes would be critical thinking and problem solving skills, and they were a little bit broad. But every department that got accredited had to show how various courses contributed to the outcomes. So somewhere in your department should be a pretty detailed list of how 6 double 00 whatever, did this. Or 6 yada, yada-- I know your grad students-- but did this. The problem is they don't have to do it for grad student courses. But it might be helpful if you looked at that, because I would imagine that sometimes the recruiters would be looking at that information, so that might be a commonality. But every program in the US that's accredited has to have, show at least how they've addressed those outcomes. So you might want to take a look at that. But yeah, if it was a little more explicit it would certainly be better. And I'm glad it's helpful. So I think we can see, you can start to see how there are better and worse learning outcomes, that if they're not measurable they're not so useful. If they are measurable, they're quite useful. So now what I'd like to do is revisit the worksheet. And what we can do is we're going to take about five minutes where you just sit quietly and write out specific intended learning outcomes. So you'll be able to upper diagonalize a matrix, you will be able to use Laplace transforms to solve this type of differential equation, whatever you've written in this column. And note that if you put it in this column, let's say you put it in the apply column, then when you write the learning outcome it's going to be you will be able to interpret data using blah, blah, blah. You will be able to model the vibration, model a car suspension system using linearization techniques. They're going to be whatever, wherever you put it in the box. These are some of the verbs, these are the verbs you're going to use in your learning outcome. Does that make sense? And I have even more verbs for you if that's helpful. Did I pass those out? You got those, right? AUDIENCE: Yeah. JANET RANKIN: So there's more verbs. It's the same thing. I mean, it's consistent. So take five minutes by yourself and do that. And then you're going to get into pairs and you're going to discuss. You're going to trade worksheets and give each other feedback. All right? So five minutes of quiet and then we'll pair up. So if you can pair up, you don't necessarily have to pair up with people that are sitting next to you. You might want to get a different perspective, so feel free to get up and move around. You can talk to somebody that wasn't in your group. So maybe if you go back and talk with Alex, maybe. We need one group of three. Oh, are you a group of three? So now just trade sheets of paper with your partner, or if you're a threesome you can figure out how to do it. And then review the learning outcomes critically. Guys, just one sec. So just review your partner's learning outcomes critically. Make sure that they're measurable, realistic, and specific. If your partner has used the word understand, help him or her work through that to get a better word. So you're going to be a critical friend here. And then you guys can discuss how to improve it. And then I'm going to pass around some flip chart paper and each one of you will write, maybe just write your name at the top, and write two or three learning outcomes on the flip chart paper. And then we can stick them up around the room, and then everybody can see everybody else's, all learning outcomes. [INTERPOSING VOICES] JANET RANKIN: Was there a question? AUDIENCE: Oh, no. I just think-- AUDIENCE: I think [INAUDIBLE]. JANET RANKIN: How are you guys doing? Doing OK? [INTERPOSING VOICES] JANET RANKIN: So you can use this zone if you need it. As you're looking at other people's learning outcomes, think about how clear, how easy it would be, or not easy, to come up with an exam question, or a project, or an assessment or measurement of whether or not the student reached, attained the learning outcome. You can put them up here, Ina and Shau. OK. And as I said, make sure that you've read everybody else's. I'm going to say a few words about them. I think we may have a record. This is like the first time ever that no one has used the word understand, which I think you're all to be commended for. So David, make sure you don't write understand. You build me up just to knock me down. AUDIENCE: [INAUDIBLE]. JANET RANKIN: You can, or you don't have to. It doesn't matter. Does anyone have any comments, either about the exercise itself, or about somebody's learning outcomes, that you see around? Again, the measurement, the thing you want to look at when you look at your own learning outcomes and those of others, is are they specific, measurable, and realistic? And would you know if a student achieved those learning outcomes? So I would say about 90% of these are very specific, which is great. I mean, there's some incredibly specific ones here. The more specific, the easier it is to measure. So that's one thing. That's a useful thing. You want to make sure you don't get so specific that there's only like one question you could ask. So that's the balance. Sometimes you can't tell that unless you're in the discipline. Do we have other comments about any of these learning outcomes, something you don't get, something that looks particularly interesting? Yes, Gordon. Perhaps you can sit down if you're done reading. AUDIENCE: I'm looking around and I'm seeing [INAUDIBLE]. I don't know how it's going to be. How are we going to test that? JANET RANKIN: Well, I guess you have to come up with a reaction that you're confident they haven't seen before. So if you can do that, then you can ask them to propose a mechanism for it, I suppose. If the person that wrote it wants to-- I know who wrote it-- but if the person that wrote it wants to talk a little bit about it, that would be great. If not-- AUDIENCE: OK, So a lot of people are going to just try to memorize each specific outcome. But it's easier to just identify your activity group and find patterns within them. So if you can try [INAUDIBLE] They can't just [INAUDIBLE]. JANET RANKIN: Right. And this is a great point. I mentioned it to one of the groups. But sometimes we'll say explain how blah, blah, blah, works. And that's a great thought, that, oh, they're going to be able to explain it, right? But many times they can fall back on a memorized explanation or sort of kind of a canned procedure. So if you really want them to be at the level of explain, you will have to kind of make some things up to get them out of, to get them away from the ability to just pull something completely from memory. So this is an example of how you might do that. Other observations or questions? AUDIENCE: I also [INAUDIBLE] something that all [INAUDIBLE]. JANET RANKIN: So perhaps the person that wrote it wants to comment on it. AUDIENCE: OK. So [INAUDIBLE]. I think what we have, just [INAUDIBLE]. So we write and compile. So you write a code and then you compile it. JANET RANKIN: Right. And this is a great example of how having people that aren't in the field read them can help clarify things for your students, who are by default not experts. So perhaps to an expert, it totally makes sense, of course, code. But to somebody that's not an expert, maybe that's confusing. Maybe the use of code twice was confusing. To an expert, it's totally readable. It's totally understandable. Perhaps to a novice it isn't. You're laughing. AUDIENCE: It's funny, because we're both [INAUDIBLE]. JANET RANKIN: Yeah, that was kind of random. Well, that's a perfect example, right? It's a perfect example of how getting maybe other people that aren't in your field to look at them can help make them more accessible to your students. OK. This is great. Oh, Hina. AUDIENCE: I had, it's on a different topic. But really, I'm talking about buzz words in learning outcomes. I think with engineering, you're really excited about telling people about real world problems. I guess that's the reason why. I guess it's more of an opinion question. As you write your learning outcomes I think in some ways it's better to be specific about the real world problems that they're going to attack. Because I've seen that on syllabuses all the time. I'm very excited about it. I think it's a way to teach people in your class. It's going to help them know what tools they're getting. JANET RANKIN: Right. I mean, real world problems in particular really do kind of encapsulate a whole pile of hopes and dreams. So it may help to be a little more specific. You can use it in your learning outcomes, but then maybe describe it in a little more detail in your syllabus, the kinds of real world problems, or why we care about real world problems. They're messy. They're ill-defined. What about them is so important? What about them is so important? Try to minimize the buzz words. But at some level, they may not know about something until they're completely through with the course, and you may have to use that word. So there's a bit of a balance. What I'm going to do before, I'll probably bring all these back to my office, take pictures of them, and put them up on the wiki so you'll have a record of them. I wanted to just as we go out here, my point is really that the hard part is these learning outcomes. And that once you write the learning outcomes, as we said before, you kind of know how you're going to test whether students got them, and you know what you're going to do to help them get them. The syllabus itself is just an articulation of that. It's a description of the course, an articulation of the learning outcomes stated, like the laundry example, with the title. This is what you're going to do. By the end of the class this is what you'll be able to do. It's kind of a promise that that's what you're going to do. It motivates students to take the class, to stay in the class, to engage with the class. It keeps you on track. So at the beginning of the semester maybe I'll be very clear, but four weeks in, and you're a little bit fuzzy, because you've been dealing with all sorts of stuff. And you kind of go, what the heck. What's this in here for? Well, if you've lined everything up at the beginning using your learning outcomes as the anchor, then you don't have to worry about that. You just follow the map. So it really, really helps you and the students get through the semester in a logical way. And it tells the students what they can expect of you, what you can expect of them. So it's motivational, structural, and it's evidence, hopefully of what they did, but certainly of what you hope to do with them. So for the post-session assignment, you'll have to write up more learning outcomes from your course. You can use these as the basis for sure. Hopefully, that table will help you as you write more of them. And I will go in and give you feedback. I want to make sure that everybody knows that you should read each other's postings. So a lot of you had great examples from the first class about how you would use these learning theories in your courses and how you teach. They were fabulous examples. Make sure you read those postings of others. That's one reason why it's up on a wiki and not just you handing it in to me. So you have the Mud cards. Fill them out if you have any more information that you want. And otherwise, we'll see next week.
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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 equations and how to teach equations in a way that promotes long lasting learning and understanding. OK. So the first example-- I'm going to give you two choices for starting the example. So this is example one for teaching equations. This is Hardy-Weinberg equilibrium. So here's one way you could start. So if a locust has n alleles and the organism is polyploid-- so that means it has, let's say, C copies of chromosomes. OK. So there's n alleles. So the allele frequencies are P1 through Pn. Then we're going to deduce the genotype frequencies. So this is the multinomial middle coefficient C. Choose K1, K2, K3, all the way to Kn. And K1 through Kn are the number of copies of each allele with K1 plus blah, blah, blah, to Kn, all equal to C. So you could actually state that and then prove it. That's option A. So now I'll give you option B for starting the same topic. So option B is from Hardy. So in Hardy's Mathematician's Apology, he writes, I have never done anything useful. No discovery of mine has made or is likely to make, directly or indirectly, for good or ill, the least difference to the amenity of the world. That's a very interesting, very bold statement. And amazingly, for such a brilliant mathematician, it's wrong. And today, what we're going to do is we're going to look at an example of the discovery by Hardy-- the Hardy-Weinberg equilibrium-- that does make a difference to the amenity of the world and helps you actually understand genetics. OK. So the option B is to quote Hardy and then point out the contradiction between what he said and what we're going to do. Now, also, in option B, you can go a bit farther. You can say, well, why would Hardy have said something like that? And indeed, why would he have said something like that in his Apology? Well, it's actually probably hard to appreciate on this side of the Atlantic, because since the Civil War, we've never had a war that just basically devastated entire countries on our own soil. Whereas in Europe, the memories of World War I are very strong. So if anyone's from Europe, you know that. In every English chapel, there's names on the wall of all the people who died in the, quote, Great War. So World War I had a very strong effect on European society. And one of the effects was on European science and people's attitude towards science. So poison gas was invented partly by German chemists. Hopper was one of them. And in the end, he committed suicide, partly maybe because of what he had done. No, sorry. His wife committed suicide. I forget if-- I should check that. But basically, people in the family were so unhappy about what had happened just there that there was a suicide. And furthermore, science after that was considered-- World War I was considered the chemists' war. So there was a wish to distance oneself from those kinds of horrible effects. And the quote from Hardy is actually a reflection of that. He himself was very strongly anti-war. He left Cambridge because Cambridge fired Bertrand Russell for protesting World War I. So Hardy left Cambridge for a professorship in Oxford and only came back basically 12 or 13 years later to Cambridge. So that statement of his partly a wish that he's hoping nothing he's done has any effect on the world because in his mind, maybe most effects are bad. Well, but actually, he did have an effect. The effect is Hardy-Weinberg. Even if you discount everything you did in number theory, there's Hardy-Weinberg, and it has an effect and we're going to look at it. So let me ask you rhetorical question. If you're a student, which would you find more engaging, more inviting? This one or this one? Who votes for that one? OK. Who votes for that one? Yeah. Now, why? So now think of-- so the question I'm going to ask you to think about just for a minute with one or two of your neighbors is why. So yeah, this is definitely more engaging. In terms of the principles we talked about last time, what makes this way of introducing it more engaging, more likely to bring students in, more likely to make them want to learn about Hardy-Weinberg. OK. So find a neighbor or two we think about the principles behind option B. OK. So I'll rudely interrupt you. And good thing I did the voice exercise so I can project all the way into the back. So what reasons for option B or against option A, which is the same thing? Yes. Can you tell me your name? AUDIENCE: Brian. PROFESSOR: Brian. I'm going to try to learn people's names. So go ahead. AUDIENCE: The reason against option A-- when you start out that way, you are activating the ability that students have to transfer material from the blackboard to their notes and then pass it to their brains. PROFESSOR: Right-- through the wave. Yeah. I'm activating the ability of the students to basically take dictation. Or as someone said, teaching is an excellent way to transfer material from the notes of the teacher to the notes of the student without it passing through the minds of either. So I'll call that not A. So that's the not symbol. Not A is-- so I'm basically asking to do dedication which doesn't necessarily lead to any kind of learning. They could, for example, copy all of that down but not really understand of it. OK. Other-- yes, can you tell me your name? AUDIENCE: Susannah. PROFESSOR: Susannah. AUDIENCE: This is right brain? So it's easier to remember and ingest it. And maybe even identify the theorems. PROFESSOR: OK. Right. And so B is the story. I'll give you an example of how stories can be so powerful, and just even the word. So a big commercial publisher-- this was several years ago-- they wanted me to write a freshman physics textbook. I had actually just put up a proposal on the web saying, we should write a freshman physics textbook that is based on the history of science. And they saw it and they thought, oh, that's great. So they flew me from England to California to talk to them. And they liked it mostly. But they said, history-- so there was the word. They liked the idea as I talked about it, but the word history really frightened them. And then, I'm still amazed-- I had this amazing insight of just two letters. I thought, oh, wait. I can actually explain to them what I mean. I said, well, actually, if you just take away that part, what I'm really talking about is that. And it was interesting. As soon as they saw the word story and they saw that history doesn't have to be names, dates, facts-- I think that's what they were seeing it as-- they saw it as story, which is actually its origin in French or Latin-- [FRENCH]. They all of a sudden were totally convinced. They said, oh, yeah. That's exactly how we should do the textbook. For various reasons, basically because I wanted to be a freely licensed book, we didn't actually sign a contract. But it's an example of how powerful story is and how people who actually spend their lives thinking about teaching and reaching students-- in other words, this educational publisher knew the importance of story. But if you present it as just dry history, it's not so interesting. So story-- yeah. So let's see. Who haven't I heard from? I haven't heard from Adrian. Yeah, and then you're next. AUDIENCE: So the fact that it's a contradiction creates some sort of tension from linear thought. And it's more of a exploratory. PROFESSOR: OK. So there's a contradiction. And the contradiction creates some kind of tension. So the tension creates interest. So every good story needs some kind of tension. So the tension here, the contradiction, is that Hardy wanted to do nothing that could harm people, basically. And so he was like, OK, well, anything could be used, even if it's for good, for ill. So let me just back off from all of that and say I'm not going to do anything that has an effect, an application. I'm a pure mathematician. His most famous book is A Course of Pure Mathematics. Well, maybe his most famous book is that book that I quoted from, A Mathematician's Apology. His most famous math book is probably A Course of Pure Mathematics. And that's what he wanted to be known for. And that's in contradiction with the fact and no, even Hardy couldn't help an application-- an actually very, very common one. It's taught in every single introductory biology course, probably. OK. So you need tension. Now, that's a general principle of learning there-- the idea of story and tension and paradox. So the one way that, as a undergraduate in physics, I learned a ton of physics, me and my friends were doing problem sets together. And a lot of problems were just grinding through math. So we didn't learn a hell of a lot from that. But we were in the library late at night, the physics library, ordering pizza, and trying to do a problem set, and we just got to making up physics paradoxes-- perpetual motion machines. We'd invent perpetual motion machines and try to get the other person to figure out what was wrong with it. Sometimes, we didn't even know what was wrong with it and we'd both try to figure it out. So from that kind of tension-- tension is almost, in a way, self teaching, because as long as the tension's there, you know you're not at the end of the story. You know there's more to do. So the same thing-- as long as there was perpetual motion going on and we hadn't found the reason, we knew we weren't done with the problem. We didn't have to ask a teacher to say, well, is this right, because it was basically self teaching. As long as the tension was there, we knew that we weren't finished. So yeah-- another general principle, stories and tension. Let's see. There was a comment over-- yes. Tell me your name? AUDIENCE: Wing Ho. PROFESSOR: Wing Ho? AUDIENCE: Yeah. So this along the line that the teacher should make themselves more human, because Hardy-Weinberg sounds like such a big name you would tattoo. That student may be driven to believe in the equation just by the matter of the fact that it's authority. But then making a story makes-- PROFESSOR: OK. So the teacher makes himself more human and makes one of the people who invented the Hardy-Weinberg equilibrium more human. So it's actually something much more easy for the students to connect to. Yes, can you tell me your name? AUDIENCE: Gregor. PROFESSOR: Gregor. AUDIENCE: Yeah. So my problem is that two different options are very, very different. And the amount of legwork it is, it's maybe more tense to me, because you can discuss it without legwork, but I think also that option A is pretty interesting, because you can approach this biological problem from more of a qualitative approach. So depending on what kind of learner you have and what kind of question you choose, A would be also very interesting. But at that point, with the background I have right now, it's very difficult to pursue that. PROFESSOR: OK. So Gregor's point, which is a good one, is that B doesn't have the same quantitative depth as A. And actually, what I'm going to show you is a way to get to this starting from here and using the principles that we're talking about so that by the time you get here, or to something like this, it actually makes sense to the students-- so that you can have both. So that's a promise. And hopefully, I'll deliver on it for you. But does that address what you're talking about? AUDIENCE: Yeah. PROFESSOR: --that B doesn't have the quantitative depth-- and that's true. AUDIENCE: Actually, I mean, you can talk about one hour on B. But if you really want to deliver an idea or a concept, that is as quantitative as option A, then you have to get to A at some point. PROFESSOR: So the point made is that if you want to get here, you do have to say this at some point. You can't just say the story for now. Well, it depends on the purpose of the class. If the purpose is the history of biology, maybe you'd continue with the story or the history of science and war. But if you want students to be able to solve problems in genetics, maybe you need to go here. Or what we'll find is I'll continue with option B moving towards A. And I'm going to show how you can get there and still have the opening of B preserved. OK so let's see. There was-- yes. Can you tell me your name? AUDIENCE: Roger. I think one of the reasons we like B is because we see A all the time. So I don't know to what extent that's [INAUDIBLE]. PROFESSOR: OK. So B is new. Right. So against A, A is very familiar and common. So in fact, yeah, I chose this-- I didn't really make a straw man here. This is how a lot of things are introduced. For example, in mathematics class, theorem proof-- so the theorem will be introduced without any of the struggle, the wondering that led to the theorem. Why would anyone even care about such a thing? Or if they cared about it, why would they come up with something like that? How can you see that? So yeah, A is seems very familiar and therefore, maybe not as interesting. Although, it may also be intrinsically less interesting for the other reasons. But I take your point, too, that it could be just familiarity. Yes. Can you tell me your name? AUDIENCE: Yeah, I'm Meg. PROFESSOR: Meg. AUDIENCE: It seems that you could really incorporate the quantitative side into B if you wanted to by having the way the proof occurred [INAUDIBLE]. And then you wind up reversing the traditional order, where you have a proof and then a theorem, because the way it happened in reality and the way it happens when you're going to do research, and the way it happens to students when they're trying to work it out themselves. And so I feel like that's much more of a natural flow [INAUDIBLE]. PROFESSOR: OK. So your point, which is an excellent one as well, is that B doesn't preclude the quantitative. What you could do is you could talk about the history, start with the history of Hardy saying that, which is not chronological, because he said that in 1940. But that's OK. It doesn't have to chronological just because it's history. Go to 1940 and then backtrack to well, how did this problem actually come to Hardy's attention, which actually turns out to be quite an interesting story. And then talk about how they solved it and what was the publication history. And all the quantitative ideas would come out in that way. And it would be almost backwards from the usual way. The usual way here is completely general. And that's probably not the way it was first figured out. And the advantage of telling about it that way is that you're preparing students themselves for a research career, because nobody comes up with theorems full blown like Athena out of the head of Zeus. You come to them from struggle and wondering. Hm, I wonder. Everything has some historical background to it. So it's called the, you could say, the genetic approach. So in biology, there's a slight, not misconception, but a saying that ontology recapitulates phylogeny-- in other words, that the organism, as it develops, say, in the womb, goes through all the evolutionary stages that it went through over the last 200 million years or whatever to become, say, a person. You're at first a fish and then maybe you're a monkey and then you're a person. So roughly speaking-- it's not exactly true. But there's a lot of truth to that for learning ideas-- that you have to recapitulate the history of ideas to really understand where we are now, the ideas today. So a great example of that is Newton's Second Law of Motion, the idea that force and acceleration are connected. So for thousands of years, millions of years, people thought force and velocity were connected. And it's actually force and acceleration. So actually, you can guide people to the understanding of force and acceleration being connected by showing them the history of how people thought it was force and velocity, because for those same reasons, those are the reasons that students will think it, too. So you're actually helping them overcome their misconceptions. So the history actually, generally speaking, helps overcome misconceptions. So I'll put that here. And the history is actually quite interesting. I think it was Punnett from Punnett squares. He played cricket with Hardy and Hardy loved cricket. And he just asked him about this problem. And Hardy said, oh. Yeah, no, it's just this, and sent off a paper to Science or Nature. So basically, it's because they were cricket playing colleagues in Cambridge, that's how a mathematician ended up interested in a biology problem. Punnett was a biologist. OK. So any other reasons? Yes. AUDIENCE: I'm going to play devil's advocate for a second here. PROFESSOR: Sure. Can you tell me your name? AUDIENCE: Paul. PROFESSOR: Paul. AUDIENCE: So when I had biochemistry the first time, the first month of the class, the teacher would tell the history about the people developing quantum mechanics. And that was the lectures. We didn't touch the material one bit. And because his philosophy was the material was in the book, you don't need to teach it. You just go home and read it. But I remember being in that class-- and maybe that's because I normally don't see things taught like that-- but everyone was outraged, because we didn't feel we learned anything. PROFESSOR: Right. OK. So the comment is that you were in an actual class that was purely taught about the history. People were outraged. They felt they weren't learning actual content. And there's two answers to that. One is that actually, I don't recommend if, say, you're teaching a physics class, that you teach it purely historically. But to the extent that you do teach the history, you can actually teach the content through the history. So it's not that the history and the equations or the ideas are separate. It's that you can use the history as a means of making the equations come alive with a deeper understanding. So it's possible that the teacher you're talking about actually didn't do that and just talked about the people who did this and did that. But he didn't really know what they did or why. So you're not forced to do that. OK. So I'll give you an example. I'll continue with this in just a moment after I take any other questions, showing you how you can start with the history, and then you can show the content. You can actually show people exactly why that would be true. And I'll show you that in one second. OK. There was a question there. Yes. Sharon. Oh, sorry. Not Sharon. Can you tell me your name? AUDIENCE: Me? PROFESSOR: No, in front. AUDIENCE: Cecilia. PROFESSOR: Cecilia. Yes. AUDIENCE: I was wondering that Option B-- it's like targeting another audience. I come to a class and I'm not very interested in the class. I need to be motivated here, Genetics or something. If I already signed up for genetics-- because if I were being in a class interested in that, I actually want to learn things. I want to know the rules. So I don't know if I want to be distracted by some story that is-- when you first said that, I was like, oh my god, if this is going to be some other big story, he's going to have the equation at the end. I'm not going to be understanding what's going on or whatever you say. And I think that could be a little bit of a turn off more, because-- try to make it look easy. You have avoided writing the equation on the board. PROFESSOR: Yeah, I'm going to do that next. AUDIENCE: OK. And also, I guess there are two kinds of stories you can tell. I don't know what story you're going to tell exactly, but this seems to be removed from the actual equation. This is more Hardy's philosophy. It's not really-- PROFESSOR: It's not directly about that. AUDIENCE: Something you did a second ago was more about the equation and more about the concepts involved with the equation. PROFESSOR: That's true. So one question is how related should the stories be? And there's a lot of freedom in that. I would say the answer is that you want it-- if it does this, if it creates interest, then it's already done something. Now, your other point was that it depends on the audience. Suppose the audience is all graduate students in genetics or people who want a crash course in genetics. Maybe actually, they want to just know the formula. And they would actually be offended if you started telling them stuff that wasn't the formula, because they feel like they're time's being wasted-- a little bit like what you were saying. So now, did you have a comment about that? AUDIENCE: No. PROFESSOR: OK. Then I'll come to you one second. So one comment about that is that actually, most of the students, at least in a big class in genetics, are going to be students who are-- for example, at MIT-- are there not because they love genetics and are going to continue in genetics. They're doing it because they're required to take introductory biology. And your job as a teacher is to actually show them that this is a really fascinating subject. It's the same in physics. It's often taught as if the only students in a physics class were the physics majors. Now, when I was an undergraduate in Standford, there were 1,600 undergraduates and 12 physics majors every year. So 1% of the student body was physics majors. So they were actually teaching in a way, generally speaking, for 1% of the student body. What about the other 99%? It was really important for them to reach them as well-- important, also, for the health and growth of the field, because the field needed people who, even if they weren't professional physicists, saw the value in the field and the value in physics. So generally speaking, it is actually a wise way to reach all of them. Now, suppose you have people who just want this. The next thing I'm going to do will show you a way of teaching to them that they'll actually learn this better, because I guarantee you, almost everyone, if you just tell them this and they take dictation, now they go away, and you say, OK, everybody, what did I write down? Most people can't reconstruct it. They'll say, well, was this a C or an N? In fact, if you look in Wikipedia, the Wikipedia entry is incorrect. Actually, I think it has an N over here. It seems plausible, but if you actually understand the equation, you think, that can't possibly be right. And I think it has an N over here. So they actually do not really have a command of the equation. So even for them, you want to not-- the story isn't the proof of it for them, but the continuation will be valuable for them. So again, I'll promise that. Now, there was a question over there. Yes. AUDIENCE: The story, you can basically use to-- and if you see you lose your students when you go on this quantitative approach, then it's maybe helpful to loosen up a little bit and basically bought you interest to see what's a big context. And the other point was-- PROFESSOR: So you're first point was that you can add stories as needed. AUDIENCE: Yeah. So depending on what it's like. And the other thing is when I make my comments related to the problem, I see it more from a various student point of view. So you point out that, depending on what the audience is, you teach differently. PROFESSOR: Yeah, that's true. Yeah. AUDIENCE: So when you give your examples-- PROFESSOR: Oh, I should tell you who the audience is. AUDIENCE: --interesting to know at a certain point what kind of audience it was. PROFESSOR: Yeah, OK. Good point. AUDIENCE: So I'm undergrad. So I don't know how it is to teach first year university. I've never had students. So how much basics or how much story you have to put in compared to how much math. So it would be very helpful to see what our target is. PROFESSOR: OK. Your comment is that it would be helpful when I'll give you an example like this to say, OK, for the particular audience, what would you do? Yeah. And I'll try to do that. That's a good point. Now, there was another question-- behind Cecilia. Did you still have a comment? AUDIENCE: Oh, I just wanted to mention that in choice A, there's a lot of jargon. PROFESSOR: OK. So there's a lot of jargon in choice A, which makes it much harder to understand. You have almost-- for example, the multinomial coefficient-- you have to keep that in your head, as well as all these subscripts, keep it all in your hand and try to manipulate that object-- very difficult. One of the points about that related is the chunking. So the chunking paper-- the idea is that these, for a student, each of these things is going to fill up one of the chunk slots, almost, because it's all new to them. You've flooded the chunking system. You've overflowed it. And they can't actually manipulate this as one object anymore, because it's far too many chunks. Now, you, as a professional in the field, are like, well, of course. It's a multinomial coefficient. What else could it be? For you, that's one chunk. For the student, that's is it C or N? K1? K? Why K? What the hell is K here? Is it commas or spaces? Parentheses? What about brackets? I've never seen something like that. Shouldn't there be as many things up here as here? So the way they look at it is completely different. Every symbol is almost a chunk. So this is just massively overflowing with chunking. So this is sort of related to the jargon. Yes. AUDIENCE: Scott. PROFESSOR: Scott. AUDIENCE: I have a meta comment. In this example, if you don't know how you're actually going to complete these, it's created a lot of tension. I'm dying to know how you're going to do it. PROFESSOR: OK. Fair enough. Would you be very offended if I gave everyone a break for a few minutes and then finished it? AUDIENCE: OK-- PROFESSOR: OK. So finish your comment. AUDIENCE: My question is in a situation like this where you have people who are making so many comments, and the other half's just dying to know, how do you know when to stop? PROFESSOR: Good question. So how do I know when to stop? Because I don't want to exhaust people's patience, but I do want to take questions. And in this case, there's the American teaching phrase saved by the bell, which is that I don't have to make the decision too hard, because it's probably time for a break anyway. But how do I make the decision? Partly, I listen to people's comments. For example, suppose I got yet another comment saying, well, we still don't know the equation. And I'm saying for the third time, oh, wait. I'll show you a way to get to the equation. I think, you know, it's probably time for me to do the equation and then I'll take more questions later. So listen for the tension in people's voice, which is also a good reason to have everyone do the voice exercise first and free the tension so that if the tension creeps back in, you know it's something you've done. OK. So I'll take more questions after the break. But it's 10:02 by that clock. 10:05, we'll start again. You can jump up and down, do jumping jacks or whatever it takes to get the blood flowing. I'll take a couple more questions and then I'll show you how to continue it to get towards that in a perceptive way. OK. So let me, as promised, continue along the lines-- not continuing the story, but continuing that approach, the alternative mirror image approach of A, which is to say, well, what would I do after telling the story? So first of all, I would try to fix some of the problems in just basically blasting people with that. So I try to make it as clear and unjargony as possible-- so as you notice, jargon up there. So imagine a gene with two flavors-- sickle cell or not. And two chromosomes-- so just like people have two copies of each chromosome. OK. So now, before I continue, what have I done just by doing that? Well, first of all, I've made something concrete. It's sickle cell or not. So right away, you can imagine it. So that's continuing the idea of a story. It's much easier to imagine a concrete situation. Either you have sickle cell anemia or you don't. Or you have a gene for it or a gene that doesn't cause it. And the two chromosomes-- so rather than having C chromosomes, I'm just restricting it to two. Now, you might think, well, that's a terrible restriction. That's a specialization. Therefore, it's bad. No. Actually, it's good for that reason, because it makes it possible to understand, in the next step, the idea behind that equation. And once you understand the idea, then that equation is not so mysterious. And furthermore, what's nice about two chromosomes? Yeah, it's a specialization, but-- AUDIENCE: Same two. PROFESSOR: Yeah, we all have two. There's some plants that have more and some plants have less, I think. But people have two. So it's already interesting to us just for that reason. OK. So then the question is how frequent are the combinations? So we want to answer how frequent are the three combinations sickle-- lowercase S is not sickle. So this is S. This is lowercase S. What happened to the fourth combination? Why are there only three? Yeah. It's the same. S, S-- this and this are the same. OK So I would actually ask that, too. Any time there's something interesting-- it should have been four. Plausibly, it could have been four, but it's actually three. Why three? OK, it's because we're actually lumping these two together. OK. So how frequent are those three combinations? Well, that's provided by this formula. P squared, q squared, 2pq-- so p and q are the frequencies of S and little S. So this is the-- OK. So now, because you've asked them about the three versus four, this two makes sense, already somehow. But how could make sense of the p squared, pq, and q squared? And so that's where I would continue with the following, which is this. So here is 0 to 1. And this is p. And that's q. So this is the frequency of sickle cell. This is the frequency of not sickle cell. So right away, you have p plus q equals 1. I'll draw this up here too. So I make a square. And I have four regions. So it's not magic. Why am I making a square? What about the problem tells me square? So I would ask the class that. I'm looking for area. And why an area? Why two dimensions? AUDIENCE: Because there are two chromosomes? PROFESSOR: Because there are two chromosomes. So this is your C. The number of dimensions is eventually going to become this thing, so you can see how we're going to get there eventually. So I'm looking for an area, because I have two chromosomes. So now, let's just look at these various areas. Here's a p squared. What's this area? There's q by p. So that's a pq. That's also pq. And that's q squared. OK. Let's look at all our pieces. Oh, we have a p squared there. We have a q squared there. Oh, and pq, pq-- 2pq. So in fact, this picture explains the entire set of frequencies here. And if you see this picture, there's nothing really new to understand here. In fact, all this picture is showing is that-- it's a picture of p plus q squared equals 1. So if p plus q is 1, its square is also equal to 1. And you break this up into the four terms, group two of them together, and you get three terms, three different kinds of terms. You get three different frequencies. So now, this square is actually very easy to use. Suppose someone tells you that sickle cell anemia, people who have both genes, are 1% in the population. Well, what fraction of the population has no sickle cell gene that all? OK. Well, we can just look at the square. The information was that this is 1%. So that means this 0.1. And that's 0.9. That's 0.9. So this here is people with no sickle cell at all. It is 0.81. And so the square actually makes everything really easy to understand. So now, the question is how do you generalize it? Well, what are some of the-- the first generalization I'll do is I'll say, suppose that there are-- well, I'll ask you. Should I generalize it to more dimensions or more different copies of the gene? What's easier to draw? More copies, because more dimensions-- I don't know how to draw in three dimensions. But I can draw three copies. That's pretty easy. So now, take a piece of paper and just on your own, draw this same figure for three copies of the gene with frequencies p, q, and r. So people have a picture? Does anyone want to describe their picture to me? Yeah. AUDIENCE: I did a three by three square with p squared and q squared and r squared diagonal [INAUDIBLE]. PROFESSOR: OK. So I do the same. I make a p, a q, and an r. And then I do the same here-- qr. OK. So there's a p squared, q squared, r squared. Now, how many guys-- so we have nine square, sub squares. Three of them are this. There's six more. Now, which of them look similar? So this is a pq. Are there any other pq's? Yeah, one other, right? pq over here. . And here is a pr. Is there a pr? Yea, another pr over here. And here's a qr and a qr. So we have p squared. And that's that guy. We have q squared, r squared, and then 2pr, 2qr, and 2pq. So that's the generalization to three copies of the gene. So it turns out that all these coefficients-- 1, 1, 1, 2, 2, 2-- those are binomial, multinomial coefficients. So now, let's generalize one more. So what have we done? There, before, we've written out this. And those are the nine terms grouped into six. So this was our C. So now, we know how to generalize. p plus q plus r to the C equals 1. If this is two chromosomes-- this is C chromosomes. So this is C chromosomes with three copies. Well, what happens if we have n copies? p1 plus p2 plus one of those plus pn to the C equals 1. And all you have to do is expand that out. You can use a square in higher dimensions. Or you can actually use formulas for math. And that's where these come from-- p1 to the 1 power, p2 to the next power, all the way to pn to the next power. And these guys are the coefficients that count the multiplicity. Yes. AUDIENCE: Why do you have to add up to 1? PROFESSOR: Oh. They have to add up to 1 because p plus q plus r, if you only have three flavors of the gene-- either you have sickle cell A, B, or C, let's say-- then these are probabilities, probability p. AUDIENCE: Oh, so parentheses says how often it's done? PROFESSOR: Yes. Frequency and probability-- so p plus q plus r equals 1. So if you square it, you still get 1. So you start from the idea that p plus q equals 1, which is what we did over here. So p plus q is equal to 1. Then you square it. It's still equal to 1. So now, you just another way of writing 1. So it turns out that Hardy-Weinberg is all just fancy ways of writing 1. Starting from this picture, we've successively complicated it. This is C equals 2, n equals 2. There's C equals 2, n equals 3. Here is the same thing again. Here is n equals 3, general C. Here's general n, general C. OK so what we've done is we've basically got here-- by stages of successive approach, one step at a time. So at every stage, it's clear what is going on. And what's the core idea? The core idea is the one you just asked a question about, which is that the frequencies add up to 1-- p plus q equals 1-- so if you square the frequencies, you still get 1. And there's nothing more to Hardy-Weinberg than that. Yes. AUDIENCE: I'm Julie. PROFESSOR: Julie. AUDIENCE: My question has to do with the original way you presented the problem-- using the word flavors instead of alleles. And I've always been fond of teaching that you should always use the probable vocabulary of your students. PROFESSOR: To switch. AUDIENCE: So you got the same thing as the-- PROFESSOR: Yeah, good question. So I used flavors instead of alleles-- so right here, imagine a gene with two flavors. So actually, probably the best way to do it is to combine the two. So you say flavors, because-- so this is a question of transmitting information to the student without noise on the channel. So if you say the word allele-- so this is again related to chunking. If you say the word allele, the problem is that now, you're expecting them to try to understand this new idea as well as this new item for taking up one of their chunks that they have available. So when I initially presented it, I would use flavors. And then I'd say, OK, now we understand the whole idea. The thing has kind of seeped-- it's not really part of short term memory anymore. It's connected to something else that they know-- for example, this. p plus q equals 1, so p plus q squared equals 1. Now, it's not taking up so many chunks anymore. Now, they're ready to hear the word allele. So I'll say, OK. Colloquially, we could say flavors, but actually, the word in the literature is alleles. So at no point, you're overloading the system. So again, it's philosophy based on, say, history and chunking. Yes. AUDIENCE: When you presented this whole second way, you made, actually, use of the equation that you wrote down in the-- PROFESSOR: Oh, over there? AUDIENCE: Yeah. PROFESSOR: Oh, no. I was just saying that for your benefit. AUDIENCE: But I thought, actually, for me, it was very useful to see what the kind of equation was. And then as you were going through the argument, I could reason out myself-- PROFESSOR: Good question, good point. AUDIENCE: --where everything was coming from. PROFESSOR: Yeah OK. That's a good point. So it actually was helpful to actually see the final goal to know where you're going. Actually, that's a good suggestion. It's actually not bad to say, OK, this is what we're going to try to understand. I don't expect to understand it now. It's full of all kinds of squigglies. So for example, maybe a really good way to do the whole thing would be to start with the Hardy story and say, we're going to talk about Hardy-Weinberg. Say, what does Hardy-Weinberg say? Well, in its full generality, it says, blah. I don't expect you to make any sense of that right now. Let's talk about what the core ideas of it are. With this, you approach it, creep up on it bit by bit, successive generalization, and then you get there. So actually, it's a good idea. Leave that on the board the whole time so you have a context and a goal, like a mountain peak you're trying to scale. Thanks for the suggestion. Yes. Tell me your name? AUDIENCE: Scott. PROFESSOR: Scott. AUDIENCE: Last week, you did present an example where you wrote down this complicated physics-- PROFESSOR: The Navier-Stokes equation. AUDIENCE: Yeah, you wrote down the Navier-Stokes equation, and then you said, yeah, I sometimes do that just to create this tension in the students [INAUDIBLE]. So given that when you complete this method-- you actually didn't use the story at all. It seems to me that if you had presented that and then you stood back and made a joke about [INAUDIBLE], it would have been perfectly satisfactory. PROFESSOR: I think that's true, too. So a lot of teaching is preference of the instructor. I, for example, happen to really like that story because I taught in Cambridge for a long time and I can imagine Hardy running into Punnett at the cricket ground. So it's interesting to me, so I can tell it with enthusiasm. And I think it says something about the relation of science on society, which I think is important for students to learn. So I might use that story. But an instructor with different purpose might do it, as you say, which is make a joke about how impenetrable this is and say, it seems really cryptic, but we'll actually understand this by the end. Don't worry. And that's a good way of-- actually, I think you're right. It's a good way of creating and releasing tension. Yes. AUDIENCE: Eric. PROFESSOR: Eric. AUDIENCE: So what level of background do we assume the audience has? Because you use sickle cell as an example, and it's obviously not critical to the example that they know what sickle cell is. Do you assume that they know that? PROFESSOR: Right. So what level of background? So if they don't know what sickle cell is, or maybe I'd use cystic fibrosis, which they're maybe even less likely to know. AUDIENCE: --presentation. PROFESSOR: Yeah. I could be adding noise. So it's a flip side. It's a two edged sword. So are you adding noise by saying sickle cell to someone who doesn't know what it is? Yeah. It would add noise. So maybe it's actually worth saying just one sentence. What is sickle cell anemia? It's a mutation of the red blood cells that makes them take a different shape and makes you unable to transfer oxygen as effectively and makes you more resistant to malaria. So actually, that part, I wouldn't say at the beginning. I'd say, well, if it's so bad-- can't transport oxygen so well-- how come it's still around? Just let people think about that for the day and come back the next time and tell them the answer. But yeah. So I'd probably say one sentence if the people haven't heard of sickle cell. When I was saying it, I was assuming mentally that they know what it is. But I think you're right. Many people wouldn't know what sickle cell anemia is, and it's worth saying just one sentence. So let me just see if there's anyone who hasn't made a comment or question. Yes. Can you tell me your name? AUDIENCE: [INAUDIBLE]. PROFESSOR: [INAUDIBLE]. AUDIENCE: Could you point to the last little bit, because I'm curious to see [INAUDIBLE]. PROFESSOR: Oh, how do you get this piece? AUDIENCE: Yeah. PROFESSOR: OK. So how do you get this whole thing? OK. So let me do that by analogy. So for example, suppose-- let me do C equals 3. I'm running out of board. Let me erase this one. And in fact, let me just use n equals 2, because it doesn't illustrate any new ideas to crank n up. But the cranking the c up, the number of copies, actually makes it important to see why you need to add them all up. So let's do p plus q cubed. So this is an organism with three chromosomes. But it's either sickle cell or not at each spot. OK. Well, let's actually just write this whole thing. You could do it by looking at a cube and seeing all the chunks, because you could actually just do it with algebra, too. So there's a p cubed. There's a ppq. So there's a ppq, pqp, and qpp. So those are all contributing to p squared q, and there's three of them. OK. Now, the three isn't what we're talking about. We're talking about this exponent here. Why is K1 all the way up to Kn equal to C? Well, let me just put that 1 in there. What does this equal? The sum of those guys is 3. That's p to the cubed q to the 0. That's also 3. What's the next term? Well, there's a 3 p cubed squared, and there's a q cubed. Well, that's q cubed p to the 0. That's 0 plus 3. This is 1 plus 2. It's always three. And it's always the C. Why is that? Well, you only have three products-- p plus q. So you have three factors. And you get to choose, when you're writing out all the terms-- there are eight of them and we've combined them into four groups. There's three here, three here, one here, and one there. You get one from each of the factors. So you get one exponent from each factor. So the total of all the exponents has to be 3. So once you understand it for 3, then this is just for C in general. Does that help? So that's a good example, actually. Someone probably would ask that question or should ask that question. And that's how I would answer it. OK. Now, Cecilia, you had another question. Yes. Did you have another question? AUDIENCE: What are we proving? For example-- PROFESSOR: Yeah. You'd think, on one hand, what are we proving? Well, we just proved p plus q plus r squared equals 1. So it looks like there's no content. AUDIENCE: [INAUDIBLE]. PROFESSOR: How has that done anything? AUDIENCE: The key fact is how you prove that the 1, the 3, the 3, and the 1 in each [INAUDIBLE]. For example, are we supposed to prove that today or break even? PROFESSOR: Well, I would say there's no less information-- well, what we've learned is that, for example, we've learned this. Why are there various products here? And then what this thing is-- we've learned intuitively what this thing does. This thing counts for the number of copies. So in the original-- n equals 2, C equals 2-- it was either there's two copies or one copy of each of these guys. So it adjusted for the number of copies. When you have p plus q plus r, there's one copy, one copy, one copy. There's two copies of that one, two of that one, two of that one. So it's that factor. It's the number of copies factor. And then in probability course, you learn how to count those factors in terms of factorials. And that has a definition in terms of factorials. But I wouldn't focus, for example, in a biology course on why it's factorials, necessarily. I'd want them to understand that this thing counts for the number of factors. But actually calculating the number of factors for general C and n wouldn't be my first goal in the course. AUDIENCE: So if I didn't have that in mind, why would I-- so the general purpose is teaching the terms in the multinomial expression. PROFESSOR: Right. So it's terms in a multinomial expansion-- so basically, it's from writing out 1 equals 1. So it seems like, oh my god, we've done nothing. What have we learned? We've just learned 1 equals 1, which we already knew. But actually, by splitting up 1 on the other side, you've actually given meaning to each of the terms on the other side. So this is the frequency of probability of having three copies of gene type A and none of the other type. This is the probability of having two copies of gene A and one of gene B. And this is the probability of having one of gene A and two of gene B. So there are three, because there's different ways of doing it. AUDIENCE: So do you have these chromosomes with different flavors [INAUDIBLE]? PROFESSOR: That's right. Because these organisms we're talking about come with multiple copies of chromosomes. Yes. AUDIENCE: [INAUDIBLE]. PROFESSOR: What is that? Yeah. So what's the phenotype in the end? And you're right. So with the sickle cell-- so actually, that is an advantage. So the point is you need to link this phenotype and the genotype. And that's true, actually. I've erased the sickle cell, but actually, the sickle cell is a good example for doing that. If you have no copies of the sickle cell gene, then you're perfectly healthy. If you have two copies of the sickle cell gene, you have sickle cell anemia. What happens if you have one copy of the sickle cell gene? Well, you do have some symptoms of it. So the question is why does that gene survive? Well, it's because you're actually more malaria resistant. At least, that's one of the theories. So somehow, I guess the malaria parasites can't eat those red blood cells as well because they have a different shape or can't invade it. So it actually gives you some advantage and some disadvantage, but they're sort of balanced. And yeah, if you have full blown sickle cell with two sickle cell genes, then you're in trouble. But that's much rarer than the 2pq, as long as q is small enough. So then you can actually continue that example. You can look at the frequency of sickle cell gene in different populations and say, OK, well, is it higher in areas where there's malaria? And test that theory. Yes. AUDIENCE: I noticed that the first one that you did was to stay mathematical with the number line. When I've, in the past, taught [INAUDIBLE], I always stick with conceptual-- p is this allele, and then you've got a phenotype in the end. I'm just wondering if you chose this because you know that you're teaching mostly to people with math rather than scientists. And if you were teaching to, say, English majors-- PROFESSOR: Good question. Yeah, so I think you're right. I did this kind of mathematically, compared to maybe how you've taught it to biology students. And yeah, I guess I'm implicitly assuming-- but I didn't say, and I should say-- that this is MIT students. I'm just, in the back of mind, thinking MIT students. But I didn't make that assumption explicit. So MIT students are perfectly happy with squaring binomials and trinomials-- trinomials, usually. And to the C power-- sort of stretching it, but it's OK. But generally, that way of doing things for them is good. If it's people, English majors, yeah, you're right. I would try to actually do even more conceptually. But one thing that's good for all of them is the picture, because once you see the picture, then you actually understand the idea in one grasp. It's really just one chunk now. Oh. It's just a square in four pieces. Oh, OK. It's just a different way of writing 1. This is one area 1 and there's four pieces. So that I would keep, no matter who I would talk to. And the question is how much would lead up to it? Yes AUDIENCE: So I think I just had an idea here that I think highlights why you can't just go in with this, because there's a disconnect between what these terms mean at the end. PROFESSOR: Why you can't go in with A. AUDIENCE: Right. So I think you have to draw a connection between, say, the areas of the boxes in the diagram and the phenotype frequency. PROFESSOR: You're right. So what you're saying is that you can't just launch in with this, because even though you could give-- say you gave really exact definitions of what all these things are. It's not clear that exact definitions are computationally productive for a student. Yeah, they may have the exact definitions, but they can't actually use it. For example, let's do an exact definition of chess. I'll tell you all the rules of chess. And that's enough of an exact definition to be able to decide what the best first move in chess it. But it's computationally useless. I still don't know what the best first move is, even if I know the rules of chess. So just telling the student all the rules that, say, define what a genotype is and what's polyploidy, doesn't mean they can actually use it in any problem. So if you want to transfer, all these things have to have meaning for them. And that's what the goal of this approach is. Now, I think the approach has been improved from your suggestions. I just wanted to show you a direction to go. But I think you've actually taken it farther and extended it and improved upon it. And the general principles among all of them are that you want to-- I would say one of the key ones is chunking. You want to not overload the chunk system. You want to somehow bring people in so they'll even listen to you. If they don't listen to you, if they don't care, the learning is going to be so much less. So all those are for that. So the dictation and jargon oppose chunking. And these all go together. OK. So now, what I want to do is give you a short answer to one of the questions that was raised earlier. How you become a good teacher? The reason I want to do that is that is exactly the same thing-- if you understand how you become a good teacher, you understand how you become good at anything-- how you become good at chess, how you become good at biology, how you become good at solving physics problems, how you become good at playing concert piano. So if that's what you want to teach your students-- to be good at those things-- well, you want to understand that in a context, say, that you're working on-- say, being a good teacher. And to do that, there is the following set of experiments. So projectors-- OK. So the way I'm going to illustrate this is I'm going to show you a chess position. And the goal is to try to remember the chess position. I'm going to give you two more seconds. OK. So everyone got the position? Now, I'm going to ask you instead of to remember exactly, to reconstruct it. How many pawns were there? That would be E. So who votes for A? OK. Who votes for B, nine pawns? Who votes for C, 11 pawns? Who votes for D, 13 pawns? Who votes for E, none of the above? OK. So let me show you. Then I'll explain why I asked you this particular question. So there are actually 11. Now, it's a very hard task. So this actually, this very tasks-- not counting the number of pawns-- I made that slight variation. But the so-called reconstruction task was given to chess players of various abilities. Grandmaster slash master is one group. What are called experts-- experts are people who are not quite chess masters, but close, in chess lingo. And then class A players, which is pretty strong tournament players. So they were given the task of looking at a position for four to five seconds. The position was knocked down, and they were asked to reconstruct it as accurately as they could. OK. So it's even harder than the task of counting the number of pawns. So the results are very striking. So by level of chess player-- so class A is the strong tournament players, experts or grand master or master. So the percent correct-- 51% of the pieces correct, 72% or 93%. So the grand masters and the masters were amazing. And in fact, for the number of pawns, I don't think they ever make mistakes on that, because pawns are one of things that you just know as a really strong chess player. So 93% percent correct-- and that's amazing. So there's an related story, which is about the memory of chess players, which is the Bobby Fischer-- yes. AUDIENCE: So I'm thinking the positions, the positions wouldn't work. It wasn't possible. They wouldn't do it as well. PROFESSOR: I'll come to that in one second. So Bobby Fischer was in a tournament and some strong master was playing in it. And Bobby Fischer went to the bathroom. And as he went to the bathroom, he happened to see that master playing a game and just continued walking. And then about six months later, he ran into him at another tournament. Fischer said to him, oh, in that position in that tournament, did you play blah? And the guy said, well, actually, I had no idea what the position was, even. Fischer said, oh, and he set up the board and said, well, see, this was the thing that you really needed to do. And by then, he sort of remembered. But Bobby Fischer remembered at one glance six months later. So what's a natural conclusion from this data? You'd say, well, naturally, grand masters and masters-- they were born with better visual memory. But in fact, the crucial experiment was then done in 1973. So this experiment was done in 1948 by de Groot. What Chase and Simon did in 1973 was that they showed positions that were random. So they redid the experiment-- they confirmed these results. And then, as you suggested, they showed just positions where the pieces were scattered arbitrarily over the board. And then everyone was basically at 12%, give or take 1% or 2%, just random variations. So what does that show? It's not that the-- maybe Bobby Fischer was an exception. But for almost everybody else, even the very strongest players, it's not that they're born with a better visual memory. It's that they've learned somehow a way of looking at chess positions that there's less to remember for them. OK. So now let's compare. Here, when the student sees this, there's a ton for the student to remember, just like when we look at a chess position, every piece is separate. But what does a chess player see? When a chess master looks at a chess position-- so I'll put the position back and I'll show you what the chess master sees. The chess masters see something very different. They see groups of related pieces together. So for example, here, the chess master sees-- this king here is not a surprising thing for this chess master. That's when you castle your king. That's where it goes. Then the rook goes next to it, and then you usually move your rook into the middle. So that's not surprising. This rook is also not surprising, the second one, because it usually comes from this corner into the middle. So all of this almost contains no information for the chess master. Here, these three pawns are very common, with the castled king on that side. But then it looks kind of strange. There's some new information there, because maybe this king castled, but then the rook went all the way to the corner. So actually, maybe black never castled. And his kind just sort of wandered into this area. What does that suggest? This suggests that the black king is really vulnerable. Maybe it's time for an attack. And in fact, I'm pretty sure this is a position from one of Garry Kasparov's games. And in fact, that is the right conclusion. The right conclusion is that it's now time to sacrifice your knight and take this pawn and draw the king out. And he actually won using that-- by sacrificing his knight. So the chess master looks at it completely differently than the novice. I'm a novice when I play chess. To me, every piece is a new bit of information. I'm way overloaded past my chunk threshold. I can't hardly remember the board at all. So your students are in exactly the same position when they're learning material that, for you, you're the chess master. So you're teaching Hardy-Weinberg. Well, clearly, you've been appointed to teach Hardy-Weinberg because you have a Ph.D. In biology. You know a lot of biology. You're the biology master. So this doesn't surprise you that much. But for the student, every single almost letter in there is news to them. So what you want to do is you want to find ways of thinking about it that you can group the ideas into chunks. So here is almost one chunk-- for example, the idea that really, it's just p plus q squared. And there's a picture for it. And once you understand that, there's another chunk. There's another idea which is well, you could actually have three kinds of flavors instead of two. OK. p plus q plus 4-- oh, I can transfer it there. Oh, and then once you understand that, you can transfer it to n flavors. Now, you can increase the number of copies of the chromosome-- so into two dimensions, three dimensions, four dimensional picture. So then, you can actually make sense of all this. You can have a way of understanding the position. And not too long ago, there's, I think, a not well enough known paper-- I'll put the reference on the website-- which shows the relative importance of symbolic calculation versus perception. So this is a perceptual mode. So much of our teaching is, let's say, left brain-- very, very symbolic. Well, there was this really interesting study done of chess grandmasters-- in fact, of the strongest chess grandmaster today, Garry Kasparov. What's the relative importance of perception versus analysis in his really strong chess playing? So the way they tested that-- there was a really good experiment-- is he would play simultaneous exhibitions all the time. So the way you do a simultaneous exhibition is there's, for example, let's say, 10 opponents. And you would just go around one component after another. They have the full time, say, three minutes, to think until you come back. But as soon as you get to a board, you just think for about five or six seconds, maybe 10, and make a move, and go to the next board, so that by the time you come around back to that same opponent, they've had their couple minutes to think. So now, he played simultaneous exhibitions against very strong grandmasters. And you can then measure his performance there. So why is that a good experiment? Well, he's now not able to do all of his calculation that he does normally. He's normally able to think for three minutes, maybe five minutes, and do a whole bunch of analysis, symbolic computation. But when he has five seconds, 10 seconds to think, mostly it's perception. Well, his chess rating effectively dropped by maybe 50 or 60 points. So 50 or 60 points, to give you an idea-- his chest rating is the highest in the world. It dropped to a level which only five people in the world are higher. So it's a very small drop. So he still plays incredibly strong chess, better than almost every other grandmaster on the planet. So almost purely with perception-- so what that shows is that the way Kasparov has become so good and in general, experts have become so good is it they look at the world different. Their perception is different. So how do you do that as a teacher? That's one question. How do you promote that in your students? Well, you want to give the ways of looking at the world that change their perception. That's why I'm so focused on the story, the tension, the human, the right brain, the pictures, the chunking, because it's those that are actually producing long term expertise, whereas this is producing what would be the equivalent of programming a chess computer. But that doesn't work. That may work for chess computers to play good chess. But it doesn't actually work for people to be able to use the knowledge later. So now, what produces that? So there's one short answer which is that for teaching, you want to change your perception of how students think. If you have a correct, new, good perception of how students are thinking, then you're actually able to make teaching judgments on the fly. You can plan chess move, your lecture, like a chess game. Your intuition is right. So you want to tune your intuition. Well, that is why I do this. I've found the single most important thing that has improved my teaching, and I highly recommend, is the feedback sheet, because for example, I learned what was confusing in question one. And in question two, I learned what works and what doesn't work. So as I see what works and doesn't work, I start to build up a more and more accurate model of you. And I start to be able to plan and reason about how to teach you and, in general, how to teach students. So I'll talk about that more. That's the idea of deliberate practice and expertise. I'm going to talk about that more in the subsequent sections in more detail and show you some of the studies around that. But the general rules is you want reflective, quick feedback on what you're doing in order to become an expert. And that's true whether you're in teaching, concert piano, physics problems, whatever it may be. So with that said, if you can fill out the sheets so that I can become a better teacher, that will be very helpful. And one announcement, which is that next week is a Tuesday. MIT is open on Tuesday, except it's Monday's schedule of classes. So we don't have a class next week. The week after that, I'm a witness in an administrative law trial. So I'm not here. So there's no class for the next two weeks. So we'll meet again in three weeks. And I'll post some readings and a short problem set for you to work on-- some readings growing out of what we've done today. OK. So if you could bring up your sheet and your index card to separate piles, that would be very helpful. And there's going to be another class coming in-- a big class, I think. So I'll just go outside and answer any questions that people I have right outside so that the new class can come in. SPEAKER: Answers from Lecture 3 to questions generated in Lecture 2. PROFESSOR: I'm going to first answer questions from before, since there are lots of questions, and all interesting. And I'm also going to do another equation example. There was lots of requests for another equation example to see how it plays out in a different field and a different way of approaching equations, not just from the entry point of a story. So I'll show you that. And then we're going to look at misconceptions in various fields and the fundamental importance of understanding that so that you can understand how to change your teaching and how to reach the students. Basically, if you can't understand where they are, you can't come to them. So questions from before-- one comment was that I don't often enough summarize the end result with the transferable lessons for later. So thanks for that suggestion. I'll make sure to do that. Another question was graduate versus undergraduate classes. We talked a fair amount about audience a bit last time in response to questions. So how do you change your teaching in response to questions, in response to the change in audience? So the particular question here is graduate verses undergraduate classes. And the sense I got from some of the questions was that somehow, it's harder to do what we were talking about last time, which is teaching equations in an intuitive way, in a graduate class than a undergraduate class. Actually, in some ways, it's the opposite. It's true that generally, in graduate classes, people just put up a ton of equations. For example, in quantum field theory, you just get a gazillion integrals with epsilons floating all over the place and then path integrals. You integrate this. And there are some 2 pis and you take a bunch of limits. And it seems like a whole bunch of methods gymnastics. But A, it doesn't have to be that way. And also there's another characteristic of graduate students which you don't have so much with undergraduates, which is that graduate students know how to read. Now, this may seem like a bizarre statement, because surely, everyone knows how to read by, say, age three or four or five or whenever they teach reading in school these days. But what I mean is that undergraduates-- generally, they have no experience on how to read a textbook because they've had so much experience with us telling them stuff everything on the board. So they have no incentive to actually read the textbook. And they don't learn how to do that. They think textbooks are read the way you read Jane Austen novels. You just read for plot. Something happened to some equation, then something else happened to some equation. And you just carry on, paragraph by paragraph, as you would a novel. That way of reading is completely hopeless for technical material. But graduate students-- not always, but generally-- have much more maturity about this. So graduate students actually can or often are closer to being able to read technical material with skill. For example, graduate students often read papers in their own field. And you know you have to read a paper differently than you would a Jane Austen novel. So because of that, you can actually teach equations very differently. What you do is you give all the long, messy, yucky parts-- you leave that for the notes, for the book, somewhere where everything is printed in a very easy to read format, rather than copying long, long, long, long strings of symbols off the board. So that connects back so what we talked about last time, which is chunks. So if you put long equations up on the board, generally, you overflow the chunk system. And once that happens, people start making mistakes. So you want to avoid doing that as much as possible. And with graduate students, it's even easier to do, because you leave all that for a type set, professionally published book or type set by yourself, but somehow printed in a clean way with no mistakes. And you can then, in class, discuss the meaning of the terms. What are the terms? Where do they come from? Why would you expect that kind of term? So I'll give you an example of doing that with an equation today. But generally speaking, all of what I was talking about last time applies perfectly well to graduate classes, even if people at first think that it doesn't. So I should say if any questions occur to you as I'm answering questions-- basically, questioning beginning questioning-- please raise your hand and ask them right now. OK. I found that the square diagram muddied the development of the Hardy-Weinberg equation. So the square diagram was this one. So the conclusion from that was the question-- which is shouldn't college students be comfortable with expanding p plus q squared? Why do they need a diagram? And the answer isn't that people aren't comfortable with expanding p plus q squared, although you will find people for whom this and this is just a symbol replacement strategy. In other words, it's something like you program in a computer whenever you see this pattern do this. But the terms don't actually have meaning for people. They don't know why those terms are that way. And if they'd mismemorize and put cubes here, they would write that down too. So the picture actually makes it clear what the meaning of the term is. So it gives, actually, meaning to what people might be otherwise comfortable with or having done from lots of practice. So I should say there's a difference between just rote matching between here and here versus this kind of understanding. And that's illustrated by the following research, which is about people's ability to multiply and add when people have brain damage. So the technical term is brain lesions. The people with brain lesions in the-- let's see if I can say this right. Some people with brain lesions in the arithmetic areas lose their ability to add, but they can multiply fine. And people with brain lesions in the verbal areas, they lose their ability to multiply, but they can add fine. Now, this seems kind of strange, right? So why would multiplication and addition to not go together when you get brain damage in the arithmetic area? So if the damage is either an arithmetic or verbal, here's what you lose. So arithmetic area damage, you lose addition. Here, you lose multiplication. And that is very bizarre, because you'd think you should lose multiplication here, too. And the reason that it doesn't work that way is because of the way multiplication is generally learned and taught. So for example, in England the way it's taught is-- this is research from Brian Butterworth in England. The way multiplication is taught in England is in the multiplication table, same is here. And people memorize it as six 9's is 54. So six 9's is 54. Here, I think I learned multiplication both places. So 6 times 9 is 54. Either way, that's a purely linguistic string. It's no surprise that when you lose verbal ability, you lose the ability to memorize linguistic strings. So the multiplication table went. So what that tells you is that most people-- and I don't know if this is true for everyone, but my guess is it depends on how you learned the multiplication table. For most people, they learned the multiplication table in a way that is not meaningful. They've learned the multiplication table purely linguistically. And that is the problem with just going down this path only, which is that-- again, this is where it's important to know where the students come from. If you know students basically have just linguistically memorized this transformation to that transformation, like the rock bands in other countries that sing in English. They know when they see English words. They know what to say. But they don't necessarily understand any of the words. That is actually very common. And this is another linguistic transformation. There's no meaning underneath it. So here to give it some meaning underneath it, some picture, it actually incorporates another brain area into the understanding, which is where addition actually has some meaning to people. So that brings up a related question, which is if people have learned multiplication in this linguistic way, how could you teach it in a non linguistic way so that people actually understand it? There are several ways. I don't know of any studies that show that after doing it this way and then brain damage, it doesn't get lost. But my speculation is that if you learn it, actually, pretty much the way I learned it, you wouldn't actually lose it. And the way is, for example, suppose you have to multiply 6 times 9. Rather than memorizing that as 54, you think about what it should be. And you reason your way to it. And we think, oh, well, that's slower. Yeah, it's slower in the beginning. But in the end, you get to the same place. But you've put it in a different part of the brain. So the way you could do this one, for example, you say 6 times 9. That's 6 times 10, which is really easy, minus 6. So 6 times 10 you know not by memorizing but from understanding the number system. So you know that that's 60 because of the way the number system works. You don't have to memorize that. That's the kind of example where I think calculators should be programmed to self destruct or at least not work for about a week if you type in a problem like that. They should just freeze as an incentive to actually think about these things before you put them into the calculator. So minus 6, and you get 54. So now, that's a bit longer the first time you do it. It's longer than memorizing it. But after you do multiplication a bunch of times like this, you actually reinforce the meaning of the number system, and you come up with the same answer. And eventually, you will memorize it, but you've memorized it in a different way. Here's another example. From the 12 times tables-- 8 times 12. Now, should you memorize that as 96? No. You should write that as 10 minus 2 times 10 plus 2. So that's equal to 10 squared minus 2 squared equals 96. And you can even draw a picture for this and show what happens to squares and areas. So again, you've come up with the same answer, but you've done it in a meaningful way. So that's one answer to why it's worth showing pictures, even if people can do the algebra. It reinforces the algebra. Is determining what constitutes a chunk simply a matter of my intuition about students' level? Well, to some extent, it is. But in the chess playing research I talked about, they actually had more objective ways of determining what a chunk was. And what they did is they put eye trackers on people. And they had the chess masters and the non chess masters and the experts look at the board, and they tracked their eyes to see what they did. So the non chess masters, their eyes wandered all over. But the chess masters looked at pieces in groups. They would go here, here, here, here, and then here, here, here, here. And the assumption was that that's a chunk and then that's a chunk. Or if they went here then to there, that somehow, these two chunks were related. So there was ways in the chess playing problem of measuring chunks. And then how do you apply that to, say, teaching physics? Well, that's a matter, partly, of intuition. And the idea is to help the students build up the chunks. So you have to be on the watch for what chunks you use. So you have to introspect. OK. How do you keep things interesting and reveal material in a time appropriate way? So the comment was that it took a while to do the Hardy-Weinberg with the story and then building up to it. So this is a question of how do you plan lecture time. What's worth doing in lecture? Is it worth spending a bunch of time on understanding the concepts? And there's two extremes to this. I'm basically towards one extreme, which is that if you don't understand the concepts, it's not even worth learning the thing, because as a student, if you don't understand the concept, you might as well just forget the material right now, because you're going to forget it soon anyway. So the only benefit from taking the course if you don't understand the material, is you just get a great on a final exam and you pass a requirement. But in terms of actually changing how you see the world, it ha no value. So there was a study done, I think, at Carnegie Mellon. Yeah, this was at Carnegie Mellon. They studied freshman physics students. So they had students who took freshman physics and students who didn't take freshman physics. And then a year later, they gave them the freshman physics final exam-- I think it was the same one, pretty much, as the students who took freshman physics took, except with numbers changed, but otherwise, it was same problems-- to see whether taking freshman physics had any effect on whether you did well on a freshman physics final a year later. And the conclusion was that it had no effect a year later. So yeah, sure, if you're give them the final the next day after the final exam that they took, they would have done pretty well, maybe 50%. Who knows what the time constant is. But certainly, by a year, it was gone. So what that tells me is that that regular way of teaching material actually did no good to the students except for passing a requirement, but no intrinsic good for their way of analyzing the world. So what that says is then you really need to find something different. And yeah, if it means it takes extra time in lecture, a bunch of time in lecture, so be it. At least people will understand something and they will change how they see the world. And that's the motive for today, which is to really understand misconceptions, because if you don't understand the misconceptions, you're not going to be able to teach in a way that produces long lasting learning. OK. So another one was how do I apply this to something really abstract? The way of approaching equations, like the proof of-- oh, this is actually one of my favorite infinite series. So the question is how do you apply it to something really abstract like this? So this is a famous infinite series. What's the sum of that? And it turns out to be pi squared over 6. Well, even that, you think, well, how can I apply it to that? And it turns out there's great stories about that one, too. I think this is a story about this problem, which is that no one knew how to do this sum from 1 to infinity. It's quite a hard sum. And so it was set as a problem, basically, for the mathematicians and physicists of Europe. And then someone produced a solution. I think someone produced a solution anonymously. And everyone basically figured out who had done it, because it had their handy mark. So the solution was by Euler. And it involved a whole bunch of trickery with polynomials and infinite degree polynomials. And it was a really sly method. So if I was going to teach this equation, I would actually teach the history of it-- how was really hard, how you could actually guess this. What are ways you could guess this? Well, you could actually approximate the sum, get a number that's one or two or three digits accurate. And then you feed it into-- does everyone know this guy? If you Google for that, you should probably find it. It's called the inverse symbolic calculator. It's a fantastic thing. I do not know how it works. And I would love to know how it works. But what it is is you feed in a number, and it will tell you all the ways of producing something really close to that. So for example, if you put in 3.141, it'll say a bunch of numbers that get near here. And one of them is pi. If you put in this to one or two or three digits, it'll probably guess for you pi squared over 6. So that's one way of getting at an answer. So part of the way of teaching it is to say, well, let's somehow get an answer with ways we can do that aren't too abstract. And then let's see if we can justify that answer. So even there, you're not lost. There's always stuff you can do. Oh, another question, which was readings-- how do you incorporate readings into a course so that students do it? So that was asked twice, actually, because I didn't answer it the first time. So one way to do readings is something called reading memos. And it's an MIT invention by Edwin Taylor, who's recently retired from the physics department. And what a reading memo is-- I'll put up the handouts. So I have often done this in my classes. I'll put up the handout for you to use that I give out and you can just copy it or do whatever with it. So a reading memo is a request to the students to write you a short memo about something that you've asked them to read. It could be the draft notes for your textbook that you're working on, which is what I often do it with. Or it could be the textbook someone else wrote and you ask the students to read chapter. And what it's not is it's not a summary of the text, because you already know what the text says. There's no point in getting a summary. What it is is students' reactions to it. So anything that questions, things that puzzle them-- oh, I didn't understand why you did this or the author did this. And then maybe three pages later, oh, now, I see-- which, if you're the author of those notes, you know that you explain the two things out of order and you should connect them. But what that does is it teaches students how to read actively, because again, I like I talked about, people just do Jane Austen's approach to reading technical material. And by getting students to read actively and formulate questions, by doing that, students learn a different way of reading, a way necessary for reading technical material. And also by writing their questions down and you seeing the questions, you actually get a view into how the students are thinking. So it's actually a way of understanding what their misconceptions are, their conceptions of the field are, and tuning your teaching. Just automatically, you'll find your teaching will impedance match to where the students are just by reading the reading memos. It has a further benefit, which is that it inverts the normal power relationship between teacher and student. So for example, most assignments, problem sets, there's the correct answer, which you know. And you're seeing whether they know the correct answer. So they're now writing an answer, worried whether they're correct or not. And then you're judging them. So normally, p set, the power hierarchy is you and then the student down here. And the student is looking up to you for validation. So now, this is not a good thing to teach. And maybe it's hard to avoid with problem sets. You have to do problems sets somehow when you teach people to do problems. But you want to minimize this as much as possible, because it's not a transferable way of dealing with the world. They can't use that when they go elsewhere. And it teaches bad habits of deference to authority. So that's normal. How does a reading memo work? Well, it's the other way around. If the student says, this is confusing, by definition, they are correct. It's confusing. They are the expert on what's confusing or not. So it inverts the hierarchy to this. And you become very interested in what the students are saying. They are the experts now. And I've had very good results with doing reading memos. And my explanation is that it's because of this inversion of power hierarchy. Now, what I mean also by good results-- two things. One is that I get fantastic feedback on the things I'm writing. The other is that I find students actually want to do reading memos after the class finishes. They say, oh, if you have more notes, can we just do some more reading memos? Great. Let's do that. And it's because it's actually-- if you write the problem sets, they'll often say, can we do more problem sets? But that requires a fair amount of work to minimize the hierarchy. But it's automatically here in the correct hierarchy. So students actually enjoy doing that and want to continue. So that's one excellent way of incorporating reading into class. So now, the problem is, what you do when you have 50 people in a class and you get 50 memos? So I've had this problem. And one thing I do is I just feel overwhelmed and I just flip through them but I don't know what to do. But another is I revise my notes based on it. But the I think correct solution is an online system. So what you want is an online system where you can post a PDF file, and then people make comments on the PDF file. So everyone gets to see an image of the page, and they can just click and make a comment. And everyone gets to see their own comments, and then when they submit them, they get to see everyone else's comments. So I actually wrote half of that system. And there's a graduate student in EECS who I think has now written a whole system independently of me. So I'm going to try it out and see how it works and try it in some of my classes this semester. So the benefit of that is that you can then see all the comments at once, rather than flipping through 50 sets of reading memos with page numbers on them. OK. How do I come up with intuition examples? How do I know if what builds intuition for me will also build intuition for the students? It's a very good question. Is it just my personal opinion or is it just the teacher's personal opinion? One of the whole themes about this class is yes, teaching does have a fair amount of art and there is a fair amount of personal opinion in it. But there's also a fair amount of science and things you can do to make it more objective. And one of them is actually to do reading memos. Any way you can to learn how students think will make it so that your intuition about the students actually matches how the students really think. That's the whole purpose of today about talking about misconceptions. Reading memos are way of understanding what students think. Oh, there they are-- the reading memos. So once you understand what students think, it's much easier to realize, just intuitively choose things that you know are going to work for them. The other way is the feedback sheet. So every time the students tell you, oh, this really helps me or this really didn't help me at all, you now have one more piece of feedback about what works for them and what doesn't work for them. So then, you can actually choose intuition examples. How do you invent them from scratch? Well, there are some general principles. One is use pictures whenever you can. Generally, that speaks to people's intuition, just because people have much more hardware for pictures than they have for equations. So I try to put myself in the position of the student, and I say, well, for example, here. I say, yeah, all these equations may well be true, but I want a way that makes me see it instantly. And that just forces me to start looking for pictures. And that tunes me, actually, towards what students need. So you can do the same. I think that was most of the questions. The other questions were similar to that. And I'll answer any that were new that I haven't answered-- but I think most of them I have answered-- at the beginning of the next lecture. So any questions that were generated by the questions? Yes, question. AUDIENCE: So your story about the physics retension issue made me question the idea of the survey courses or topics in class, where it's like, I don't really need to learn this. I just want you to-- been exposed to it so that they remember enough to know to go back to it if you need it someday. That research, that that has any prayer of working. PROFESSOR: Right. So the comment was what I'd said about even a year of intense freshman physics-- maybe it was a semester. I forget if it was just mechanics. But it was either one semester or a whole year freshman physics did no good towards long term understanding and change of understanding of freshman physics. Well, what does that say about these big survey classes, where you're not expected to understand? So in freshman physics, at least they have the expectation that you're supposed to understand everything. Now, what about the courses where they start with the expectation that you're not going to understand most of it? That one is going to be totally hopeless. Then I think that is basically true. And maybe you could justify back in the day-- let's say, 400 years ago, even when books were around but there was no web. People need to know what books are out there. So this is a traditional thing in Cambridge, for example. They just give people big huge reading lists. So then you go back. Later, you're like OK, these are the key books in the area. So you know one place to go for a reading list. That's kind of obsolete now with the web. If you want to find out something-- for example, suppose there was an equation I didn't know about. Let's say the Black-Scholes equation. Would I say I wonder if I took any classes about Black-Sholes-- maybe, let me go flip through all my course notes? No. You just type it into some web search and see what shows up. And that is much more likely to be more relevant than some notes that you might have had or not had. So yeah, I think the survey courses are completely pointless. Now, that doesn't mean that introduction to a field is pointless. But it means that the way to do the introduction has to be very different. You can't just scatter a bunch of topics at people. What you have to do is figure out-- and we're going to talk about this when we talk about course design. You have to figure out what are the core reasoning ideas special that that field has to offer to the world? For example, history-- what's special about history? Well, historians have a sense of how to evaluate the validity and reliability of evidence and contradictory evidence. That's something you don't get in many other fields. For example, it's in between a science and a pure literature field. In just straight literature, reading novels, there's historical evidence and things, but generally, you're reading in a different way. In math, it's hard to know where contradictory evidence comes in, although there are ways of teaching math which I like which do that. But generally, history has something new to offer, which is it's a messy world. You have noisy evidence. What do you do? Well, that's something that an intro survey course could actually teach. And that, somebody could transfer, even if they don't remember when did the Magyars invade Europe and all the random stuff that would be in a survey course. So those would be grist for the mill hung off big ideas. So I'm going to talk about that when we talk about course design. But yeah, you're right. That course design is completely hopeless the general big survey. Other questions? Yes. Yes. Could you tell me your name? AUDIENCE: Meg. PROFESSOR: Meg. And what was your name? AUDIENCE: Amy. PROFESSOR: Amy. Thank you. AUDIENCE: I was [INAUDIBLE]. I was just thinking about how for me, even if I understand something really well at the time, and I know that I'll use it again, it takes me-- having a test in front of me will not actually reflect whether or not I'm going to understand it, given the short period of time to remind myself. And so I'm wondering, if testing people out of the blue a year later is actually capturing whether the people who take this and perform faster [INAUDIBLE]. PROFESSOR: Yeah. That's a good question. So maybe it was a slightly unfair test, because they were just tested out of the blue. AUDIENCE: If they had been using it all along-- if, in that period of time, they had taken courses that built on that material-- then, they would be reinforcing it all the time if they can. PROFESSOR: Right. So if they had taken courses that used freshman physics throughout, maybe they would have remembered the freshman physics better. And I'm sure that that's true. So I can tell you one story from my graduate time. So I did a Ph.D. In physics. And I had to do the qualifying exam. And to do the qualifying exam, you have to study a whole bunch of undergraduate physics and then take the exams on it. And then you take a bunch of courses in various fields. Now, the only thing, basically, I remember from all the electromagnetism is one thing, which is I understand pretty well the index of refraction. And why is that? That's because I was really pissed off-- sorry for the camera. I was really annoyed about the following thing. So this actually goes back to what I was talking about about contradictions, which is that you're always told in relativity that speed of light equals C. And that's the great postulate of relativity-- that the speed of light does not change, dammit. And that's what Einstein said. And there's all these thought experiments with trains and lightning bolts and people throwing rocks from the train at different speeds. And it's C. It's C. It's C. And then, somewhere later in an electromagnetism course, they say, the speed of light in a medium with an index of refraction N is C over N, where N is typically around 1, maybe a little bigger like 1.001 for air maybe, 1.33 for class. And how do those fit together? So that really annoyed me. And I wanted to get to the root of it and say, well, how could it be that you could have a speed of light that's always C, yet it looks like the speed is some lower number, V? So I worked out a whole bunch of stuff about how electrons scatter radiation and all the scattered radiation adds up and makes it seem like it's slowing down the light. And because of that, I actually understand this. And also because of that, which may be not so good, every time there's an electromagnetism problem I have to do, I always try to fit into a scattering problem. And if I can't do that, then I just can't do it at all. And that's despite taking two years of electromagnetism-- two years as an undergrad and one year as reviewing as a grad student. So the point I'm trying to make by that is that most stuff disappears. And the way to really make stuff stay is you really have to struggle with something. And that's the most efficient way to make something stay. And that's not what happens in your traditional class, and even less in a survey class. AUDIENCE: I'm just also asking-- it's almost impossible sometimes when you see material for the first time-- you might have to see it three or four times before anything comes, you're really going to understand it. And if the class is set up so that you will only see it one time because that's all the time you have, then you have to take more classes on it. PROFESSOR: Right. So maybe you need to see something a few times to really understand it. So if a class doesn't give you that chance and then you have to take Thermo Two and then Thermo Three. AUDIENCE: Hopefully the class gives you a chance. Say that it's building on this. PROFESSOR: Right. So you should try to build that into the class. So there's a name for that, which is called the spiral curriculum. And there's a lot of sense to that. So I'll just put a quick-- so you show the idea in its crude form. And then you spiral back to it in a more sophisticated way. But you'd like to do that soon, before the connection is gone, before this is actually wafted away. You want to spiral back to it. And so you'd like your class to do that. So what that shows is that you should start with the big ideas and then you should refine them, because if you start with all the little details here, you'll just flood the chunking system. And actually, there will be no memory of it here and then you have to start over from scratch here. You had a question. Could you tell me your name? AUDIENCE: Roderigo. PROFESSOR: Rodrigo, yeah. AUDIENCE: Regarding the reading memos, do you think good to also ask the students questions about the readings? The reason why I am asking is because I know of a class that's actually implemented that PDF annotation system. And a lot of the comments that they get are too trivial, so to speak. They don't ask those type of questions. [INAUDIBLE]. PROFESSOR: So the comment is that a class that actually implemented the online annotation system, they found that the comments are too micro level and not broad enough about what's really confusing or interesting. And so I haven't tried the online system myself yet. What I do know is that on the paper, if you do it on paper, you get really insightful, detailed comments. Now, I don't know what variables are different. One might be that the online system just encourages-- because everyone does things quick online-- it encourages quick clicking. So it may be that it encourages less depth of thinking, whereas writing it on paper, I actually find I get a mix of a whole bunch of not trivial comments, but small comments like, some typos, that equation isn't quite right. But then I get things like, I don't see the picture here or I don't understand why you did this now. And that, to my mind, is it useful comment-- or a question like that can't be right because of the following counter argument. So on paper, I get a lot of interesting things. So it may be that online isn't as good. And that's one reason I want to try it and see. So it may be that they need to go back to paper despite it being harder. Another is also what is the material? If the material is really boring, you're going to get really micro comments. So it helps to actually have written interesting material or give people interesting stuff to read that people are likely to think about and make comments. Question. Can you tell me your name? AUDIENCE: Brian. PROFESSOR: Brian. Yes. AUDIENCE: I notice when you talk about equations, you like to give them the name that they're most commonly [INAUDIBLE]. Do you find that students get more understanding of concepts like this when they're given a name based on discovery versus terminology based on use of the equation? I think, from my mind, if you want to describe the relationship between stress and strain in material, do you want to call it Hooke's law? Or do you just want to call it the constitutive equation for solid material? PROFESSOR: Right. OK. So the question is what about naming equations? Should you name them by who made them or by how they're used or what they are? So another example of that is, for example, the fluid mechanics equation. Should you call them Navier-Stokes or should you call them fundamental fluid equations? And there's a tension there. First of all, I'll say that you do want to give a name. The big win is giving the thing a name, because that makes a unit of thought for the students. So that's the first order bit, the first order term. The second order term is how should you name it? And there, there's something which is that you want a name that's common. So if, for example, people look it up elsewhere, they're likely to find more stuff about it. On the other hand, you want a name that's intuitively meaningful. So there's a tension. There's not often a right answer to that. And you can go either way. So for example, I wouldn't probably use constitutive equation for the solid, because I have to think, what the hell does constitutive mean? So myself, it doesn't mean anything to me. There's another word. I'm trying to remember what it is. Epistemology-- people just use it like it's just a plain, obvious word. But every time I hear it, I have think, what the hell does that mean? And then I put the translation into the sentence they're saying. And then I can sort of parse what they're saying. So constitutive equation, to my mind, that's a word that is meaningful to the experts and not so much to the students. So I would maybe call it the ideal spring equation, because it is the ideal spring equation. It's just the proportionality is slightly general because you have tensors. But otherwise, it is the ideal spring equation. So it connects to something. And you can say, OK, who discovered? Hooke. So we often called it Hooke's Law. So then they have both And there's no harm in doing that. Question. AUDIENCE: Because you mentioned a little bit before [INAUDIBLE]. If you're trying to teach something many times, then maybe it's better to tell the end, and then-- PROFESSOR: Yeah, the spiral. AUDIENCE: Yeah. And I was just thinking because of the whole work-- so there are these really topics like the ideal gas law. And is it OK if I were to lie or say to students just for the sake of simplicity, for example, do you want to say whether you can actually use that equation in some cases or is it not logical always, and then you go-- PROFESSOR: Could you tell me your name? AUDIENCE: Cecilia. PROFESSOR: Cecilia. Yeah, thank you. That's an excellent question as well. So the question is, how much should you lie, if at all? For example, if I'm recommending teaching the big ideas in the overall approach first, that's almost necessarily going to involve some amount of lying, because the truth is complex and messy, the full truth. And basically, you do want to lie. Some people hate it. But there's actually a really good book that follows this principle. And even if you're not interested in the typesetting system, you can see how it's played out in this book. It's called The Tech Book. So that's the manual for the tech typesetting system, which I use and many people in math and physics use. Now, the reason it's interesting is that Knuth actually tells you in the preface, I'm going to lie to you. So what he does is he has three levels of statements. There's statements that aren't marked with a-- so there's that sign. On the road, it means slippery, icy, something like your car might fish tail basically, danger. So there are statements without one of these, which may have some lies in it, not the full truth so that you just get the idea-- what are the fundamental concepts that tech uses. When he starts to get into some gory details, but not super gory. He puts one of those bends. And when he has a super gory details, there's two of those bends. And then he says, look, don't read any of these things unless you've been working with tech for a year and are pretty competent with it. Don't worry about that. You'll be able to do what you need to do just by reading this and maybe the single bend sections. So that's an example where the lying was put to really good use. And yeah, you should lie. In fact, you have to lie. There's no way to avoid it. And in fact, everything is a lie. It has to be, because to understand the universe, our brains are a constituent of the universe. So there's no way to understand the full universe, because that would involved packing more than our brain capacity into our brain. So just there is a pigeon hole principle proof that you have to lie to understand the universe. So you have to say some stuff that isn't quite true. And where the art is is in choosing what is a useful lie. So what you want to do is develop the art of skillful lying. And that's a mark of a really, really good teacher-- skillful lying. Yes. AUDIENCE: How do we know that you're not lying to us now? PROFESSOR: I probably am. The question is how do you know that I'm not lying to you now? And I probably am, the reason being that I've now practiced lying so much, I don't even know when I'm lying and when I'm not. So no, I'll give you an example. I am definitely lying to some extent. For example, there probably are situations where you don't want to any lies at all-- for example, teaching people how to manipulate the machines in the intensive care unit. Maybe the first thing you need-- if you have a half an hour to teach them, you'd better teach them, memorize these damn things and don't mess it up or you'll kill somebody. Maybe there are situations where lying is less important and lying is more important. And I haven't talked about those. So right away, I have lied to you, just because that. And I've skipped details because I wanted to get the big idea across. So lying actually comes very naturally to me because I think the most important thing is the big details. So just by saying big details first, you're automatically lying. And I'm recommending it highly to you, too. Rodrigo. AUDIENCE: I have a comment on the lying. As a student, I have students, a lot of friends, who told me that they don't like it when teachers lie. But what I think the differentiating factor is whether they tell you that they're actually lying. And if they do, then it's totally fine. And if they don't, and a couple months from there, they just told you that everything was not really true, the students might get upset. PROFESSOR: Right. So the comment is that students often don't like when they find out that they've been lied to. But it's OK if you tell them that you're lying to them. And there's a general principle there which you can use in all your teaching, which is that whenever you do anything slightly nontraditional-- whenever you do anything like that-- it's really important to tell the students what you're doing and why, because they will go with you. They'll go along with you if you explain to them the motive for it. So you tell them, look, this is a really complex subject. There is no way to understand the whole subject at its first glance. You need to develop high level structures first. And then you can put in the details underneath that. So I'm going to tell you just the high level structures first. And there will be some untruths in that, but I'm not going to tell you things that are completely false that you have unlearn. I'm going to tell you things that you have to refine your understanding of or that aren't the whole truth. So together with the idea of lying, you want to minimize stuff that you tell them that they have to unlearn and make it so that you're telling them stuff that they can keep using. It's just not the full story. OK. So what I'm going to do is I'm going to give you just a quick example of another equation and how you could teach it. Then we'll take a short break and then we'll do misconceptions. OK. So the other equation that I want to explain is this one. So I did a biology example before, so I chose a physics equation this time, which is the wave equation. OK. Now, also to vary it, with the biology example, I introduced it with a bit of history. With the wave equation, I'm going to actually introduce it with a different approach, which is not to talk about the history but actually to get the students to try to construct the equation. OK. So the question is what the hell is the wave equation? Well, the wave equation describes-- so here's some string. And here's your coordinate x. And you want to know how does the height of this piece here, the height being f of x as a function of time as well, change with position and with time? So you want to figure out an equation for that behavior of that string, that stretch between two points. For example, this might be a guitar string, and these are the two ends of the guitar string. And it' under tension. And you want to know, how does it move? OK. So we're going to construct the equation. So let's say this is a differential equations class for people in physics who are learning mathematical methods. And they want to learn how to construct differential equations. So maybe they're engineers, physicists. And they're, say, juniors. So they have some mathematical sophistication and some knowledge of forces in physics and some knowledge of differential equations. So I'm going to write down the rough form of the equation, and we're going to try to figure out all the pieces and fill in the missing pieces. So I'm going to give the general form of the thing. So there's some derivative of f with respect to time-- one or two derivatives, we're not sure-- is equal to something here. And then there's some constant here or maybe here. So there's a whole bunch of question marks to fill in. And what we're going to do is reason about what they are. OK. So the first question is to figure out this guy. How many time derivatives do we need? So what this equation describes is the motion of this point. Now, why does the point move? Well, you could ask the class, and eventually, they'll come up with, well, there's some forces on it, because the string's under tension, so there's forces on the point. So here's our point. And here's, say, the string going through it. This is a blow up now of this region. OK. So now, what are the forces on this guy? Well, there's a force from that piece of the string and a force from that piece of the string. OK. Because there's force, what's going to happen to the thing? Is it going to have a velocity or an acceleration? Anyone? AUDIENCE: Acceleration. PROFESSOR: Acceleration. OK. How many derivatives does acceleration have? Two. OK. So we're going to put two derivatives here. So you need two derivatives. So this, and now, we're going to say this is some kind of acceleration. And maybe there's masses in there and stuff, which would all be slurped into these constants. So we have an acceleration. And now, we have to decide what generates acceleration. This side is generating acceleration or the force. So we want to decide, for example, one or two derivatives. Generally, most equations either have one derivative or two. Some really nasty ones have four. But generally, it's one or two. So we'll just choose. Is this one or two? OK. And then the last thing we're going to do is to figure out the constant. OK. So now, one or two derivatives-- one way to decide that is to make something that has just one derivative in it. So if here is my string and here is the point-- so f has a non-zero-- so it has a non-zero df/dx. But the second derivative is 0, because it's a straight line. OK. So we know the second derivative is 0. Let's see what we can figure out what the force or the acceleration should be. OK. Well, here is a point. There's going to be a force on it from that end and that end. And what's the net result of these two forces? 0. So when the second derivative is 0, in this case at least, we would like the force to be 0, which means that force has to be connected to the second derivative of position. So we got that. Now, the next problem is to work out what goes here. The first thing is what's the sine. Should it be plus or minus? OK. So I'm going to ask you that. So find a reason whether it should be plus or minus here and a reason. So find a neighbor or two, and we'll take a vote-- plus or minus, intuitive reason for it. I can't use this example to decide. Let's vote. Everyone have their votes ready? Who votes for plus? So about-- who votes for minus? OK. That's great. So we have a diversity of opinion. So right away, what that shows you is that it's worth actually discussing that point in class, because if you just tell people, they'll write down something you tell them, but they won't have actually internalized it. It'll just be something that maybe contradicted what they said, thought, or not. And then they have to remember, was it what I thought or was what I not thought? Or was it what I not thought or what I thought? So they don't actually understand why. So it's actually worth going through the discussion. So what example could you use to decide? Who haven't I heard from? Yeah. Could you tell me your name? AUDIENCE: Mike. PROFESSOR: Mike. AUDIENCE: Well, if you're at the top of the arc, and you don't it's going to want to be pulled down. It's coming down. PROFESSOR: OK. So let's use an arc like this. We can't use this one, because it's 0, and 0 doesn't have a sign. So it doesn't help us. So the next most complicated thing is this-- so an arc like that. Here's my point. So the force is downwards, or the acceleration is downwards. So we should have negative acceleration. And what's the second derivative with respect to position here? That's also downwards, because the arc is like that. So the derivative has the same direction-- the space derivative and the acceleration. So it should be plus. So we got the sign right. And to emphasize the importance of getting the signs, there's an interesting comment from Feynman back in the '60s at some conference about quantum electrodynamics. He was commenting about how crazy the whole process that he invented for solving quantum electrodynamics is. And his reason that is so crazy-- he said, well, we do the first order term, and then we calculate the second order term and add it to the first order term. But it's very worrying that when we calculate the second order term after the first order term, we don't know whether the second order term is positive or negative. We just have to calculate it and see what it's going to be. But we can't predict ahead of time whether it's positive or negative. And what that speaks to is the importance that physicists attach to knowing the sign of an effect. Is a plus or is it a minus? And so that's fundamentally important. You want to make sure you get that right. Now, why did I do the sign of the effect after this, after the order of the derivatives? Because the most important thing is what do the terms mean. So this is a curvature. This is an acceleration. So here, let me write that-- curvature. Until you know what the terms mean, there's no hope of figuring out what the sign that connects them is. But the sign is the very next thing you do. And it's really important. It's a plus. And now, we need to put in one more thing, because the dimensions are totally bogus. This is a position divided by a times squared. This is a position divided by a length squared. So we need to actually multiply by something to make the units come out correct. So we need something that puts a time squared here and a position squared over here. So then the units will work out-- or a position squared over time squared, which is C squared, which is some speed squared. So speed squared is position squared over time squared. And that makes all the units work out. So there you go. There you have the wave equation. So now, if you're going to actually formally derive it, that's all fine. But I would do this first so that people know where every single term in the equation comes from. And now, I've been a bit sloppy. These are really partial derivatives. But that's, again, an example of lying. I wouldn't worry about whether it's a partial or total derivative at the beginning, because that's not the fundamental idea. The fundamental idea is that it's curvature on this side connecting to acceleration on this side, and they're connected by a positive sign and by something that has dimensions of speed squared, which turns out to be the speed at which the wave moves. So that's a way to introduce yet another equation, not related to the history, but actually connecting to the intuitive approach. Any questions about that?
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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: -- 10 seconds to answer the first clicker question. As you can see we're having another competition today, so see if you can beat out Justin's recitation and get the most correct today. So, this is a topic from Friday, which is asking us which of the following molecules are free radical species. OK. So it looks like most and you got it, that it's o h. Let's quickly go over why that is just in case you weren't one of that 74%. So, if we're looking at o h versus c o, what are we looking at to decide if we have a radical here? Valence electrons. So we have 6 plus 1 in terms of o h, that's 7, versus having 4 plus 6, which is 10. Specifically, what do we look for in terms of valence electrons? Odd numbers. So, we have a radical here with o h, whereas we have an even number of valence electrons for c o, so that's not a radical, in terms of what we know how to think about so far, which is using our Lewis structures to determine that. All right. So that's a little review from Friday, let's move on to today's lecture. In terms of what we're going to be talking about today, we're going to finish up with talking about polar covalent bonds, and finish up with discussing the idea of polar molecules. And then we'll move on to start talking about well, we know how to draw our Lewis structures, how can we use those Lewis structures to think about the actual shapes of molecules. So, first thinking about our polar covalent bonds. What we had established is if we have a case where we have a covalent bond, but there's unequal sharing of electrons between the two atoms, namely because they have different electronegativities, what we see is we have a polar bond or a polar covalent bond where one of the atoms is going to pull away more of the electron density from that bond, giving it that partial negative charge where the other atom is going to end up with a partial positive charge. In terms of thinking about well, what do we call a covalent bond versus a polar covalent bond versus a purely ionic bond, it turns out that the distinction is a little bit fuzzy, it's not a clear line, but in terms of talking about things in general, and especially in thinking about problems for this course, what we're going to say is if you have an electronegativity or chi difference between the two atoms that is greater than 0 . 4, and less than 1 . 7, and we're talking about the Pauling electronegativity scale here, which is what's used throughout your book. If you have a difference in electronegativity anywhere in that range, we'll have what we call a polar covalent bond between those two atoms. So, for example, if we compare the chi values for hydrogen versus carbon, we see they only have a difference of 0.4 . 0.4 is not greater than 0.4 . So we would not call this a c h bond, we would not say it's polar covalent, we would just call it covalent. In contrast, if we talk about a carbon oxygen bond, now we have a chi difference of 0.8 , so that is greater than 0.4 , so we'll call carbon oxygen bond polar covalent. We can extend this idea of talking about these polar covalent bonds to thinking about an entire molecule. I'm sure you've all heard of the term a polar molecule versus a non-polar molecule, and essentially, when we talk about having a polar molecule, what we're saying is that there's is a net non-zero dipole moment within the whole molecule. So essentially, what we have to do is combine all of the individual bonds to think about the molecule as a whole. So, let's take an example of carbon dioxide here, or c o 2. Very soon in the lecture you'll learn how to predict shapes based on Lewis structures. So you will quickly see that this is a linear molecule. We can think about the 2 bonds within it. There's two carbon oxygen bonds, and we know that there is a dipole in that bond, which is going toward the oxygen. Remember, in chemistry we always draw the arrow, we're always interested in what those electrons are doing. So if we draw in the bond dipoles, and you can draw these into your notes, you want to draw an arrow towards the oxygen in each case. But we're not just interested here in thinking about do we have a polar bond, we actually want to know in general do we have a polar molecule. So, looking at c o 2 as a whole, do you think we have a polar or a non-polar molecule here? Non-polar. OK, very good for all of you that said non-polar. The reason it's non-polar is we simply have two equal vectors in opposite directions, so they cancel each other out. The net effect is that we have a 0 dipole moment in the c o molecule. In contrast, we can look at h 2 o. H 2 o actually has this bent shape, and again, we'll see very soon and how to predict that h 2 o has a bent shape, that water is bent. And again, we do have dipole moments -- we have bonds where there is a polar bond and we should point our arrow toward the oxygen, because electron density is being pulled from the hydrogen atoms to the oxygen. And again, we can cancel some of these vectors out, but we still have a net dipole moment that is going to be going up if we think about combining these 2 vectors here. So, what we would say, which is what we know from everything we've always heard, which is that water is a polar molecule. So, this is a great way to think about molecules in general. It's very easy to think about if those vectors cancel each other out or if they're additive when we're talking about just single bonds, so we have two atoms or if we have just three atoms, as we do in these cases here. But it turns out, a lot of the molecules we consider are actually much, much, much larger than just a couple of atoms. So we can't often think about canceling all the different vectors out. And in general, when we're talking about these really large molecules, whether they're large organic molecules or we're talking about proteins, which can have thousands of atoms in them, instead of adding up all those individual polar bonds, what we do is we talk about the number of polar groups that it will have in the molecule. So we can think about a protein, for example, as having a lot of different polar groups or one that has not too many polar side groups. That's one way you hear people talking about proteins, and that has a lot to do with their solubility in water, and also about how the protein's going to fold. But let's look at a little bit about less complicated example than a protein with thousands of different atoms. Let's instead consider a couple of vitamins here, which instead have a few dozen different atoms in them. Specifically, let's look at vitamin A and vitamin B9. Does anyone know another name for vitamin B9? I don't think I hear it -- does anyone say folic acid. It's also sometimes called vitamin M. It's one of the vitamin B9 vitamins, there's actually other forms of vitamin B9, but this is one of the big ones that you hear about. Hopefully you're all very familiar with folic acid, because it's an important vitamin to take, especially if you're a woman, especially if there's any chance you ever might become pregnant in any kind of near future time, by accident or on purpose. And the reason for this is because folic acid deficiency in pregnant women leads to neural tube defects in babies, and your brain develops very, very, very early in pregnancy, the brain of an embryo. So it turns out that a lot of women don't realize they're pregnant while the fetus' brain is developing, and if these women are not getting enough folic acid, you can end up with spina bifida, which is an absolutely devastating and very preventable disease or other neural tube defects. So hopefully, if you're not familiar with folic acid, you'll become familiar at least within the next 10 or 15 years or so before you're thinking about maybe, on purpose, becoming pregnant at some time. Vitamin A you probably all heard about growing up, getting enough of your carrots so you can see at night. Vitamin A is important for eye health. But just looking at the structure of these two molecules, knowing what we know so far about general chemistry, we can already say a lot about how these are going to function the body or how they'll be treated in the body. And specifically, let's take a look at which of these two molecules has more polar bonds and see what that tells us. So take a look at your two structures and tell me which has, between folic acid and vitamin A, more polar groups within the vitamin. And this should go pretty quickly since you don't need to have an exact number, we're just looking and glancing and seeing which has more. So let's take 10 seconds on this. All right, great. So most you said folic acid or vitamin B9, so let's switch back to our notes and take a look at why. So vitamin B9 has more polar bonds. It's very easy to see, once I highlight, that it's not even a close call at all. We have a bunch of different polar bonds in B9 versus just one in vitamin A. Remember, the c h bond we're not calling polar covalent because it only has an electronegativity difference of 0.4 between the two atoms. So what we know is that vitamin B9, folic acid, is more polar. Remember, we also just said that water is polar, so would we say that B9 is water or fat soluble? It's going to be water soluble -- everyone knows the saying, "like dissolves like." B9 is going to be very water soluble. This actually is very important if you think about taking your vitamins. This means if it's water soluble, any vitamins that are water soluble, and now you should to be able to look at the structure of any vitamin and tell me, for example, that vitamin C is water soluble, that folic acid is water soluble. If you take these vitamins, what happens is that they're very quickly and easily excreted into your urine. So, for example, some people like to take mega doses of vitamin C. What really happens is that you have a mega dose of vitamin C in your urine, it doesn't stick around long in your body. So it doesn't help and just take all of your vitamin C at once. That's why it's important to eat balanced meals throughout the day, you need to be getting a constant supply of these water soluble vitamins. The same is for folic acid, you can't just take it once a month, you need to be taking it regularly in order that you keep the stores up in your body, otherwise you're going to excrete it, you're going to get rid of it very quickly. In contrast, if we think about vitamin A, is this going to be water soluble or fat soluble? Yup, so this is the fat soluble vitamin. Vitamin E is another big one that's fat soluble that gets a lot of press in terms of being an important vitamin to take. We can think about what it means when a vitamin is fat soluble instead of water soluble. Well, now it's not going to get excreted out of our body so quickly, so we actually can build up amounts of vitamin A vitamin E, for example, but that can also pose a problem if you think about biologically what's happening. And up here I just show two different supplements that I found on the internet. This is One A Day vitamin, it has about 100% of what you need of everything. And then this vitamin is one that I found that's supposed to really help you with your eye health, if you have bad vision instead of glasses, they suggest that you try this vitamin here. And what you can see is that it has five times your daily value of vitamin A, and 13 times what you need of vitamin E. Is this a good idea? No. Basically, you are just building up more and more and more of these fat soluble vitamins into your body. And there have been studies that have come out recently trying to look at the health benefits of vitamin E, and in some of these studies they give these mega doses, and instead what they find is increased bleeding in these patients, an increase in overall different types of death that can be happening. You want to take your vitamins, it's very important, but you don't want it just build up, and you want to think about is the vitamin that I'm going and getting 800 times what I need a day, is that a water soluble vitamin or a fat soluble vitamin, and using your chemistry, you should be able to very quickly take a quick look at that structure and figure out what kind of a vitamin that is. All right. So that's just one reason we would want to be able to think about polarity is thinking about whether something is very water soluble or not. Let's think about some other things that have to do with polarity and then can tell us a lot of other information. So let's move on to talking about the shapes of molecules. And the shapes of molecules is very important for a number of different properties when we're thinking about chemical reactions and reactions that take place in the body. When we talk about shapes of molecules, we're talking about the geometry of that molecule. And the geometry influences all of its different properties, including things like melting point or boiling point, it's reactivity. We just saw when talking about polar molecules, that it influences whether or not the molecule itself is polar or apolar. It's also really important when we talk about biology. Shape is particularly important when you think about enzymes having an active site where a molecule needs to fit perfectly into the active site, it does that because of its shape. A quick example that we can think about is with sucrose. Does anyone know what sucrose is? It's just table sugar -- sucrose is that crystal in sugar that we hopefully use in most of our sugar intake as sweeteners. A lot of times we use corn syrup, which is often not sucrose and which tends to be not always as good for us, although anything in too much excess is obviously a non-ideal situation. But in order for us to use the energy from sucrose, sucrose is made up of two individual sugar monomers. It's made up of a monomer of glucose and one of fructose, a 6 and a 5 membered ring, so in order for our body to use it, we need to break it into with its monomers. And if we that, we can do it by what's called hydrolysis -- sucrose and water breaks down into its two monomers, now our body can use it. The problem is this process takes on the order of 10 or maybe 100 years in order for us to get enough of the sugar broken down. That's why we need an enzyme molecule, and the enzyme in our body is called sucrase. Sucrase breaks down sucrose, it catalyzes that reaction, so it happens very quickly. And the key is, and here's a ball and stick shape of what sucrose looks like. It needs to fit exactly into that active site. When it does, it combines into the enzyme, the enzyme can then catalyze the hydrolysis or the breaking apart of those two individual monomers into its separate pieces. And then it let's go of the glucose and the fructose, and we get our enzyme back again, and something else combined into that active site. So, this is one very simple example of why molecular shape is very important. A lot of times you're thinking about small molecules interacting with proteins or interacting with other molecules, and you want to think about the shape of that molecule to think about how that interaction is going to take place. So what we're going to use to think about molecular shape or molecular geometry is what's called valence shell electron repulsion or vsper theory. And this theory is very convenient for us to be thinking about because we are now masters of drawing Lewis structures, and vesper theory is based on Lewis structures, and also the principle that when we have valence electron pairs, they're going to repel each other. This make sense any time you have negatively-charged electrons, you want to get them as far away from each other as possible because they're all negatively charged. And the other principle is that the geometry around that central atom, which you've identified as the central atom in the Lewis structure, is going to be such that it minimizes the repulsion between those either bonding electrons or the electron pairs. So when we talk about vsper there's a special nomenclature that we do use, and in vsper we have a -- does anyone know what a means in vsper theory? It's a central atom. What about x? So this is actually the bonding, any bonding atom you call x. And then we have e, which is equal to lone pairs. So e is a lone pair -- e is not a lone pair electron, so if you have two electrons, that's one lone pair. All right. So there are some guidelines that we use when we're coming up with these vsper geometries. The first thing we talk about is the steric number, which we use to predict what that geometry will be. When we're talking about steric number, all we're talking about is adding together the number we have of bonded atoms, plus the number of lone pairs. So, essentially were just adding x to e and that gives us our steric number. So, for example, if we look at this molecule, which is a x 2 e, what is the steric number here? Yeah, it's 3, so we have a steric number of 3. Something that I want to point out is that we could have a different molecule, which is also a x 2 e, but in this case we have a double bond between the central atom and one of the x atoms here, and what I want to point out is in vsper theory we treat bonds as bonds, we don't worry about if they're single or they're double or they're triple. So again, we call this a x 2 e, this again, has a steric number of 3. So what's important in terms of vsper theory is the number of atoms bonded to a central atom. What's not important is the types of bonds that we're dealing with. A few other guidelines I want to mention is first of all, to think about resonance structures. So on Friday we were talking about the chromate anion, and we said that here are two of its resonance structures here, but that we could actually draw four more resonance structures, and I just want to point out when you take a molecule that has many different resonance structures and you want to draw its vsper, it's Lewis structure and its vsper geometry, you can take any single one of those resonance structures, it doesn't matter, you'll all end up with the same geometry. And lastly, if we're talking about a molecule that has more than one central atom, which is what we're very, very often doing, you need to deal with each one separately. So for example, with methanol here, we have a carbon, which is a central atom, and we also have an oxygen, which is a central atom, you need to talk about the geometry separately, we don't talk about the geometry of the entire molecule. So, let's go ahead and look at some of these vsper examples, and Professor Drennan is going to help demonstrate what some of these are with some models here. So the first case we're going to talk about is without any lone pairs -- this is the most straightforward case. And our first case that we can have is a x 3, which is going to have a linear shape, and that will have a bond angle of a 180 degrees. The next case that we can talk about is trigonal planar or a x 3. Now, you do need to know these geometry, you need to know the names. A lot of them are very easy to remember. I expect not too many of you will get linear incorrect. Also, trigonal planar, pretty easy. Trigonal, it has 3 atoms around the central atom and the molecule is planar. PROFESSOR: So, what is the bond angle for this geometry? 120. PROFESSOR: All right. Great. So, next we can think about a x 4, and this is what's called a tetrahedral geometry. If you're trying to remember this name you can think of each of those bonding atoms as being in the corner of tetrahedron. One thing I want to point out is you're seeing some notation we haven't used before in this class where you have a wedge coming out at you, and you also have a dashed line to one of those bonds. Any time you see a wedge in a Lewis structure, it means that it's coming out at you -- it's either coming out of the screen or it's coming out of your paper. And any time you see a dashed line here, which might be easier to see in your notes, that means that the bond is actually going into the page or into the screen or into the blackboard. PROFESSOR: And what are the angles for a tetrahedral? PROFESSOR: All right, so they're 1 0 9 . 5. Pretty close. So that's as far away as those bonds can get from each other would be 109 . 5 degrees. The next case we have is a x 5. That is what we call trigonal bipyramidal. Again, we have trigonal, because we have those 3 bonds in the center, and if you can picture putting walls between all those bonds, you can see there's a pyramid on the top and a pyramid on the bottom, so bipyramidal. PROFESSOR: And there are 2 sets of angles here, what are the angles for equatorial atoms? 120. And then for axial? 90. PROFESSOR: And again, axial, those are just the ones in the axis, and equatorial if you put your globe around the equator. So the last case we have is a x 6. A x 6 is what we call octahedral -- you can picture each of those bonded atoms as a corner of an octahedron. PROFESSOR: And what are the angles in this, this one set of angles? 90. PROFESSOR: Great. So, 90 degrees. All right. So this is all straightforward when we're just thinking about different molecular shapes that don't have any lone pairs in them, but once we start having lone pairs, we now need to think about how those lone pairs are going to affect the geometry. But before we do that, let's talk about some examples of molecules without lone pairs. So the first case is what we saw at the beginning of class, carbon dioxide, and we said that that was linear, and now we know why it's linear -- it has a formula of a x 2, and an s n number of 2. And that's a bond angle of 180 degrees. So, you can tell us about borane, borane is a x 3. PROFESSOR: And what is the geometry? And the angle? 120. Feel free to yell out very loudly, we want people in OpenCourseWare to hear the answers coming from the room. PROFESSOR: All right. So the next case we're going to look at is c h 4 or methane, and let's do a clicker question here to make sure everyone is remembering these geometries. And this should be very quick, and try to not turn the page back one, and tell us in 10 seconds here what the geometry is. 98%, that is a new record. Very good. Tetrahedral. PROFESSOR: All right, while I'm holding carbon dioxide in one hand and methane in the other hand, I just want to do an additional plug for the energy debate tonight. So there will be representatives from both presidential campaigns that are going to be there to answer questions. And I'm not sure of the exact format, I'm not sure whether all audience members are going to be able to ask questions or not. I'm not sure if I'm going to be there, I'm supposed to be giving a talk actually, ironically, about energy at the same time. But if I can't make it and you're allowed to ask questions, I'd like you to ask a question for me. So, my lab looks at enzymes that remove carbon monoxide and carbon dioxide from the environment. And I have an energy initiative grant to do this, because MIT recognizes that an important part of any energy initiative is thinking about lowering pollutants and greenhouse gases. So, I heard on Thursday at the debate when Governor Sarah Palin was asked directly if she really doesn't think there's any connection between global warming and any man-made activities, that she said, well, you know, it could be just regular temperature fluctuations, and she doesn't want to point any fingers. So, she's not quite sure. I also learned on Thursday that she will head the energy initiative in a McCain/Palin White House. So, if you go to this debate, I would like you to ask what is your plan for sequestering carbon dioxide and other greenhouse gases. You can ask both candidates, and please let me know what the answer is, because I think that's a really important question -- here we have carbon dioxide, here we have methane, we need to be thinking about greenhouse gases. PROFESSOR: You can also throw in the question that you do know this is tetrahedral and 109 . 5 degrees -- show your chemistry knowledge here. All right, and let's keep showing that we know our geometries, let's look at p c l 5. PROFESSOR: And what is geometry here? Yup. And the angles? 120 and 90. PROFESSOR: So, 120 equatorial, and 90 axial. And last we have s f 6. PROFESSOR: So, what is the geometry here? Octahedral. And angle? 90. PROFESSOR: All right, so there's our set of examples, one for each for those shapes or geometries that have no lone pairs. And now let's think a little bit about once we do have molecules with lone pairs. And the biggest point to keep in mind when we're comparing lone pair electrons with bonding electrons or electrons in bonds, is that when you have electrons in bonds they have less spatial distribution than lone pairs. So that's just a way of saying that electrons and bonds, they take up less space. So, when we think about lone pair electrons, they're taking up more space and this means that they're going to experience more repulsion in thinking about the lone pair electrons with either other lone pair electrons or with bonding electrons. So let's think a second about the order that this will be in. The biggest repulsion we would feel is if we have two different lone pair electrons or lone pairs. They're going to have the most repulsion. In the middle is if we're talking about a lone pair with a bonding pair. And then the least repulsion is going to be between two bonds or two bonding pairs of electrons. So, let's look at an example of where this comes up. So the first example I'm going to talk about is a molecule that has the geometry of a seesaw shape, and once we get to having Professor Drennan actually show you that shape, you will never forget. See-saw, that's going to be one of the easy geometries to remember. But the first thing that we need to consider in terms of the shape, which starts, if you picture first of all, and these models sometimes don't quite stay together, we could actually have two different possibilities. The first is the idea that we could have an axial lone pair, and the second is the possibility that we could have an equatorial lone pair. And we could consider both, and we should be able to use our vsper principles to think about which one is actually going to happen. So if we think about having an axial lone pair here, that would mean that these lone pair electrons are going to be within 90 degrees of three different loan pairs, and actually we'll see later, it will be more than 90 because they're actually going to push them down. But in terms of considering how many bonding electron pairs they'll repel strongly, what we care about is anything within 90 degrees initially. So, what we would see with the axial lone pair is that we have three lone pairs that we're going to very strongly repel. Now would you tell me, if we think about an equatorial lone pair, how many different bonding electrons or a bonding pairs that will strongly repel if we have an equatorial lone pair? So you might need to look at your notes to actually compare these two before you submit your answer up here. All right. Let's take 10 seconds on that. OK, strong showing with the clicker questions today. It's going to be a tight race for clicker competition. It's correct that the equatorial is, in fact, has two, only two that it strongly repels. So, actually it's probably easier to look at Professor Drennan's model here to see that. So it's where the equatorial lone pair is that is the actual seesaw geometry. PROFESSOR: So, how many of you had seesaws in playgrounds when you were growing up? Oh, a lot of people. I know that they're not considered totally safe anymore, so some of them are going away. So this is seesaw. Now you should never forget it. PROFESSOR: All right, so seesaw, we've got seesaw. Just to point out a few more shapes. We took one out, we replaced one bond with a lone pair in terms of seesaw. If we have a x 3 e 2 now we have two equatorial lone pairs. This is called the T-shaped PROFESSOR: T-shaped. PROFESSOR: See, these aren't too hard to remember the geometries, the geometry names. And we also can think about if we have a x 4 e 2. So now in order to get our lone pairs as far away from each other as possible, they're going to be on the axial position, and this is called square planer. PROFESSOR: So, this is easy to remember, square, also planar. PROFESSOR: All right. So, let's talk a little bit now about what the actual angles are between bonds when now we have these lone pairs present. And remember, what we said is when we have a lone pair of electrons, they actually are going to have more repulsion than if we just have a c h bond or a bonded pair of a electrons. So essentially what that means is that in molecules that have lone pair electrons, so for example, if we look at n h 3 or ammonia versus methane here, and Professor Drennan is showing those models to you here. What you end up seeing is actually that the bonding angle in n h 3 between the n h bonds is smaller. And you can't see it with these models because these are not actual lone pairs. But if they were actually repelling each other, they would be pushing those bonds as far away as possible, and instead of being 109 . 5, you'd see that the angle is now 106 . 7. So it's a smaller angle between bonds, because you have more repulsion from those lone pairs pushing those bonds down. We can also think about the influence of atomic size in terms of this effect. So first of all, what happens to atomic size as you go down the periodic table? Good, it increases. So we have size increasing as we go down the periodic table. Phosphorous is right underneath nitrogen on the periodic table, so phosphorous is going to be bigger than nitrogen. In terms of picturing what happens with those lone pairs, when we have a larger atom, the orbitals are also going to be larger, they can take up more space. That means these electrons, this lone pair electrons are going to take up more space. This means they're going to push away those bonding electrons even more. So would you expect the bond to be larger or smaller for p h 3? STUDENT: [INAUDIBLE] PROFESSOR: It's going to be smaller. So, what we see is that angles between bonded atoms actually decrease as you go down a column on the periodic table. So the actual angles for p h 3 are now going to be really quite a bit smaller, they're 93 . 3 degrees. All right. So let's create a list for ourselves in terms of all the geometries that we can have now that we're dealing with lone pairs. One thing I want to point out in terms of remembering the names of the different geometries, when you're naming a geometry, the geometry name, for example, when we looked at square planar, we're only actually naming where the bonds are, where actual atoms are. The name doesn't really depend on the lone pairs. And also, when you draw a geometry, you don't always have to draw the lone pairs in, but you have to remember that the lone pairs are very much affecting the angles within your molecule and also the actual shape. So, for example, let's start talking about different types that have lone pairs in it. First of all, thinking about a x 2 e, so we have one lone pair, and the geometry here is bent, and I want you, thinking about lone pair repulsion, to tell us what you think the angle between these bonds are going to be? And let's take 10 seconds on that. OK, good. 75%, that's not bad. Let's think about why. It's going to be less than 120 degrees, because we know that normally in a trigonal planar situation, we have an angle of 120 degrees. And since lone pairs are going to cause more repulsion, we're actually pushing down these two bonds here closer together, so what we end up seeing is that they're going to be less than 120 degrees. And it depends on the actual molecule what the exact angles are going to be, so you never have to learn the exact angles in terms of lone pair electrons, you just need to be able to tell us if the bond angle is going to be less than 120 degrees. So let's look at another example here, a x 3 with one lone pair, e. This is called trigonal pyramidal. Again, you have trigonal because there's three atoms bonded to the central atom, and this looks like a pyramid here. PROFESSOR: So, what would the angle of this be? STUDENT: [INAUDIBLE] PROFESSOR: I think I heard it. PROFESSOR: I think so, too. Less than 109 . 5. Let's take another example, a x 2 e 2. This is also bent. Tell us what the geometry -- we told you the geometry, tell us what the bond angle is going to be between these bonds. So, let's take 10 seconds on this. Let's re-poll F 4. So 10 seconds again. All right. In about 10 seconds, Darcy, in 10 seconds we'll hit the next side. OK, so it's less than 109 . 5 degrees now. So some of you wrote that it was less than 120 degrees, and we can think about if we switch back to the class notes the difference here. So even though, so this is now going to be less than 109 . 5, if we looked at the case we had bent in the first case, that started with 120 and one of those bonds was replaced with a lone pair, but now we have two lone pairs here, so now what we're going to see is that the two bents are not, in fact, equal. The bent where you started with the tetrahedral shape is actually going to be less than 109 . 5, and Professor Drennan can maybe show us that with the models here. PROFESSOR: It's a little had to see, actually, between the two, but I think it's easier to see on this, that if you consider the sort of starting place for the two, if you look at the bottom set and you put orbitals, so it also depends on your starting geometry. It can be bent if take off these bonds on the top, but it depends on what the starting geometries were. And so you will go with your starting geometry, if it was 109 . 5, it'll be less than that. If it's 120, then it's less than that. PROFESSOR: So, this is a good case of where, even when you have the geometry the same, you need to think about how many lone pairs you're dealing with in your molecule, and think about what would the geometry be if I just pictured those lone pairs in there as bonds and then push them even closer together so that it's less than that actual angle. OK, let's talk about a x 4 e, and that is what we call the seesaw shape. PROFESSOR: And so, what difference in angles are you going to have? STUDENT: [INAUDIBLE] PROFESSOR: Yup. PROFESSOR 2: So, two sets again, one is less than 120, and one is going to be less than 90. So, let's look at a x 3 e 2, you'll notice there's a lot more combinations we can get to once we're talking about lone pairs. This is called T-shaped, and we had briefly discussed that but not mention what the angles would be. PROFESSOR: So, what angle would you have here? Yup. PROFESSOR: All right, great. Less than 90 degrees. So, we'll start a new page here and talk about a x 2 e 3, and we'll let you tell us what this geometry is. Yup, so I hear a lot of linear out there. And you need to keep in mind we give the names based on where the actual bonds are, not where the lone pairs are, even though the lone pairs, of course, affect the structure in the geometry. PROFESSOR: The angle? STUDENT: 180. PROFESSOR: Yup, 180. PROFESSOR: So, it's exactly 180 here, not less than. PROFESSOR: So, for a x 5 e, what we have here is called square pyramidal. This is square because of the four square at the bottom of your pyramid, then you can picture the pyramid as you go to the top. PROFESSOR: And so what angles would you have here? Yup, less than 90. PROFESSOR: So, less than 90, great. All right, there are more. So, our next combination that we can think about is if we have four bonded atoms and two lone pairs, a x 4 e 2, that's going to be called square planar. PROFESSOR: So, again, we have square and it's planar. And so what are your angles here? Yup, exactly 90. PROFESSOR: So next let's look at a x 3 e 3. This is what we call T-shape, this is also T-shape. PROFESSOR: And angle here? PROFESSOR: Yup, I think we heard less than, it's less than 90 degrees. And let's look at a x 2 e 4, and again, you guys can tell us what this geometry is, first of all. Yup, it's linear. So, we have lots of different ways we can get to a linear molecule, some of which have lots of lone pairs, some of which have no lone pairs at all. This is linear and a 180 degrees. All right. So we'll do just a few examples to show you some actual molecules that have these geometries. We won't go through all of them, because as we saw, there's lots of different combinations we can have once we start getting into lone pairs. The first we'll look at is water here, and water is a good one to look at because we had seen that at the beginning of class here when we were talking about all polar molecules. When we talk about the formula type of water, what when you say the formula type is? STUDENT: [INAUDIBLE] PROFESSOR: A x 2 e 2, that's correct. And what is the geometry of water? It's bent. So what we saw at the beginning of class is water is bent and now you can see why and should be able to predict that yourself. I just want to mention that if you look, and depending on the edition of the book you have, in one edition it's called bent, in the other edition it's called angular. We'll call it bent, but just remember that bent and angular, they're the same geometry. Let's look at another example, s f 4. And here it is drawn here. What is the formula type for s f 4? I think I heard it, a x 4 e. And the geometry? Seesaw, good. All right, we never get seesaw wrong, that's the easy give-away one. All right, what about b r f 3? This is the shape here. What is the geometry for b f 3 -- the original geometry, not the new geometry here? What was that? STUDENT: T. PROFESSOR: T-shaped, that's right. B f 3 is T-shaped. All right, so let's look at xenon f 2. That's going to have a formula type of a x 2 e 3, if we actually look at the Lewis structure here. So if you think about what the geometry is, and you should just be able to look at this and see, what would you say the geometry of xenon f 2 is? STUDENT: Linear. PROFESSOR: Linear, good. All right. And let's try one more here, which is a x 4 e 2, or a xenon f 4 -- very explosive, but a good example. STUDENT: Square planar. PROFESSOR: Square planar, all right, great. So, in general, when we think about vsper theory, even though it doesn't talk about or tell us anything about the energies of these different shapes, it's very useful in making a first approximation and coming very close to thinking about what the actual shapes of molecules are. So one thing I want to point out in terms of what you're responsible for is you should be able to fill out one of these charts with only seeing the labels up here -- tell us the formula type, tell us the steric number, tell us the geometry, you should be able to draw the Lewis structure, you already know how to do that, and also talk about the angles. So, as you finish up your problem-set, you can, I think now, get through just about all of it. You can do now this part with the geometry. So, we'll see you on Wednesday -- we're going to get out a little bit early today.
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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 back to 8.701. So in this video, we'll talk about the nuclear shell model. We've already seen an interesting empirical model to describe nuclear binding energies-- the liquid drop model. But it comes short in the description of all aspects of the nucleus. So let's see what we can find here. First of all, you probably remember shell models from atomic physics. And shell models are very successful in describing hydrogen, for example. The question is, can this also work for the nucleus? After all, the nucleus is a many-body system, compared to hydrogen, where you have a proton and an electron circling around. There's no analytic solutions, like the Schrodinger equation. There is no dominant center for a long-range force, like the proton has been the dominant center. And we have short-range forces with many pairs of interacting nucleons. And I can continue the list of difficulties. On the other hand, the interactions kind of average out and result in a potential which depends only on the position, but not on the timing of the nucleus. And that leads us, then, to what we call a nuclear mean field. So on average, our proton and our neutron inside the nucleus sees a specific potential. And we can use that, then, parameterized as potential with a harmonic oscillator, and use that model, then, in order to describe our nucleus. So this works, actually, surprisingly well. But before we go there, we'll look at experimental evidence for closed nuclear shells. So again, here is our plot of the binding energy. And you see that there are those areas here that seem to be some sort of higher binding energies. And it turns out those happen at so-called magic numbers. Magic numbers are 2, 8, 20, 28, 50, and 126. So the question now is, how can we explain this? Where does this come from? So again, the experimental evidence is numerous. We find that the number of stable isotopes or isotones is significantly higher for nuclei with a proton-- or neutron, or both-- numbers equal to one of those magic numbers. The nuclear capture cross-section, meaning the likelihood to capture a proton or a neutron, are high for nuclei where exactly one nucleon is missing from a magic number. But it's significantly lower for nuclei with number of nucleons equal to the magic number, meaning that there is this concept of a closed shell. We either just add a nucleon to close it or you have to pay a higher price. The energy of excited states for nuclei with a proton or neutron number equal to the magic number are significantly higher than for other nuclei. And these are all experimental observations. And the excitation probabilities of the first excited states are low for nuclei with a proton-- or neutron, or both-- numbers equal to the magic numbers. Quadruple moments-- we haven't discussed those at length, but you can think about them as deformations of the nuclei. They almost vanish for nuclei with proton or neutron numbers equal to the magic numbers. So those are more kind of sphere kind of objects. Here's a plot which shows or points out the double magic numbers-- as in, those are a nucleus where both the proton number and the neutron number are laying on the magic number. So calcium here has two of those, with 20 protons and 20 neutrons, or 20 protons and 28 neutrons. And there's alphas. Those are specifically interesting object of research. There was some historic confusion in this, and it came from the fact that while the experimental data pointed to nuclear magic numbers of 2, 8, 20, 28, 50, and 126, if you just think about a flat bottom potential, just a flat potential, you find magic numbers which are 2, 8, 20, 40, 70, and 112. And those are typically not in agreement. So therefore, it seemed like that this shell model kind of worked, but not really. We found agreement here, but then disagreement in the higher part of the magic numbers. So something was missing. And so what was missing was the spin-orbit part of the discussion. We alluded to this in the nuclear force. What you have to do is, beyond three-dimensional harmonic oscillator, you have to add the spin-orbit coupling to the Hamiltonian. And when you do that, you change the orbit such that the magic numbers agree with the experimental data. So you see here, the potentials for proton, which has also the Coulomb repulsion added, and the nuclear potential, and then you see that the spin-orbit coupling slightly changes the potential. All right. As a comparison here, the nuclear and atomic shell models, just for an example. And you see we call them shells because we see that the energy gaps between individual shells are quite large, much larger than within the shell. And the same-- this is for the atomic model. And for the nucleus, you see very similar. So it's not that extended, but still larger gaps in energy when you go from one state to the next.
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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 what do we do? We are going to sum over final states the probability to go from i to final at time t0 to first order. Since the sum of our final states is really a continuum, this is represented by the integral of the f i t0 1, multiplied by the number of states at every little interval. So this will go rho of Ef dEf. So this is what we developed about the number of states. So I'm replacing this-- I have to sum but I basically decide to call this little dN, the little number of states in here, and then I'm going to integrate this probability, so the number of states over there, and therefore the dN is replaced by rho times dEf. So then this whole transition probability will be 4 integral, I'm writing now the integral, VfI squared, sine squared, omega f i t0 over 2 Ef minus Ei squared, rho of Ef, d of Ef. And you would say at this moment, OK, this is as far as you go, so that must be Fermi's golden rule, because we don't know rho of Ef, it's different in different cases, so we have to do that integral and we'll get our answer. But the great thing about this golden rule is that you can go far and you can do the integral. Now I don't even know, this VfI also depends on the energy, how am I ever going to do the integral? That seems outrageous. Well, let's try to do it, and part of the idea will be that we're going to be led to the concept that we already emphasized here because of this suppression, that only a narrow band of states contribute, and in that narrow band, if the narrow band is narrow enough, in that region VfI high be approximately constant in a narrow enough region, and rho will be approximately constant. So we'll take them out of the integral, do the rest of the integral, and see later whether the way we're doing the integral shows that this idea is justified. So I'll just-- you know, sometimes you have to do these things, of making the next step, so I'll do that. I'll take these things out, assuming they're constant enough, and then we'll get 4 VfI squared, rho of E, what should I put here? E sub i, is that right? Because if it's all evaluated at the initial energy Ei, if only a narrow band will contribute, I'll put an h squared here so that this will become omega fi, and now I will integrate over the sum range of energies the function sine squared omega f t0 over 2, over omega fi squared dEf. So I just took the thing out of the integral and we're going to hope for some luck here. Whenever you have an integral like that it probably is a good idea to plot what you're integrating and think about it and see if you're going to get whatever you wanted. Look, I don't know how far I'm going to integrate, I probably don't want to integrate too far because then these functions that I took out of the integral are not constants, so let's see what this looks like, the integrand, this function here. Well, sine squared of x over x squared goes to 1, you know when-- this we're plotting as a function of omega fi. Why? Time is not really what we're plotting into this thing, we're plotting-- we're integrating our energy, Ef, omega fi is Ef minus Ei, so omega is the variable you should be plotting, and when omega goes to zero, this whole interval goes like t0 squared over 4, and then sine squared of x over x squared does this thing, and the first step here is 2 pi over t0, 2 pi over t0 and so on. And now you smile. Why? Because it's looking good, this thing. First what's going to be this area? Well, if I look at this lobe, roughly, I would say height t0 squared with 1 over t0, answer proportional to t0. This whole integral is going to be proportional to t0. The magic of the combination of the x squared growth, t0 squared and the oscillation is making into this integral being linear in t0, which is the probability the transition [INAUDIBLE] is going to grow linearly is going to be a rate, as we expected. So this is looking very good. Then we can attempt to see that also most of the contribution here happens within this range to the integral. If you look at the integral of sine squared x over x squared, 90% of the integral comes from here. By the time you have these ones you're up to 95% of the integral. Most of the integral comes within those lobes. And look what I'm going to say, I'm going to say, look, I'm going to try to wait long enough, t0 is going to be long enough so that this narrow thing is going to be narrower and narrower and therefore most of the integral is going to come from omega fi equal to 0, which means the f equals to Ei. If I wait long enough with t0, this is very narrow, and even all the other extra bumps are already 4 pi over t0 over here is just going to do it without any problem, it's going to fit in. So another way of thinking of this is to say, look, you could have argued that this is going to be linear in t0 if you just change variables here, absorb the t0 into the energy, change variable, and the t0 will go out of the integral in some way, but that is only true if the limits go from minus infinity to plus infinity. So I cannot really integrate from minus infinity to plus infinity in the final energies, but I don't need to because most of the integral comes from this big lobe here, and if t0 is sufficiently large, it is really within no energy with respect to the energy, Ei. So our next step is to simply declare that a good approximation to this integral is to integrate the whole thing from minus infinity to infinity, so let me say this. Suppose here in this range omega fi is in between 2 pi over t0 and minus 2 pi over t0. What does this tell us that omega fi is in this region? Well, this is Ef minus Ei over h bar so this actually tells you Ef is in between Ei plus 2 pi h bar over t0, and Ei minus 2 pi h bar over t0. All right, so this is the energy range and as t0 becomes larger and larger, the window for Ef is smaller and smaller, and we have energy conserving. So let's look at our integral again, the integral is I, that's for the integral, this whole thing, will be equal to integral dEf sine squared of omega fi t0 over 2 over omega fi squared. So what do we do? We call this a variable, u, equal omega fi t0 over 2, so that du is dEf t0 over 2 h bar, because omega fi is Ef minus Ei, and Ef is your variable of integration. So you must substitute the dEf here and the rest of the integron. So what do we get from the dEf and the other part? You get at the end h bar t0 over 2 integral from-- well, let's leave it, sine squared u over u squared du. So look at this, the omega fi squared, by the time you get here omega fi goes like 1 over time, so when it's down here we'll give you a time squared, but the dE gives you 1 over time so at the end of the day we get the desired linear dependence on t0 here, only if the integral doesn't have t0 in here, and it will not have it if you extend it from minus infinity to infinity. And there's no error, really, in extending it from minus infinity to infinity because you basically know that n lobes are going to fit here and are going to be accurate, because there is little energy change if t0 is large enough. If t0 is large enough, even a 20 pi h bar and a 20 by h bar here, that still will do it. So we integrate like that, we extend it, and we get this whole integral has value pi, so we get h bar t0 pi over 2, that's our integral, I. So our transition probability, what is it? We have it there, over there, we'll have the sum over final states, i to f of t0, first order is equal to the integral times this quantity, so that quantity is h t0 pi over 2, so it's 4, what do we have, VfI squared, rho of Ei over h squared, then h bar t0 pi over 2. So your final answer for this thing is 2 pi over h bar VfI squared, rho of Ei t0. So let's box, this is a very nice result, it's almost Fermi's golden rule by now. Let's put a time t here, t0 is a label, not to confuse our time integrals or things like that, so we could put the time, t, here, is 2 pi over h bar VfI squared rho of Ei t. From here we have a transition rate, so a transition rate is probability of transition per unit time, so a transition rate would be defined as the probability of transition after a time t, divided by the time t that has elapsed, and happily, this has worked out so that our transition rate, w is 2 pi over h bar VfI squared rho of Ei, and this is Fermi's golden rule, a formula for the transition rate to the continuum of final states. You see, when I see [INAUDIBLE] it almost seemed you still have to integrate, there is a rho of E and let's integrate [INAUDIBLE] but the interval has been done and it says transmission amplitude squared evaluated at the state initial and final with the same energy and final state, and the rho evaluated at the energy of the initial state. You don't have to do more with that. So we have this formula, let's look at a couple more things. Do units work out? Yes, this is transition per unit, this is 1 over time, this is energy squared, this is 1 over energy, and this is an h bar, this will give you 1 over time, so this thing goes well. How about our assumptions? This was calculated using some time t, we used to call it t0. How large does it have to be? Well, the larger it is the more accurate the integral is, but you don't want to take it too large, either, because the larger it is, the transition probability eventually goes wrong at first order of perturbation theory. So this argument is valid if there is a time, t0, that is large enough so that within this error bars, rho and the transition matrix elements are constant so that your integral is valid. But this t0 being large enough should be small enough that the transition probability doesn't become anywhere near 1. That will happen in general or if VfI is sufficiently small, so when VfI is sufficiently small, this will always hold, and in physical applications this happens and it's OK. So that's our presentation and derivation of Fermi's golden rule, and we will turn now to one application and we will discuss.
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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: Mendel's second law-- this thing over here about a three to one ratio about a single trait being controlled by a pair of alleles, and those alleles being distributed independently of each other to the offspring, the stuff you always learned about Mendel-- that's often referred to as Mendel's first law. Mendel, by the way, didn't call it Mendel's first law. It's considered-- you don't write that in your own papers or something like that, right? So Mendel did actually observed some other things beyond this independent segregation of the alleles for a single trait. Mendel began to cross his peas together and try to make combinations. He had rounds and wrinkles. He had greens and yellows, talls and shorts. He started making combinations. How about a plant that was wrinkled and yellow? Green was the normal color. Round was the normal shape. But he had yellows. He had wrinkleds. How about making a combination? So he begin to make plants that bred true for different pairs of phenotypes. So for example, he had a pure breeding line here that was both round and green. That was the stuff he could pick up at the market. But he made a line here that was both wrinkled and yellow. What is the genotype of this round, green strain? At the round gene-- the gene for roundness- what is its genotype? Big R, big R. It's a pure breeding strain. It was homozygous-- we're just testing our words here-- homozygous for the big R allele. Green is controlled by a different gene. It has an allele big G. It's pure breeding for this. And we call it big G, big G. Convention, when we use capital letters, it tends to mean that the associated phenotype is dominant. OK? It'll make you think that the allele is dominant, but it's the associated phenotype that's dominant. Now, wrinkled-- what was wrinkled? Wrinkled was a homozygous wrinkled-- little r, little r. And what was the genotype at the yellow locus? Little g, a little g is how we'll denote it. Geneticists use four or five different kinds of notations. We're going to use this notation today of big R's and little r's and little g's. But you'll get used to other kinds of genetic notations. So when we cross these guys together, this F0 generation here of these two pure breeding parental strains, we get an F1 generation. The F1 generation-- what is it phenotypically? What did it look like? Round and green, yep-- round and green. What was in genotypically? Big R, little R, big G, little g. Now we could self these plants. And our head would hurt with the nine to three to three to one ratio. So instead, why don't we cross these plants back to the wrinkled, yellow strain to make our life easier? Little r, little r, little g, little g-- and when we cross that back, what is the segregation that's going to happen? Well, what are the possible gametes that could emerge from this parent? We could get a big R and a big G. We could get a big R and a little g. We could get a little r and a big G. We could get a little r and a little g. Those are the four possibilities that could be contributed by this parent. What could be contributed by this parent? Little r and little g-- that's it, right? No other options. So that's what our little Punnett square looks like here of our options-- parent number one, parent number two. What will this be? This will be big R, big G over little r, little g, big R, little g over little r little g, little r, big G over little r little g, little r little g over little r little g. In other words, this will be round and green. This'll be round and yellow. This'll be wrinkled and green. And this'll be wrinkled and yellow. And what will be the ratio of these? One to one to one to one, provided that those gametes were all equally frequent, provided that that parental plant made each of those four types in equal proportions. That one to one to one ratio in the gametes will necessarily translate into a one to one to one ratio in the phenotypes observed in the next generation. When we cross back-- not by selfing-- but when we cross back to the parent that has the recessive phenotypes, we'll often call this a backcross or a test cross. If I haven't done a backcross or a test cross where I crossed back to the wrinkled, yellow parent, I would instead have had to make a square here that had 16 boxes in it, and I would have had to add up to 16 boxes to figure out how many were round and green-- nine out of the 16. How many were round and yellow? Three out of the 16. How many were wrinkled and green? Three out of the 16. And how many were wrinkled and yellow? One out of the 16. But for the purposes of using the white board up here, I did the test cross or the backcross, because it's simpler. But you can also do the four by four matrix and figure out what it looks like. That was news. It didn't have to be that way, right? Maybe it was something else. This is pretty cool. What it tells you is not just is it the case that here the alleles segregate independently. It's a random coin flip which one you get. It tells you no correlation between the two traits. They're independent. This is called independent assortment. Mendel's second law is the law of independent assortment. All right. So Mendel publishes the paper. 1865, it comes out. Here's a copy of Mendel's paper. It's translated into English from the original German. So I got Mendel's paper here. There's no Punnett squares. It's kind of messy notation. Look at that. Look at all that-- big A, little b. He's got A's, B's, C's. He's got three factor crosses running around in here. Mendel really goes to town. It's a beautiful paper here. But it just goes on and on. It's incredibly hard to read. Look at this. It takes a lot to read this thing. But in a way, it's simple. It's nothing so sophisticated. Oh yeah, here. Look at this-- long, green, inflated, constricted-- all this kind of stuff. It's pretty cool. You should look it up. It's online. You can find Mendel's paper. What happens to Mendel's paper? Nobody reads it. It sinks like a stone. It's this cool paper and nobody reads it. Nobody reads it for a lot of reasons. Scientific communication wasn't so big in those days. Oh, well. Mendel also ends up getting promoted to become the abbot of the monastery, and that's pretty much the end of his scientific career, in my opinion. He got too many administrative duties, doesn't do more science there. Also, he has some poor choices. The next plant he works on is hawk weed. Hawk weed turns out to have really weird genetics that totally leads him astray. Basically, this is the one important paper Mendel ever publishes. It's an incredibly important paper. It's so important because it contains the clue to what Darwin, living at exactly the same time, wished he understood, which is what the basis of genetic variation is. Wouldn't it be great if Darwin had read Mendel's paper? Darwin actually owned a copy of Mendel's paper. He received a copy of Mendel's paper. In those days, the way they printed books, there were folded pages and you had to slit the page to read it. Darwin never slit the pages of the copy of Mendel's paper. So we know he's never read Mendel's paper, but he has one in his library. He had Mendel's paper. He had the answers sitting there on the shelf, but never read it. The stuff sinks like a stone. Nobody really pays much attention to it. And Mendel goes on, dies that's it. [LAUGHTER] PROFESSOR: Until the end of the 1800s, right at the beginning of the 20th century, along comes cytology-- looking at cells in the microscope. Microscopes began to get good in the late 1800s. And cytologists began to see in their microscope that when cells divided, these funny structures began to appear-- these long, thread-like things. And the German chemical industry being developed at that time had invented all sorts of dyes. And cytologists began experimenting putting dyes on these cells. And the dyes let them see really clearly these funny things that were condensing out when cells divided. And they had no clue what these funny things were other than that they took up dyes. And so in the absence of any clue what were, they called them chromosomes, meaning colored things. That is what chromosome means. They called them colored thing Chromos colored bodies, colored things. That's all they knew. And they observed that these chromosomes, these colored things, did really interesting choreography. When cells divided, when they underwent mitosis, what would happen is that the chromosomes would line up along the midline. And they would have, at that point, these funny X-like structures. And I'll draw four of these chromosomes lining up like this. And what would happen during mitosis? The cell would divide, and each of the two cells would get one half of the X. So if you started with four of these X's, you ended up with four like this. There you go. That was the chromosome being somehow tugged apart. Now, anything that gets tugged apart when a cell divides-- that's kind of interesting. Then those chromosomes would disappear. You couldn't see them again for a while until the cell was ready to divide again. And when the cell is ready to divide again, darned if those single lines hadn't turned into X's. Somehow the cell had turned the single lines into these two pieces, these X's, and they were ready to divide again. And that was mitosis-- the process of ordinary cellular division. But there was another process. Folks observed meiosis. That's what happens when you make gametes. So when gametes get made, the choreography was a bit different. Instead of all the chromosomes lining up on the midline as individuals, they lined up as pairs. They line up as pairs. When the cell divides, you end up with now only two X's, not four X's. Then what happens is those cells divide again. And you end up with those straight lines-- but not four of them, only two of them. The first step gets called meiosis number one-- meiosis I. The second is called meiosis II. The second step looks just like mitosis, doesn't it? Chromosomes are lined up along the midline. They separate it. It just looks like mitosis, ordinary cell division. But that first step is special. That first step says, somehow, the chromosomes come in pairs, and the cell picks one from each pair and gives to its gametes-- its sperm or its egg-- one from each pair. And what do you think happens on fertilization? Well, you had one from each pair, one from each pair, it comes together and it restores a pair now. And you know what folks said? They said this sounds just like what that dead monk was talking about. Pairs, particles of inheritance-- particles that come in pairs and you give to your gametes one of the two pairs. These colored things must be the basis of inheritance or genes or something. Wow, because it fits Mendel's model beautifully. It explains the first law. What about the second law? What about Mendel's second law? How could it explain the big R and the big G being inherited independently of each other with no correlation? What would that have to mean? They're on different chromosomes. The genes are on different chromosomes. Because if big R is on one chromosome and big G is on the other chromosome, then it's a coin flip whether or not the big R might be here and a little r might be there or the big G might be there or maybe it's over there. It's a random draw which way it's going to go. So it perfectly explains Mendel's second law. Unless-- what happens if big R and big G are on the same chromosome? Then they're going to go together. I'm not going to have independent assortment. I may have totally dependent correlated assortment. If big R and big G are on the same chromosome, they're going to be inherited together. Mendel's second law is going to be wrong. So why did Mendel find big R and big G going together? Maybe was lucky and he picked traits on different chromosomes. But what about his next trait? Lucky again? Lucky again? Mendel studied seven traits. How many chromosome pairs do peas have? Turns out, seven. But anyway, what happens? What's going on? Mendel's second law can't be right if the chromosome theory is right when the genes are on the same chromosome. They would be dependent. They won't be independent one to one to one to one. So which is it? Is Mendel right, my hero? Or is this chromosome theory right? Because they can't be both perfectly right. So which is it? Oops, we've run out of time. [LAUGHTER] PROFESSOR: Next time.
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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: All right. Then let's go back to our discussion of what happens to classical and quantum mechanical magnetic moments when they are exposed to magnetic fields. I just want to remind you of what we did last class and what we want to wrap up today. This is rapid adiabatic passage. I mentioned to you and explained it to you that rapid adiabatic passage is a powerful way to manipulate a classical and quantum system, and what we discussed is that when a spin points in the up direction and you sweep the resonance of an oscillating magnetic field-- the frequency of an oscillating magnetic field, through the resonance, you create an effective magnetic field in the moving frame which will rotate. And the atom, when the change is done adiabatically, will follow the rotation and therefore invert the spin. So it's a perfect, very robust method to invert population in spin systems. What I want to pick up today is the question, how slow is adiabatic. We have to fulfill an adiabatic condition and we have already an idea already of what the adiabatic condition is but now we want to derive it. That we had this picture of a spin which is rapidly precessing. It always precesses around the direction of the effective magnetic field, and the condition adiabadicity is that the rotation of the effective magnetic field has to be much slower than the precession. Let me just make it clear by a counter example. If the atom precesses around the magnetic field and the magnetic field would suddenly jump, then the atom would now start precessing about the new direction of the magnetic field and it would have completely changed its angle relative to the magnetic field. It would have lost its alignment with the magnetic field. So you clearly see that the condition is the direction of the magnetic field must not jump, and the only other time scheme is the frequency of the Larmor precession, so our condition for adiabadicity is the rotation of the effective magnetic field has to be slow compared to the precession frequency. And so we want to now derive from that condition the conditions for adiabadicity. And just as an outlook to make it interesting, what I will derive for you in the classical picture is now something which you will later encounter as the Landau-Zener parameter. But Landau-Zener sweeps, we talk about it later today is a quantum mechanical version of rapid adiabatic passage, but we now get classically a result, which will feature the Landau-Zener parameter. So with that, let us write down what we want to look at. It is adiabatic condition, and to write it down in words is that the Larmor frequency, omega L, which is given by the effective magnetic field, has to be much larger than theta dot. Now things in general are rather complicated. If we are far away from resonance, you change the frequency, but the effective field is not changing a lot. The critical moment is really when we have the real field, we add the fictitious field and that cause a rotation. The critical moment is when we are near resonance. So in other words, we have to fulfill an inequality. The left side has to be larger than the right side, but the left side is actually smallest in the vicinity of the resonance, and the angle theta dot is actually largest near resonance. That's when it sort of quickly goes to 90 degrees. So therefore, if we want to find the condition of adiabadicity, we can derive it by looking at the region around the resonance. So the effective magnetic field is the real field minus the fictitious field caused by the rotation by the transformation into the rotating frame. So we have the magnetic field at an angle theta with respect to the z-axis. I've just written down the z component for you, and the transverse component is-- so this is this component and the transverse component is the amplitude of our drive field B1. So we can just read it form the diagram. The resonance data is 90 degrees, and the correction angle is whatever we have of the effective z field over B1, and that means that the derivative, the angle theta dot, the angular velocity at which the magnetic field rotates, there's a time derivative because we sweep the frequency. So therefore, theta dot is nothing else than omega dot, the sweep rate of the frequency divided by gamma B1. But gamma B1 is nothing else than the Rabi frequency. And on resonance, the Larmor frequency is just a Rabi frequency, because on resonance-- sorry for repeating myself-- the fictitious field has canceled the bias field, and the only field left is the rotating field, but the rotating field is the Rabi frequency with a gamma factor. So therefore, we have the adiabatic condition that omega dot over omega Rabi has to be smaller than omega Rabi or to say it inverts the change delta omega of your drive frequency, the change delta omega in one Ravi period has to be smaller than the Rabi frequency. So omega has units of frequency. Omega dot is a derivative that's in units of frequency squared, and this is to be smaller than the Rabi frequency squared. You find that actually quite often if you do an adiabatic change of your trap frequency, things are adiabatic as long as the change of the trap frequency in one period of the trap frequency is smaller than the trap frequency, and you find something else that the derivative of your trap frequency has to be smaller than the trap frequency squared. So these are adiabatic conditions, when you tighten up magnetic or optical confinement for atoms. So this is very, very genetic. The small rate of the frequency you change has to be smaller than the relevant frequency squared. As I said, we come back to that when we do the quantized treatment of rapid adiabatic passage and we encounter that combination in the Landau-Zener parameter, which takes us to our next topic. We want to now talk about quantized spin in a magnetic field. And one of the first things I will be telling you is that everything we have learned about the classical magnetic moment you don't have to unlearn or re-learn, it exactly applies to the quantum spin. Before we look at the Hamiltonian and the standard Hamiltonian for two-level system, let us first look at something more general, which is Heisenberg equations of motion. So we want to write down the differential equation, the equation of motion for expectation values. So for an atom in a magnetic field, our Hamiltonian is simply the same Hamiltonian, which involves the gyromagnetic ratio, the angular momentum operator, and the magnetic field. And you know from quantum mechanics that Heisenberg's equation of motion for any operator are simply that the time derivative of the operator equals the commutator with the Hamiltonian. In some cases, if you have an explicit time derivative of the operator, you have to edit, but we are talking here about the angular momentum operator which has no explicit time dependents. So we are interested in the operator which describes the magnetic moment, but the magnetic moment is nothing else than the gyromagnetic ratio times the angular momentum. So therefore, the relevant commutator is the commutator of the Hamiltonian with the angular momentum operator. Just to remind you, the Hamiltonian was proportional to z. So what we are talking about is commutators, not surprisingly, between the angular momentum operators and those commutators are just this cyclic commutation involving the epsilon tensor. So if you just put that in component by component, you realize immediately that the operator equation for the time derivative of the magnetic moment is nothing else than the cross product, the vector product, of the operator of the magnetic moment times the magnetic field. To commence, this is exact, but also it looks exactly like the classical result. Well, you would say it's an operator equation. Usually operator equation some of them are pretty useless because you can't calculate the operators, but in this case, we can immediately take the expectation value, so we can get some immediately meaningfully equation namely that the expectation value follows the same equation, and this tells us that whenever we have a quantum system, it has a magnetic moment. In an applied external magnetic field, the result is simply rotation, precession, and this is exact. It's not a classical approximation. It's an exact result for quantum mechanics. So the way how we derived it makes it obvious that it's an exact result, which is valid not only for spin-1/2, but it is valid for any spin. If you have a magnetic moment corresponding to a spin of 10 H bar, this spin follows the same equation of motion as spin-1/2. Of course, a special case is valid for spin-1/2, but spin-1/2 is isomorphous to a two-level system. Any two-level system can be regarded as spin-1/2 system, therefore, this geometric interpretation that the dynamics of the quantum system is just a precession rigorously, exactly applies to any two-level system. It's also valid, and this will be relevant for atoms, if we have composite angular momentum. For instance, we will encounter the total angular momentum F of an atom which has components from the orbital motion of the electron, the spin of the electron, and the spin of the nucleus. But if we have such an angular momentum, F, the creation of motion is it will precess around a magnetic field. Well the small print here is, unless the B field is so strong that it de-couples the components or that it breaks up the coupling of the different parts of the angular momentum. In other words, we have simply assumed here that magnetic moment is gamma times angular momentum, and that requires that the angular momentum are coupled in a certain way. If we don't fully understand what coupling of angular momentum is, we really talk about that when we talk about atomic structure. So as long as the spin state coupled to one total spin, this total spin will just precess. This picture of precession will also be valid for a system of N two-level systems coupled to an external field, and this will be the example of Dicke superrradiance, which we will discussed towards the end of the course. So very simple result, but very powerful, and this is your permission whenever you encounter any of the systems to see a vector precessing in your head. This is exact. So we've talked about Heisenberg equation of motion for general spin, but-- you have a question, Nancy? AUDIENCE: What did you mean by N two-level systems here? Are we talking about coherent systems or non-coherent systems? PROFESSOR: We talk about in two-level systems and to be more specific the coupling comes because they all talk to the same magnetic field. So we have in two-level systems connected to the modes of the electromagnetic field. We start with the symmetric state. The coupling is symmetric and that preserves the symmetry of the atomic state. In other words, we will have a situation where the angular momentum is the maximum angular momentum we can get in two-level system, and the dynamics of this two-level system, the description of Dicke superradiance has the geometric visualization of this precessing motion. I know I'm not explaining it exactly. I want to sort of whet your appetite for what comes later and also sort of prep you that some of the simple pictures will really carry through the course. OK, so this is for very general spin. Let's now talk about features of, yes, the most important spin for us, namely the two-level system, which is spin-1/2. Well, the most generic system is an electron in a magnetic field B times ez. And well, why don't we start with a clicker question. So the question is, what is the level structure of the electron in the magnetic field? It's a two-level system, and you have two options A and B. One option is the upper state is spin up. The ground state is spin down or the opposite. So in other words, and tell me whether an electron is in the lower state when the speed is up or when the spin is down. You would say it's a stupid definition, but we talk all the time about an electron is in the spin down state or the spin up state, which is the lower energy state of the electron. So I think by exchanging the better recent number of responses has considerably gone up. OK. The answer is the electron is in the ground state. Spin down is the lowest state for the electron. To say it in words, of course, a compass needle wants to be aligned with the magnetic field. So you want the magnetic moment to be aligned with the magnetic field, and that means the magnetic moment has to point in the plus e direction, which up but the electron has negative charge, the gamma factor is negative, and that's why for the electron, the vector of the spin and the vector of the magnetic moment are opposite. So just try to find some main-mode technical thing. The electron lives in the basement. It wants to be down, spin down. This is the lowest state for the electron. However, if you have a system which has a positive gyromagnetic ratio, which would correspond to, well, positive charge, nucleus if it has somewhat normal magnetic moment, then in that case, the spin up state is less energetic than the spin down state. So let me just write that down. So the correct answer is this one, and it involves that gamma is negative. The gamma is a gyromagnetic ratio, the ratio between magnetic moment and angular momentum, and for negative charges, it's negative. For positive gamma, the situation is inverted. So let's just use for a moment the result we got from Heisenberg's equation of motion. We know the classical result. We have already derived for the classic spin, the classical result for the expectation value of the magnetic moment. But now I want to sort of relate it to something quantum mechanical because we know that the classical solution equals the quantum mechanical solution. So if you have a two-level system, the z component of the magnetic moment is the difference between spin down and spin up. And because of conservation probability, p up and p down is unity, we can also write that as 2 times-- let me now introduce e for excited state, just I know it's hard to keep track of spin up, spin down. I want to make sure that I mean now the excited state, so the excited state for the electron is spin up. So we have this condition, so therefore, the excited state fraction of a two-level system is related to the expectation value of the magnetic moment in that wave, and now we want to use the classical result we derived. We derived the classical result when for t equals 0 all the spins were in the ground state, and by using the result we derived previously, we have the 1/2 from the previous line and then the magnetic moment, mu z. We found an expression which involved the Rabi frequency, and the off resonant Rabi frequency are sine square generalized Rabi frequency times time over 2. So the two factors of 1/2 and 1/2 cancel, and what we find now for the quantum mechanical system using Heisenberg's equation of motion is that if you prepare a system initially in the ground state, the excited state probability, the fraction, the excited state oscillates with a Rabi frequency, and this is, I think, the second time in this course and not the last time that we see the Rabi, that we obtain, the Rabi transition probability. But let's go further. We have now discussed the classical spin. We have sort of done classical quantum correspondence with Heisenberg's equation of motion. We know that this in general, it implies Rabi oscillations, but now we want to go deeper into the quantum domain by talking about the spin-1/2 Hamiltonian. So in other words, we want to go beyond expectation values. We want to talk about the wave function itself. So the Hamiltonian, which we will use for major parts in this course, it's one of the fundamental Hamiltonians in physics. Of course, there's the harmonic oscillator. There is a hydrogen atom, but then there's this Hamiltonian, which is a two-level system with splitting omega naught. And then we make, which is often a simplification, where we use a pure exponential, so a single frequency in complex notation where the complex exponential e to the i omega t is the drive term. So this is one of the simplest Hamiltonian for this kind of system. It's a two-level system with a splitting, and now it is driven and the simplest drive term is not cosine omega t or sine omega t as we will see. The simplest drive term is e to the i omega t. So this is now our Hamiltonian. and since it is so important, let me ask you a clicker question whether this Hamiltonian can be exactly realized in nature or it is an approximation. For instance, that you always have cosine omega t as a drive, and cosine omega t is e to the plus i omega t and e to the minus i omega t. And then maybe with the rotating wave approximation, you throw away a term. So I sort of wanted to ask you, is this an idealization that we have a coupling which is a simple, complex exponential, this nature always more complicated or is there a simple way to realize this Hamiltonian in nature? So what do you think? Stop display. What's funny about that? AUDIENCE: Equal to. AUDIENCE: There are two [INAUDIBLE]. PROFESSOR: Oh, they couldn't make up their mind. The system should reject those votes. Anyway, it means most of you anticipate what I want to derive to you that we can actually exactly get this Hamiltonian and indeed, this is the Hamiltonian I will derive for you in the next few minutes, which is the Hamiltonian of spin-1/2 in a magnetic field coupled to a rotating magnetic field. So we start out with fields which are real, real field, no imaginary numbers, no complex numbers. These are real fields, but when we write down how the real fields couple to spin-1/2, we get e to the i omega t and e to the minus omega t without any approximation. So since at least 80% of you know the result, just regard it as an exercise to introduce how we spin matrices in a nice way, also this will help you how to do pre-set number 1. So the Hamiltonian is the spin coupled to a magnetic field, and if we express that by angular momentum operators, the gyromagnetic ratio it involves the operator for the spin in the z direction. So the two-level system has a splitting of H bar omega, so it's plus 1/2 minus 1/2 H bar omega, and that means the diagonal part, the non-driven part is simply given by the Pauli spin matrix sigma z and omega naught is the energy splitting, which is proportional to the applied magnetic field. And up-down, and excited and ground are the eigenstates of this Hamiltonian with energies plus minus H bar omega naught over 2. So this is the same Hamiltonian, but now add a real rotating B field B1. So the drive Hamiltonian H1 is the same magnetic moment but now coupled to a time-dependent rotating field. The amplitude of the rotating field is the Rabi frequency divided by gamma, and we assume that the field is rotating in the xy plane. So it's ex ey cosine omega t sine omega t. And now I put in two minus signs here for convenience. If you want, I've just shifted-- it's just a definition. I've changed the definition of the amplitude by a minus sign. So this is the Rabi frequency, and the magnetic moment divided by gamma is nothing than the spin. Magnetic moment is gamma times the spin, so therefore, we are back to the spin operators and the spin operators, if I factor out 1/2 H bar, are now the Pauli spin matrices sigma x and sigma y. And if you look at those spin matrices, then you'll realize that we go complex in our Hamiltonian, not because we have approximated a real field cosine omega t by some e to the i omega t, but because when we have a rotating field and we write down it in Pauli spin matrices, we get imaginary units form the sigma y spin matrix. So that means we have now for this system rewritten the coupling term took you to the rotating field H1 as 0 0 e to the plus i omega t e to the minus i omega t. And therefore, the Hamiltonian is the famous two-level Hamiltonian with omega naught and the Rabi frequency, which I wrote down at the beginning of this chapter. So we'll leave that here, but we will use it even more in 8.421. This is the famous dressed atom Hamiltonian. It is the starting point to calculate eigenstates and eigenvalues, not just in perturbation theory. You can go to oscillating fields at arbitrary strengths, so you can solve exactly in the dressed atom picture using this Hamiltonian, the problem of a two-level system plus one mode of the electromagnetic field no matter what the drive term, what the strengths of the electromagnetic field in the spin mode is. So this describes the two-level system plus one mode of the electromagnetic field with arbitrary strengths. And as I said, we talk about some things here but others I explored in 8.422. Questions? Yes, Will. AUDIENCE: We refer to the eigenstates and eigenenergies of this Hamiltonian as dressed in states in the same way as we refer to address states in a fully quantized fiction? Do we refer to both cases as dressed states? PROFESSOR: Yes. That's a good question. The question is now, what are the dressed states, and Will, I think you are referring that there are two ways to talk about the coupling of a two-level system to one mode of the electromagnetic field. It is this same classical picture where we introduce and let me say an analog amplitude of the electromagnetic field which drives it. And then there is a fully quantized picture where you first quantize electromagnetic field and you couple to photon number states. Actually the beauty of it that the two solutions are exactly the same. So in other words, how to say, if you couple an atom to one mode of the electromagnetic field. We have two ways how we can solve it. One is, we introduce a coherent electromagnetic field, and there is an exact unitary transformation which tells us if we have the quantized field in a coherent state, we can do unitary transformation, and what we get is exactly this Hamiltonian. So therefore, this is also-- you may not recognize it-- this is actually the quantum description of the electromagnetic field when it is in a coherent state. The other option is, we use the dressed atom picture maybe following some work of [INAUDIBLE] and others where we assume this single mode of electromagnetic field has in photons, and then we solve it for this photon number state. So in other words, these are the two ways how we can relatively, easily treat the problem. Either we assume the quantum field is in a coherence state or it's in a flux state. But since the dressed atom picture in the standard way assumes that the photon number, N, is large, there is a correspondence that in the limit of N of N being large, the flux state description and the coherent state description fully agree. And you pick what you want. If you introduce the electromagnetic field explicitly with it's quantum state, you get the dressed atom picture as a solution of a time independent problem, whereas here with a coherent state description, the coherent state oscillates, cosine omega t, with a time-dependent problem. And actually I should say whenever I get confused in one picture, I look in the other picture and it becomes clear. I generally prefer where we have N photons, it's because we can discuss everything in a time-independent way, but for certain intuitive aspects, this is also variable, so in the end, you have to learn both. And in your homework, you will actually write down the general solution for this Hamiltonian as an exercise. Nancy. AUDIENCE: I think I'm confused a little bit. So in the flux state picture, the dressed states can be exactly part of an independent matter of coupling between a lesser photon on any excited state. So like we can write eN as 1 and g or something like that. PROFESSOR: Yeah. You couple a photon field with N photons and energy in H bar omega to N minus 1 H bar omega. AUDIENCE: But in this one, is there like a direct photon number thing, because we haven't quantized the field yet? So what do the dressed states mean at this point? PROFESSOR: Well, the fact is that if you start out with a coherent state, your photon field is not in photons, it's a laser beam. The laser beam or the coherent state is in a quantized description, a superposition of many flux states. So therefore, the number of photons in a coherent state fluctuates or has a large Plutonian statistics, and if you take one photon out or not, it doesn't make a big difference. For instance, for those of you who know how the coherent state is state defined, the coherent state is defined as when you act on the coherent state with an annihilation operator, you get the eigenvalue times the coherent state. So that tells you you have a fully-quantized description of your laser in terms of a coherent state. You take one photon out and what you get? the same state back. And this may immediately justify that what we write down here is simply the coherent state with its amplitude and the amplitude of the coherent state would be B1, the amplitude of the drive field. And we don't really need other states because a coherent state has the property. You to take a photon out and you still have the same state. So therefore, we don't have to keep track of the coherent state. It's there all the time. But what I'm saying can be formulated more exactly when we use the appropriate formulas. But this is sort of the bridge. That's why we do not have to keep track of the photon state. It's because the coherent state has those wonderful properties. Other questions? OK. So this is the famous Hamiltonian. And of course, if it's the famous Hamiltonian, we want to solve it. As I said, the general solution is left to the homework, but I want to sort of show you parts of the solution to tell a story. And the question is, well, how do we solve this Hamiltonian? The answer is, we do exactly what we did in the classical problem. We transform to the rotating frame. In other words, this Hamiltonian is best solved by doing-- you can actually solve it directly. You can just put in a tri wave function and solve it. But I want to sort of bring out the big idea here which is analogous to what we have done in the last few classes, namely we have involved rotating frames. So what solves this Hamiltonian is a unitary transformation, and the unitary transformation is this one. And so this unitary transformation, let me first write it down, it transforms the Hamiltonian to the time independent one. We have now time independent of diagonal matrix elements. Our diagonal matrix element have changed. Delta is now the detuning of the tri frequency from the energy splitting of the two-level system. In particular, when we are on resonance, the diagonal matrix elements have disappeared. This is the result of the unitary transformation, and let me just show you this transformation over here. Can be actually written as an operator involving the z component of the magnetic field. And what I just wrote down for you is actually the quantum mechanical operator, the rotation operator for performing at rotation around the z-axis. So by selecting the rotation angle to be omega t, that's how I can generate the unitary transformation, and this unitary transformation makes the Hamiltonian time independent. So in other words, everything is in the classical system. We just go to a frame which rotates with a [INAUDIBLE], and we find the time-independent problem. So now this Hamiltonian can be easily solved. And you will find as a special case when you start with an amplitude, initially you start in the ground state, then the excited state amplitude square is the Rabi oscillation, something we discussed 40 minutes ago, but before, we got it from the classical quantum mechanical correspondence using the Heisenberg equation of motion and here it comes out by explicitly solving for the wave function for the dressed Hamiltonian. Questions? I want to say a few words now about rapid adiabatic passage, but this time by emphasizing the quantum mechanical aspects. In other words, we have a clear understanding what happens classically. We have a clear understanding what happens in the adiabatic limit, but I just want to sort of in the next 10 minutes use what we have already learned, combine it with the quantum mechanical Hamiltonian and tell you that, well, when you are not fully adiabatic, you actually have transition probabilities between the two states. So I want to sort of bring in the concept of transition probabilities to the case of-- what I want to say is rapid adiabatic passage when it's no longer adiabatic, but what this just means when we sweep the frequency and we're not in the adiabatic limit. So how do we describe it quantum mechanically? We start out with a Hamiltonian, which has-- we use our rotating framework for convenience that allows to write down exactly the same Hamiltonian in time-independent picture. So the Hamiltonian has two parts, a diagonal part and an off-diagonal part. So if I show the energy as a function of detuning delta-- well, maybe I should times 2 over H bar, just normalize it so then it becomes just the straight line at 45 degree y equals x. So the unperturbed-- the Hamiltonian without drive has a level crossing at detuning 0. Then we add to it the drive term. Well, let me just write down, not the drive term but the full Hamiltonian. So the full Hamiltonian with the addition of the drive term has delta minus delta and now it has the coupling with the Rabi frequency. That means that on resonance, the degeneracy between the two levels is split by the Rabi frequency, and if I now show you the energy eigen levels of this two-by-two Hamiltonian, it will asymptotically coincide with a dashed line through this and through that. So in other words, I'm just reminding you that a non-diagonal matrix element turns a crossing into an avoided crossing. So when we take the frequency omega and we sweep the detuning, so we change delta and do a sweep of the frequency omega at a rate omega dot, then we sweep through the resonance and in one limit, we have rapid adiabatic passage or in general, we realize the Landau-Zener problem of a sweep through an avoided crossing. So what I'm formulating here is it's the so-called Landau-Zener crossing or the Landau-Zener problem, which is the quantum mechanical description of you take a system by changing an external parameter. Here, we sweep the frequency of the rotating field, but by changing the external parameter, we sweep the system through the avoided crossing. And it has the two limiting cases that when we go through this crossing very, very slowly, the adiabatic field then tells us we stay on one of these adiabatic solid curves, and this is the case of rapid adiabatic passage, which we discuss in the classical limit. But it is also the other solution if you would sweep through it very, very fast, you're in the diabatic limit, you follow the dashed line and you start up here and you wind up there. The Landau-Zener problem is actually a problem which you find it in all it text books, but to the best of my knowledge, there is no simple, elementary derivation which I could give you in a few minutes. And if the mathematical problem is a nice, mathematical demonstration of an exact, solvable model, but to my knowledge explicitly deriving it is not providing additional insight. It's one of the cases where the result is more insightful and much simpler than the derivation. So what I want to give you is, therefore, simply the textbook result. So in the adiabatic limit, you stay on the solid line. If you do not cross the avoided crossing very, very slowly, you'll have a non-adiabatic probability to jump from one level to the other one. And this non-adiabatic probability is expressed as an exponential function which involves the Landau-Zener parameter. And the Landau-Zener parameter in this exact solution is omega Rabi squared times this new rate, d omega dt or d delta dt minus 1. This square should go outside the brackets, so therefore, what we find is from the exact solution that the Landau-Zener parameter is a quarter times-- and this should now look familiar, the omega Ravi frequencies squared over omega dot. And when we discussed the limit of adiabaticity classically, I hope you remember I gave you the argument by looking at the adiabatic condition that the adiabatic case requires omega dot to be much smaller than omega Rabi squared. So here very naturally what appears in the quantum mechanical problem is just the ratio of the two quantities we compared when we looked for the limit of adiabaticity. So therefore, the probability for a non-adiabatic transition is simply involving this ratio omega Rabi squared over omega dot. So in other words, we know already from the classical argument, but here we confirm it, adiabaticity require that this inequality is met. OK, I could stop here, but since we are using sort of diabatic sweeps in the laboratory, as long as I've been involved in doing quote "atom science," I want to sort of go one step further and teach you a little bit more about this formula and try to provide insight, and often insight is also provided when you apply perturbation theory. So I know the adiabatic case is very simple, but I want to look at the diabatic case and then look at transition probabilities in a perturbative way. This is actually the way how we often transfer population in the laboratory. So I want to understand better the way how we transfer population from one curve to the other one. So if we do a fast sweep, we call it diabatic. So in other words, if we have this crossing and we go really fast, well, what happens is this is the crossing between spin up and spin down. If you go much, much faster than the Rabi frequency, the spin has no opportunity to change its orientation. So therefore, the wave function, the spin has to stay up or down and that means the system just goes straight through the crossing. Because spin up has positive slopes, spin down has negative slope. Being adiabatic, staying on this lower adiabatic curve would actually require the system to go from spin up in this part of the adiabatic curve to spin down in the other part. And to flip a spin cannot be done faster than the Rabi frequency, so if you sweep fast, that's what's happening. So we have two trivial limits, one is the adiabatic limit or just the adiabatic curve and nothing happens. The other limit is the infinitely-fast limit and nothing happens again when we look at the diabatic basis, which is spin up and spin down. But now let's be almost adiabatic, and this is a problem which we really want to understand physically and intuitively because that means the system spins the small timelier resonance, and there is a small probability to make a transition. So if you had one of your hyperfine states, pick your favorite hyperfine state, you'll rapidly sweep the frequency. You will find, unless you sweep it infinitely fast that there's a small probability in the other hyperfine state. And that's what you want to calculate now, and I want you to understand how would you estimate and calculate the small probability. So let's now estimate the result, namely for the small probability in perturbation theory. And actually what I'm calculating for you here is, if you've used evaporation-- I know half of the class is doing that-- if you apply an i F field, you don't have to make it so strong that you and the adiabatic limit. You are exactly in this limit. The atom will slosh several times through the resonance in an almost diabatic way, but there is a finite spin flip probability and that's how you evaporate atoms. And I want you now to fully understand the derivation, what is the probability of ejecting atoms in the almost diabatic limit with i F spin flips. That's a limit where 90% of the BEC experiments operate. So I hope everyone realize it's an important question and also I hope everybody understands the question because now I have bigger questions for you. The first question which I will ask you, should we calculate that transition probability by using perturbation theory for an incoherent transition or for coherent transition? Let me just explain you what I mean and then I ask you for your opinion. Coherently, we simply say in perturbation theory, we start with our population in state 1. We have to do the coupling Hamiltonian time dependence of the population in state 2 and that means if we integrate this equation for a short time, we find an amplitude a2. And the probability to be in the state 2, which is the amplitude squared, is proportional to the Rabi frequency squared times the effective time squared, the effective time of tribing the system. Coherent processes are always quadratic in time. If we do it incoherently, well, the way how we describe incoherent processes are Fermi's golden rule, which we've all seen. And the probability in Fermi's golden rule is very different. Well, it is proportional to the Rabi frequency squared, to the matrix element squared, but Fermi's golden rule gives us a constant rate, and for constant rate, the probability is rate times time. So now it is linear in time and then because of the delta function in Fermi's golden rule-- I'm missing a symbol so I use gamma here. It has nothing to do with the Landau-Zener probability. This is just the density of states. So I hope you know now what is the difference between coherent or incoherent. The most important part is that things are linear in time for an incoherent pulses rate equation and at least for small times quadratic in time for coherent pulses. So now we come to this process where we take atoms from spin up to spin down. We evaporate with a weak, course, almost adiabatic with weaker f drive, so we are closer to the diabatic limit. And so if you think about this problem, I want you to tell me if this process, the perturbative transition close to the diabatic case is that should we use when we apply perturbation theory, the coherent picture or the incoherent picture. In other words, is the dynamics of the quantum system, when we go relatively quickly for the Landau-Zener crossing, is that a coherent or an incoherent process? I could see where it doesn't matter, but it does matter. So I think it's an open question. Let me give you the answer. It is coherent, and you can see it in the following way. What is the source of incoherent here? We have a Hamiltonian. The Hilbert space is by two-by-two. There is no coherence which can be lost. There is no spontaneous emission to other states. There is no reservoir. We don't have a small system which couples to a bigger system, and then the small system-- we do that on Wednesday-- has to-- tomorrow, Wednesday-- has to be described by density matrix. We have the none of the physics which would give arise to incoherent physics. It is coherent. But maybe I'm oversimplifying. Is somebody who said it's incoherent who wants to maybe press me harder and tell me why you think it is incoherent. Well, one possibility is-- and this is why we often use Landau-Zener sweeps in the lab. We have fluctuations of the resonance frequency. And when we go and sweep through it, we don't know exactly when we hit the resonance. And if you would take an ensemble of systems and you go through the resonance at different times, you will get an ensemble of wave functions, which has different phase factors in it, and in the end, you will actually need a density matrix to describe it. But this is now an experimental imperfection which I haven't assumed here. So in other words, what you should do is the following. Whenever you sweep through Landau-Zener crossing, you start with the ground state and what you get out is a superposition of ground and excited state. And the Hamiltonian determines absolutely every aspect of the amplitude in the ground and the excited state including all phase factors. In other words, if you do a Landau-Zener crossing in a way that you prepare 50-50% of the atoms, they are always face coherent and you can use this process as a beam splitter in an atomic thermometer. Fully coherent. We have five minutes, so if we assume now, I hope we all agree that the amplitude is obtained in this coherent wave, then, of course, the question is, but what is the effective time? When we sweep the resonance, we are far away. Nothing happens. Nothing happens. Nothing happens. We go through the resonance, everything happens. Nothing happens. Nothing happens. Nothing happens. So this is the effective time when the wave function really changes and we create the coherent and mixture in the second state. And it is this effective times squared which determines what happens, what our transmission amplitude is. So therefore, the question is, what is the effective time in the Landau-Zener crossing? I can give you three choices. One is, the effective time is, well, we change omega and the effective time is, how long, what is the time until we have a detuning, which is equal to the Rabi frequency. Another possibility is that the effective time where we coherently drive the system is 1 over the Rabi frequency, just the Rabi period. Or another choice, how I can construct time out of the two frequencies where it's omega Rabi and omega dot. These are our two elements. So another possibility how I can get time is, omega dot is frequency squared, and the square root of 1 over omega dot at least fulfills the dimensional criterion that this is time. So my question for you is, what do you think is the effective time during which we drive the system coherently? I have to tell you before I made up the problem, I do not know the answer. But I can also tell you that there is only one answer which is correct. All right. I have to say I expected A to be the correct answer, but I convinced myself it's only B. And the answer is the following, and I know I have to stop, but I only need three more lines. The effective time is not the time until you are detuned, the effective time is which I can call the dephasing time, the time during which everything is coherent. What happens is, we change the frequency delta omega, and the delta omega is, of course, omega dot times delta t. So if we change the frequency by sweeping it in such a way that we are detuning and now with that detuning, if we let the system evolve, we would get a phase shift of pi. That's sort of the maximum where everything adds up coherently. If we would go longer in time, the frequency has changed to the point that what we add to the amplitude of the other state is no longer in phase to what we have added before. You can sort of look at it like this. You have a little bit of phase a2, You build up. You build up. You build up by adding amplitude with the same phase. But now you are sweeping, and this is the criterion where you start to pile up things with the wrong phase and then the phase eventually becomes randomized and you're not effectively contributing. This equation defines the window delta t during which we effectively add substantial amplitude in the second state, and it involves delta t squared, so neglecting factors of unity, the result of this is that, indeed, delta t is 1 over omega dot plus. And indeed, if we say the probability for coherent drive is Rabi frequency squared times delta t squared, what we now obtain is the Rabi frequency squared over omega dot, and this is our Landau-Zener parameter. So in other words, if we check with the exact result, the Landau-Zener probability, 1 minus p non-adiabatic-- the Landau-Zener problem is e to the minus 2 pi over gamma or gamma pi over 2 over the Landau-Zener parameter, if I do an expansion of the Landau-Zener parameter for a small value of the exponent, the exact result is 2 pi times gamma, and this is indeed proportional to omega Rabi squared over omega dot. So in other words, what I've shown you is that coherent time evolution with this weird effective time, what I motivated physically, exactly that produce the limit of small gamma from the exact result for the Landau-Zener crossing. OK. I know time is over. Any questions? OK. So today was officially our Monday class, so we meet again tomorrow, the same place, the same 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 want to do is talk about something at a much higher scale than what we've thought about through most of this semester. And that's probably by design. Over the course of the semester, we started with kind of enzyme kinetics or molecular binding kind of events, and we slowly built our way up the larger and larger scales. Now there's always this question about whether we're claiming that we really understand how the higher levels of organization result from the lower level interactions. And I'd say, we definitely don't understand all of it. So you shouldn't come away with that as the notion. But at least one thing that I think is fascinating about this area of systems biology is that much of the framework that we use to understand, let's say, molecular scale interactions or stochastic gene expression, so these dynamics at the smaller scale, much of those ideas and such certainly transport up to these higher scales or translate up to the higher scales, where, in this case, we're using kind of master equation type formulas to try to understand relative species abundance. And so I think part of what I like about this topic of neutral theory versus niche theory and so forth in ecology is that you can just see how very, very similar ideas, that we applied for studying stochastic gene expression, can also be used to try to understand why it is that some species are more common than others when you go and you count them, in this case, on an island in Panama. Now, the subject is, by its nature, less experimentally focused than much of what we've done over the course the semester. And this is really a topic the tends to be a combination of mathematical theory with kind of careful counting of species in some different areas and trying to understand what that means. But it's an area that there have been a number of physicists involved in over the last 10 years. And I think that it's fascinating, because it does get to the heart of what we are looking for from a theory, what kind of evidence do we use to support a theory or to refute it. So I think there are a lot of very basic issues about science that come up when we start thinking about this question of neutral theory in ecology. And since it's, for many of us, a totally new area that we don't know very much about, you can come to it with maybe fresh eyes. And you don't have the same preconceptions that you would have for many other models that you might be more familiar with in the context of molecular cell biology. So the basic question that we're going to try to talk about today is just the question of why is it that, when you look out at the world, you see that there are some species that seem to be abundant and some that seem to be rare? Are there other patterns that are somehow universal? And what kind of sort of lower scale processes might lead to the patterns that we observe? And I think that this paper that we read is-- I mean, it's not that it's. Well, can somebody say what the actual scientific contribution of this paper was? Yes? AUDIENCE: They did a calculation. PROFESSOR: They did a calculation. But it's a little bit more specific than that. What is it? AUDIENCE: They came up with the closed form equation? PROFESSOR: That's right. Basically, there was a model of this neutral theory in ecology that we're going to explain or try to understand. You can simulate the model, but then there are possible issues associated with convergence or something of those. Although it's hard to believe that that's really such a concern. But you can simulate that model. What they did is they just showed that you could get an analytic-y kind of expression for it. It's not a super analytic expression, but, at least, it's not a straight up simulation. You kind of numerically do something, integrate something, as compared to doing the stochastic simulation. So it's not that that, in and of itself, is what you feel like-- it's not what we necessarily care so much about. But I think that it's still just a nice, short description of the model and the assumptions that go into it. And you get a little bit of a window into the debate that's going on between these two communities of kind of the neutral theory guys and the niche theory community. So there's only one figure in this paper. And it's an example of the kind of data that we want to try to understand. So there's a particular pattern in terms of the relative species abundance. And we want to understand what kind of models might lead to that observed pattern. But given that there's just one figure in the paper, we have to make sure that we understand exactly what is being plotted. And what I've found from experience-- and, actually, even the answer to the email question that was sent out, I think, was incorrect on one of these things. So we'll talk about that some more. So beware. We'll figure it out. But I think it's actually surprisingly tricky to understand what this figure is saying. But first of all, can somebody describe not what the figure is saying but just what the data is supposed to be? Where do they get the data? Anything that's useful? AUDIENCE: They were on an island ecosystem. PROFESSOR: There's an island. It's called BCI, Barro Colorado Island. AUDIENCE: [INAUDIBLE]. PROFESSOR: So it's a 50 hectare plot. Does anybody know what a hectare is? AUDIENCE: It's a lot more than a square meter. PROFESSOR: It's a lot more than a square meter, yes, indeed. Yeah. Is this an English unit of measure? This is the kind of thing that I have to Google. But it's one hectare is equal to 10 to the 4 meters squared. That's a good thing to memorize. I AUDIENCE: Exactly or approximate? PROFESSOR: I think it's exact. I think I think it's an exact. AUDIENCE: Then it's a metric unit. PROFESSOR: Yeah, so apparently it is a metric unit. So the idea is that if you take a 100 meters by 100 meters, this is a hectare. And there's 50 of them. It's about like a half a square kilometer to give you a sense of what we're talking about. And what do they do on this plot? AUDIENCE: They count a certain number as canopy trees. So the trees that are, like, really big. PROFESSOR: And how do they decide which trees to count? Did they count every tree? AUDIENCE: No, just the ones that like formed the top layer. PROFESSOR: I think that the way that they decide-- OK. Does anybody remember how many trees were counted? AUDIENCE: [INAUDIBLE]. PROFESSOR: So there are 21,457 trees in this 50 hectare plot. They identify the species for each one of these 21,000 trees. And they assign them. And they found that there were 225 distinct species. So this is really quite an amazing data set. Because I can tell you that I would not be able to do this. This was highly skilled biologists that can distinguish 225. If they can identify these 225, that means they have to be able to identify other ones as well. And they did it for 20,000 trees. And indeed, Barro Colorado Island is one of the major Smithsonian research institutes, where they've been tracking. They do this like every five years or so, where they do a census, where they count all of the trees. And they're also tracking many other-- it's not just trees. They're doing everything there. AUDIENCE: Is there only plants? PROFESSOR: What's that? AUDIENCE: Is it only plants? PROFESSOR: No. So actually, I visited BCI, and it seemed like they were studying all sorts of things. And there were nice looking birds there. AUDIENCE: No, I mean in this census. PROFESSOR: In this census, it's only trees. And the way that they decide which of the trees to do, it's the ones that are more than 10 centimeters DBH. Anybody can guess what DBH might mean? It's actually diameter at breast height. So what they do is they walk up to the tree with a ruler, and then, if it's larger than 10 centimeters, then they count it. You need to have some threshold at the lower end, otherwise you're in trouble, right? And there were plenty of trees that satisfied this requirement here. Then what they do, for all of these trees, it's assigned to some species. The basic goal of this branch of biology or ecology is to try to understand the pattern, from this sort of data, where it comes from. Or first describe it, and then once you have a description of it, then you can try to understand what microscale processes might lead to the pattern. And the pattern is what's plotted in figure 1. It's the only figure in the paper. I have reconstructed a rough version of it, here, for you on the board. But if you want a more accurate version, you can look at your paper. Now, we want to make sure that we understand what the figure is saying. So we will ask the following question. What is the most common number of individuals for a species in this data set? The most common/frequent number of individuals for a species to have in this data set. Now, it's maybe worth just saying something a little bit more. So you notice that they were not trying to count the total number of species, altogether. And in general, all of this field of relative species abundance, to try to understand them, what you do is typically take one trophic level. So some of the classic studies were of beetles in the Thames River. The idea is that it's some set of species that you think are going to be interacting, maybe competing, with each other, in some way, in the sense that they're maybe eating related things and being eaten by related things. And so in this case, these are the trees in Barro Colorado Island. And you can imagine that this is useful. The fact that it's trees instead of something else means that you can actually track the individuals over time. And when you go to the island what you see is that all the trees, they're wrapped by some tag. And presumably, they have some system to tell you which species that is so that they keep records of everything. But the question is, what's the most common number of individuals for species in the data set? Do you understand what I'm trying to ask? And we're going do approximate, so we'll say. Or this, can't determine. We want to know, what is the mode of this distribution of the number of individuals for each of these species? Do you understand the question? I'm going to give you 20 seconds to look at this. AUDIENCE: Should we just hold a blank piece of paper? PROFESSOR: Oh, we don't have our-- ah. AUDIENCE: [INAUDIBLE]? PROFESSOR: You know, the TA always lets me down. All right, yeah. So you can do A, B, C, D, E. Are we ready? AUDIENCE: [INAUDIBLE]? PROFESSOR: You can just do this if you're not. But given this was the only figure in the paper, and that this is a basic property of the distribution, I'm sure that you figured that out last night, anyways, right? Especially since it was one of the questions in the [INAUDIBLE]. So you presumably already thought about this question, right? OK. Yes? AUDIENCE: Yes. PROFESSOR: Ready, three, two, one. I'd say we got a lot of B's. So it seems like B is the most. So this, we'll put a question mark here. Can somebody verbally say why their neighbor said that the mode of the distribution is around 30? Yeah? AUDIENCE: The tallest bar. PROFESSOR: The tallest bar there is around 30. That's a very practical definition. So that's normally what we mean by the mode. There is a slight problem in all of this, which is that this thing is plotted in a very kind of funny way. So if you look at the figure, what you'll see is that it's number of individuals. And down here, it says, log2 scale. Now, when we say the mode, what we're wondering about is that, if you just take the most typical kind of species of tree that's there, how many individuals do we think there should be there? Of course, typical is hard to define. We can talk about mode, median, mean, et cetera. But the most common number of individuals for a species of the data set ends up not being 30. It ends up being 1. And we will try to reconstruct this right now. Because you have to do a little bit of digging to figure out what is being plotted here. But it's not the raw data. The problem here is that this is on this log scale, where the bins here are growing kind of geometrically or exponentially, whatever, as you move to the right. So over here, this thing only contains one real bin. And actually, we're about to find it's half a bin, which is even weirder. Whereas out here, this is maybe 30 bins. So the number of species that we're going to put in this bin is everything between around 20 something up to 50 or so. The number of kind of true bins that end up in each of these plotted bins is going to grow geometrically as we move to the right. So this is a very funny transform of the data. And indeed, I think it's always nice to just, in life, you always plot the raw data first. And then what you can do is then you can do funny. There's a reason to plot it this way. Because this is where they get this idea that this might described as described as a log normal. The idea is, if you take a log of the data, then you get something that looks like a normal. But you always plot the raw data first. So let's try to figure out what the raw data looked like. And now what we're going to do is we're going to have real scalings, honest to goodness numbers. Now the number of species you get still. So this is asking, how many different species do we see with one member or with two members or with three, four, et cetera? And I don't know how far we're actually going to be able to get. But in this one figure, in our paper, they tell us what the histogram means. So the first histogram bar represents what they call phi 1 divided by 2. Phi 1 was the number of species observed with one member, which means that even this first plot bar is not the number of species observed with a single individual. It's half of that. You can argue about the consistency of how these things should be, but that's what this thing's plotted. And it looks like it was nine, here, so this should be 18. So I'm going to put up here, here's a 20 and here's a 10. Right, so here is an 18. Now, what do they say? This bin represents phi 1 divided by 2 plus phi 2 divided by 2. So they took the number of species where they saw just a single individual plus the number of species where they saw two individuals, and they added those and they divided by 2. That's this number. We're not going to go through this whole process, because it's a little bit tiresome. But I've already done it for you. So I'm going to plot a few of things to get you there. And so I calculated it was 19, 13, 9, 6. It becomes ill-determined once you get out here, in the sense that we don't have enough. It's not uniquely specified going from that to that as it has to be. But I calculated it. It's around 5, in here, for a few. And somewhere in here, it's going to go into 4. And then this might go down to 3, and then deh, deh, deh. Now, if you look at this and the rapid rapid fall-off, do you think that you're going to find any species that have more than 20 individuals? We're going to vote. So you see this falling-off? So let's say that I've just showed you this, and I haven't yet calculated the rest, do we think that there's going to be any species with more than 20 individuals? Greater than 20 individuals, question mark? 1 is yes. 2 is no. It's going to be yes, no. Ready, three, two, one. So we got some 2s. So I'd say that most people are saying, no. Look at this fall-off. They're not going to be any species with more than 20 individuals. Although we already know that there are many species with more than 20 individuals. So this plot is useful for something. You can see that there are. And we know exactly the number of species that have more than 20 individuals, roughly. So those ones are all in these. So you can see that there are hundreds of species with more than 20 individuals. And indeed, it looks like there were two or three species that had more than 1,000 individuals or 1,500 or whatever the cutoff there was. So this distribution starts out rather high but then falls quickly. And out here, it's going to be very, very sparse. So there's going to be a bunch of numbers in here where there's not any species in the histogram. And then out there, there's going to be one, right? And indeed, you have to go really far out. Because there's one species out there that has a couple thousand. And indeed, the mean number of individuals per species has to be around 100. We know how to calculate a mean. This divided by this is just short of 100. So the mean number of individuals in a species is around 100. The mode is one. And the median? Well, ready? We decided this was the mode. Where is the median going to be? Is it going to be A, B, C, D? Ready, three, two, one. Indeed, this tells you pretty clear where the median is. This thing is indeed around the median. Because you can say, oh, it's about the same numbers to either side. So the median is around here. And I told you where the mean was, again. You guys remember? Ready, three, two, one. Mean, uno. Mean. So this is a very, very funny distribution. I guess I want to highlight that. And I think it's not at all what you would have expected somehow. At least, if you had described this measurement process to me, if you told me that you went to this island and you counted 20,000 trees, I don't know how many species I would have guessed. But OK, 220, it's reasonable. Well, I would have guessed it would have looked something like this on a linear scale, maybe, right? You know, that there would be a bunch of them around 50 to 100 and some would go couple hundred, some of them. So I guess I would have thought that the mean, mode, median would all be kind of a more similar thing. But this is just not the way the world is. It's not just on BCI. People, for hundreds of years, have been studying these distributions. And things that look like this, with extremely long tails, this is what people see. Now you can argue about exactly how fast it falls off and whether it's different on a mainland or an island. But this basic feature, that rare species are common, this seems to be just that's what you always see. This is the thing that you have to remember, rare species are common. And I think that this is the basic, surprising thing in this whole field. And the ironic thing is that even after spending all this time reading about theories to describe these distributions, it's still very possible-- and I would say, based on the statistics, this year and past years, it's not just possible, but it is the standard outcome-- is that after reading this paper, you do not realize that the distribution looks like this. You somehow still think that it looks-- you kind of still think it's like a linear scale, where the typical species has this, where the mean, median, mode are all about the same thing. So I guess always plot the raw data in an untransformed way. There are theoretical reasons why it might be nice to plot it like this. But be very careful about what you're doing. Because then you're left with a mental image of a histogram that looks like this. And that's very, very dangerous. Yeah? AUDIENCE: Why does it matter [INAUDIBLE]? [INAUDIBLE] the aggregate data in bins like that. And I mean, sure, exactly one species is the mode, but do you really want the--? PROFESSOR: I understand what you're saying. It's just that there's a qualitative aspect to the data, which is that most species are very rare. And this is something that I think is surprising. I think it's deep. And it's something that you do not get realized. AUDIENCE: Most species have more than 16. I mean, it depends what you mean by rare. PROFESSOR: Yeah. AUDIENCE: Look at the way that the distribution is away from trend. AUDIENCE: That's a good point. But the species density is clustered around the low numbers. PROFESSOR: Right. AUDIENCE: But actually most species have more than 30. PROFESSOR: Maybe the surprising thing is that just if you take-- the mean is 100. And so I would've thought that, if you plot number of species as a function of the number of individuals, given those numbers, I would have guessed, OK, here's 100. I would have guessed-- here's 50, so just to highlight that this is 150. So linear scale, I would have guessed it would look something like that, maybe larger than Rudin or something. AUDIENCE: What would that look like in a log2 scale? It would look like It's like the log of [INAUDIBLE]? So it goes up really fast and then-- PROFESSOR: So this thing would be kind of like shoom. I mean all the weight would be in. It would be like all here plus a little bit on each of these. AUDIENCE: But yeah. I don't think it's actually that different. The only thing that's different is the tail on the left. PROFESSOR: And the tail on the right. AUDIENCE: Yeah, it's a little bit longer. PROFESSOR: No, it's lot longer, right? Because this thing, all of the weight is between 50 and 150, which means that all of the counts are basically going to be these two, basically. Because this thing comes out either way. So in this case, if you take that histogram put it on this kind of scale, you end up with two bars up high, nothing outside. So it's a very different distribution. And it's not to say that this is a ridiculous thing to do. It's just that. But the problem is that your mental image of what the distribution looks like ends up being incorrect, in the sense that you have a qualitatively different sense of what's of what's going on. And if you go up to 10 species, here, and 10 is way down here. If this is what it looked like, there would be essentially no species with fewer than 10 individuals. But if you come over here and you add it up here. It's like a mean of 6 times 10 is 60 out of 200. A quarter or a third of the species on this plot of land have fewer than 10 individuals. And 10 is really a very small number. Well, rare species are common. I think it's a true description of the observed distribution here and elsewhere. And it's not something that you appreciate or realize when you plot it in that way. AUDIENCE: But you can get this information from that plot. PROFESSOR: No, I agree. You can get it. You can get it. But it was only 10% of the group got it. Right, the fact that you can get it-- right, it's possible. But you don't get it. That is a practical statement. Yeah, I'm not dead set against this distribution. It's just that it makes everybody think something that's not true. So if you think that that's OK, then I can't help you. It's OK, but it's just you have to be careful is my only statement. And I very much want you to take away. Because I this is an accurate description of the data. Rare species are common. And one of the readings-- I think it was in this paper, maybe it was a different one that I was reading. Even Darwin, when talking about this, commented on this fact that rarity of species is somehow a typical event. AUDIENCE: And common species are rare. PROFESSOR: And common species are rare, that's right. This distribution is hugely, hugely skewed. These are the measurements. It's good to look at them in both of these ways. Because you can't even plot the data on a linear scale. So that's a good reason for doing it. But I think it's good to have both of these pictures in mind. What we want to do is to talk about two classes of models that give something that's essentially this log normal distribution. So on a log scale it looks normally distributed, approximately. And those two models are going to be kind of a niche-based model and a neutral model. Can somebody, in words, explain what they maybe see as the difference between this niche and a neutral kind of approach? Yeah? AUDIENCE: [INAUDIBLE]. PROFESSOR: Every species is--? AUDIENCE: [INAUDIBLE]. PROFESSOR: In which one? AUDIENCE: In niche. PROFESSOR: In the niche theory, the species are different. So it seems like a ridiculous statement. Do you believe that species are different? We can vote, yes or no. Ready, three, two, one. Yeah. Well, somebody's been convinced by the neutral theory. It's clear that species are different. And the question is which patterns in the data do you need to invoke differences in order to explain? And I think that one, maybe, theme that's come out of this relative species abundance literature and the debates between the neutral and the niche guys is just that this distribution is less informative of the micro scale or individual kind of interactions then you might have thought. Because multiple models can adequately explain such a pattern. In all areas, we have to remember that you make an observation, and you write down a model that explains that observation. So what you do is you write down a model. And writing down a model, what that means is that you make some set of assumptions. And then you look to see what happens in that model. And if the model is consistent with the data, that's good. But it doesn't prove that the assumptions that went into the model are correct. And this is a trivial statement. And I've said it before. You have to tell yourself this or remind yourself of this kind of once a month. Because it's just such an easy thing to forget about. Now, the niche models indeed assume that the species are different. And that's reasonable. Because we think it's true. But then, of course, there are many different ways of capturing those differences. And then you have to decide whether the assumptions there are reasonable or whether they're necessary, essential. In the context of the niche models, we're going to think about the so-called broken stick models. So basically, you get log normal distributions when there's some sort of multiplicative-type random process that's being added together. You get normal distributions when you have sums of random things going together. This is the central limit theorem. But when you have multiplicative kind of errors or random processes coming together, you get log normal distributions. And I want to highlight that that does not necessarily have to tell you so much about the biology of it. Because a classic situation where you get log normal distributions is if you take a stone and you crush it. You can do this experiment at home. And then you measure the mass distribution of the resulting fragments. And the distribution of mass is log normal. Just take a stone, grind it under your boot or hammer it, just kind rub it right in. You'll get you'll get some distribution of fragments. For each of the fragments, measure the mass, and, indeed, you end up getting a log normal distribution. Because there's some sense that what's happening is that you take a larger mass, you break it up randomly, and then the resulting fragments, at some rate, each of them you break up randomly. and the small ones are maybe kind of less likely to get broken up as the big ones, so then the small ones can still get even smaller. But then there's going to be, at some rate, some very large ones. So such a process ends up-- I mean it's not biology. This is just something about the nature of the breaking up of this physical object. And indeed, the basic idea behind many of the niche models that give you a log normal distribution is equivalent to crushing a stone and measuring the resulting distribution. I'll describe what I mean by that. Typically, the broken stick models, they say there's some resource axis. This is a resource axis. And this could be, for example, where you're getting food from. Now, we're going to have to divide up this resource access among some number of different species. And what we're going to assume is that the number of individuals in the species is proportional to the length of the resource axis that it's able to capture. And I want to make sure I find my notes. I want to highlight this. This comes from MacArthur in the 1950s. MacArthur and it's 1957. So we imagine there's this homogeneous resource axis. We're going to break it up into N segments. And the abundances are proportional to the length. And the idea is that, if you just break this up randomly, so let's say you just draw N minus 1 lines randomly, or N minus 1 points randomly here. Now you have N species with N different abundances. The question is does that give a log normal? We'll say N minus 1 random points. Do you understand what I mean. You sample uniformly once, sample uniformly twice. You do that N minus 1 times, and now you have N and deh deh. And then we say, OK, the first species has this many individuals. The second has this one. The third is this one, et cetera. The question is does random points, does that lead to a log normal? Yes and no. Let's think about this for 10 seconds. N minus 1 random points, log normal distribution, ready, three, two, one. So I'd say that we have a majority are saying no. Can somebody say why that is? AUDIENCE: [INAUDIBLE]. PROFESSOR: Because it's something else. That's fair. But can you say qualitatively why it is that this is not going to work? AUDIENCE: You can't have very long gaps. PROFESSOR: Right. That is it's going to be very unusual that you get a very long gap. What about the other end? AUDIENCE: Also a very long tail. PROFESSOR: Now I'm a little bit worried. I think that that's true, right? Well, I'm going to say that you're not going to get this super long ones. I think that the distribution might still be peaked at short values. No? AUDIENCE: No. PROFESSOR: Random? If we were just traveling along this resource axis, at a rate that's kind of exponentially distributed, like Poisson rate, we just dropped points, that's something very similar to this random-- AUDIENCE: It said we're limited in the number-- PROFESSOR: No. Is that not true? AUDIENCE: Your sample [INAUDIBLE]. PROFESSOR: I'm a little bit worried that I might be-- now, I'm not 100% confident. Depending on how I look at this, I get different distributions. Yeah? AUDIENCE: But I think the first thing that he said, where you just say, I'm going to pick N minus 1 points-- PROFESSOR: Yes. AUDIENCE: --is a different thing than going along the axis and exponentially dropping ones along. PROFESSOR: I agree it's different. AUDIENCE: I don't think that would be the idea simulated, because you would be very likely to just get this giant thing at the end when you're finished. AUDIENCE: What you could do, you could go on to draft N plus 2 points. PROFESSOR: No, I think-- AUDIENCE: These scales that are your two end points [? are doubled. ?] PROFESSOR: Because I think that the probability distribution does grow. I think that I'm going to side with you. So we've decided that there are not going to be as many short sticks, and there's not going to be as long sticks as compared to a log normal. Do we agree with that? At least we agree that it's not going to be a log normal. So you're not going to get this huge variation of some very long sticks and some very short ones. Now, the question is how would you change this sort of model in order to generate a log normal? And the answer is that what you have to do is you have to what is called some niche hierarchy or so some hierarchical breaking. Just like what led to the stone giving you a log normal is that you have to have some successive process of breaking things. So this is what they call some hierarchy model. And then they key thing is that it's sequential. You have your resource axis. First, you have some rule for breaking it up. It could be that you just sample uniformly or some other probability distribution. And the way that you might think about this is via-- just everything up on the board is so nice and useful. I feel bad getting rid of it. This thing is not true, so I don't mind erasing it. So let's imagine some bird community in the forest. And we're going to think about where is it that the birds are getting their grub or their food to eat. First, well, now the axis is somehow vertical. You could divide them up into the ground foragers as compared to the tree foragers in terms of where they're getting their food. And you say, oh, well, how much of the food is on each side? Oh, well, we'll say 30% is on the ground, 70% is on the tree. This is along the stick. You cut the stick in some way, or you break the stick in some way. But then within the tree foragers, you'd say, well, the resources might be separated. And this is really like speciation, a species is in the niche, the species are focusing on different niches. So you'd say, oh, some are going to focus on the trunk, some will focus on branches. And again, this part of the stick is now broken or divided among different resource locations with some amount. But then also, you're going to get speciation in different directions here, because there's both the surface-- I don't know if you guys have ever eaten grubs-- but there's the surface grubs, and then there's also the sub-bark grubs. And so you kind of do this process multiple times, where you kind of pick different branches and break them to divide up the niche. And then you end up with a log normal type distribution. And this is a similar process to the crushing of the stone, because the idea is that there's sequential breaks of the stone. So the stone first breaks into maybe simply two or it could be three. First, there's one breaking. And then one of them is broken more. So given this process, you end up getting a log normal distribution. Yeah. AUDIENCE: But you also have a distribution of like how far. Because I guess there are two questions. Like when you break your stick, you assume, somehow, that you uniformly break it. PROFESSOR: Yeah. A lot of work has gone into the question of how it is you should break the stick. Given that you have this tree foraging stick. On a practical level, what they do is they ask, well, what probability distribution gives you the best agreement with the data? Is it uniform? Or is it, oh, it's broken like this? And in some cases people say, well, it's actually tilted on one side. Well, in the context of a succession and some other environments, there's an idea that, if a species first gets somewhere, they can kind of monopolize a larger fraction of the resources then if it's divided kind of an equally at the beginning. And that's going to effect where this probability distribution is going to break each one. But there's always this question about how constrained are the notions and so forth. And I'm agnostic on that point. AUDIENCE: But you also need distribution for how many times it breaks [INAUDIBLE]. PROFESSOR: Yes. It's just that, if you do this process, it's like a central limit theorem type result. So you have to do it enough times so that you get to some limiting distribution. And then you could keep on doing it. In the end, we always say that species abundance is proportional to the size. So we're going to scale, ultimately, to get the correct number of individuals. It's just that you have to do it some reasonable number of times so that the randomness kind of washes out, and you end up approaching that limiting behavior. Does that make sense? And indeed I just want to mention a major result in this field. These niche type models successfully explained or predicted another pattern that had been observed, which is the so-called species area relationships. So this is just saying that, here, we looked at 50 hectares, and we asked how many species where there. 225 species in 50 hectares. Now, the question is, if instead of looking at 50 hectares, we instead looked at 500, do you think of that the number of species we observed would have gone up, stayed the same, or gone down? Up, same, down, ready, three, two, one. Up. Up. If you look at a larger area, you expect to see more species in a larger area. And people really do this. They look in some area, going from, say, they take a meter, and they count all the species. And then they go and here is 100 meters, and they count all the species. And they ask, how many species do you see as a function of the area? And what people have found is that the number of species you observe it is proportional to the area to some power, where Z is around a 1/4. And of course, the area goes as some r squared. If you wanted to, you could say it goes as the square root of the radius, whatever. But the number species in some area, it grows, but it grows in a manner that is less than linear. Does that make sense? It definitely makes sense that's less the linear. Because linear would be that you sample a bunch of species here, and then you look at another identical plot, you get some other species. And they were saying that, oh, that you really don't expect any of those species overlap. That would be a weird world. So it very much make sense that this is less than 1. Of course, it didn't have to be this power law. But one thing that has been discovered, around the world, is that power laws are very interesting. But once again, many different microscopic processes can lead to power laws. The niche models have successfully predicted or explained why it might have this scaling. But it turns out that neutral models can also predict it. And may just be that lots of spatially explicit models will give you some power law type scaling that looks kind of like this. So once again, it's a question of how convinced you should be about microscopic processes based on being able to explain some data. And I think the best cure for this danger, of assuming that the microscopic assumptions are correct, because the model is able to explain something, is that, if you find some other very different set of microscopic assumptions that also explain the patterns, then it becomes clear that you have to take everything with a grain of salt. And that's I think part of what's been very valuable about the neutral theory contribution to this field. AUDIENCE: Does this just come from-- you assume that all the individuals are uniformly distributed and then [INAUDIBLE]? PROFESSOR: There are multiple derivations of this, so it's a little bit confusing. The neutral models, that I have seen, that lead to these patterns, they basically have the individuals randomly, either with sex or without sex, kind of diffusing around, and then they divide, deh-deh. And then you can explicitly just do the different spaces and see that you get a scaling. It seems to be a surprisingly emergent feature of many of these models. And once again, it may be something that tells us less about biology than it does about math or something. Any other questions about this, the base notion of this niche hierarchy type models? So I want to spend some time talking about this neutral theory in ecology. The math, in particular the derivation of this particular closed form solution, is not really so interesting or relevant. But I think it's very important to understand what the assumptions are in the model and maybe also something about the circumstances in which we think that it should apply. So the basic idea is that we have, what we hope, is some metacommunity that is large. And then we have an island. So this has to do with this theory of island biogeography. We have an island over here. And in the context of the nomenclature of this paper, they are some community size, size j here. This tells us about the number of individuals. And they're distributed across some number of species. Now, the neutral theory, the key thing is that we assume that all individuals are identical. And once again, it's not that the neutral theorists believe that this is true. It's that they think that it may be sufficient to explain the patterns that are observed. And when we say that all individuals are identical, what we mean is that the demographic parameters are the same, birth, death rates. And it's even a stronger assumption, in some ways, than that. It's assuming that the individuals are the same, the species are the same, and that there are no interactions within the species as well. So there's no Alley effect, or no specific competition. So the birth, death rates are going to be independent of everything, which is an amazingly parsimonious model. And it's kind of amazing you can get anything out of it. And then we have a migration rate m. It's either a rate or a probability, depending on how you think about it. Rate or probability m. And can somebody remind us how we handle that? AUDIENCE: Both just in a community? PROFESSOR: Yeah. AUDIENCE: At some probability that is proportional to the distribution of the species in the metacommunity? PROFESSOR: Yeah, that's right. AUDIENCE: --transfer an individual from the metacommunity to the island. PROFESSOR: Perfect. AUDIENCE: We do stick to the island to make sure that number of individuals. PROFESSOR: Right. So what we're going to do is we're basically going to pick a random individual, here, each cycle. This is kind of like a Moran process. We're going to pick an individual here. And we're going to kill him. And then what we're going to do is, with probability m, replace that individual with one member of the metacommunity at random. So the rate coming from here will be proportional to the species abundance in the metacommunity. And with a probability of 1 minus m, what we're going to do is we're going to replace that individual with another individual in the island. Now, the math kind of gets hairy and complicated. But the basic notion is really quite simple. You have a metacommunity distribution, which is going to end up being the so-called Fisher log series in this model. This describes the species abundance on the metacommunity. But then on the island, we're just going to assume that there's birth, death that occurs over here at some rate. But we don't even have to hardly think about that. From the standpoint of, say, a simulation or model, we just run multiple cycles of this, where we have j individuals. And we always have j individuals, because it's like the Moran process. At every time point, we kill one individual, and we replace it, with somebody either from the same community or from the island. And you can imagine that in the limit of m going to zero, what's going to happen on the island? Yeah, so you'll end up just one species, just because this is just random, like genetic drift. It's ecological drift where one species will take over. Whereas if m is large, then somehow it's more of a reflection of the metacommunity. Are there any questions about what this model is looking like for now? AUDIENCE: Could we talk about the Fisher log series? PROFESSOR: Yeah. AUDIENCE: So we would put it on the same axis as the [INAUDIBLE]? PROFESSOR: Yes, this is a very, very good question. So we'll do this in just a moment. Because this is very important. I want to say just a couple things about this model. So when I read this paper, what I imagined is that it really looked like this. This was Panama, and that, 30 kilometers off the coast, there was this island, BCI, Barro Colorado Island. But that's not maybe an accurate description of what the real system looks like. Does anybody know where BCI is? AUDIENCE: It's in Panama. PROFESSOR: Hm? AUDIENCE: Panama. PROFESSOR: So it is in Panama. But it's not off the coast of Panama. I guess that was my original. AUDIENCE: It's in the canal. PROFESSOR: Yeah, it's in the canal. So it's an island that was created when they made the Panama Canal. So this thing was not always an island. It's been an island for 100 years. And it's in the middle of a canal. And they actually have cougars that swim back and forth from the mainland. But it does make you wonder whether this is-- it's much more strongly coupled to the mainland then I imagined when I read this paper at first. I don't know what that means for all this. But certainly, you expect this to be a more or less appropriate model depending on this. Because, of course, if you went and you sampled 50 hectares here, you wouldn't believe that it should have the same distribution. You'd believe it should be more like the Fisher log series. And there's some evidence that things are tilted in a way that you would expect. And we'll talk about that. It's tricky. And of course, you have to decide in all this stuff, oh, what do you mean by free parameters? And actually, it seems like people can't count. And we'll talk about this in a moment, too. Because, of course, constructing the model, there's some sense of free parameters that you have there. Because we could have said, oh, it's just going to be the Fisher log series, or we could have said, oh, it's going to be island. Or we could have said, oh, there's another island out here. And then that would be another distribution. And not all of these things introduce more free parameters, necessarily, because you could say, oh, this is the same migration rate, or you could do something. But they are going to lead to different distributions, and you have that freedom when you're trying to explain the data. There are a lot of judgment calls in this business. But let's talk about Fisher log series, because this is relevant. So the model is very similar to what we did for the master equation in the context of gene expression and the number of mRNA. So was the equilibrium or steady state distribution of mRNA in a cell, was that a Fisher log series? Yes or no, five seconds? Was the mRNA steady state probability distribution a Fisher log series? Ready, three, two, one. No. No. What was it? It was a Poisson. And you guys should review what all these distributions are, when you get them, and so forth. So what was the Difference why is it that we have some probability, P0, P1, P2? This could be mRNA or it could be number of individuals in some species with some birth and death rates. What was the key difference between the mRNA model, which led to this distribution becoming Poisson, and the model that we just studied here, where it became a Fisher log series? And I should maybe write down what the Fisher log series is. So this is the expected number of species with n individuals on the metacommunity. Here is the Fisher log species. There was some theta X to the n divided by n. So what's the key difference? Yeah. AUDIENCE: I think that the birth and death rates are both proportional [INAUDIBLE]. PROFESSOR: Right, the birth and death rates are both proportional. AUDIENCE: In the Fisher log series. PROFESSOR: In the Fisher log series. So what we have is that b0-- and what should we call b0 in this model? AUDIENCE: [INAUDIBLE]. PROFESSOR: Well, right now, we're thinking about the metacommunity. AUDIENCE: Speciation. PROFESSOR: Speciation. b0 is speciation, which we're going to assume is going to be constant. In this model, do we have speciation on the island? No. The assumption is that the island is small enough that the rate of speciation is just negligible. So speciation plays a role in forming the metacommunity distribution, but it doesn't play a role in the model. So this is speciation. But then what we assume is that b1, here, is equal to some fundamental rate b times n, but it's b times, in this case, 1. So more broadly, bn is equal to some birth rate times n. This is saying that the individuals can give birth to other individuals. Now, we're not assuming anything about sexual reproduction necessarily or not. We're just saying that the kind of rates are proportional to the numbers. So if you have twice as many individuals, the birth rate will be twice as large. This is reasonable. This is Pn and this is Pn plus 1. So this is d of n plus 1 is equal to some death rate times n plus 1. So each individual just has some rate of dying. It's exponentially distributed. This again makes sense. What was the key difference between our mRNA model, from before that gave the Poisson, and this model that gives the Fisher log series? AUDIENCE: So with the mRNA, it's with a standard like a chemical equation where there's some fixed external input. But then the degradation is according to the amount that you have. So death is proportionate [INAUDIBLE]. PROFESSOR: Perfect. In both cases, the death rate is proportional to the number of either mRNA or individuals. However, in the mRNA model, what we assume is there some just constant rate of transcription, so a constant rate, per unit time, of making more mRNA. So just because there's more mRNA doesn't mean that you're going to get more mRNA. But here, we assume that the birth rate is proportional to the number. So that's what leads to the difference. And so this is one of the few other cases that you can simply solve the master equation and get an equilibrium distribution. And it's the same thing we do from just always, where we say, at steady state, the probability fluxes or whatever are equal. So you get that P1 should be equal to P0. and then we have a b0 divided by d1. And more broadly, we just cycle through. The probability of being in the nth state, it's going to be some P0. And then basically, it's going to b0 divided by d1, b1 divided by d2, b2, d3, dot, dot, dot, up to bn minus 1 dn. And indeed, if we just plug in what these things are equal to, we end up getting-- there's P0, the fundamental birth over death to the nth power. And then we just are left with a 1 over n. Because we're going to have a 2 here and a 2 here, and those cancel. A 2 here and 3 here, and those cancel. And we're just left with the n at the end, finally. So this x, over there, is then, in this model, the ratio of the birth and death rates. So which one is larger? Is it A slash 1? Is it b is greater than d? Or is it b slash 2, that b is less than d? Think about this for five seconds. Do you think that birth rates should be larger than death rates or death rates should be larger than birth rates or do they have to equal? Ready, three, two, one. So we got a number of-- it's kind of distributed, 1 and 2's. Well, it's maybe not that deep, not deep enough. Can somebody say why their neighbor thinks it's one or the other? People are actually turning to their neighbor. A justification for one or the other. AUDIENCE: So if this problem where b over d is greater than 1, then this distribution is not normalized. PROFESSOR: Right. So if b over d is greater than 1, so if x is greater than 1, then this distribution blows up. Then it gets more and more likely to have all these larger numbers. But then if b is less than d, shouldn't everybody be extinct? No. Can somebody else say why it is that it's OK for b to be less than d? If birth rates are less than death rates, shouldn't everyone be extinct? AUDIENCE: Because there's a rate b0. PROFESSOR: Because there's a rate b0, exactly. So there's a finite rate of speciation. So it's true that every species will go extinct. But because we have a constant influx of new species, we end up with this distribution that's this Fisher log series. Now, if you plot the Fisher log series, it looks a bit like this. But let's think about it a little bit. Does the Fisher log series, does it fall off, A, faster or slower than this? Fisher falls, A, faster-- this is in this direction-- or, B, slower? AUDIENCE: Faster or slower than what? PROFESSOR: Than the island distribution. Because you can see that this falls off pretty rapidly. Ready, maybe? Three, two, one. I saw a fair number of people that don't want to make a guess. Indeed, it's going to be faster. Can somebody say why? Yeah. AUDIENCE: [INAUDIBLE]. PROFESSOR: Is it going to be because of the 1 over n? I mean the 1 over n is certainly relevant. Without the one over n, then we just have sort of a geometric series. And the log normal is not just a geometric series either. AUDIENCE: [INAUDIBLE] Whereas this has a very long tail. PROFESSOR: That's right. So this falls off. This would be kind of exponentially, and this is faster than exponentially. And indeed, this make sense based on the model. Because this community, the reason that it has some very, very abundant species is partly because it gets migration from the abundant species here. This falls off pretty quickly. But those frequent species still can play a pretty important role in the island community, because the migration rate is influenced by large numbers. And the other thing is, of course, that the rare species are going to often go extinct. I mean the distribution on the island is some complicated process of the dynamics going here, plus sampling from here. But there's a sense that it's biased towards-- it's not just a reflection of the metacommunity, because the migration rate is sampled towards the abundant species. So the migration of these species ends up playing a major role in pushing the distribution to the right. So you have much more frequent, abundant species on the island as compared to the mainland. AUDIENCE: [INAUDIBLE]? PROFESSOR: Yeah. AUDIENCE: [INAUDIBLE] measurement of the distribution on the-- PROFESSOR: Well, I'm sure they have. I think the statement that there's a faster fall off on mainlands than on the islands I think is borne out by the data. But I don't know if trees on the Panama side of the canal are actually better described by a Fisher log series as compared to this, though. AUDIENCE: I guess my question was the abundant species that we see on the island, is it just the result of diffusive drift? PROFESSOR: Well, this also has the diffusive drift. AUDIENCE: But in the sense that what really pushes. PROFESSOR: Well, I mean I think you need both, the diffusive drift and the migration. But I think that the fact that the migration is from the mainland, and it's biased towards those abundant things, I think is necessary or important. AUDIENCE: I guess just in terms of distinguishing between the niche and the neutral models, as applied to the mainland, does the niche model predict also a log normal? Because it seemed like, in the discussion earlier, the neutral also predicted log normal [INAUDIBLE]. PROFESSOR: That's a good question. In this whole area, I mean it's a little bit empirical. The fact that the niche model kind of predicts this, or this broken stick thing predicts a log normal, they didn't say anything about islands there, right? I guess even Fisher's original log series, he used it to describe-- I think maybe that was the beetles on the Thames. But his original data set, where the Fisher log series was supposed to described it, as it was sampled better and better, it eventually started looking more and more like a log normal anyways. I mean it's easy to see the frequent species, because you see them. This tail can actually be very hard to see, because you have to find the individuals. It's a good question of to what degree each of the models really predicts one thing on one place and another. There's always tweaks of each model that adjust things. So I think it's a bit muddy. But the one thing that I want to highlight. So there's a lot of debates, then, between these different models. And each of the models have some fit. They have red and black. There's one that kind of goes like this. And another one that kind of goes like that. And they're not labeled, because they look the same. And you can argue about chi squareds and everything, but I think it's irrelevant. They both fit the data fine. And the other thing, just the sampling of kind root n sampling, if you expect to see 10 species, then if you go and you actually do sampling, you expect to have kind of a root n on each one. I mean the error bars, I think, around this are consistent with both models. So I'd say that the exercise of trying to distinguish those models based on fit to such a data set I think is hopeless from the beginning. And then you can talk about the number of parameters. And if you read these two papers, they both say that they have fewer number of free parameters. And it is hard to believe that there could be a disagreement about this. But then, you know, it's like, oh, well, what do you call a free parameter? And then what they say, any given RSA data set contains information about the local community size j. So they say, given that, it's not a free parameter, because you put that in. That's the number of individuals. And then outcome is your distribution, right? And you say, OK, well, all right, that's fine if you don't want to call that a free parameter. But then when you fit the log normal to this distribution, the overall amplitude is also to give you the number of individuals in the metacommunity. So if you don't call j a free parameter in this model, then you can't call the amplitude a free parameter when you fit the log normal, at least in my opinion. I think that they both have three. Because if you fit a log normal to this, you have the overall amplitude. That's the number of individuals. And then you have the mean and the standard deviation or whatever. From that standpoint, I think they're the same. Yeah. AUDIENCE: But I mean how do you fit the log normal when you don't impose? Do they impose the amplitude? I mean it's still a parameter. PROFESSOR: No, that's what I was saying. It's a parameter. I mean the normalized log normal, you integrate, and it goes to 1. But then you have some measured number of individuals in your sample, and then you have to multiply by that to give you. AUDIENCE: But is that what they do when they do their fit? PROFESSOR: Yeah. AUDIENCE: Or do they keep that amplitude as also a parameter? PROFESSOR: I think that you can argue whether this is a free parameter or not. But I think that you can just put it as the number of individuals, and it's not going to affect anything. You could actually have it be a free. But this gets into this question about what constitutes a free parameter or not. And actually, there is some subtlety to this. But I think, at the end of the day, the log normal is not going to look like this. You have to. You basically put in the number of individuals that you measured. AUDIENCE: So when you calculate [INAUDIBLE]? PROFESSOR: Huge numbers of pages of has been written about comparing these things. At some point, it comes down to this philosophical question about what you think constitutes a null model. And this gets to be much more subtle. And I think reasonable people can disagree about whether the null model that you need to reject should be this neutral model or if it should be a niche-based model. Or maybe it's just that there's some multiplicative type process that's going on and gives you distributions that look like this, and you need other kinds of information to try to distinguish those things. And in particular, I'd say that it's really the dynamic information in which these models have strikingly different predictions, and then you can reject neutral-type models. Because that neutral models predict that these species that are abundant are just transiently abundant, and they should go way. Whereas the niche-based models would say, oh, they're really fixed. And indeed, in many cases, the abundant species kind of stick around longer than you would expect from a neutral model. Of course, the neutral model is not true in the sense that different individuals are different. But it's important to highlight that even such a minimal model can give you striking patterns that are similar to what you observe in nature. And so I think we're out of time. So with that, I think we'll quit. But it's been a pleasure having you guys for this semester. And if you have any questions about any systems biology things in the future, please, email me. I'm happy to meet up. Good luck on the final.
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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: I think for the physics educators, I would suggest then to introduce the kind of big question we are trying to solve in the very beginning of the lecture. Then people will come into your lecture and then knowing that we are trying to solve this specific question. Then we work on the theoretical calculation using our physics intuition and physical laws we learned from other class to solve this question. And then we check if we have answered the big question which we actually introduced in the beginning. And finally, I would like to stress that comparison between, say, analytical calculation and the demo is actually very, very important. OK? Because that gives you a chance to verify your theoretical assumption to see if our calculations actually make any sense. And finally, I usually summarize the kind of take-home message, or the conclusion from this course so that people can walk out of the classroom and knowing that this is actually what they learned from this lecture. So actually, I think that is actually a applicable to all kinds of different physics course. If we actually try to do that way, then that would make the physics course much more enjoyable. In general, teaching a course, the lecturer has to show that he is so interested in this course. It is true when the lecturer actually knows about the course very well and knows the excitement about this course very well. So why don't you just show it? Right? Because that will excite a lot of interest from the students in the classroom. And also, that will make your lecture much more enjoyable, not only for the students, but also for yourself because you feel really happy when you're teaching in the classroom.
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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've emphasized in the first lecture, you know, that there's a lot of stuff that happens just in your ordinary life. I saw two examples of this. Yesterday's Boston Globe, just on the front page there was a discovery about ìHeart Cell Discovery Raises Treatment Hopesî. Scientists announced yesterday the discovery of cells in the heart that can create new muscle cells raising hopes that doctors may find dramatic new ways to treat heart disease. The team showed that the cells, which are similar to stem cells, can be expanded from just a few hundred in the laboratory dish up to more than a million. And these can be guiding into becoming the pulsing muscles that power the heart. So when we were talking about those yeast dividing and saying how one cell becomes two, this is a general principle throughout life that cells come from other cells and they divide. And we'll see the relationship to that with DNA replication as we go along. In the case of yeast, as I said, they're just always the same. Your progeny are always the same. But in something like our own cells we start out as a single fertilized cell but somewhere along the way the cells have to become specialized. So the very early ones are the embryonic stem cells. They have the potential to become any cell in the body. But at some point, at one of these cell divisions the cells are going to have to start to become more specialized. And, for example, this one might be a lineage that would lead to heart muscle or to becoming a nerve or something. And at that point it loses its ability to become any cell in the body. And in many cases by the time you get out ultimately to the final cell that's making up the muscle or the nerve or something it has no capacity to regenerate. So that's why, for example, spinal cord injuries are so damaging because nerves at this point cannot be regenerated. Or heart disease, you get a damaged heart we're stuck. This is why this result is exciting. Because there seem to be at least a few cells in the heart that have the capacity to regenerate more heart muscle. Now, this is early on. It hasn't been rigorously shown to be a stem cell. But there's an example from the front of yesterday's paper about something we were virtually alluding to in class. There was also an article about AIDS testing. Again, you know, we're talk more about the HIV-1 virus. And then today on the front page of the Boston Globe yet again is ìRomney Draws Fire on Stem Cellsî. And you can look at this. But, you know, he's sort of trying to straddle, I guess, between being supportive of research on the one hand and the concerns of the conservatives and the religious right on the other hand, and he's drawing fire from both sides. But it's an issue that is in our society today. You're going to be expected to make decisions on it, to know about it and understand. I'm just trying to drive home that what we're talking about isn't taking place in a vacuum. Nobody emailed me an idea as to what happened here. I showed you this little movie. This is water that is cooled below the freezing point but hasn't formed ice crystals, but if we put a little bit of this pseudomonas syringae in it then somehow that super-cooled water turned into ice. And I told you it was a protein on the surface. Nobody had any ideas. So why don't you turn to whoever is close to you and you can talk about it for 30 seconds and see if anybody can come up with an idea as to why. All right? I won't look. You know, just go ahead. Talk to somebody and come up with an idea. OK. Well, let's see. Did we manage to get any ideas? Anybody got the courage to try and guess what that protein might be doing? Pardon? It's a nonpolar molecule. It's not disturbing the bonds. It's an interesting idea. Do you have an idea then, are you able to extend that as to why then the ice would start to form? I mean it's certainly true that nonpolar bonds sort of interfere with the water. That's something we've talked about. Let's see. Any other ideas? Yeah? That's a version of the same idea, I think, hydrophobic because you think it wants to repel the water and push it together. That's interesting. You're sort of getting closer on these. Yeah? There it is. If you were to design a protein that basically could bind water molecules in a lattice that mimicked what you found in ice then the water molecules coming up and binding to these little pockets in the protein would present then a little field of stable water molecules that looked to the next water molecule like it was part of an ice crystal. And that's indeed how that bacterium does that trick. It's called the ice nucleation protein. And they do things like take this bacterium and they put it into things like when you're doing snowmaking, you put this in and then you spray the super-cooled water, and this makes it go into ice crystals and then it helps you get nice snow for ski resorts and things. That's at least one of the areas where it's used. OK. So I'm just going to show you this movie again. These are just baker's yeast, saccharomyces cerevisiae, a kind of single-celled yeast that's used in baking bread or making beer. And here we're seeing cells divide. And this particular kind of yeast has a way of doing, it kind of buds the daughter off from the side. Some double and then split down the middle. But you can see what's going on. There's a lot of cell growth going on. And the issue that we're going to address now is where does the energy come that's needed to do that? You know from your own experience that to build things, to make things takes energy. You cannot put up a bridge, you cannot put up a building, you cannot build a computer chip without somehow putting energy in. You're taking a bunch of matter in the universe and ordering it in a very specific way making new contacts that didn't be there. It's an energy-requiring process. And I'm going to talk today about where that energy comes from. And then I want to tell you a little bit, just a very brief historical thing along the way, because a point I've emphasized here is biology is an experimental science. And many of the greatest discoveries weren't because somebody had the idea and then went out to prove it. Very often we didn't even understand how it worked. And somebody was investigating a phenomenon, found some peculiar things, and then began to get insights. And the insights were what then led to a fundamental increase in our understanding. And this little bit of history involves some names that you see on the MIT buildings around here. One is Lavoisier who is a French scientist. And he was studying what happened when grapes were converted into wine, a good topic for a French scientist to be studying. So, in essence, what he was studying was glucose being converted to two molecules, excuse me, of -- -- ethanol and two molecules of carbon dioxide. This transformation, there's C6H12O6. Remember, carbohydrates have that composition. And so he was studying that. He managed to figure out that's what happened to the sugar when you were making the wine. And at that point he got beheaded. That terminated that part of his investigation. But this problem was then picked up by Lois Pasteur who, again, his name is on one of the MIT buildings. He worked in France as well. There's a Pasteur Institute in Paris. There's a nice museum in Lille in Northern France that has a lot of this. But he grew up in Arbois which is a town in sort of Eastern France that, as you can see from the little picture of the village, winemaking was a major industry. So he was interested in that probably from when he was a small, small kid, although probably not dressed like that. But anyway. So one of the issues that he took on, which was a real problem for the wine growers in his little town and in France in general was sometimes wines would go bad. They'd come out sour and couldn't be drunk and then you'd lose all the profit that would have come from that wine. So there was a lot of interest in trying to figure out how to prevent wines from going bad. And so Lois Pasteur started to study this. And he discovered that there was this conversion that had been figured out now of two ethanol and two carbon dioxide. So this was a conversion. And we now refer to it generally as ìa fermentationî. But what he discovered with this conversion occurred -- -- if yeast were present. That the rate of this conversion varied as the number of yeast, so it went faster if there were more yeast. And the yeast stopped growing -- -- when the sugar ran out. So what he discovered here was a correlation. He hadn't proven anything. He just saw that if you watch sugar go to ethanol there were yeast around, if you had more yeast it went faster, and when you ran out of sugar the yeast stopped growing. There was something connected here. So he came up with the idea that the yeast were responsible for this conversion that was happening when you made wine. And it was further helped out in this because he discovered an alternative -- -- conversion in which C6H12O6 went instead to give two molecules of CH3CHOH. This molecule which you know, galactic acid, it too has C6H12O6 on both sides of the equation but it's a different molecule. And what he found was that this is the lactic acid you know as what's in yogurt. It makes yogurt sour. Or if you exercise really hard and your muscles are sore that's because you accumulate lactic acid in your muscles, and I'll tell you why that is in the next lecture. But what the other thing that Pasteur realized was when you got this alternative conversion you didn't have yeast present, you had some other organism. And so that was a huge advance just of practical value to the winemakers because they knew they had to have yeast in there to get wine and there problems were coming when some other organism that wasn't yeast got in there and it did something different with the sugar and made it into lactic acid instead of making it into ethanol and carbon dioxide. So there was Pasteur working away on a practical problem and it was, you know, a really major advance to the winemaking industry for him to do this, but it also then sort of unexpectedly led to another issue. And that was why were the yeast doing this? Because one of the things that Lavoisier had noticed and Pasteur noticed was that you did this conversion. The two ethanol plus two carbon dioxide. But you could account for virtually all of the carbon and hydrogens and oxygens that started out as sugar and seemed like virtually of them showed up in the ethanol and the carbon dioxide. So why was the yeast doing this? And the idea began to develop out of that was that rather than being used to make biomass, in which case you would have expected to see a whole lot of mass in the yeast cells and no so much up here, that instead most of this sugar was being used to make energy and that somehow the cell was getting the energy necessary to all that synthetic work involved in cell division by carrying out this conversion. And there's a fundamental relationship then between chemical energy and whether a reaction can proceed. And I'll just take it through in sort of your typical introductory chemistry reaction, A plus B going to C plus D. You know, there are certain classes of reactions that will go almost to completion. Probably an overstatement to say it's to go to completion, but it's effectively over here. Those are termed irreversible reactions, and there are certainly some of them. If I have hydrogen and oxygen and I light a little match, you pretty much go all the way to making water with a great big boom and no hydrogen or not much hydrogen and oxygen left on the other side. However, most reactions that one finds in nature don't have that quality. Instead they are going forward at some rate and back at another. And they reach eventually an equilibrium that's characterized by what's known as an equilibrium constant which is the product of the concentrations of the products over the product of the concentration of the reactants. And that's a characteristic of every particular chemical reaction. And we really have to worry about this in biology because if everything was irreversible that would be fine, but in order to do all this synthetic work you have to deal with a lot of reactions that aren't going to go to completion. And nature has had to figure out a way of doing that, just the same way that bridges and buildings don't spontaneously assemble and engineers and others have had to work out ways of putting all of those things together. So at some level you see the same kind of problem. Now, there's a way of expressing this energy associated with a chemical reaction that can be used to directly calculate whether a reaction is going to go and how far it will go. And a person who did this work is another person who's on one of the MIT buildings. It was [Willard? Gibbs who was a faculty member chemist who worked at Yale in the 1980s, excuse me, 1800s, and he came up with an expression that's now known as ìGibbs free energyî. And what's important about this way of talking about the energy change associated with the chemical reaction is it considers not only the internal energy of the system but also the change in disorder. Or another way of saying that, for those of you who've run into the laws of thermodynamics, it combines the first and second laws of thermodynamics. And you have to consider both of those if you're going to consider whether a reaction will go. And you cannot measure an absolute free energy but you can measure a change. And this is the equation. It's the change associated with a chemical reaction is equal to the change associated with the chemical reaction under some set of standard conditions times RT times the log of the concentration of the products multiplied together over the concentration of the reactants. So if we could just go to the same example we were just thinking about, the energy change with that reaction that we were considering would have been this. So this is the energy change -- -- associated with the concentrations -- -- the reactants and products that we're considering. This is the energy change under standard, or the term standard conditions where everything, each reactant, each product is present under one molar concentrations. So not something you'd ever find in most cases, but it's a frame of reference. And then this is the universal gas constant -- -- which is two times ten to the minus third kilocalories per mole per degree Calvin, the temperature in absolute. This is the temperature in degrees Calvin. And the temperature for most biology, most life is around 25 degrees Centigrade, so that's equal to 298 degrees Calvin, which is about equal to 300 degrees Calvin. So for most -- And since the range in which life can occur on an absolute temperature scale is really pretty small, it sort of fluctuates in only very minor ways around 25 degrees Centigrade, then for most of the biological reactions we'll be thinking about this RT number is about 0.6 kilocalories per mole. Now, biochemists actually have a special form of free energy they use, which we put a delta G prime. And in this case the delta G prime is equal to delta G prime under a set of standard conditions plus RT natural log of C products over the reactants. But the assumption is made that the reaction is in water which, I mentioned the other day, is 55 molar. Yeah? This is the degree Celsius. I've just expressed it in degrees Calvin. Sorry. My mistake. Excuse me. Because I was wrong is why. OK. Thanks for catching that. All right. So water is very concentrated. And so under these conditions the other convention is then you can set the hydrogen ions and water molecules to one. And you don't have to think about them when we're doing this. This is a convention that biochemists do. Now, this free energy, the delta G that gives free energy is a thermodynamic -- -- property. And I'll just share with you the same visual image I've had since I was an undergrad, which I think is not a bad way of thinking about it trying to understand what happens, that if we have a plot of the free energy as a function of what happens as the reaction goes along so that we have A plus B here and C plus D down here. When you go from reaction to products, the way I've drawn it, some kind of energy is given off in this kind of reaction. And if you know that you will know then that the reaction will be able to go forward because it's able to give off energy just the same way hydrogen and oxygen give off a lot of heat and stuff, and you know that reaction really goes a long way to completion. So it's kind of as if you were out here on your spring break on your skis already to go down the black diamond hill, you know, you can sort of see what would happen. Now, because it's a thermodynamic property it doesn't matter what route you take to get from the reactions to the products. So if you go down the double diamond slope or you go down the bunny slope you still end up with the same amount of energy coming out of the reaction. And that's important because if that wasn't true you could make a perpetual motion machine and you'd be very rich. The second thing that's important is that the free energy will tell you what would happen if the reaction went but it will not tell you whether it can go. If I did a demo here and I brought some hydrogen and some oxygen and I mixed them together in a vessel in the front of the class we could all sit here waiting for it to explode. But the likelihood is we would sit here for a very, very long time and not see an explosion, right? And the reason is that in order to get that hydrogen and oxygen close enough together we had to give them some extra energy and push them so they overcome repulsion and stuff. So if you were out here on your skis again getting already to go, but in fact you got off at the wrong stop on the ski lift and you were there, even though there would be energy getting down from here it's not going to happen at any discernable rate given the sort of little bounce in energy you have in your normal lives. So what we're doing when we do hydrogen and oxygen is by putting a match into it or something we're giving it enough energy that actually a few of the molecules get up here, they drop down, then they give up so much energy and heat that all the rest of them get pushed up and the thing goes. But that's sort of not a bad way of thinking about it. And we're going to talk in a minute about what determines how fast reactions go, not whether they go or not. And then, of course, at that point we're going to have to worry about this issue. But before that what I want to show you is that there's a direct relationship between this Gibbs free energy and the equilibrium constant. So we have this, well, what we could do is you have the reaction over there. So let's consider that reaction has come to equilibrium. And that means there'll be no further energy change. So we'll just set the delta G to zero. And that would mean then that delta G prime zero is equal to minus RT concentration C over D over concentration of A over B. You'll recognize this. That's the equilibrium constant, right? I'm sorry. There's a natural log in here. I didn't get it in. OK? So which is equal to minus RT the natural log of the equilibrium constant or the natural log of the equilibrium constant is equal to minus delta G prime zero over RT. Or another way of saying that is the K equilibrium is equal E to the minus delta G prime zero over RT. So if you think back to consequences of an equilibrium constant, if the reaction is going to go almost all the way then there are going to be mostly products, very few reactions, so the K equilibrium will be large. So if a reaction is going to go a long way then the equilibrium constant will be large. And in order for an equilibrium constant to be large then this delta G is going to have to have a large negative sign. So if the reaction -- -- is favorable then K equilibrium will be large and the delta G prime zero will have, at least within the scale of an activation energy, a large negative value. And let me give you a couple of examples. When we talked about carbohydrates, I briefly told you sucrose was what we call a disaccharide, two sugars joined together. What do we do when we join two things together pretty much usually in nature? You split out a molecule of water. So we take a molecule of glucose, a molecule of fructose, both carbohydrates, stick them together and we get table sugar. If we want to reverse that reaction we have to put in a molecule of water and we can run it the other way. We get glucose plus fructose. The K equilibrium for that reaction is 140,000. The delta G prime zero is minus seven kilocalories per mole. So that's an example of what I was just telling you, a fairly large negative value. If we think about a reaction that's not favorable, here's acidic acid. That's what makes vinegar sour. And the hydrogen ion can come off here to give you a hydrogen ion and the negative ion of acidic acid or acetate ion. The equilibrium constant for that one is, what is it, I think two times ten to the minus five. So only a little tiny bit of the acidic acid actually ionizes. And the K equilibrium constant then, excuse me, the delta G prime zero is plus 6.3 kilocalories per mole. So buried in this example is not showing you that a reaction that's unfavorable will have a positive free energy associated with it, whereas one that's favorable will have a negative free energy. This is also sort of telling you why you don't die when you put salad dressing on your salad, because if acidic acid ionized as thoroughly as sulfuric acid and you put an equivalent amount of sulfuric acid on our salads none of us would be here. It's only a little tiny bit that's going, and so that's what's happening. So what this really sets us up for is this fundamental problem in biology, and that is that this reaction here, you can see what it would go, this one doesn't go, but most of the reactions that you have to carry out in biology demand an energy input because they just won't go. We could sort of force this a little bit. We could raise the concentration of the reactions and it would give us a little bit more product, but that's not a useful solution to all the things. So this was a really fundamental problem that had to be solved in evolution in order for life to ever exist. And I'll give you just an example. If we consider taking a couple of molecules of glutamate, which is one of the amino acids we talked about, a couple of molecules of amino and making it into a couple of molecules of glutamine. Now, this is an amino acid needed for making proteins. This is an amino acid needed for making proteins. The cell has to have both of them. Glutamate has two methylene groups and then are carboxyl group that's one of the acid amino acids. And glutamine the side chain -- -- is now amid. The delta G from zero associated with this reaction is plus seven kilocalories per mole, so it's as unfavorable almost as that one we're looking at. In fact, it's worse than the one we're looking at over there. The reason that this is sort of pushing the thing uphill energetically is that the electrons here actually distribute themselves back and forth. So you can kind of think of the molecule as going back and forth between these two forms. And that makes it more stable. And when you stick on the amine group to make the amid it cannot do that, and so you're actually pushing everything energetically uphill. So how does a cell accomplish this? There's energy available. If we consider what happens with C6H12O6 going to two lactate the delta G prime zero associated with that is minus 50 kilocalories per mole. So the cell has got a lot of energy out of making even that simple conversation of a sugar molecule into two lactate. But it somehow has to figure out how to use that energy in order to drive these unfavorable reactions. And the solution, which is really one of the secrets to life, is to use coupled reactions -- -- with a common intermediate. And if you look outside a cell, as Lavoisier did or Pasteur did, this is what you'd see. But if you could look inside the cell and see what's happening when that conversion is being made you'd discover that the full reaction looks like this. It's the sugar molecule plus two molecules of ADP plus two molecules of inorganic phosphate are going to give two molecules of lactate plus two molecules of ATP. What's ATP? It's a ribonucleotide. That's ADP. And what happens when you make ATP is an extra phosphate gets added onto that end of the molecule. So by having yet another phosphate on here you've got a whole role of negative charges. This is a molecule in which the various parts are not happy to be together because all these negative charges would like to push apart so when you break the bond of ATP then energy is released. So using ATP is a way of sort of storing chemical energy so you can use it in some other kind of context. And so by burning it, by carrying out the reaction in this way a cell is able to not only make a molecule of sugar, glucose into two lactate, it's able to generate ATP along the way. And the delta G prime zero for this reaction is minus 34 kilocalories per mole. So even though it's taking out some of that energy and putting it in ATP, this is a reaction that goes very, very efficiently. Then instead of trying to carry out just that reaction, what the cell is actually doing is taking the two glutamate plus the two molecules of ammonia plus two ATP. And then this is converting it to two glutamine plus two water. I think I failed to put that in here so you can correct it back there. Plus two ADP plus two molecules of inorganic phosphate. And so the Pi very commonly used in biochemistry to denote just inorganic phosphate ion. So what's happen here then are these two reactions going on. This reaction now, because ATP is involved, is now favorable, and the delta G for this reaction is minus nine kilocalories per mole. So by having an ATP hydrolyzed as part of the reaction mechanism, this reaction that used to be unfavorable is now favorable. And then the kind of cute thing then is if you sum this all up, the ATPs and the ADPs are on both sides of the equation so they just drop out. And what you're left with is C6H12O6 plus the two glutamines plus two ammonias going to give two glutamines, excuse me, two lactate plus two glutamines plus the two waters. And the delta G prime zero for this is minus 43 kilocalories per mole. So this is not, you can think of it as using energy in the form of ATP like this a little the way we use money in our society. I do some work at MIT. I don't get given food to eat or TV to watch the Super Bowl. Instead I get given money, then I go to the store, I give them the money, I end up with the food or the stuff. And if you're watching it from the outside you see me do work at school and then food, TV or whatever shows up at home. But what's happening is the money is serving as a common intermediate in those transactions. And that's what basically ATP is in the cell. It's energy money. And in making ATP the cell has to take this ribose with an adenine on it, I think I didn't put the adenine on here I realize. The adenine is sitting on the ribose now. There are two phosphates, both of which have a negative charge on them. And to create that third bond it has to push it together. It's a very sort of an intrinsically unstable molecule. When you break the bond it will give you energy back. And that's one of the really amazing secretes to life, and that's the underlying principal of why it is that life can go forward. Now, the second issue that we need to quickly address here is -- -- not only can a reaction go, which is what thermodynamics tells us, but how can fast can it go. And this epitomizes the problem that all chemical reactions face because literally every chemical reaction that you carry out involves bringing a couple of entities together. And as they get closer and closer and closer they don't want to be there so you have to sort of push them together in some kind of way or make sure they have enough energy to get together. And that's what we see represented here. And that's a special term called the activation energy. It's given the term delta G with a double-dagger. And that is what -- It's the size of that activation energy that limits how fast chemical reactions can go. So the solution you use in chemistry, most of you, is you use a catalyst. And the catalyst doesn't change the outcome of the reaction. It just changes how fast you get there. So there are many reactions you've heard about in chemistry. Just stick the thing at 500 degrees centigrade, put in a piece of platinum, and now the reaction will go a whole lot faster. By heating it up molecules have more energy. So if they have more energy they can get closer together just from that. And then what the platinum surface would do is allow the molecules to both stick and that would bring them in proximately and also help them come together. Well, you cannot raise the temperature in a biological system, but still you have to overcome this. But the principal then, what you have to do when you carry out a catalyst, what any catalyst would do is that it lowers this activation energy. And if you lower the activation energy then enough of the molecules, just at whatever condition they're in will have enough energy to be able to go. It won't change the size of the drop. It just changes how fast you reach that final equilibrium. And there are two forms of biological -- Two molecules that are biological catalysts. One of the molecules you know is enzymes. Enzymes are made of a protein. We spent a bunch of time working at that. One of the things I showed you the very first day, this is a thing made by the anthrax bacterium, anthrax lethal factor. What it actually is, it's a protein and it's an enzyme that's able to catalyze the cleavage of certain peptide bonds in proteins in our body. And in particular it goes after molecules that are involved in signaling processes inside of cells. And if we don't have those then we die. More recently it was discovered that RNA can be a catalyst. And these are called, if you have an RNA that's a catalyst it's called a ribozyme. And these seemed pretty exotic for a little while they first discovered the idea that a piece of RNA could serve as a catalyst in a biological system, but it eventually turned out that the ribosome, which we'll talk about in some detail which is the protein synthesizing machinery that creates those peptide bonds between each of the amino acids to make the proteins. It's a big conglomeration of RNA shown in gray and a bunch of different proteins that are shown in yellow, but the actual formation of the peptide bond, the thing that makes all proteins is actually catalyzed by a piece of RNA. And so the ribosome is actually a ribozyme. And it's ironic that that sense that a piece of RNA is catalyzing the bond that makes proteins possible. So we'll finish this up and get in then to glycolysis which is the most evolutionary ancient of these energy-producing systems on Monday. OK?
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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 we've now drawn pictures of the interaction of a person jumping off a cart in both the ground frame and a reference frame moving with velocity, vc, which is the final speed of the cart. These were pictures in the reference frame moving with vc, and these are the pictures-- momentum diagrams-- of the person and the cart in the ground frame. Now, we would like to apply the momentum principle to solve for vp and vc. We are given that the velocity of the person in the moving frame, u. We'll treat this as a given quantity. And we'll express the velocity, vc, and so we want to find vc and vp in terms of u and p and mc. Now, we'll use the ground frame first, and our assumption here is that the person jumped horizontally and that there are no external forces in the horizontal direction. So all of the vectors we've drawn are horizontal, and there's no external forces in the horizontal direction. Then we know that the initial momentum of the system is equal to the final momentum of the system. In our initial picture, nothing is moving. And in our final picture, we have m cart v cart plus m person v person is 0. Now, recall that we also showed that the velocity of the person in the ground frame is related to the velocity of the person in the moving frame by adding the relative velocity of the two frames. The relative velocity of the two frames is the velocity of the cart, vc, so what we have is vc plus u. So this is how vp is related to u. And now, I can use this start equation to write it as 0 equals mc vc plus mp times vc plus u. And by adding terms, I get that mc plus mp vc equals minus mpu. Or the velocity of the cart is equal to minus mpu divided by mc plus mp. Now that I have the velocity of the cart, I can just substitute that in here to find the velocity of the person, which, remember, was the velocity of the cart plus u. So now, I've solved this problem in the ground frame. Next, I'll do the same analysis in the moving frame.
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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: My watch says five after 2:00, so why don't we get started? This is going to be a strange lecture. I'm expecting some wise guy in the back to say, all your lectures are rather peculiar! But I have a phone call from Europe coming in sometime around 2:15, 2:30. Somebody's coming to visit us next week, and we're going to set up a seminar. So I don't know exactly when it's going to come in, but I have to be in my office. So what I will do is adjourn at quarter after, 20 past the hour. And then so I don't keep you here sitting restively wondering when I'm going to come back, we'll adjourn until 3 o'clock. So you seem to have been taken longer and longer breaks during intermission. So I'll give you a full 3/4 of an hour so you can get it out of your systems. And then, we'll take shorter breaks from here on in. OK this is going to be pretty much our last lecture on symmetry. And we will begin, at least in half of the next class on Tuesday, to begin talking about physical properties, and in particular tensor properties. But I'd like to say a little bit about the derivation of space groups and take a look at how this information is tabulated for you by the kind folks who prepare the international tables. And there are really very strict parallels between what we did in two dimensions. And that's the reason why we did it so thoroughly and systematically, except that there are three dimensions. And because we are taking 32 point groups and putting them into 14 lattices, there are a lot more combinations to be considered. And there are a lot more of them that are unique. Also, they're a little bit more difficult to visualize and depict. So there are some new conventions in preparing a representation of three dimensional symmetries on a sheet of paper or page of a book which, of necessity, has to be two dimensional. So we'll go over some of those conventions. But the main reason for going a little bit further is that there is another surprise lurking in there for us as soon as we begin to combine translation in three dimensions with a symmetry element. But let me go through in some detail the monoclinic space groups. So we have at our disposal to decorate with symmetry elements two lattices. And they are a primitive monoclinic lattice with two translations, a and b, that define an oblique net, and a third translation, c, at right angles to that net. And then, either two choices for a double cell-- and I'll draw these in projection because I can do both of them with one diagram. Either this is a and b and the extra lattice point is in the center of the cell halfway up from the base. This would be body centered. And that's represented by I for innenzentriert. In German, that's German for body centered. Or the alternative would be to pick a cell like this, in which case the extra lattice point gets caught in the center of one of the faces. So this depiction here would be a side centered lattice. And it could have the extra lattice point in the middle of the face out of which b comes. And in that case, the symbol for the lattice is B. Or alternatively, without changing any of the nature of the specialness of the lattice, the extra lattice point could be in the middle of the face out of which a comes. And that would be an A lattice. With the first, setting with the c axis unique, there is no c lattice because the oblique base of the cell would not give us anything new if it were centered. All right, so there are those two lattices. And then, there are three point groups. And they were 2, m, and then a combination of a twofold axis perpendicular to a mirror plane. So it looks as though in principle there are going to be 6 combinations. In point of fact, there are a lot more because of the surprise that we have in store for us. Before proceeding further, though, let me ask rhetorically the question. Which lattice type would you pick as the standard one? And here, there's a major decision that has to be made. Should the labels on an axes give a unique symbol for the lattice? In other words, if you have the double cell for a monoclinic crystal, should you define the cell to make it body centered always, or to make it A centered or B centered always? Or should the labels be defined by the relative lengths of the axes? So those are the two possibilities. And there's no right or wrong way. You pays your money and you makes your choice. And on pragmatic grounds, this is the convention that's followed. And it's purely a pragmatic decision because lattice constants have been determined for literally tens of thousands of materials. Sometimes, the lattice constants are known and the space group is not. But it would be terrible if the label that you applied to the axis could be C attached to the longest one, the shortest one, or the intermediate one. It would be impossible to calculate any database for lattice constants displayed by particular materials. So there's all this-- this is what the decision was made. This is the decision that was made many years ago. And for monoclinic crystals, the labeling of the axes is determined by the c axis being the unique axis. That is, the axis that is along the twofold axis or perpendicular to the mirror plane. And then, the labels b and c are defined by magnitude of b greater than the magnitude of a. So I inherently drew this is in the proper fashion. So the labels are applied to monoclinic crystals in that fashion. The price you pay for that is that the symbol for the lattice for a monoclinic crystal that has the double cell could either be A, B, or I depending on the relative lengths of the axes. If these were the two shortest, it would be a body centered lattice. On the other hand, if this and this were the shortest pair, then it would be a side centered lattice. So the symbol for the lattice type will bounce around depending on the dimensions of the translations. So that's a complication in the space groups. In the two dimensional plane group, that issue never came up. We just took B greater than A for the rectangular and the oblique nets. OK, so let me do quickly a couple of the monoclinic space groups. Let's take a twofold axis and put it into a primitive monoclinic net. And this is exactly the same as our plane group, P2, with the twofold axes extended parallel to the c translation. So we would have twofold axes in all of these orientations. And that's simply P2, little p in two dimensions, with the twofold axes extended normal to the plane of the plane group. So this is p2. This is capital P2. And that is way we distinguish plane groups from space groups. So the symbol for the lattice is a capital symbol for the space groups. So I didn't have to use any new combinations theorems. I simply say there's a third translation perpendicular to the net. So everything protrudes out of the plane group into a third dimension. Let me now combine a twofold axis this a and this b and this c with a lattice that is not the primitive lattice but one of the double cells. And I'll take it, for convenience, to be the body centered flavor because it's easier to draw. And the operation that we've added to the lattice, if this is a combination of a twofold axis, is the symbol A pi. And I know what happens if I combine A pi with a translation that is perpendicular to the rotation axis. I get a new rotation axis, B pi, that's halfway along the perpendicular part of the translation. And that's where all of these additional twofold axes came in. But if a lattice is a body centered lattice, there is now a translation that goes up to the centered lattice point. So what's that going to be? What is T followed by A pi followed by T, where T is 1/2 A plus 1/2 of B? And that is a part that is perpendicular to the axis, and then a third component, c, which is parallel to the axis. As with all these theorems, what you do is you draw it out once and for all and see what it turns out to be. So let's do that. Let's say that this is the translation that goes up to the point 1/2, 1/2, 1/2. Here is the parallel part-- the perpendicular part, rather. This is 1/2 of A plus 1/2 of B. And then, we have a part that's parallel to the rotation operation, A pi. And that is 1/2 of c. So let's just do it. Here's my first object number one. It's right handed. I'll rotate 180 degrees to get a second one that's right handed. And then, I will translate it up to here. And here sits the third one. It's also right handed. How do I get from one to three? It's not really clear how I do that. And again, I'll draw it in projection because this is looking pretty messy in three dimensions. So here's my first one. I rotate 180 degrees to get a second one. The chirality is all the same. And then, I translate up to this point. So if this one is at z and this one is at z, this one here will sit at z plus 1/2. And this is the third one. Anybody got any idea? Can't be reflection. They're of the same chirality. It can't be translation because the orientation is different. Yes, sir? AUDIENCE: Then it has to be some form of rotation. PROFESSOR: Right, has to be. But how do we get the object up to a different elevation? Well, we've stumbled headlong over a new type of operation, just as we discovered the glide plane when we started putting mirror planes into a lattice in combination with translation. The only way I can get from the first to the third is to rotate 180 degrees about exactly the same point that I would have before, namely at 1/2 of the part of the translation that is perpendicular to the operation. But then before putting it down, I've got to take a second step. I've got to slide it up by a translation component that is parallel to the axis about which I've rotated. So to complete my theorem, A alpha followed by a translation that has a perpendicular part plus a parallel part is a new type of operation that involves rotating through 180 degrees. And this is located at 1/2 of T perpendicular. But it has a translation component which I'll call generically tau. And tau is equal to the parallel part of the translation. So if I separate out this new sort of operation in all of its grandeur, what it does is it takes a first object, rotates us 180 degrees, does not yet put it down, first slides up by the amount tau. Doing the operation, again rotates it 180 degrees, doesn't yet put it down, slides it up by tau. Doing it again moves us to another one up here. So what we've generated is a helical spiral of objects about the central axis about which we're rotating. And this is a new two step operation that's called a screw axis. There is a persistent rumor that it derives its name from the effect that it has on 360 quizzes. But that is just an ugly rumor. There's nothing to that at all. So this is a new two step operation that we'll call, in general, A alpha tau. And what it will do is to generate a screw like distribution of objects. For those of you who have come in late, since you've been demonstrating a predilection towards longer and longer breaks, we're going to adjourn now and have a 40 minute break so you can get it out of your system. And we will resume at exactly 3 o'clock sharp. And we will launch into an elucidation of the nature of screw axes. And again, for those of you who were not here at the very start of class, the reason is that I have a phone call that had to come in about this time of day and no other time. So rather than keep you shifting, wondering whether I'm coming back, we'll start at three when I should be done with my business. OK, so that's it. You missed all the important discussion of what's going to be on the next quiz, and nobody will tell you because they want to keep class average down. All right, so sorry for the longer than usual interlude, but we'll start again promptly at 3:00.
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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 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. Hi. I'm Brian. We're going to be going over problem number 13 of the fall 2009 final exam. I like to think of the theme of this problem as polymers. So that's the overarching concept. But I'm going to give you some things I think are important to review and to know before actually meaningfully attempting this problem. So there's three things that I that I sort of thought were very important. The first one is polymerization processes. So we learned in class and in the book as well that there's two ways that we've covered that you can create a polymer. And that is through either addition polymerization or condensation polymerization. And I'll talk about that more in a second. The second thing you want to know is basically molecular weight and how it applies to polymers. And you want to know about elastomers for this problem. If you don't know about elastomers then you'll have trouble with the last part of the problem. So that's kind of what I would review. And then once you've got that under your belt give the problem a try. So we're going to go and we're going to start the problem now. We're given in the problem the 6-aminohexanoic acid, which I've drawn here. And we're also given at the beginning of the problem the structure of nylon 6. This is the actual reaction that's used to create nylon. So what I'm going to do is I'm going to show you-- or ask the question of how do we create nylon from 6-aminohexanoic acid. And that's actually going to answer part A and part B to this problem. So to create a polymer-- in this case we're talking about the polymer nylon 6-- you actually have to add together many mers. So merge is the term we used to denote the single unit which is repeated n times to create the poly mer-- polymer. So in this case the mer, the individual unit of nylon 6, is 6-aminohexanoic acid. And that's this. That's the same thing right there. So we're going to add these together and we're going to start creating our long chain of nylon 6. Remember, nylon 6 a polymer is a really, really long chain. Think of it as spaghetti, very long spaghetti. It isn't just two molecules that react. So let's just start somewhere. It's got to begin somewhere. You have to have two molecules at some point in the beginning reacting with each other. So let's look at our two molecules of 6-aminohexanoic acid-- I have to keep looking back at how to say that-- so let's take a look and we have to think about now our two possible processes or routes towards polymerization. There's addition and there's condensation. Now addition, as you probably read and heard, requires there to be a double bond present between two carbon atoms. The reason for this is because usually have a free radical coming in, which will then initiate. That's why oftentimes the free radical molecule or free radical provider is called the initiator. You need an initiator to come in to break that double bond and then create another free radical. Which then lets the process continue along in a chain like that. If we look at our molecule here, we don't actually have any double bonds between two carbon atoms. We have a double bond with an oxygen but that's not sufficient for an addition-type reaction. For a condensation reaction, what that basically implies is two mer units, or two things coming together to react and they give off a byproduct. So that's how I remember condensation, very often the byproduct is water. You can have other byproducts, like HCl or any other sort of small molecules can be given off. But most times what we'll see in 3091 will be water given off in condensation, so it's sort of intuitive to call it condensation. In this problem, if we look at this, we have a molecule that has the Hs on one side and has Os on the other side, we can actually almost see this immediately as being a condensation reaction. What's going to happen is we're going to have this O reacting with the positive end of this molecule, which has Hs. And we're going to have an H2O liberated and given off. We're going to have the new molecule formed. Let me draw the molecule for you. Let me write it out. So we've just begun forming our polymer and this is a polymer which has an n of 2. It has 2 mer units building it. Notice I've kept the ends here. The positive and the negative here. In reality what's going to happen is we have a big vat of this 6-aminohexanoic acid and they're going to keep reacting. So we might have another one of these molecules float over and react with this O. Or conversely, we could have another one of these molecules come over and react with the H. So this thing is going to keep building. And the way these polymerization processes begin is the control of some parameter of the system. It could be temperature, pressure you can add something in to sort of initiate it. But that's how this is going to start. So we've actually sort of answered part a and b. We've come to the conclusion that this is obviously a condensation reaction. Because we're going to give off-- I left that out there, see if you guys caught it-- H2O. So we're losing this O and two of these Hs and we're forming water. And that's the byproduct. And a byproduct immediately should tell you we've got condensation going on. So condensation polymerization, part A. Part B, we've actually already done to sort of logic out the answer to part A. This is the answer to part B. Part B is just the reaction. I'll read it very specifically. Write the reaction that converts two molecules of 6-aminohexanoic acid into a dimer of nylon 6. So we have two molecules forming the dimer. You are going to have trimer, and it goes on and on. The most common mistake on this problem was to give us the full repeating units. Let me just show you what that would look like. And then we'll move on with the problem. So the most common mistake on this problem was to not have these end units. And instead say, oh it's created this huge, long chain. So I'm going to write it as follows. And this is the symbol to imply that it repeats. Likewise over here. So people would give us this. And that's not correct. And the reason that's a problem is because not only do you lose points on this part, you also lose points on the next part. So part C is now asking us to calculate the molecular weight of a molecule given a certain n. n is the degree of polymerization. So in this case, n equals 2, we've got a dimer. In part C, we have n equals-- surprisingly-- 3,091 You'll see that's a recurring theme in this class. So we have n equals 3,091 And the question is, how do we approach the molecular weight problem? Well we're told that we have 3,091 one of these mer units making up this polymer chain. So let me just write this down. We're going to form an equation. If we take the mass of the full chain and divide it by the mass of a single mer unit we should be able to extract the number of units in the chain. So let me write that out for you. So the total molecular weight of the molecule, divided by the weight of a particular mer unit. OK so we're going to put that as mer. And so this is easy. Let's first identify what our mer unit is. And this is why it was easy get tripped up on this problem. Because if you made a mistake on the part B, you have trouble on part C. What is our mer unit? So in our mer unit, what we're talking about is a chain which has 3,091 of these linked pieces here. So let's look at this. What do we have? Here's our mer unit. Let me circle it. We have 6 carbons. We have 11 hydrogens, 1 nitrogen and 1 oxygen in this mer unit. We're not going to take the full, complete end unit. We're not counting this O. We're not counting those 2 Hs. We're dropping off 18 grams per reaction into water. So this is our mere unit. Let's calculate the mass of our mer unit. So it's going to be-- let me write this here-- we're looking for molecular weight. We're going to divide that by the mass of the mer unit. Let me just write out the full line of what it would be. Approximately 6 times 12 plus 11 times 1. So this is 6 carbons, times the mass of carbon. 11 hydrogens times the mass of hydrogen. We're going to have 1 oxygen times the mass of oxygen. And we're going to have the nitrogen as well. So let me add that down here. 1 nitrogen times the mass of nitrogen. I rounded. That's OK. As long as you get the idea right. So now we can easily just solve for the molecular weight. And we'll find that molecular weight in this problem will be equal to 3.5 times 10 to the fifth grams per mole. There's another way to do this problem, actually. And what you can do is you can calculate n times the number of this-- this particular molecule-- and you can subtract off 3,090 orders. 18 grams. 3,090 times eighteen. You subtract that off. Think about a little bit why it's 3,090 and not 3,091. So that's just maybe a brain teaser for you. That is the answer to part C. And now part D asks us if we're able to convert this into an elastomer. So this is where it's really required that you understand what an elastomer is and what that implies and what the structure looks like. So an elastomer-- and I sort of paraphrased this from the book and added my own thing to it-- is basically a randomly oriented amorphous polymer with some cross-linking such that you have the ability to move the chains but not slide them completely over each other. So in order to have an elastomer you must have cross-linking. In order to have cross-linking, what must you have? Well from lecture and from the book we know that to cross-link you must have double bonds in your carbons. So you must have 2 carbons-- going back over here-- which are double-bonded together. And why is that? Because what happens is those double bonds will be broken by some initiator element-- many times we'll have sulfur-- will come in, break the double bond and then you'll have links forming between things. So let me actually give you real-life example of this so you can see what I'm talking about. We're going to show disulfide bridges bounding and creating cross-links in a molecule. Something we're very familiar with is rubber. And we're also very familiar with car tires. The big difference between those two things is that rubber has vulcanized Which means we're actually putting sulphur into it to create a different type of polymer. We're looking at a cross-linked, much stronger rubber which we use on our car tires. So polyisoprene looks something like this. You should be familiar with this notation by now. We have this denotes it repeats. We have our double bonds, we have our carbons and hydrogens. This is polyisoprene. And this is what we're going to add sulphur to vulcanize rubber. We're going to create vulcanized rubber. sulphur generally actually looks, it's a ring of 8 sulphur atoms. We'll then add it to rubber, maybe heat it up mixed. And you'll make what you're used to on car tires. So what's actually happening here is that you can see we have this double bond between carbons. The sulphur's going to come in, it's going to react with this double bond, and is actually going to pull it off. And it might form a chain like that. And then this sulfur will then connect to another one of these polyisoprene molecules that was originally double bonded. And I'm not going to draw the rest of it out. And it's going to react, kill this double bond, and it's going to connect it. So basically what's happened in this process of forming sulfide bridges is you've got this polymer, polyisoprene, connecting to this polymer, polyisoprene, and you have cross-linking going on. And that's what we would need to have in our case for the nylon 6 in order to get an elastomer, And because nylon 6, because our mer unit, 6-aminohexanoic acid doesn't have double bonds, we can't break them. We can't network. We can't cross-link. And therefore we can't make an elastomer. So the answer to this problem-- in a very long-winded manner-- is no. You cannot create an elastomer given that mer unit.
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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: interpretation of the wave function. --pretation-- the wave function. So you should look at what the inventor said. So what did Schrodinger say? Schrodinger thought that psi represents particles that disintegrate. You have a wave function. And the wave function is spread all over space, so the particle has disintegrated completely. And wherever you find more psi, more of the particle is there. That was his interpretation. Then came Max Born. He said, that doesn't look right to me. If I have a particle, but I solve the Schrodinger equation. Everybody started solving the Schrodinger equation. So they solved it for a particle that hits a Coulomb potential. And they find that the wave function falls off like 1 over r. OK, the wave function falls off like 2 over r. So is the particle disintegrating? And if you measure, you get a little bit of the particle here? No. Max Born said, we've done this experiment. The particle chooses some way to go. And it goes one way, and when you measure, you get the full particle. The particle never disintegrates. So Schrodinger hated what Max Born said. Einstein hated it. But never mind. Max Born was right. Max Born said, it represents probabilities. And why did they hate it? Because suddenly you lose determinism. You can just talk about probability. So that was sort of funny. And in fact, neither Einstein nor Schrodinger ever reconciled themselves with the probabilistic interpretation. They never quite liked it. It's probably said that the whole Schrodinger cat experiment was a way of Schrodinger to try to say how ridiculous the probability interpretation was. Of course, it's not ridiculous. It's right. And the important thing is summarized, I think, with one sentence here. I'll write it. Psi of x and t does not tell how much of the particle-- is at x at time t. But rather-- what is the probability-- probability-- --bility-- to find it-- at x at time t. So in one sentence, the first clause is what Schrodinger said, and it's not that. It's not what fraction of the particle you get, how much of the particle you get. It's the probability of getting. But that requires-- a little more precision. Because if a particle can be anywhere, the probability of being at one point, typically, will be 0. It's a continuous probability distribution. So the way we think of this is we say, we have a point x. Around that point x, we construct a little cube. d cube x. And the probability-- probability dp, the little probability to find the particle at xt in the cube, within the cube-- the cube-- is equal to the value of the wave function at that point. Norm squared times the volume d cube x. So that's the probability to find the particle at that little cube. You must find the square of the wave function and multiply by the little element of volume. So that gives you the probability distribution. And that's, really, what the interpretation means. So it better be, if you have a single particle-- particle, it better be that the integral all over space-- all over space-- of psi squared of x and t squared must be equal to 1. Because that particle must be found somewhere. And the sum of the probabilities to be found everywhere must add up to 1. So it better be that this is true. And this poses a set of difficulties that we have to explore. Because you wrote the Schrodinger equation. And this Schrodinger equation tells you how psi evolve in time. Now, a point I want to emphasize is that the Schrodinger equation says, suppose you know the wave function all over space. You know it's here at some time t0. The Schrodinger equation implies that that determines the wave function for any time. Why? Because if you know the wave function throughout x, you can calculate the right hand side of this equation for any x. And then you know how psi changes in time. And therefore, you can integrate with your computer the differential equation and find the wave function at a later time all over space, and then at a later time. So knowing the wave function at one time determines the wave function at all times. So we could run into a big problem, which is-- suppose your wave function at some time t0 satisfies this at the initial time. Well, you cannot force the wave function to satisfy it at any time. Because the wave function now is determined by the Schrodinger equation. So you have the possibility that you normalize the wave function well. It makes sense at some time. But the Schrodinger equation later, by time evolution, gives you another thing that doesn't satisfy this for all times. So what we will have to understand next time is how the Schrodinger equation does the right thing and manages to make this consistent. If it's a probability at some time, at a later time it will still be a probability distribution.
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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: Lewis structures. So I tell students who have sort of no background in chemistry before they come into this class, and there always are some, that there are some topics in 5.111 that having no experience with the topic is actually a good thing, and Lewis structures is one of these things. I think if you've seen it before, you're like, oh yeah, that's easy. I know how to do that. You don't practice and you get on an exam and you're like, oh wait a minute. I forgot how to do this. And the students who haven't seen it before know all the rules and just get brilliant, perfect scores on the exam. And the other students are like, man. I forgot how to do my Lewis structures. So here are Lewis structures. We're going to go over them. So I really like Lewis structures because they're relatively simple, and they work. Like 90% of the time, they are the correct structure. And I'm a big fan of simple things working. If you wanted to get from 90% to 100%, you'd have to use Schrodinger's equation, but you can get 90% just with the simple Lewis structures I'm a big fan of that. I love it when simple things really work pretty well. OK. So when Lewis structures the key to this is thinking about the electrons being shared so that you get a full valence shell. And having the electrons distributed in such a way that all of the atoms have the number of electrons that make them happy, which is usually eight electrons, which is an octet, that noble gas configuration. So every dot in a Lewis structure represents a valence electron. And we can then look at some atoms and put dots around them to indicate the number of valence electrons. So we also have to know how many valence electrons atoms have. And so why didn't you just practice with a clicker question. And here's part of the periodic table up here if you need it. All right. I'm told 10 seconds. Everyone was crazy fast. Yes. So seven is the correct answer. You could look at the periodic table and sometimes with these it's a counting thing. So this is one where you want to always double check if things don't make sense. All right. So we can put seven electrons around fluorine, and we'll have two fluorines here. They'll both have seven electrons around them. And now I'm going to jump to another slide, but I'm going to show you the seven again in case you haven't written them down. If you don't want to, they're not in your handout, but that's probably OK. So when you bring them together, you can bring them together in such a way that they can all share. And so if we put in green, then, one of the fluorine's seven. And then we put in blue the other fluorine's seven, you can see that they can share two in the middle and both are very, very happy. So just thinking about this really simple idea, how many electrons will give you an octet, will give you eight? And how can you put things together in such a way that allows for that to happen? Now there are a few elements that do not want eight in their valence shell, and hydrogen is one of them. It just has that one S. So it only wants two, that's all it can handle. So this is an exception, hydrogen is going to want two electrons. Hydrogen loves to interact with things, though. It interacts with lots and lots of things. And here hydrogen with its one valence electron is interacting with chlorine with its seven valence electrons, and they are sharing two electrons forming a bond together. So when we're talking about Lewis structures, we're talking about different kinds of electrons. So we're talking about bonding electrons, the electrons that are involved in the bond, and also lone pair electrons. So chloride in HCl is going to have two bonding electrons, one was its, and one came from hydrogen. And it's also going to have six lone pair electrons, or we could say three lone pairs. So when we say a lone pair, that indicates two electrons there. So it has one, two, three, four, five, six or one pair, two pairs, three pairs. Now there are rules to Lewis structures, and here is the complete rule. In your handout, this wouldn't fit on one page. It's on two pages. And these rules, if you do work these problems, you will remember these rules, and they become pretty easy. But it's important to work Lewis structure problems so that the rules become really familiar to you. And it takes time to work Lewis structure problems, so don't wait to the last minute to start this problem set. There's a lot of Lewis structure problems on it, which means it's not difficult, but it's going to take some time. All right. So let's briefly go over these rules. First what you want to do is draw in the skeleton structure. Just put the atoms down. Hydrogen and fluorine are always going to be terminal atoms. Don't put them in the middle of a molecule. That gets chemistry professors really upset to see hydrogen in the middle with lots of bonds to things, so don't do it. And typically the element with the lowest ionization energy goes in the middle, and there are some exceptions and we'll see some of those exceptions. But that should be your first guess. You want to count the number of valence electrons. If there's a negative charge, you need to count that in or if there's a positive charge you need to subtract that from the total. Then you want to figure out the total number of electrons needed, so everyone has their full valence shell. You need to subtract these two to get the number of bonding electrons. And here are some of the things that it's really easy to make math mistakes here, so if your structure makes zero sense at the end, go back and check your math. Assign to bonding electrons to each bond. If any remain, you want to think about whether you have double or triple bonds. And there's only certain kinds of atoms that can have double and triple bonds. So be careful where you're putting your double and triple bonds. If any valence electrons remain, those are loan pairs. And then lastly, you want to figure out the formal charge on all of the atoms in your structure to make sure that this is a valid structure, and we're going to talk about formal charge. So first, let's just try an example. And your sheet has two examples on the same page. We have HCN and we also have CN minus. So we're going to do HCN first, so don't fill your entire page because you're going to have to write things for CN minus as well. But before we start, we need to figure out which atom is likely going to be in the middle. And so why don't you tell me what you think on the clicker question. OK, 10 more seconds. Yup. So here again we want to have one that has a lower ionization energy, and you also want to consider other things like hydrogen can't be in the middle. OK. And it was written that way, but sometimes they're written in a way that is not as straightforward. OK. So I'll put that up. All right so we'll go through the rules and we'll try to work this out. So first I can write-- I'm going to start-- I guess I'll write over here. So number one, we're just going to write HCN with C in the middle. So that's the first thing we're going to do. Next we're going to consider the valence electrons. And you can just help me out by yelling things out. How many valence electrons does hydrogen have? AUDIENCE: One. CATHERINE DRENNAN: What about carbon? AUDIENCE: Four. CATHERINE DRENNAN: Nitrogen? AUDIENCE: Five. CATHERINE DRENNAN: And you can always check me on my math. How much does that equal? AUDIENCE: 10. CATHERINE DRENNAN: Excellent. There's nothing like adding simple numbers in front of 350 people to really put the stress in one's day. OK, so to have a complete full valence shell, what do I need for hydrogen? AUDIENCE: Two. CATHERINE DRENNAN: What about carbon? AUDIENCE: Four. CATHERINE DRENNAN: Eight to be complete. Nitrogen? AUDIENCE: Eight. CATHERINE DRENNAN: Eight. And I think we add this up, we should get 18. How's that? All right. So for four, now we're going to subtract these numbers from each other to tell us how many bonding electrons we have. So we have 18 minus 10. And we should have eight to bonding electrons. For five I'm going to now assign two per bond. So I'm going to put one here. Another here. Another here. Another here. So I've assigned two per bond. And now I see if I have any left. Do I have some left? I've used four, I had eight. So yes, I have four more. And if you have more bonding electrons, then you are supposed to assign those bonding electrons. And think about whether it's allowed to have double or triple bonds. Can hydrogen be involved in the double bond? No. Carbon nitrogen? AUDIENCE: Yes. CATHERINE DRENNAN: Yes, or a triple bond. So I have four more, so I'm going to put one, two, three, four. So I'm going to have a triple bond between my carbon and my nitrogen. And now I'm good. So now I want to see do I have any extra electrons. So for this, I had 10, I used eight, so I have two left. So I'm going to assign those two as a lone pair. Should I put them on hydrogen? What about carbon? No. Carbon already has its eight, because it has four bonds. So then I'm going to put it on nitrogen. And then the only thing left is formal charge, which we haven't talked about yet. So before we do formal charge, I just want to do the same thing with CN minus. And I will say if you want to draw triple bonds, that's fine too. You don't have to indicate the dots as bonds. It's perfectly fine to write out a triple bond. So I could have written this as well. OK, so let's look at CN minus. So how many valence electrons does carbon have again? Four. Nitrogen? Five. Am I done? AUDIENCE: No. CATHERINE DRENNAN: No. I need to add one because there's a charge on this molecule of minus one. So now so I have 10 again. Three I'm going to figure out how many electrons I need to complete my valence shell. How many from carbon? AUDIENCE: Eight. CATHERINE DRENNAN: Eight. Nitrogen? Eight so now I have 16. I will subtract. Now 10 from 16 and get six bonding electrons. And I'm going to assign first just two of them. So I'll assign one, two here. And then we-- this is to assign. Six said, do you have any left over? I had six, and I only used two, so the answer is yes. I have four more. So I can put those in one, two, three, four. Again, we're going to have a triple bond. And then we ask are there any electrons left? So we had 10, we used six of them. And so we're going to have now four more here. So now I can assign-- this only has-- this is not complete for carbon. So I can put a lone pair on carbon, and I can put a lone pair on nitrogen. And now they have their complete octet. And we get to assign formal charge. So I could write it this way, or I could have written it with a triple bond here. And don't forget to put the charge on the end. All right. So now let's consider formal charge, because we're never done with our Lewis structures until we've considered formal charge. So formal charge is a measure of the extent to which the atom has really lost or gained an electron in the process of forming this covalent bond. So as we'll talk about, it's not the same as oxidation number where you have like sodium plus one, but again, there's some differences in how many are brought to the table and what it ends up with in the end. So there's an equation which you'll have to learn, but if you do enough of these problems, it'll be stuck in your brain and you can't purge it even if you want to. And so formal charge, FC, is equal to the number of valence electrons, symbol V, here, minus the number of lone pair electrons minus half of the number of bonding electrons. So at least this equation makes sense. If you forget what they mean, you can probably think about it and it'll come back to you. So in doing these formal charges, you want the formal charges to add up to the charge on the molecule. So if we had HCN, that's a neutral molecule so the sum of all of the formal charges must be zero or you did something wrong. If it's CN minus, the sum of the formal charges has to add up to minus one or you did something wrong. So this is a good way of checking your math. So always remember that the sum needs to add up to the total charge on the molecule. If you remember that, that's a really good check to make sure you didn't make some kind of weird math mistake and add things wrong and have an appropriate number of loan pairs or something going on. OK. So let's calculate formal charge now on our CN minus molecule up here. So the formal charge now on carbon here. So how many valence electrons do carbon have? AUDIENCE: [INAUDIBLE] CATHERINE DRENNAN: Four. How many lone pairs does it have? AUDIENCE: [INAUDIBLE] CATHERINE DRENNAN: It has one lone pair, two lone pair electrons. This is the total number of lone pair electrons here, and then half the number of bonding electrons. So it is expanding electrons and half of that is three. And so that should add up to a charge of minus one. You can also, instead of thinking half the number of bonding electrons, you can also just think about number of bonds if you want to to do this. All right so to see if you have the hang of it, let's do a clicker question. OK, 10 more seconds. Most people got it right. I can't actually read the number, but that was very good. It always seems wrong to put zero as the answer, but that is, in fact, the answer here. So if we look at this again, we have five valence electrons on nitrogen. We have two lone pair electrons, and we have six bonding electrons. Half of six is three, and so that's zero. Or you could have said, well, if this is minus one, and the charge is minus one, then that had to have been zero, otherwise Professor Drennan did something wrong, and that's just not possible. So the answer there would be zero. And you can see that the total formal charge is minus one, and the total charge of the molecules also minus one. So again formal charge does not equal oxidation number, it's something special. It tells you kind of in this arrangement of atoms in the molecule, did this atom kind of come up with a little bit more at the end or a little bit less, depending on where it started from and where it is. And where you want in these structures for the formal charge to be low. So we can use a formal charge to decide between Lewis structures so that the structures with the lowest absolute values of this formal charge are more stable structures. So if you have really high formal charges, that means that molecule isn't really very stable because you want low charges. You want lower energy. You want things to be in a more neutral and happy state. So we want to figure out which ones are going to have low charges. So let's look at another example, Thiocynate ion, and it has a carbon, sulfur, and nitrogen in it, and it has a charge of minus one. So I might tell you the ionization energies for carbon, sulfur, and nitrogen, and then ask you which atom do you think is going to be in the center of the molecule? So what do you think? What's in the center of this molecule based on those numbers? Just yell it out. AUDIENCE: [INAUDIBLE] CATHERINE DRENNAN: Sulfur. So sulfur has the lowest ionization energy, and I told you that's usually the thing in the center. But you can start with that. It's always good to start with that, but then you want to check the structure and make sure that a structure with sulfur in the middle has the lowest formal charge. So let's take a look at that. So we can draw structure A with sulfur in the middle, and then calculate the formal charges on that. And if we do that, we see for the nitrogen here, nitrogen has five valence electrons, it has four lone pair electrons, and it has half of four bonding electrons, so it would have a formal charge a minus one in this particular structure that has sulfur in the middle. We can look at carbon. Carbon has four valence electrons, Four lone pair electrons, and half of four bonding electrons, so it has a charge of minus two. Then we can look at sulfur. Sulfur has six valence electrons, zero loan pairs, and half of eight bonding electrons, so it has a formal charge of plus two. So overall, this does equal the minus one. So it's a valid structure. But is that the lowest energy one? We also could put carbon in the middle or nitrogen in the middle. So let's look at what this does for us. So with the formal charge on nitrogen now, we have five minus four lone pair electrons, minus half of four bonding electrons, minus one. So that's the same. Now we can look at carbon. We have now just no lone pair electrons. It has four minus zero minus four, half of eight bonding electrons, or zero. And for sulfur, six minus four lone pair electrons, and half of four bonding electrons, or zero. Next, structure C, five minus zero lone pairs minus half of eight, plus one. So that one's different. Carbon, we have four valence electrons minus four lone pairs minus half of four, bonding minus two. And for the sulfur, six minus four lone pairs electrons, half of three or zero. So now with the clicker, tell me which is most stable. All right, 10 seconds. I think this can be 98%. That's what I'm thinking. I'm feeling good. Well, close. Yeah. What? No, no. Sorry. It should be B. Yeah, it should be B. Yeah. Sorry I actually-- Yay. There we go. Yeah, B is correct. So if we just look at it over here, it has the lowest number of formal charges, so the answer is B. OK. Let's start with this simply. Who wants to tell me why one, how they could look at this and realize one was not the correct answer? I think this is on. Give it a try. AUDIENCE: One's not correct because if you look back at your atomic radius chart, this is pretty much doing the exact opposite of that CATHERINE DRENNAN: Yeah, so helium definitely not the biggest atom there is. OK so six got a lot of attention, and so did two. And ionization energy, electron affinity, and electronegativity are definitely connected to each other, but there is a clue that electron affinity would not be the correct. Does someone want to say what you might have noticed? AUDIENCE: For electron affinity, it increases and then stops at the noble gases because noble gases do not want electrons. So in this particular chart, all the noble gases are like the highest-- are the highest ones in the relative area, which would mean that electron affinity would be incorrect. CATHERINE DRENNAN: Yep. That's right so the noble gases were the clue. So [INAUDIBLE]. Yep. And so if electron affinity also is not high at the noble gases, they're also not electronegative. Noble gases just don't want extra light. They don't want to lose electrons, they don't want to gain electrons, they just want to be left alone. So this trend is for ionization energy. And because noble gases want to be left alone, they don't want to lose any of their electrons. Great. So this is good to be thinking about, because we're finishing up now for the handout from last time. And we're going to be talking about electronegativity again. We never move very far away from a lot of these topics. They just keep coming back. So we just keep reviewing them. All right, I don't need to microphones, although, I don't know, I kind of like having this one. Anyway. So let's take out the handout from last time, and let's finish it up. We were talking about formal charge, and we had looked at examples where we calculated formal charge, and then we looked at which structure would be lowest in energy, and that was the structure where you had the least separation between the charges on them, so the smallest absolute numbers. If you have formal charges of zero, that's fantastic. That's what the molecule wants. Minus one, plus one, if you must, but when you start having plus two minus two, that's a lot of charge separation, so that's less favorable. So we're going to have more examples of that as we go along. But now, what-- if you had calculated formal charge and they're all the same, how do you know which structure is correct? So what if you have-- and this is-- just some people who are having trouble. This is the top of page four from the handout. You have two valid Lewis structures that have the same formal charge, how do you know where it goes? And the answer is that the negative formal charge should go on the most electronegative atom. And so that's why we are sort of talking about electronegativity again. And so electronegativity-- remember electronegativity is high when the electron affinity is high, meaning that the atom wants to get an electron, has a high affinity for electrons, and also a high ionization energy, which means it doesn't want to give up its electrons. So that's something-- it likes to have electrons, and so you want to put a negative value, which indicates there's more electrons on something that's electronegative. So negative on negative over here. And so that's what you're looking for. That's how you're going to make a decision. So let's look at an example. So here is a molecule, and we're going to look at two possible Lewis structures of this with similar formal charges, and decide which has the correct structure. So first let me give you a couple hints that can be useful in problem sets and in exams. When you see CH3, that's a methyl group, and that's going to be terminal so you're going to have these three hydrogens associated with that carbon, and that's going to be at an end of the molecule. So it could look like one of these two things. So we have the carbon, we have it attached to three hydrogens, a carbon attached to three hydrogens, and then attached to something else, this nitrogen here. And this structure where it's just kind of written out in a line or a chain of atoms, what we call chain molecules sometimes. Often the atoms are actually written in the order in which they're attached, so that's definitely true here. CH3, three hydrogens attached to the carbon, so they're attached to the atom that came before in the chain. The nitrogen is also going to be attached to the carbon. Even though it follows the hydrogen, you're not going to have hydrogen in the middle of a bond. It's not going to be bonded to two things. Hydrogen is always terminal, so even though nitrogen is here, it's got to be attached to the carbon. So we have three hydrogens and then a bond between the carbon and the nitrogen. Now we have a hydrogen after the nitrogen, and by this rule it should go on the nitrogen. But you might want to double check that that's true. And then we have an oxygen. Again, the oxygen is going to have a bond with the nitrogen. You're not going to have a bond with hydrogen in the middle. Hydrogen's always terminal. So the only real choice we have here is we can put the hydrogen on nitrogen or we can put the hydrogen on the oxygen here. And so we can use this rule about electronegativity and formal charges to figure out which of these structures is right. So in this particular case, all of the formal charges are zero on all of the atoms, except there's one minus one charge. And in this structure, the minus one charge would be on the oxygen, and in this structure the minus one charge would be on the nitrogen. And if everything else is zero then you have the sum of your formal charges, minus one, equal to the charge on the molecule, which is minus one. So both of these are valid structures. Both have low values of formal charge, which is right. So it's going to be the structure that has the negative charge on the most electronegative atom is the right structure. And so here you need to remember some of your rules about electronegativity. And in terms of electronegativity, we see that oxygen has greater electronegativity than the nitrogen. And so that's where we would want to put our negative charge-- on the negative charge goes on the atom that's the most electronegative. And that would generate the lower energy structure. So if you're given a table of electronegativities, which you often are, you can look it up and validate that that's going to be the correct place. And in fact, experimentally we know that that's the right structure. So that works. So if you have two structures, identical formal charges, valid structures, then the last step is to think about where that negative charge should go, and pick the atom that is the most electronegative. All right we have one more thing we need to talk about in Lewis structures before we start violating various rules that we've learned, and that is that we need to talk about resonance structures. And so to explain to you what a resonance structure is, it's really helpful to start with an example, so that's what we're going to do. And we're going to consider ozone, which is three atoms of oxygen. And we have the ozone layer, which protects us from UV damage and is very valuable, and we should not destroy it with chemicals being released into the environment. And because you don't have complete say over that, always wear sunscreen. OK. So let's build up these Lewis structures, and then consider what's meant by a resonance structure. So here we have part of the Periodic Table that you're going to need, and we need to think first about the valence electrons. So oxygen has six valence electrons and there are three oxygens, so that's 18. To get a full octet for each of the oxygens, three oxygens, an octet is eight valence electrons, so that would be 24. To figure out the number of bonding electrons, we're going to be subtracting our octet electrons from our valence electrons. So 24 minus 18 is six. And then our next step is to assign those bonding electrons two at a time, two per bond. So let's take a look at that. We can put one bond here between these two oxygens, one bond here between these two oxygens, and then ask do we have any more? And we do. Because we have six bonding electrons, we used four, so we have two more bonding electrons. So we need to make a double bond. But now we have the question of where to make that double bond. Am I going to put it between these two oxygens or am I going to put it between those two oxygens? And so I could say put it there, but I could also, in structure two, put the double bond over here. All right. We'll come back to that question in a minute. Let's first figure out if we have any remaining valence electrons, and we do. So we had 18 and we've only used six, so we have 12 left. So we're going to put those in as lone pairs. And so I can put them in over here. One set here, one, two, three, four, five, six. And I can also do that over here. One, two, three, four, five, six. And now we have two structures, so we need to think about formal charges to see if that can help differentiate structure one from structure two. OK. And let's look at that. Be sure everyone's ready. And that is a clicker question. So I'll put that back up. All right. Let's just take 10 more seconds, and we'll talk about what the right answer is, and a little bit of a trick for doing these, perhaps, a bit faster. OK. So that was pretty good. So let's look at why that's the right answer. And we'll take a look at that over here. So let's do the calculations. So you have to remember the equation for formal charge for sure and once you do enough problems it should stick in your head pretty easily. So if we look over here-- so this is the formal charge on oxygen A. There were six valence electrons, there are four lone pair electrons, minus half of the bonding electrons. There are eight bonding electrons, so that's two. So that's a formal charge of zero. For oxygen B over here, we again have six valence electrons, and we have two lone pair electrons. We have six bonding electrons, so half of six is three, which is plus one. And on oxygen C over here, six minus six lone parallel electrons, half of one bond, so half of two is one, minus one, overall the charge here is neutral. And that's good because it's a neutral molecule. So to do this structure faster, you have to realize that oxygen A over here is the same as C over there. So you can just copy down what you had for C is now A. B is exactly the same in the two structures, so it's the same as what you calculated. And now this C was the same as this A, so we can put that same value that we calculated for A into C. So they're both the same. Same formal charges, which structure is correct? And the answer is both of them. And in fact, you need both of them, so there's data-- chemists love data-- and the data is that the bonds are actually equivalent in the molecule. So it isn't that there's one double bond and one single bond, there's just one kind of bond in ozone, and it's between a single and a double bond. And so this is how one would draw that kind of thing. You would have structure one here and structure two here. You would put them both in brackets, and you would put this special double headed arrow between them and that indicates that both of these structures are needed to describe the properties of the molecule. You do not have a stationary double bond and a single bond. You have kind of a mixture between these two, and that's what a resonance structure is. So let's take a little bit more of a look at this. So this is just what I had before, experimental evidence is that the bonds are equivalent. There isn't a double and a single, there's something in between, a one and a half. And this is called a resonance hybrid, and it's of length of these two structures. So this structure is blended with that structure. So some of you are aware of hybrids from biology, and now with cars. So a mule would be an example. And if you're thinking about what a mule is, you don't walk out into your barnyard one day and see a donkey one day and you go out the next day and you see a horse. A mule is a hybrid between a donkey and a horse. So if you were a chemist, you would do this. You would put the donkey in brackets, and the horse in brackets, and put your double headed arrow. And if you see this, you'd say, oh yes, a mule. So that is what this is. Both structures are needed to describe ozone. One structure isn't enough. You need both of them. They're in resonance with each other. And so what's true about the electrons in ozone is that they're delocalized across all of these bonds, so there isn't like a double single. All of those electrons are delocalized. They're shared over this set of atoms. And you can have two resonance structures, you can have three, you can have four. It depends on the molecule in question. And in all those cases, those electrons would be delocalized. So just to sum that up, resonance structures, two or more, same arrangement of atoms-- and that's important. It's the electrons that are different. And this isn't in your handout, out but just think about this for a second-- because it was in your hand out a minute ago. Are these resonance structures? No. They're not resonance structures. The atoms are in different positions. So one of these structures is right one of these structures is wrong. With resonance structures, they're both correct and both needed to define the structure. So pay attention. This is a common mistake. Pay attention. Ask yourself, are the atoms in a different position? That's not resonance. You're just looking-- atoms are the same, formal charge are the same, you're just looking at whether you have different arrangements of electrons.
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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 That is the largest value it under a Creative Commons license. Additional information about our license and MIT OpenCourseWare in general is available at ocw.mit.edu. And your equation is equal to E Psi. And we saw that these energies, the binding energies of the electron to the nucleus, were quantized. Well, these energy levels then were given by minus this Rydberg constant R sub H over n squared. And this n was our quantum number. It is a principle quantum number. And we saw that n, the smallest value was 1, 2 and ran all the way up to infinity. Well, today what we are going to do is solve this equation, or we are going to look at the results of solving this equation for the wavefunction Psi. Now, Psi, in general, is the function of r, theta, phi, and also time. But we are going to be looking at problems in which time does not have an effect. In other words, the wave functions that we are going to be looking at are what are called stationary waves. We actually are not going to be looking at the wave function as a chemical reaction is happening. We are either going to look at it before or after, but not during. And, in those cases, we are looking at a wave function. And the atom is just stable and is sitting there. The time dependence here does not have a consequence. And so, therefore, the wave functions that we are going to be looking at are just a function of r, theta and phi. And we are looking at what is called time-independent quantum mechanics. Later on, actually, if you are in chemistry, in a graduate course in chemistry, is when you look at time-dependent wave functions. We are going to look at time-independent quantum mechanics, the stationary wave. Now, it turns out that when we go and solve the Schrˆdinger equation here for Psi, what happens is that two more quantum numbers drop out of that solution to the differential equation. Remember, we said last time how quantum numbers arise. They arise from imposing boundary conditions on a differential equation, making that differential equation applicable to your actual physical problem. And so, when we do that, we get a new quantum number called l. And l is, I think you already know, the angular momentum quantum number. Absolutely. It is called the angular momentum quantum number because it indeed dictates how much angular momentum the electron has. It has allowed values. The allowed values of l, now, are zero. Zero is the smallest value of l, the lowest value of l allowed. 1, 2, all the way up to n minus can have. It cannot be any larger than n minus 1. Why can't it be larger than n minus 1? Well, it cannot because the angular momentum quantum number, at least if you want to think classically for a moment, dictates how much rotational kinetic energy you have. And remember that this energy, here, is dependent only on n. This energy is the sum of kinetic energy plus potential energy. If l had the same value as n, well, essentially, that would mean that we would have only rotational kinetic energy and we would have no potential energy. But that is not right. We have potential energy. We have potential energy of interaction. So, physically, that is why l is tied to n. And it cannot be larger than n minus 1. And then, we have a third quantum number that drops out of that solution, which is called M. It is the magnetic quantum number. It is called that because indeed it dictates how an atom moves in a magnetic field. Or, how it behaves in a magnetic field. But, more precisely, m is the z-component of the angular momentum. l is the total angular moment. m dictates the z-component of the angular momentum. And the allowed values of m are m equals zero. You can have no angular momentum in the z-component. Or, plus one, plus two, plus three, all the way up to plus l. Again, m is tied to l. It cannot be larger than l, because if it were then you would have more angular momentum in the z-component than you had total angular momentum. And that is a no-no. So, m is tied to l. The largest value you can have is l. But, since this is a z-component and we've got some direction, m could also be minus 1, minus two, minus three, minus l. We have three quantum numbers, now. That kind of makes sense because we have a 3-dimensional problem. We are going to have to have three quantum numbers to completely describe our system. The consequence, here, of having three quantum numbers is that we now have more states. For example, our n equals 1 state that we talked about last time, more completely we have to describe that state by two other quantum numbers. When n is equal to one, what is the only value that l can have? Zero. And if l is zero, what is the only value m can have? Zero. And so, more appropriately, or more completely, that n equals 1 state is the (1, 0, 0) state. And, if we have an electron in that (1, 0, 0) state, we are going to describe that electron by the wave function Psi(1, 0, 0). Now, what I have not told you yet is exactly how Psi represents the electron. I am just telling you right now that Psi does. Exactly how it does is something I haven't told you yet. And we will do that later in the lecture today, sort of. That is the energy, minus the Rydberg constant. But now, if n is equal two, what is the smallest value that l can have? Zero. And the value of m in that case? Zero. And so our n equals 2 state is more completely the (2, 0, 0) state. And the wave function that describes the electron in that state is the Psi(2, 0, 0) wave function. And then, of course, here is the energy of that state. Now, when n is equal to 2, what is the next larger value of l? One. And if l is equal to one, what is the largest value that m can be? One. And so we have another state here, the (2, 1, 1) state. And if you have an electron in that state, it is described by the Psi(2, 1, 1) wave function. However, the energy here is also minus one-quarter the Rydberg constant. It has the same energy as the (2, 0, 0) state. Now, if n is equal two and l is equal to one, what is the next larger value of m? Zero. And so we have a (2, 1, 0) state. Again, that wave function, for an electron in that state, we label as Psi(2, 1, 0). And then, finally, when n is equal to two, l is equal to one, what is the final possible value of m? Minus one. We have a (2, 1, -1) state and a wave function that is the (2, 1, -1) wave function. Notice that the energy of all of these four states is the same, minus one-quarter the Rydberg constant. These states are what we call degenerate. Degenerate means having the same energy. That is important. Now, this is the way we label wave functions, but we also have a different scheme for talking about wave functions. That is, we have an orbital scheme. And, as I said, or alluded to the other day, an orbital is nothing other than a wave function. It is a solution to the Schrˆdinger equation. That is what an orbital is. It is actually the spatial part of the wave function. There is another part called the spin part, which we will deal with later, but an orbital is essentially a wave function. We have a different language for describing orbitals. The way we do this, I think you are already familiar with, but we are going to describe it by the principle quantum number and, of course, the angular momentum quantum number and the magnetic quantum number. Except that we have a scheme or a code for l and m. That code uses letters instead of numbers. For example, if l is equal to zero here, we call that an s wave function or an s orbital. For example, in the orbital language, instead of Psi(1, 0, 0), we call this a 1s orbital. Here is the principle quantum number. Here is s. l is equal to zero. That is just our code for l is equal to zero. This state, (2, 0, 0), you have an electron in that state. Well, it can describe that wavefunction by the 2s orbital. Here is the principle quantum number. He is s, our code for l equals zero. Now, when n is equal to 1, we call that a p wave function or a p orbital. This state right here, it is the 2p state. Because n is two and l is one, this state is also the 2p wave function and this state, it is also the 2p wave function. Now, if I had a wave function up here where l was equal to two, we would call that d orbital. And, if I had a wave function up here where l was three, we would call that an f orbital. But now, all of these wave functions in the orbital language have the same designation, and that is because we have not taken care of this yet, the m quantum number, magnetic quantum number. And we also have an alphabet scheme for those quantum numbers. The bottom line is that when m is equal to zero, we put a z subscript on the p. When m is equal to zero, that is our 2pz wave function. When m is equal to one, we are going to put an x subscript on the p wave function. And when m is equal to minus one, we are going to put a y subscription on the wave function. However, I have to tell you that in the case of m is equal to one and m is equal to minus one, that is not strictly correct. And the reason it is not is because when you solve Schrˆdinger equation for the (2, 1, 1) wave function and the (2, 1, -1) wave function, the solutions are complex wave functions. They are not real wave function. And so, in order for us to be able to draw and think about the px and the py wave functions, what we do is we take a linear combination of the px and the py wave functions. We take a positive linear combination, in the case of the px wave function. The px wave function is really this wave function plus this wave function. And the py wave function is really this wave function minus this wave function. Then we will get a real function, and that is easier to deal with. That is strictly what px and py are. px and py are these linear combinations of the actual functions that come out of the Schrˆdinger equation. pz is exactly correct. pz, M is equal to zero, in the pz wavefunction. Now, you are not responsible for knowing that. I just wanted to let you know. You don't have to remember that m equal to one gives you the x subscript, and m equal to minus one gives you the y subscript. You do have to know that when m is equal to zero, you get a z subscript, because that is exactly right. Now, just one other comment here. That is, you see we did not put a subscript on the s wave functions. Well, that is because for an s wave function, the only choice you have for m is zero. And so we leave out that subscript. We never put a z on there or a z on there. It is always 1s, 2s, or 3s. Well, in order to understand that just a little more, let's draw an energy level diagram. Here is the energy again. And for n equals 1, we now see that the more complete description is three quantum numbers, (1, 0, 0). That gives us the (1, 0, 0) state, or sometimes we say the 1s state. There is one state at that energy. But, in the case of n equals 2, we already saw that we have four different states for the quantum number. Four different states. They all have the same energy. They are degenerate. Degenerate means having the same energy. They differ in how much angular momentum the electron has. Or, how much angular momentum the electron would have in the z component if it were in a magnetic field. The total energy is the same because only n determines the energy. It is just that the amount of angular momentum, or the z component of the angular momentum, differs in 2s, 2py, 2pz, 2px, but they all have the same energy. In general, for any value of n, there are n squared degenerate states at each value of n. If we have n is equal to three, then the energy is minus one-ninth the Rydberg constant. But how many states do we have at n equal three? Nine. Here they are. Just like for n equal two, we have the 3s, 3py, 3pz, and 3px. And there are the associated quantum numbers for that. But now we have some states here that I have labeled 3d states. Well, when you have a d state, that means l is equal to two. All of these states have principle quantum number three and angular momentum quantum number two. And they differ also by the amount of angular momentum in the z direction. They differ in the quantum number m. For example, we are going to call the m equal minus two state the 3d(xy). We are going to put xy as a subscript. For m equal minus one, we are going to put a subscript yz. For m equals zero, we are going to put a subscript z squared. For m equal one, d(xz). And for m equal two, x squared minus y squared. Again, for m equal minus two, minus one, one and two, those wave functions, when you solve Schrˆdinger equations, are complex wave functions. And what we do to talk about the wave functions is we take linear combinations of them to make them real. And so, when I say m is minus two, is the 3dxy wave function, it is not strictly correct. Therefore, again, you don't need to know m minus two. You don't need to know that subscript. Or m equal one, you don't need to know that subscript. But for m equal zero, this is a z squared. Absolutely. These wave functions are linear combinations. This one is not. It is a real function when it comes out of the Schrˆdinger equation. You will talk about these 3d wave functions a lot with Professor Cummins in the second-half of the course. You will actually look at the shapes of those wave functions in detail. So, that's the energy level diagram, here. All right. Now let's actually talk about the shapes of the wave functions. I am going to raise this screen, I think. What do these wave functions actually look like? Well, for a hydrogen atom, our wave function here, given by three quantum numbers, n, l and m, function of r, theta and phi, it turns out that those wave functions are factorable into a function that is only in r and a function that is only in the angles. You can write that, no approximation, this is just the way it turns out. The function that is a function only of r, R of r, is called the radial function. We will call it capital R, radial function of r. It is labeled by only two quantum numbers, n and l. The function that is a function only of the angles, theta and phi, we are going to call Y. This is the angular part of the wave function. And it labeled by only two quantum numbers, but they are l and m. Sometimes we call this angular part, for short, the Y(lm)'s. There is a radial part, and there is an angular part. The actual functional forms are what I show you here on the side screen. And, in this case, what I did is to separate the radial part from the angular part. This first part, here, is the radial part of the wave function. And here on the right is the angular part of the wave function. And I have written them down for the 1s, the 2s, and the bottom one is the 3s, although I left the label off in order to get the whole wave function in there. I want you to notice, here, that the angular part, the Y(lm) for the s wave functions, it has no theta and phi in it. There is no angular dependence. They are spherically symmetric. That is going to be different for the p wave functions. The angular part is just one over four pi to the one-half power. And it is the radial part, here, that we actually are going to take a look at right now. Let's start with that 1s wave function, up there. If I plot that wavefunction, this is Psi(1, 0, 0), or the 1s wave function versus r. Oh, I should tell you one other thing that I didn't tell you. That is that throughout these wave functions, you see this thing called a nought. a nought is a constant. It is called the Bohr radius. I will explain to you where that comes from in a little bit later. But it has a value of about 0.529 angstroms. That is just a constant. Let's plot here the (1, 0, 0) wave function. If I went and plotted it, what I would find is simply that the wave function at r is equal to zero, here, would start out at some high finite value, and there would just be an exponential decay. Because if you look here at the functional form, what do you have? Well, you have all this stuff, but that is just a constant. And the only thing you have is an e to the minus r over a nought dependence. That is what gives you this exponential drop in the wave function. What this says is that the wave function at all values of r has a positive value. Now, what about the Psi(2, 0, 0) wave function? Let's look at that. Psi(2,0,0), or the 2s wave function as a function of r. What happens here? Well, we are plotting essentially this. All of this stuff is a constant. And we have a two minus r over a nought times an e to the minus r over 2 a nought. That is what we are really plotting here. And, if I did that, it would look something like this. We start at some large, positive value here. And you see that the wave function decreases. And it gets to a value of r where Psi is equal to zero. That is a radial node. And in the case of the 2s wave function, that radial node occurs at r equals 2 a nought. And then the wave function becomes negative, increases, and gets more and more negative, until you get to a point where it starts increasing again and then approaches zero. This part, here, of the wave function is really dictated by the exponential term, the e to the minus r over 2 a nought. This part of the wave function is dictated by this polynomial here, two minus r over ao. If you wanted to solve for that radial node, what would you do? You would take that functional form, set it equal to zero and solve for r. And so you can see that 2 minus r over a nought set equal to zero, that when r is 2 a nought, that the wave function would have a value of zero. That is how you solve for the value of r at which you have a node. Now, this is really important here. That is, at the radial node, the wave function changes sign. The amplitude of the wave function goes from positive to negative. That is important because at all nodes, for all wave functions, the wave function changes sign. And the reason the sign of the wave function is so important is in chemical bonding. But let me back up for a moment. Many of you have talked about p orbitals or have seen p orbitals before. Sometimes on a lobe of a p orbital, you put a plus sign and sometimes you put a negative sign. You have seen that, right? Okay. Well, what that is just referring to is the sign of the amplitude of the wave function. It means that in that area the amplitude is positive, and in the other area the amplitude is negative. And the reason the amplitudes are so important, or the sign of the amplitudes are so important is because in a chemical reaction, when you are bringing two atoms together and your electrons that are represented by waves are overlapping, if you are bringing in two wave functions that have the same sign, well, then you are going to have constructive interference. And you are going to have chemical bonding. If you bring in two atoms, and the wave functions are overlapping and they have opposite signs, you have destructive interference and you are not going to have any chemical bonding. That is why the sign of those wave functions is so important. So, that is Psi(2, 0, 0). What about Psi(3, 0, 0)? That is the last function, here, on the side walls. And let me just write down the radial part, 27 minus 18(r over a nought) plus 2 times (r over a nought) quantity squared times e to the minus r over 3 a nought. And now, if I plotted that function, Psi(3,0,0), 3s wave function, I would find out that r equals zero, large value of psi finite, it drops, it crosses zero, gets negative, then gets less negative, crosses zero again, and then drops off. In the case of the 3s wave function, we have two radial nodes. We have a radial node right in here, and we have a radial node right in there. And, if you want to know what those radial nodes are, you set the wave function equal to zero and solve for the values of r that make that wave function zero. And, if you do that, you would find this would come out to be, in terms of units of a nought, 1.9 a nought. And right here, it would be 7.1 a nought. The wave function, in the case of the 3s, has a positive value for r less than 1.9 a nought, has a negative value from 1.9 a nought to 7.1 a nought, and then a positive value again from 7.1 a nought to infinity. So, those are the wave functions, the functional forms, what they actually look like. Now, it is time to talk about what the wave function actually means, and how does it represent the electron. Well, this was, of course, a very puzzling question to the scientific community. As soon as S wrote down is Schrˆdinger equation, hmm, somehow these waves represent the particles, but exactly how do they represent where the particles are? And the answer to that question is essentially there is no answer to that question. Wave functions are wave functions. It is one of these concepts that you cannot draw a classical analogy to. You want to say, well, a wave function does this. But the only way you can describe it is in terms of language that is something that you experience everyday in your world, so you cannot. A wave function is a wave function. I cannot draw a correct analogy to a classical world. Really, that is the case. However, it took a very smart gentleman by the name of Max Born to look at this problem. He said, "If I take the wave function and I square it, if I interpret that as a probability density, then I can understand all the predictions made by the Schrˆdinger equation within that framework." In other words, he said, let me take Psi and l and m as a function r, theta, and phi and square it. Let me interpret that as a probability density. Probability density is not a probability. It is a density. Density is always per unit volume. Probability density is a probability per unit volume. It is a probability per unit volume. Well, if I use that interpretation, then I can understand all the predictions made by the Schrˆdinger equation. It makes sense. And, you know what, that is it. Because that interpretation does agree with our observations, it is therefore believed to be correct. But it is just an assumption. It is an interpretation. There is no derivation for it. It is just that the interpretation works. If it works, we therefore believe it to be accurate. There is no indication, there are no data that seem to contradict that interpretation, so we think it is right. That is what Max Born said. Now, Max Born was really something in terms of his scientific accomplishments. Not only did he recognize or have the insight to realize what Psi squared was, but he is also the Born of the Born-Oppenheimer Approximation that maybe some of you have heard about before. He is also the Born in the Distorted-Wave Born Approximation, which probably none of you have heard before. But, despite all of those accomplishments, psi squared interpretation, Born-Oppenheimer Approximation, Distorted-Wave Born Approximation, he is best known for being the grandfather of Olivia Newton-John. That's right. Oliver Newton-John is a singer in Grease. Two weeks ago in the Boston Globe Parade Magazine, which I actually think is a magazine that goes throughout the country in all the Sunday newspapers, there is a long article on Olivia Newton-John and a short sentence about her grandfather, Max Born. So, that is our interpretation, thanks to Max Born. Now, how are we going to use that? Well, first of all, let's take our functional forms for Psi, here, and square it and plot those probability densities for the individual wave functions and see what we get. The way I am going to plot the probability density is by using some dots. And the density of the dots is going to reflect the probability density. The more dense the dots, the larger the probability density. If I take that functional form for the 1s wave function and square it and then plot the value of that function squared with this density dot diagram, then you can see that the dots here are most dense right at the origin, and that they exponentially decay in all directions. The probability density here for 1s wave function is greatest at the origin, r equals 0, and it decays exponentially in all directions. It is spherically symmetric. That is what you would expect because that is what the wave function looks like. You square that, you get another exponential, and it decays exponentially in all directions. That is a probability density, probability of finding the electron per unit volume at some value r, theta, and phi. And it turns out it doesn't matter what theta and phi are because this is spherically symmetric. What about the 2s wave function? Well, here is the 2s probability density. Again, you can see the probability density is a maximum at the origin, at the nucleus. That probability density decays uniformly in all directions. And it decays so much that at some point, you have no probability density. Why? Because that is the node. If you square zero, you still get zero. r equals 2 a nought. You can see that in the probability density. But then again, as you move up this way, as you increase r, the probability density increases again. Why? Remember the wave function? It has changed sign. But in this area, here, where it is negative, if I square it, well, the probability density still is going to be large. Square a negative number, you are going to have a large positive number. That is why the probability density increases right in here, and then, again, it decays towards zero. You can see the radial node not only in the wave function, but also in the probability density. Here is the probability density for the 3s wave function. Once again, probability density is a maximum at r equals 0, and it decays uniformly in all directions. It decays so much that at some value of r, right here, the probability density is zero. Why? Because the wave function is zero. You square it, and you are going to get a zero for the probability density. And then the probability density increases again. Why? Because you are getting a more and more negative value for the wave function right in this area. Square that, and it is going to increase. And then, as you continue to increase r, probability density decreases. It decreases again, so that you get a zero. You get a zero because the wave function is zero right there. This is our second radial node. But then, the probability density increases again. It increases because the wave function increases. Square that, we are going to get a high probability density, and then it tapers off. So, the important point here is the interpretation of the probability density. Probability per unit volume. The fact that the s wave functions are all spherically symmetric. They do not have an angular dependence to them. And what a radial node is. If you want to get a radial node, you take the wave function, set it equal to zero, solve for the value of r, and that gives you a zero. Now, so far, we have talked only about the probability density and this interpretation. We have not talked about any probabilities yet. And, to do so, we are going to talk about this function, here. It is called a radial probability distribution. The radial probability distribution is the probability of finding an electron in a spherical shell. That spherical shell will be some distance r away from the nucleus. That spherical shell will have a radius r and will have a thickness. And the thickness, we are going to call dr. This is not a solid sphere. This is a shell. This is a sphere, but the thickness of that sphere is very small. The thickness of it is dr. And, to try to represent that a little bit better, I show you here a picture of the probability density for the (1, 0, 0) state. This is kind of my density dot diagram. And then, this blue thing is my spherical shell. This blue thing, here, has a radius r, and this blue thing has a thickness dr. And so, I am saying that the radial probability distribution is going to be the probability of finding the electron in this spherical shell. That spherical shell is a distance r from the nucleus and has a thickness dr. Now, I want to point out that the volume of the spherical shell is just the surface area, here, 4 pi r squared, times the thickness, which is dr. Not a very thick spherical shell. The radial probability distribution is the probability of finding that electron in that spherical shell. It is like the probability of finding the electron a distance r to r plus dr from the nucleus. Why is that important? Well, because if I want to calculate a probability, what I can do then is take the probability density here, Psi squared for an s orbital, which is probability per unit volume, and I can then multiply it by that unit volume. In this case it was the 4pi r squared dr. And the result will be a probability, because I have probability density, probability per unit volume times a volume, and that is a probability. Now we are getting somewhere in terms of figuring out what the probability is of finding the electron some distance r to r plus dr from the nucleus. In the case of the s orbitals, I said that the Psi was a product of the radial part and the Y(lm) angular part. Remember that the Y(lm) for the s orbitals was always one over the square-root of one over 4pi. The Y(lm) squared is going to cancel with 4pi, and you are just going to have r squared times the radial part squared. For a 1s orbital, if you want to actually calculate the probability at some value r, you just have to take Psi squared and multiply it by 4pi r squared dr. However, in the case of all other orbitals, you cannot do that because they are not spherically symmetric. And so, for all other orbitals, you have to take the radial part and multiply it by r squared dr. I will explain that a little bit more next time. Okay. See you on Monday.
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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: And let me I assume, for example, that I'll put the state alpha beta in. Alpha and beta. What do I get out? So you have this state, alpha beta. What do you get out? Well, state comes in and is acted by beam splitter 1. So you must put the beam splitter, 1 matrix. And then it comes the mirrors. And lets assume mirrors do nothing. In fact, mirrors-- the two mirrors would multiply by minus 1, which will have no effect. So lets ignore mirrors. And then you get to beam splitter 2 and you must multiply by the matrix of beam squared 2. And that's the output. And that output is a two-component vector. That gives you the amplitude up and the amplitude down. So I should put BS2 here, BS1 over here, alpha beta. The numbers move away, 1 over square root of 2 and 1 over square root of 2. Commute in matrix multiplication. Then you multiply these two matrices. You get 0, 2, minus 2, and 0, alpha beta. And you put the 2 in so you get beta minus alpha. So here is the rule. If you have alpha and beta, you get, here, beta and minus alpha here, or a beta minus alpha photon at the end. Good. So let's do our first kind of experiment. Our first experiment is to have the beam splitters here. D0-- detector D0 and detector D1 over there. And let's send in a photon over here only-- 1 or input 0, 1. Well this photon, 0, 1, splits here. You act with BS1-- the matrix BS1. You get two things. You act with the matrix BS2, and it gives you this. But we have the rule already. If you have an alpha beta, out comes a beta minus alpha. So it should have as 1, here, and minus 0, here, which is 0. So you get a 1, 0. So what is really happening? What's really happening is that your photon that came in divided in two, recombined, and, actually, there was a very interesting interference here. From the top beam came some amplitude and gave some reflected and some transmitted. From the bottom beam, there was some transmission and some reflection. The transmission from the top and reflection from the bottom interfered, to give 0. And this, too, the reflection from the top and transmission from the bottom, were coherent and added up to 1. And every single photon ends up in D0. If you would put the beam-- well, Mach and Zehnder were working in the late 1800s, 1890s. And they would shine light. They had no ability to manipulate photons. But they could put those beam splitters and they could get this interference effect, where everything goes to D0. So far, so good. Now let me do a slightly different experiment. I will now put the same thing, a BS1 and a beam going in, mirror, mirror, BS2 here. But now, I will put a block of concrete here on the way. I'll put it like this. So that if there is any photon that wants to come in this direction, it will be absorbed. Photon could still go like this, but nothing would go through here. And here, of course, there might be D0 and D1. And here are the mirrors, M and M. Now the bottom mirror is of no use anymore because there is a big block of concrete that will stop any photon from getting there. And we are asked, again, what happens? What do the detectors see? And this time, we still have a 01. Now I would be tempted to use this formula, but this formula was right under the wrong assumption-- that there was no block here. So I cannot use that formula. And certainly, things are going to be different. So I have to calculate things. And we're doing a quantum mechanical calculation. Well, up to here, before it reaches here, I can you do my usual calculation. Certainly, we have BS1 acting on the state, 01, and this is 1 over square root of 2, I think, minus 1, 1, 1, 1. Yup, that B is 1, acting on 01. And that gives me 1 over square root of 2, 1 over square root of 2. So, yes, here I have one over square root of 2 amplitude. And here I also have 1 over square root of 2 amplitude. OK. Now that's the end of this amplitude. It doesn't follow. But on the other hand, in this branch, the mirror doesn't change the amplitude, doesn't absorb. So you still have 1 over square root of 2 here. And now you're reaching BS2. Now what is the input for BS2? The input is a one over square root 2 in the top beam, and nothing in the lower beam because nothing is reaching BS2 from below. This is blocked. So yes, there was some times when something reached from below, but nothing here. So to figure out the amplitudes, here, I must do BS2 acting on 1 over the square root of 2, 0. Because 1 over square root of 2 is coming in, but nothing is coming in from below. And, therefore, I get 1 over the square root of 2, 1, 1, 1, minus 1, 1 over square root of 2, 0. This time, I get 1/2 and 1/2. OK, we must trust the math. 1/2 here and 1/2 there, so 1/2 a column vector, 1/2, 1/2. OK, let me maybe tabulate this result, which is somewhat strange, really. So what is strange about it is the following. In the first case, where the interferometer was totally clear, nothing in the middle, everything went to D2. And nothing went into D1. But now, you do something that should block some photons. You block some photons in the lower path, and yet, now you seem to be able to get something into D1. There is an amplitude to get into the D1. So by blocking a source, you're getting more somewhere. It's somewhat counterintuitive. You will see by the end of the lecture in 10 minutes, that it's not just somewhat counterintuitive, it's tremendously counterintuitive. Let's summarize the result here-- the outcome in the blocked lower branch case and the probability for those events. So photon at the block-- the photon can end in three places. It can end on the block. It can end on the D0. Or it can end on D1. So photon at the block-- well, the amplitude to be here is one over square root of 2. The probability should be 1/2. Photon at D0, probability amplitude, 1/2, probability, 1/4-- photon at D1, probability, 1/4. You could put another table here-- outcome all open, probability. And in this case, there's just photon at D0. That's 1. And photon at D1 was 0.
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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: So yeah, but today, what we're going to do is talk about evolution. And in particular, we're going to complete our discussion of evolution in the presence of clonal interference when multiple mutant lineages are competing in the population a the same time. And then, we'll move on to try to think about evolution on these so-called rugged fitness landscapes. So such ruggedness occurs when there are interactions between the mutations within the organism. So a so-called epistatic interactions can, perhaps, constrain the path of evolution. So we'll complete our discussion of Roy Kashoney's paper on the equivalence principle and the presence of clonal interference. Then we'll say something about how clonal interference slows down the rate of evolution. It slows down the rate at which the population can increase in fitness because of this competition between different beneficial lineages. And then, we'll discuss this paper by Daniel Weinreich, which I think, for many of us, had a real important effect in just terms of getting us to think about evolution in a different way. Any questions before? Yes. AUDIENCE: Question about the exam. PROFESSOR: Yes. In case you've tried to forget, we have an exam again next week. So research and education indicates that the more exams, the better. So if that makes you feel any better in terms of process, then use that. So next Thursday, 7 o'clock, we will announce the room later. Did you have a question about that? AUDIENCE: Is it everything from the start or [INAUDIBLE] PROFESSOR: Yeah, so it will be weighted towards the material that we did not test through exam one. But you can expect to get two plus or minus 1 questions on material that was covered in the first part of the class. OK. Any other questions? OK. All right, so coming back to this discussion of the equivalence principle. So the last figure of the paper-- it's a figure of four-- illustrated some alternative underlying distribution. So what we wanted to know is-- all right, so they're going to be some beneficial mutations that these E.coli can get in this new environment. And they're going to distributed according to some probability distribution. So what we, in principle, want to know is this distribution of effects of beneficial mutations. So the probability of distribution, P beneficial, is a function of s. And what was, kind of, the scale of beneficial mutations that they observe in the paper. All right, how much better do the populations get after these few 100 generations? Hm? AUDIENCE: 1%. PROFESSOR: All right, so [? order of ?] percent. A few percent, right? Precisely. And indeed, they had three different models for some of the underlying distributions and how they might behave. So they had the exponential uniform and the [INAUDIBLE] kind of delta function. And in all these cases, are those the only three possible underlying distributions? Those are just, kind of, typical distributions. What you want, in this case, is you want these three degree distributions to, somehow, be qualitatively different so you can drive home the point that you can, in principle, describe the results of their evolution experiment via wildly different underlying distributions. OK. Now in all these cases, you have to specify the mutation rate, as well as the mean selection coefficient. All right, so the mean of this distribution. OK, now the question is, can we get any intuitive insight into why there were the patterns that they observe in terms of the mutation rate that they had to assume and the mean selection coefficient. There were some region of parameter space for those two distributions, for each of those distributions, that were, kind of, consistent with their data. So in particular, there was a mutation rate-- mu-- for the exponential distribution, a mutation rate for the uniform, and a mutation rate for the delta function. And also, there was a mean s associated with the exponential, mean s for the uniform and, again, a mean s for the delta. AUDIENCE: Were they trying to minimize the parameter numbers? PROFESSOR: Right. OK, so in this case, they wanted to compare these three distributions. Each of them were specified by two parameters. You could come up with other underlying distributions that might, for example, have a larger number that might be specified by a larger number parameters. But then, it's harder to compare the quality of the fit and so forth. So they chose these three distributions just because this is, somehow, the rate that these new mutations will appear in the population. And then, this tells us something about how good those mutations are. Right. Now some of you have the paper in front of you. And that's OK. But based on our understanding of how the clonal interference, kind of, manifests itself in terms of leading, eventually, they have the log of the fraction. So the fraction starts out 50-50. Log F1 over F2. So like, cyan and yellow, say? All right. So this thing starts out here. And then, one side gets beneficial mutation. So it, kind of, comes up. So they measure the slope, for example, of the lineage that is taking over the population. So they want to know, well, which of these distributions and associated parameters will be able to explain the range of different trajectories that they saw? So the question is, can we order these things? And why? All right, which one of these should be the largest, second largest, third largest, and so forth? OK. Now it's OK if you just-- well, if you had the paper in front of you, you could just read it off. But ultimately, you're going to have to be able to explain why it is that one is larger than the other. OK? So what I want to know is, for example, at what order should these things come in? All right. So what I want to do is let you think about it for a minute. And then, we're going to vote by putting our cards from high mutation rate to low mutation rate among a, b, and c. Yes? AUDIENCE: So the means are constrained to be the same? PROFESSOR: So it's really going to be some range of parameters on each of these. So the question is, why is it that in the range of parameters that are consistent with what they observe experimentally, that these things have some order? Are there any other questions about the question? I'll give you 30 seconds to think about what the mutation rate should, kind of, be in this situation. AUDIENCE: The answer is highest to lowest mutation rate. PROFESSOR: Right. So you're going to put the highest mutation rate up high, the lowest mutation rate down there, and the middle one in between. Yeah. All right. Do you need more time? All right, let's see where we are. And it's OK if you're confused or don't know what I'm trying to ask. But let me see where the group is. Ready? Three, two, one. All right. So I would say it's, pretty much, all over the place, whether people are voting something in reality or not. OK, right. So the situation is that we have the data, which is shown in figure 3A, which is a bunch of these things that, kind of, look like this. All right. So we have some times. We have some slopes. We want to know how can we understand that data that we get out. So what we're going to do is we're going to take a model in which we say, all right, we're going to start with this population that's all identical. And then, we're going to allow some mutations to accumulate. And we're going to let them compete against each other. And then, see what happens to the other. So there's going to be some, again, distribution of slopes and so forth. All right. To what degree does this sort of data constrain that underlying distribution? Between something that looks like an exponential, something that has a uniform distribution, and something that is a delta function. So that's the exponential. This is the delta. And this is the uniform. Yeah. AUDIENCE: So we measure the slope at what time [INAUDIBLE]? PROFESSOR: Yeah, it could. That's right. So this thing could turn around at various times. So I think that there are a number of different ways that you could argue about the right way to do this. In practice, I think it's not going to be very sensitive because there's a minority of them that will actually be turning around, for example. So you could, for example, just say all of the trajectories that cross some point, I mean, measure the slope. And I think that would be sufficient. AUDIENCE: But if you don't see this fraction turn over, you could still have clonal interference? PROFESSOR: If you don't see the fraction. AUDIENCE: Like, in the sense-- PROFESSOR: That's right. So even if you don't see these things like, the flatten out, for example, then you could still have clonal interference because the slope might still be steeper than it would be in the absence of clonal interference. All right, what I'm going to do is I'm going to let you discuss with a neighbor for one minute. And then, we'll, maybe, discuss as a group just because I want to make sure that everybody gets a chance to try and verbalize their thought process. And if we discuss in a group, then only a few of us get to. All right, so one minute. Try to discuss it with your neighbor. And then we'll reconvene. [SIDE CONVERSATIONS] Yeah, and we're going to discuss the means in a moment. So indeed, these distributions will not end up having the same mean s. AUDIENCE: What are you controlling? PROFESSOR: What we're controlling is that we're asking about what range of parameters for each distribution will adequately fit the data. AUDIENCE: I know [INAUDIBLE]. Does it depend on what you get? PROFESSOR: The data will be, basically, the initial slopes here and when they deviated from a 50-50 mixture. AUDIENCE: OK. PROFESSOR: All right, so that's what [INAUDIBLE] is those histograms. AUDIENCE: Yeah. OK. [INAUDIBLE] PROFESSOR: All right, so it seems like we've quieted down, which means that we all agree on the answer. Is that-- no? OK, well I think that this is, actually, pretty tricky. So that's fine. I just want to see where we are, though. All right. Reconfigure your cards. Your best guess for the orders of the mutation rates between exponential uniform and delta. All right, ready? Three, two, one. OK, so we're migrating towards some things. OK, great. And can somebody verbalize the answer that their group got? AUDIENCE: So our answer is A, B, and C. PROFESSOR: OK. AUDIENCE: [INAUDIBLE] exponential [INAUDIBLE]. We can not see most of the [INAUDIBLE] lower selection coefficient mutations. PROFESSOR: OK. AUDIENCE: So we're actually underestimating the mutation rate from the data. PROFESSOR: Underestimate. OK, no, I can see what you're saying. OK, yeah, so the idea is that you're saying that we don't see an awful lot of the mutations here, which means that the true mutation rate, the underlying mutation rate is, somehow, much larger than you would have thought based on the mutations that you actually see here or something. And there's maybe another. OK, so they're different. All right. OK, it's certainly along-- yeah, sometimes it's true. And then, of course, there are different ways of saying this. Yes? AUDIENCE: Yeah, same answer but a slightly different way of thinking about it. If you're just randomly sampling any of these distributions, then your sample drawn from the exponential distribution. It's going to be low selection [INAUDIBLE] more often than it's going to be for the other ones. Like the delta is [INAUDIBLE]. PROFESSOR: That's right. Yeah, yeah, yeah. AUDIENCE: [INAUDIBLE] every time the uniform. It's going to be equally likely to be a high selection coefficient as opposed to a low selection coefficient. But with the exponential distributions, your most likely to be a low selection coefficient. So you want more mutations. PROFESSOR: That's right. You, somehow, need more mutations of that exponential in order to sample out there. Right? So which one is going to have a more clonal interference? Which of these distributions will end up having the most clonal interference after you fit the data? Yeah? AUDIENCE: The one with the highest mutation. PROFESSOR: The one with the highest mutation, right? Kind of has to. Of course, and even though some of those mutations are going to be loss, still it's going to have the most colonel interference there. And indeed, if the underlying distribution were modeling as a delta function and, in their [? fit ?], what they got was that this might be around 5 and 1/2%, I think. Yeah, so [? 5 to 5 and 1/2 ?]. OK. So here, this guy was around 0.055. Between 5 and 5 and 1/2%. So what they're saying is, all right, well you could explain all of our data just by assuming that there's some mutation rate where, periodically, some individual gets a beneficial mutation that is a 5, 5 and 1/2%. And that could, in principle, be used to explain the base features here, how long have to wait before anything happens, and the slope when something starts happening. So the histogram that they plot is actually, somehow, this initial this initial slope once you start seeing it deviate from 50/50. OK? All right. But their point is it that that does not prove that the underlying distribution is a delta function with some mutation. And indeed, to explain the data with a delta function, you don't actually don't need any clonal interference. Right? You just say, OK, well somebody gets a mutation. It's 5%. And eventually, it's going to spread. And that's what we see. If you want to explain the later dynamics of flattening out and so forth, then you have to allow the other lineage to get a mutations, as well to cause a flattening. But as far as the base dynamics of when you leave the 50-50 in the initial slope, you don't even really need to have any clonal interference to explain their data with a delta function underlined. And that's why you also can get by with a very low mutation rate because you don't really need much in the way of competing lineages. Yeah? AUDIENCE: But what if the slopes are [INAUDIBLE]? PROFESSOR: Yeah, yeah. No, right. So you're not actually going to get the true distribution of slopes. But their argument is that a lot of that could just be noise and measuring the slopes and so forth because, if everything is a delta function, then you would start out by just getting one slope, unless you offer multiple mutations on a lineage. And then, things could get more complicated. But yeah. In this case, all of these guys would have the same slope. But that's, at least, a reasonable first order approximation to the data. However, as you move to these distributions in uniform and exponential, you're going to need more and more clonal interference to, kind of, explain the data. So you'll need higher and higher mutation rate. What's interesting is that you also have a lower and lower mean s. OK? AUDIENCE: Can you just explain why you need to explain the data? PROFESSOR: Yeah, sure. And I think that drawing these underlying distributions is really helpful. So first, we're going to draw the delta function. That, kind of, makes sense that you can fit everything just by assuming 5, 5 and 1/2%, right. So what we're going to do is draw the various P. So I drew those distributions. But they weren't necessarily to scale. i.e, they didn't necessarily have the proper mean selection coefficient. What we can do here is we can draw this is the mean s of the delta function, which was 5, 5 and 1/2%. All right, we got this guy here. Now the question is can we describe the data using a uniform distribution with the same mean selection coefficient? All right, so we're going to have you vote yes and no. And if you say no, then you have to say what's going to go wrong. All right? The question is can we just use the same mean selection coefficient for our uniform distribution. AUDIENCE: [INAUDIBLE]. PROFESSOR: Right. So what we're going to do is we're going to measure the experiment where we would have, say, 96 different evolutionary trajectories where we measure the time that it takes for something to happen and then the initial slope. And we're going to take a histogram of those things, and compare it between what we would get in the model with what we got experimentally. All right, the question is can we use the same mean s for a uniform as we did for the delta function. And if you say no, you have to say why not. A is yes. B is no. Ready? Three, two, one. All right, so we have a bunch of no's. Maybe a few yes's. All right but then some of the no's, it's incumbent on you [INAUDIBLE]. So I don't know. So yes, so one of the no's, why not? AUDIENCE: So if we have [INAUDIBLE] if it has the same average, then there are going to be outliers that are more better for the [INAUDIBLE]. PROFESSOR: That's right. So the problem here is that, if we use the same mean selection coefficient pretty for the uniform, then what we're going to end up with is something that comes out twice as far. AUDIENCE: Wait. But the delta function is just a uniform distribution with 0. AUDIENCE: Yeah. AUDIENCE: So you could always just fit it with [INAUDIBLE]. PROFESSOR: Yeah, we're assuming it's a uniform distribution that starts at 0 and then goes out to some [? amount ?]. Yeah. So the idea is that-- I mean, whatever. That's the model. And I think it's a reasonable model because we know that there are a lot of mutations that have little effect. So it makes sense for a distribution to start at 0, if you're going to have something like a uniform. Right? And the problem with such a uniform distribution is that because there's going to be some clonal interference, what that means is that you're going to be, kind of, weighted out here. And that means that you'll, kind of, see mutations that are out here around 10%, instead of around 5%. So what you actually want, then, is something where the mean selection coefficient is around half of what you had as the delta function. So you want something that really looks more like this. And that's actually why, if you look at the data or if you look at this figure, then the area that works for the uniform distribution has a mean/coefficient of 3%. So this thing comes out to around 6% here. So just beyond the s correspond to the delta function. So this is the delta. And this is the uniform. So here is around 6%. And the mean of this is [? at half ?] of that. Right? What's happening is that there's some clonal interference. And you only need a modest amount of clonal interference because if you sample from a uniform distribution just a few then, the most fir one will be around here. And it'll already be relatively peaked. And of course, there's also this issue that you have to survive stochastic extinction. So that amplifies the effect further. So really, if you just have, say, two mutations sampled from the uniform that survives stochastic extinction, you're already going to get something that's peaked around there. Does that make sense? Yeah. AUDIENCE: But what if we [INAUDIBLE]? PROFESSOR: So then, the question is what exact mutation rate do you need. And basically, you need a high enough mutation rate that you get some mutations and that you can have some clonal interference. And the question is, what prevents you from having a mutation rate that's too high, maybe? AUDIENCE: My question was why do you need clonal interference to explain the data. PROFESSOR: Ah, right. AUDIENCE: I mean, of course, you have these [INAUDIBLE] focused on what happened early on. PROFESSOR: That's right. Yeah. Right, OK. So in there it comes down to how peaked this distribution of slopes is because there are very few shallow slopes. And of course, then, it gets into questions about quality of data and so forth. And that's more subtle. But certainly in principal, in the absence of clonal interference, you would have some fair number of shallow slopes. It would be under represented just because of the stochastic extinction business. But still, you need some to explain the, sort of, peakiness of that distribution of the slope distribution. Now the exponential is interesting because it's in a very, very different regime. So I just want to show this is mean s for the uniform. And actually, if you look at the figure of the mean s for the exponential, it's down there around 1%, which is a little bit surprising because what this is saying is that, if you look at this distribution, this initial slope, kind of, extends down here. And you have something that falls off, dramatically, though, right. OK. So how is it possible that you could use such a distribution that's peaked over here that's so far over on the left and still explain the same data? AUDIENCE: [INAUDIBLE]. PROFESSOR: That's right. So it's true that I've drawn this, kind of, around 0. But the exponential and principle goes to [? infinity ?]. It's just that it falls off exponentially. But what this means is that you're sampling pretty far out on the exponential in order to get the same mean effect. So you're actually going out to five or six times this characteristic s. So you're talking e to the minus 5. Right? That means there's a lot of clonal interference that has to be happening in order to explain this, right? But why is it that it's way over here? I mean, why not just use an exponential with an s that's more like 3, 4, 5, 6%? What's that? Well OK. But, you know, I mean, this is just a model. I can do whatever I want because, actually, they fit their data with these models. So realistic just means that it explains their data. So there's nothing a priori wrong with saying, oh, here's an exponential with a characteristic fall off of 5%. That's, in principle, fine. I mean, it doesn't work, for some reason, but we have to figure out why. Yeah. AUDIENCE: [INAUDIBLE]. PROFESSOR: OK, right. So you're likely going to get a low selection coefficient, OK. Is that the problem? AUDIENCE: [INAUDIBLE]. PROFESSOR: OK, that's true. But I guess the question is why can't we use? AUDIENCE: [INAUDIBLE]. PROFESSOR: Two mutations? Two clusters? What do you mean? AUDIENCE: [INAUDIBLE]. PROFESSOR: OK, the time between two. OK, that's an interesting statement. Although, the region or parameter space that they claim works is actually a region where we have a really high mutation rate. Orders of magnitude higher than the other two distributions. So in that sense, the time between mutations being established is really small because they're saying that, oh, if you want to fit the [INAUDIBLE] exponential, then you have to assume lots of clonal interference. So that means the time between successive establishments or mutations is really, actually, very short. So that's actually the regime where they claim works. So the question is why is it that we can't go to this other regime? In this figure that they make, why is it that the allowed mean coefficient doesn't extend out to over there? Yeah. AUDIENCE: [INAUDIBLE]. PROFESSOR: Right. AUDIENCE: Because the exponential is [INAUDIBLE] that [INAUDIBLE]. PROFESSOR: That's right. Yeah, that's right. The point is that, to explain their data, you need something to cause this distribution to get more peaked over here, which means you need to have a fair amount of clonal interference. But that means you need to have a high mutation rate. But once you have that high mutation rate, if you have an exponential that comes out here, then you would actually sample way out here, as well. So the only way to get a peaks distribution around here is to have it so that the exponential is really suppressing those really good mutations. But you have a lot of clonal interference that, kind of, pulls things out. Now there still is a fair range of parameters that work here. You know, it goes from, say, half a percent up to, maybe, 1 and 1/2% in terms of this mean selection coefficient. And the mutation rate that works then changes. So as you got a larger mean selection coefficient, the mutation rate that is compatible goes down because you need less clonal interference. So that's why, if you look at this figure for the region that works in terms of the mutation rate for the beneficial mutation and the mean s, there's some region that looks, kind of, like this that works for the exponential. And this is, actually, a big range. So this is, actually, a factor of 100 in mutation rate that would be compatible. And then, a factor of, maybe, three in mean selection coefficient. So there's some range of parameters that work. And you can understand why it is that this thing has to be shaped the way it is. Does that make sense? Yeah. AUDIENCE: [INAUDIBLE]. PROFESSOR: Right because the uniform is out here. And then, the delta function is here. So this is delta, uniform, and exponential. And you're just saying that it all, [? kind of ?], following. Is that-- yeah. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, well maybe some power law fall off would do that. AUDIENCE: [INAUDIBLE]. PROFESSOR: There's nothing there's nothing magic about these three regions. And it's not that we're claiming that the mean selection coefficient cannot be in here. It's just that, if you pick these three underlined distributions, you get this range of different values that would work. So if you chose other underlying distributions, you could get other blogs. But it's true that there is a general trend that, the higher the mean selection coefficient, the less clonal interference you need or want to explain the data. AUDIENCE: [INAUDIBLE]. PROFESSOR: Right. Why can't it go higher or lower? Yeah. I don't know. A factor of 100 isn't enough for you? Yeah, it's a good question. I'd have to think about it to figure out which-- because there's going to be a different effect on each side, presumably. Yeah, and all of these distributions, there's some floor just because we know that you have to get these beneficial mutations within the first few tens of generations. Otherwise, the mutation wouldn't have gotten a chance to spread when it did. So that means we know that there's going to be a lower bound on the mutation always because, if it's too low, then we wouldn't have gotten the mutations in time. And now, in terms of why it can't be higher, yeah, I'd have to think about it. Are there any other questions about this paper? I think it's a challenging paper, kind of, conceptually/mathematically. But I think it's interesting because it does get you to think about this process of clonal interference in new ways. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah. AUDIENCE: So is that enough? Like, is this condition rate known [INAUDIBLE]? PROFESSOR: Oh, I see. Oh, that's a good question. Yeah. Yeah, so I think that the mutation rates that-- first of all, this is not a per base pair mutation rate. This is the rate that you get beneficial mutations. So I'd say, the numbers are not ridiculous. But it's not that you can just take the known per base [? pair of ?] mutation rate and say, oh, well it has to be here because an awful lot of them are deleterious. And it's very sensitive because the mutations that are occurring around here don't really matter for evolution because they tend not to survive stochastic extinction. They're not going to survive clonal interference. Yet, it could be a big part of the distribution. Right? It could be that the majority of the mutations-- for example, in the exponential-- is there. So you can actually have very different distributions that don't really change the evolutionary process but would change the rate of beneficial mutations. So I think the numbers are not ridiculous. But it's hard to constrain, actually. What I want to do is, before talking about these rugged fitness landscapes and the Weinreich paper, let's just say something about the rate of evolution. And when we say rate, we're referring to the change and the mean fitness of the population with respect to time. All right, so this is the rate of evolution. So this is the change in the mean fitness. Delta. Delta mean fitness divided by delta time. So what we want to do is just start by thinking about a situation where we assume that we're not all running out of new mutations that are good for us. All right, so what we can do is just assume that, at some rate, mu-- so there's at rate mu beneficial, we'll say. We sample from some probabilities distribution of beneficial mutations. And then, something happens to cast an extinction. Maybe clonal interference and whatnot. But someone of them will fix. And then, that increases the fitness. And then, for now, we'll just assume that the fitness is add. But for small S's, it doesn't matter whether we're thinking about [? fitness as ?] adding or multiplying. You guys understand what I just said there? So if you get a mutation that has affect S1 and maybe the right way to think about this is that, if you get a mutation s2, then the [? fitnesses ?] perhaps , should multiply as the [? mu ?] model. But this is, of course, for a small s1 and s2, this is around 1 plus s2. So for small S's for short times, maybe we don't need to worry about this because this is, for s1 and s2, much less than 1. Yes? AUDIENCE: My intuition would be that, if you already had a 5% increase i [? fitness ?], it would be harder to get [INAUDIBLE]. PROFESSOR: That's right. So eventually, that certainly is going to be the case that we're going to run out of these beneficial mutations. But for the first few thousand generations, it's roughly linear. So eventually, it does start curving over. But maybe not as fast as you would have thought. And at the very least, this is a good [? no ?] model. And then, we can, of course, complicate things later. But for now, we'll just assume that you always sample from the same probability distribution of beneficial mutations, just for simplicity. The question is, how fast will the fitness of the population increase with time? OK? Do you guys understand the question? All right, let's start by thinking about the regime where mu b N is much less than 1. So very low rates of mutation relative to the population size. Now you might recall-- for clonal interference. So clonal interference not relevant. What we found last time was it required that the time that it took for mutation to fix had to be much less than the time between successive establishments of these beneficial mutations. And this was 1 over s log N's. You should be able to [? derive ?] both of these. This and that and this, and the next step, as well. All right, so you can ignore clonal interference if this is true. So let's say that we can ignore clonal interference. So for small population sizes or in the limit of low mutation rates. What we want to know is the rate of evolution. How will it scale with various things? I'll go ahead and have us vote for-- OK. How does it scale with, [? particular ?], both, the mutation rate and the population size? It's proportional to what? Holding another thing is constant. Holding, for example, the distribution and n constant. OK? Do you understand the question? Yes. AUDIENCE: [INAUDIBLE]. PROFESSOR: OK. So this is just the rate of beneficial mutations. And this is just to the 0th power. i.e, it doesn't defend. AUDIENCE: Oh, OK. PROFESSOR: Linearly, it's squared. All right, ready? AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, assuming that clonal interference is not relevant. So assuming, for a small mutation rate, how does it scale in mutation rate? OK, ready. Three, two, one. All right, so we got a lot of B's, which is nice. So this is saying that, indeed, if mutation rate is small, then there's going to be some rate that mutations enter into the population. They may or may not survive stochastic extinction. But if they do, and that doesn't depend on mutation rate, then they can fix. And in this low mutation rate regime, they don't compete with each other. In which case, if you double the rate of these things entered into the population, you'll double the rate that they get established. And you'll double the rate that these beneficial mutations will fix in the population So you double the rate. OK? All right, now as a function of n. Does it go as N to the 0 or other? We got A, B, C, D. All right, I'll give you 15 seconds to think about it. Ready? Three, two, one. OK, so now we're getting more disagreement. All right. And I'd say, largely, A's and B's. All right, there's enough disagreement. Let's go ahead and spend just 30 seconds. Turn to your neighbor. [SIDE CONVERSATIONS] Per individual, yeah. [SIDE CONVERSATIONS] All right, let's go ahead and re-vote just so I can see where we are. How is it that the rate of evolution in this regime, no clonal interference, how is this going to scale the population size? Ready? Three, two, one. OK So, I'd say that we have not really convinced each other of anything. All right, so A's B's? Somebody explain the reasoning. Yeah. AUDIENCE: [INAUDIBLE] mutation rate [INAUDIBLE] larger the population, the larger we're going to get [INAUDIBLE] population. PROFESSOR: OK. All right, so if you double the size of the population, you'll double the rate the new mutations enter into the population. AUDIENCE: And the other important thing is that the fixation time just goes, like, [INAUDIBLE]. PROFESSOR: OK, right. So then there's the fixation time business. So how is that relevant? Yeah, somebody that said A, what was your partners reasoning? Or where you convinced by this argument? Or confused. AUDIENCE: Well if it's [INAUDIBLE], then the [INAUDIBLE]. PROFESSOR: Ah. Right, so if it's a nearly new [INAUDIBLE] mutation, then it's true that the [INAUDIBLE] would be 1/N but-- AUDIENCE: But if N is very large then, sometimes, s will not [INAUDIBLE]. PROFESSOR: OK, right. But now you're invoking N being large, which I don't think we necessarily want to do. I mean, I guess there are a couple things to say there. One is that the mutations that are really nearly neutral will not have a very significant effect on the fitness. And within that regime, I think, yeah, you'd have to check what happens there, right? But I think that, in most of these cases, small population, we're often saying population [? must be ?] 10 to the 4 or so. In which case, the nearly neutral mutations are not very relevant. Right? So indeed, over a broad range of conditions, in this situation, it's going to scale as N. So the rate, so far, it's going to equal to there's a mu b times an N. And that's basically because the rate the new mutations enter into the populations is mu N. And the rate that they get established is just mu N's. This is, indeed, what this calculation is telling us. Now the question is, how much of a fitness gain will we get? How is it going to scale with this probability distribution. So this probability distribution will give us some function of s. So this distribution has to be relevant. Do we agree? So we want to know in what way is it relevant. Is it relevant via the mean, via the mean squared, the mean cubed-- I don't know-- the mean squared, or other? This one is harder. So it's worth spending, I'll give you, a full 30 seconds to think about it. So the question is, how is it that the probability distribution will enter into the rate of evolution in this situation? OK? Question. No? AUDIENCE: Can [INAUDIBLE]? PROFESSOR: I'm sorry, multiple what? AUDIENCE: Is it possible that it's entered in multiple situation? PROFESSOR: Oh, yeah. You know, this is why I gave you flashcards. You can put up any combination you want. Another 15 seconds. This one is trickier. Let's go ahead and vote. Ready? Three, two, one. All right, so we got a lot of A's and B's. Some, maybe, C's and D's. All right, so yeah. We're pretty far. OK, we're all over the place. All right, so there's going to be some distribution. There might be, for example, probability distribution function of s, function of s. We want to know it's going to something that's going to fall off in some way. Maybe exponential. Maybe something else. Now to keep track of this, what we need to do is remember that the probability of establishment goes as s. The probability of establishment. And the [INAUDIBLE] model is equal to s. In other models, it might be 2s. But it's around s. And then, the question is, how much of a benefit will you get if you do establish? All right. And that is, again, going to go as-- and whatever the delta fitness is actually, again, equal to whatever acid is that you sampled. And the given mutation that appeared, you sample somewhere. It has to, both, survive and well- if it survives, then it gives you an s. So what that means is that, actually, you end up averaging s squared of the distribution to determine the rate of evolution. If it was just a delta function at s, then it's just an s squared. So then you think, oh, it's going to be mean of x squared. Right? But if it's a delta function, then the mean of s squared and the mean s squared are the same thing. But in this situation, it's just useful to play with some different distributions and see how it plays out. But in this case, it is, indeed, mean s squared. So the rate of evolution in the limit of no clonal interference is, actually, rather simple. It just goes as mu N. But then, you have to take the expectation of s squared of whatever this distribution is of underlying mutations. OK? But of course, this is going to break down at some point, as everything does. And can somebody remind us why it is that it's going to break down? Clonal interference. Perfect. And we know exactly where clonal interference is going to start being relevant here. In particular, what we can imagine drawing is something that's [INAUDIBLE] the rate of evolution as a function of N. So rate is a function of N. And we might even want to plot in a log, log scale. So we'll say the log of the rate. The log of N. And at the beginning, what do I draw here for small n? A line with slope 1. OK. If i had depended on n squared, then what would I be drawing here? A line with slope 2. Yep. Don't mess that up because it's really easy to. OK, so it's a line with slope 1. Right? So at the beginning, for small population size, if you double the population size, you should double the rate of evolution. But this can't go on forever, and it won't. So it's going to curve over somewhere. all right. AUDIENCE: How can the rate go down [INAUDIBLE]? PROFESSOR: the rate goes down relative to what it would have if it had continued. And that's because you're wasting some beneficial mutations. With clonal interference, what's happening is that you have a great mutation over here but also a great mutation over here. And only one of those great mutations can win. So with clonal interference, you're, somehow, wasting some of the beneficial mutations that you acquired. AUDIENCE: Even though you're always taking the maximum because what you're doing is you're [INAUDIBLE]. PROFESSOR: Right because the alternative would have been to take the sum of them, which is what happens over here. If they're not competing, then you got one. And you get the other. And you just get higher and higher. With clonal interference, you indeed take the maximum. But that's lower than sum. And that affect just gets worse and worse as you get to larger population sizes. So although it's relatively simple to calculate the rate of evolution in the limit of small populations, the rate of evolution in the case of clonal interference is, actually, a very hard problem. Hard, well, experimentally, theoretically, and in all ways. I just want to say a few things. There have been some really, I think, interesting studies occurring over the last 10 years trying to get at this regime. So the question is how should it behave. And there's a paper by Desai, Fisher, and Murray when they were all at Harvard. Since then, Daniel Fisher, the [INAUDIBLE], has moved to Stanford. Desai went to Princeton but then came back to Harvard. So he's now Harvard faculty. So this is current biology in 2006, '07, '08. It's 2007. So they have, kind of, a simplification of this where they asked let's just assume that we don't have a probability distribution of beneficial mutations. Instead, let's just assume that there's some mutation rate to acquire beneficial mutation that's exactly s. OK? So you say, all right, well that sounds super simple. Of course, it's not true. But even that problem is hard. But what they can do in this regime is that then it's nice because everything's, kind of, discrete because then the population is going to described by a series of-- so this is abundance as a function of the fitness. So this is, maybe, the bulk minus fitness relative 0. Here, this is s, 2s, 3s, 4s, and maybe there's a little bit here at 5s. OK? So what happens is that there's going to be some equilibrium distribution of this front or the nose of the population. And at some rate, these guys get mutations where some individual, kind of, comes like this because it gets a mutation that's beneficial. But that doesn't actually affect the dynamics very much because all these guys are growing exponentially. What's really relevant are when individuals that have several or more of these beneficial mutations than the rest of the population, when these guys get a mutation that they can move forward. And it's this dynamic that is, kind of, pulling the population here. So you can actually do analytic calculations on this model that you would not be able to do if you did a full distribution of mutations here. But it's actually still complicated and hard. And I do not claim to have gone through the full derivation because the full thing is they get that the velocity in this model approximately equal to. By then, it's an s squared. All right, so you might sneer at this model and say, oh, well, you know, this is an oversimplification, blah, blah, blah. But even this is hard. And you get a complicated expression. And it's describing something fundamental that's happening in these populations with clonal interference. The important thing is that you can-- more or less, this term is going to be the dominant one in many cases, especially because we're interested in the large end regime here. So what you see is that, for large N in this model and they did experiments in this paper that are consistent with it, they find that the velocity-- the rate of evolution-- this is the rate. You see her the s squared term that we already talked about. Right? And there's no probability distribution, so it's just an s squared. But what you see is that it ends up going as log N for large N. So this is when there's a lot of clonal interference. So maybe this thing goes as log N. But I'd say, maybe, this is not an open and shut case because real populations are more complicated than this because it's not that every mutation has magnitude s. But I think it's a reasonable first order model. And I think this is a nice set of calculations to make sense of things. Are there any questions about that calculation or why this thing curves off [INAUDIBLE]? Yes. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah. AUDIENCE: [INAUDIBLE]? PROFESSOR: Sure. OK, so the question is why is it that it's average s squared instead of average s squared? So what is it that's useful to say? I'm trying to think of a nice probability distribution where those things will have very different-- AUDIENCE: [INAUDIBLE]? PROFESSOR: For a delta distribution, they're the same thing. So if we wanted to do an intuitive explanation, maybe, [? probably ?]-- I'm trying to think if we had two delta distributions. So the mean s would be here. Yeah, OK. So imagine that-- I know that this is not a real beneficial mutation so maybe but OK. But it's something that's small. You know, negligently small here and magnitude s over here. So the mean s is, indeed, equal to this s 0 over 2. Right? Whereas, the mean of s squared is, then, s squared 0 over 4. Are we looking at something different? Oh, no. OK, I think I'm going to have to come up with a better example to answer. So maybe after class we can come up with one. I don't want to take five minutes finding a good explanation. Yeah? AUDIENCE: [INAUDIBLE]? PROFESSOR: Yeah, so this is the rate of evolution in the regime of large population size. AUDIENCE: OK. PROFESSOR: And I don't know how small you have to go before [? funny ?] [INAUDIBLE]. AUDIENCE: That's only for the [? large ?] [INAUDIBLE]. PROFESSOR: Yep. Yeah, and, of course, you could imagine that taking various limits is complicated here. But the important thing is that the rate goes as log N for, in the case, a clonal interference because that's the key thing to remember, besides the fact that they actually had to work to analyze this model. OK, so what I want to do is, in the last 10, 15 minutes, to talk about this [INAUDIBLE] paper because I think it's pretty. And I mean, it's an elegant example of how, if you look at a problem in a different way, then you can get, I think, really interesting insights using a minimal amount of, like, measurements. I mean, basically, how many measurements are in this paper? AUDIENCE: 32. PROFESSOR: Basically, 32. Right? So what they are doing is they're analyzing mutations in the enzyme beta lactamase or the gene encoding beta lactamase, which confers resistance to beta lactam drugs like, in this case, cefataxime OK, so this guy gives resistance to these beta lactam drugs that are like ampicillin or penicillin. In this case, cefataxime. but not all of the versions of this gene or the enzyme, actually, can break down this new drug. So all 32 versions of the enzyme that they study break down, for example, ampicillin. But they had widely varying levels of resistance or ability to break down this drug, cefataxime. So they wanted to try to understand something about what happens if you start out with this base version of the enzyme. You know, if you just look up what's the sequence for beta lactamase. What's going to be the sequence? And that we're going to call the minus, minus, minus, minus, minus. And we might, reasonably, want to know how does it get to the version of the enzyme that has all five of these [? point ?] mutations? Does anybody remember what those five mutations were? Like, what kind of mutations are they? AUDIENCE: [INAUDIBLE]. PROFESSOR: They're all [? point ?] mutations. And are they all protein coding mutations? No. So actually, this one here is actually a promoter mutation that increases expression by a factor of two or three. Whereas, these things here are indeed protein coding and change the amino acid. Protein coding. The amino acids in the end. Well they each change one amino acid in the resulting protein. So what Weinreich in this paper was trying to understand is what is the shape of these fitness landscapes? And what does that mean about the course of evolution or the repeatability or predictability of evolution? And I just want to stress this is the Weinreich 2006 because this version of the gene/enzyme is, essentially, unable to break down cefataxime at all. So E-coli that has this version of enzyme, it's almost as if they don't have any enzyme at all. Whereas, this version of the enzyme is able to break down this drug, cefataxime, at very high rates. And indeed, the way that these things are quantified in this paper is we have what's known as the MIC or the Minimum Inhibitory Concentration. And basically, you just ask-- oops, inhibitory concentration-- what's the minimum amount of the antibiotic that you have to add to prevent growth of the bacterial population after 20 hours starting from some standard cell density, OK? So it's a very easy experiment to do because, 96 well [? plate ?], you just have many wells. And you just go down a concentration, maybe, by a factor of a root 2 each time. So then, you go across 12 or 24. And you get over a broad range of antibiotic concentrations. And what you should see is that this is, maybe, dividing by route 2 each time. So you get growth here. You get growth here. Growth here but then no growth, no growth, no growth. And the concentration and that you added here is the MIC. What you'll see is that, depending on the version of the enzyme that the bacteria have, the growth will occur up to different concentrations. AUDIENCE: Why would you [INAUDIBLE] PROFESSOR: Why would you? Why is that-- AUDIENCE: [INAUDIBLE]. PROFESSOR: Oh, I'm just telling you what they actually did experimentally in this paper. You could do a factor of 2. Or it's just a question how fine of a resolution you want. How to do root 2? OK, so this is like a mathematicians question, all right because your point is going to be that-- AUDIENCE: [INAUDIBLE] route 2. PROFESSOR: OK, so it's true. Root 2, it's an irrational number. It's the first proof in a analysis textbook. It doesn't matter, OK? Our error in pipe heading is a percent, which means that if you do 1.41, that's fine. Yes. So don't be paralyzed by petting a root 2. OK? AUDIENCE: Is it always very sharp, this changing in growth? PROFESSOR: You know, biology and the word always should never be used in the same sentence. I'd say that it's a reasonable [? assay. ?] It's, typically, sharp. It happens, though, that you get growth here. And then, you get stressed out because you don't know what to do. I mean, the important thing is that you do the experiment multiple times and you have some reasonable rule for treating these things. Yeah, it can be more complicated though. All right, what we're going to use in the context of this paper, though, is we're just going to assume that this MIC is a measure for fitness. The mapping from MIC to fitness is, actually, very nontrivial. Something my group has spent a long time thinking about. But for the purpose of this paper, just when you hear MIC, you can just say think of it as fitness. OK, [? higher ?] MIC, he assumes that it could be selected for by evolution. So there are, in principle, 2 to the five different states. Different versions of this gene. So what he did is he constructed each 2 to the 5 versions of the gene, put them into the same strain of E-coli, and then measured the MIC of each of those 32 strains. And that was all the measurements in this paper, basically because everything else is just analysis of that resulting fitness landscape. But what's exciting about this is just the ability to have an experimental fitness landscape because we've talked about fitness landscapes for years. But then, it tends to be much more like what you saw in Martin Novak's book, right? That you can think about these fitness landscapes. And you can do calculations of what should happen on them. But this is a case where we can actually just measure something akin to a fitness landscape, and to try to say what it means. So you can ask questions about how rugged is the landscape. How many different paths can you take from this version to this version? So first of all, how many peaks were in this landscape that he measured? AUDIENCE: [INAUDIBLE]. PROFESSOR: One peak. This is important. This was the one and only peak. And when you read this paper, you come away thinking, oh yeah, this is a really rugged landscape, right? Many of the paths were not allowed by so-called Darwinian or selective evolution. But it's easy to forget that, actually, it's not that. It's a moderately rugged landscape because there was still one peak. In particular, if you just assume that the population starts at that minus, minus, minus, state, starts travelling uphill in fitness, gets a mutation and goes uphill, is there any possibility for it to get trapped in a non-optimal? No because there are no other peaks. So there's only one peak. That's the same thing as saying that you can take any path you like going uphill, and you will always get to the same final location. You will never get stuck anywhere. I mean, it does not mean that you can take any old path that you want. Many of the paths may be blocked in the sense that they may go downhill and so forth. But at any location you're at, there's always, at least, one path going up in fitness, up in MIC. Doesn't matter which path you take. You will always be able to get to the peak of this landscape. OK? So it's not too rugged of a landscape, in that sense. Yeah? AUDIENCE: Wasn't it, [INAUDIBLE]? PROFESSOR: Right. So some paths are not allowed in the sense that some paths decrease fitness, locally. But what I'm saying is that you can take a different path that goes up in fitness. And you'll still get to the same peak. AUDIENCE: Right, but you started in that [? wrong path ?] and you [INAUDIBLE]. PROFESSOR: Well no, no. The thing is that you can't take that path because that path goes down in fitness is the claim. So the statement from this paper is that, if we just assume that the only mutations that can fix in a population are mutations that increase fitness, than it does not matter which of those beneficial mutations you take because you will always end up reaching the peak. So there are 120 possible trajectories. Can somebody say how we got 120? AUDIENCE: [INAUDIBLE]. PROFESSOR: Right. So this is 5 factorial. What he found by analyzing the resulting fitness landscape is that 102 were selectively inaccessible. Oh, I don't know how many-- is that the right way to spell that? Yeah. And what he's assuming is that the only mutations that can fix are beneficial mutations. OK, right. In particular, if you have two states, there's a mutation that you could acquire. From here, let's say you get this mutation, and that just leads to the same MIC. He assumes that that is inaccessible. All right? You know, and of course, like all things, you can argue about it. What he's saying is that it won't fix in reasonable times, which is fair if it's really a neutral mutation. In particular, if you have 10 to the 6 bacteria, and if the mutation rates are the same everywhere and some mutations lead to a significant increase in fitness, then a neutral mutation would be unlikely to fix. So what he found is that there were only, then, 18 of the 120 paths off this fitness landscape that had monotonically increasing fitness values. And indeed, if you analyze those trajectories, what he found is that, actually, 18 [? isn't ?] maybe even over [INAUDIBLE] because only a few of those trajectories would likely occupy majority of what you might call the actually observed paths just because of the statistics of when those paths branch and so forth. So the argument from this paper is that we can measure fitness landscapes. And from it, we can say something about the path of evolution, perhaps. Other people have since gone and done, actually, laboratory evolution on a different antibiotic resistance gene-- again, Roy [? Kashoney ?], actually-- to confirm that these landscapes, at least in some cases, can inform laboratory evolution. So there is a sense that maybe evolution is more predictable than you would have thought. We're out of time. But if you have any questions, please, go ahead and come on up and ask them. All right? Thanks.
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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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RAFAEL JARAMILLO: All right, so today is the last lecture of new material. We're going to talk about reactions between gases and condensed phases. So we're going to talk about that, and then, we'll make it specific to oxidation. And then, on Friday, I'll do an extended practice problem, I suppose, on reactions between gases and condensed phases. So in general, some reaction-- I'm going to write this really generally. aA plus bB going to cC plus dD. And these are not just for ideal gases. So the last time we've seen reaction like this in 020, it's been for ideal gases reacting. And we're going to move away from that and do this more generally. So let me just quickly remind you of how we get to the action coefficient. We have a general expression for the change in Gibbs free energy with these unconstrained internal variables, and we can re-express this in terms of the reaction extent, remembering that DNI over mu I is a constant for all I reaction extent. I don't need to have an I there. Just xe. And reminding us that mu of I, the chemical potential of a given species is its reference chemical potential plus [INAUDIBLE] activity. So this is still completely general. We take that expression and we write d of g equals-- let's see, mu of I not mu of I plus rt log a of I to the power mu of I. And this is all times dt. And this is d G0 plus RT log product a of I mu of I. Again, whole thing times dt. And of course, this is 0 at equilibrium, because at equilibrium, Gibbs free energy is optimized. So the coefficient is 0, and we get, at equilibrium, this product, a of I mu of I, which we call the equilibrium constant, equals e to the minus delta G0 over RT. So that's just a reminder. But this is why I wanted to take three minutes for that because we're going to make this specific to metal oxidation. That's what we're doing. So equilibrium constant for metal. And what you're going to find is that it simplifies a lot. So that's good. Metal oxidation. As we started the last lecture, we have some metal reacting with a modal of oxygen gas going to MzO2. Remember, z here has to do with the fact that we don't know what the oxidation state of that metal is. So potassium would be 1, iron has a couple of different oxidation states. Copper would be 1 or 2-- so there's different oxidation states. So this is to do this generally. So all right, now, we're going to write the equilibrium constant for this reaction. So we have the activity of the oxide MzO2. Sorry, the writing is getting a little bit small there. The activity of the oxide over the activity of the metal to the power of z times is the activity of O2 gas. The products of the numerator, reactants, and the denominator. The metal is z modal of metal. So we have that activity to the power of z. And straightforward to write down. Everyone with me so far? All right, this oxidation reaction at equilibrium. We're going to come back to that. This is the metal and oxygen and the oxide all coexisting at equilibrium. And now come the approximations. We're going to treat O2 as-- how are we going to treat oxygen, anybody? How will we model oxygen? AUDIENCE: Constant pressure? [INAUDIBLE] question. RAFAEL JARAMILLO: Not constant pressure, but we have a-- AUDIENCE: It's a gas. RAFAEL JARAMILLO: --as a-- what kind of gas? AUDIENCE: Ideal gas. Ideal? RAFAEL JARAMILLO: Yeah, because it's the only model for gases that we know. [INAUDIBLE] So an ideal gas-- the activity of an ideal gas is just as partial pressure by atmosphere for 1 atmosphere reference statement. Good. That's easy. And we're going to assume that condensed phases are pure. That's big. If we assume the condensed phases are pure, that's equivalent to saying that their composition is unchanging, all right? That's saying that this metal does not dissolve any oxygen, and it's saying that this oxide is what? What kind of a compound is always at fixed stoichiometry? AUDIENCE: Like, perfect crystal? RAFAEL JARAMILLO: Sorry? AUDIENCE: The perfect crystal? RAFAEL JARAMILLO: No, well, not perfect. No, not a perfect crystal, actually. Be something else. We learned this just two days ago. AUDIENCE: [INAUDIBLE] ? RAFAEL JARAMILLO: Mm, yes. It's a pure oxide, meaning that stoichiometry is ideal. It's a line compound. It doesn't deviate from that. So if we can assume the condensed phases are pure, what is the activity of a pure phase? We've got to think back. What's the activity of a component in its reference statement? AUDIENCE: 1? RAFAEL JARAMILLO: Right. Thanks, that's recalling material from the middle of the semester, so it goes back a little bit, but the activity of the material in its reference date is 1. So now, I have some really handy approximations. The activity of oxygen gas. We're treating it like an ideal gas, so it's just Po2 by atmosphere. And the activity of the metal and the activity of the oxide are both 1. So it's equivalent to assuming that there's vanishing oxygen solubility in the metal and that the oxide is a line compound. So with those approximations, we have the equilibrium constant is Po2 by atmosphere, so minus 1, and this is going to be equal to the E minus delta G0 over RT. So you see this simplified quite a lot. Simplified quite a lot. All right, I want to share some slides on binary metal oxygen systems to convince you that they do tend to be pair of materials. So let's look here. I grabbed a couple. OK, so here's the tin oxygen system. And you could see that an oxygen-- well, it's just a gas here. Tin down here is a metal until it becomes a liquid. In the liquid phase, it dissolves some oxygen. You can see that the liquid phase here has a solution region. So you can dissolve oxygen in liquid tin, but what about in solid tin? I know oxygen is off. So you see solid tin will just coexist with this line compound tin oxide. So this is what we're talking about, a line compound of an oxide and a metal that doesn't dissolve any oxygen. Here's another example, copper oxygen. Again, copper. It's written here as a parentheses as if it's a solid solution, but really, there's vanishingly small oxygen dissolution and copper. There's some. There's some, but it's so small as to be invisible on this part. I'll tell you that when you're making a component for cryogenics or for space applications, you want oxygen-free copper. You can buy it from McMaster, and you want it, because oxygen is magnetic when it freezes, and so a lot of times you want non-magnetic copper. So you have to get oxygen-free copper, but it's there at the parts per million level. It doesn't show up at all on the spot. So again, oxygen not dissolving in copper and these oxides being line compounds. Here's more examples. This is a little counter example. This is manganese. Manganese actually does dissolve some oxygen. So you see it dissolves 1% or 2% of oxygen. So maybe the activity of manganese should be not quite 1. Maybe it should be 0.99. And making these oxide similarly to not quite long compounds. They have some solid solution region, so that's a counterexample. Another example a titanium oxide oxygen system is titanium dissolves a lot of oxygen. So this is an even counter counterexample. This is a titanium, so funny system. It dissolves a lot of oxygen into solid titanium, but its oxides here are line compounds. So what to make of it? It's good to understand where the approximations are and where they might not. Questions on these faith diagrams before I go back to the board? AUDIENCE: Does this dissolving all owe to the molecule or just oxygen in general? RAFAEL JARAMILLO: I'm sorry, I didn't quite-- could you repeat? AUDIENCE: Going back to the solution, is this like dissolving oxygen as auto form or other forms of oxygen? RAFAEL JARAMILLO: It almost certainly disassociate. So this would be dissolving oxygen atoms. Well, I mean, how do you dissolve oxygen atoms? You expose the material to oxygen gas, and there's some catalytic process by which the oxygen gas splits into oxygen atoms at the surface and the oxygen atoms diffuse into the metal. It's almost certainly not dissolving O2 molecules. Metals that do dissolve oxygen are the basis for a lot of interesting technologies, but we won't have time for that. So let's move back to-- sorry, any other questions on these phase diagrams and interpreting them or applications and so forth before I move back? All right, so we have this really simple expression for the equilibrium constant, and now we want to evaluate delta G0 equals delta H0 minus T delta S0 for metal oxidation. So we're going to evaluate that. So we're going to start with enthalpy. Enthalpy delta H0 equals delta H0-- let's say 298 plus 298 temperature D-- temperature and you can have a delta Cp for the reaction. So this is general, but here's the-- and you know what's coming is an approximation. So I'm going to tell you for most metal oxides, delta H0-- let's say 298-- is large. These are energetic reactions. Why would the enthalpy of formation of an oxide be large? Does somebody have the sense for why that should be? I'll give you a hint. It's large negative. It's large in the negative direction. What reaction are we talking about here? AUDIENCE: Oxidation? RAFAEL JARAMILLO: Oxidation. So what kind of reaction is an oxidation reaction other than just-- you could say it's an oxidation reaction, but what's actually happening? Maybe somebody who hasn't replied yet today. What's happening on the atomic level during an oxidation reaction? AUDIENCE: Does it include a favorable transfer of electrons? RAFAEL JARAMILLO: Yeah. Yeah, this is due to-- so thank you to the two of you. Due to exothermic nature of electron transfer during metal oxygen bond formation. And I could-- I get to ionic. So it's an ionic bond formation. You have very electronegative element, and that's oxygen, and less electronegative elements, those are metals. And that bonding mechanism is electron transfer, and it's very exothermic. So as a result, delta H's tend to be very large and negative. And this allows us to neglect temp dependence. Neglect temp dependence. We'll just say that it's whatever it is at 298, because that's a convenient reference point. And these are tabulated. The enthalpy of oxidation reactions at standard condition, those are tabulated. So that takes care of the enthalpy. What about the entropy? Entropy delta S is delta S at some reference temperature, 298 plus integral. OK, so who knows-- who has a guess on what we're going to do? We're going to approximate this as what? What am I looking for? I want to justify what approximation. AUDIENCE: Aren't entropy values usually much smaller? So if the enthalpy is already so big, we could assume entropy is 0? RAFAEL JARAMILLO: I see. So you're right that entropy values tend to be much smaller, but the thing is they're multiplied by temperature. So they do matter quite a lot, because the temperature number can get large. So we're not we're not going to neglect it completely, but I'll tell you-- I'm sorry? AUDIENCE: Neglecting its temperature dependence? RAFAEL JARAMILLO: Yeah, we're going to neglect its temperature dependence. So I'm going to write this before I write this. It's just the why we get to do that. So let me write neglect the temperature dependence, and now, somebody please tell me why we might be able to do this? And it's not just that we really like to make our lives easier. It turns out to be a really good approximation in many cases. What about this oxidation reaction? What's the dominant entropy change? What's the dominant contribution to delta S? AUDIENCE: Could it be when the phase change happens? So when the temperature is constant? RAFAEL JARAMILLO: The phase change meaning from metal to metal oxide? AUDIENCE: Yeah. RAFAEL JARAMILLO: That's true, but maybe tell me what part of that? AUDIENCE: [INAUDIBLE] condense it? RAFAEL JARAMILLO: Yes, thank you. That's what I'm looking for. The reaction entropy is dominated by-- I didn't see who said that, but thank you so much. By condensation of O2 from the gas. That's it. So here's metal. There's a chunk of metal, and then I have O2 gas. Gas is very high entropy, as you know. O2. And this reaction pulls the oxygen out of the gas and makes a metal oxide. So the reactants have much, much higher entropy than the products, simply due to the fact that the reactants included a mode of gas and the products don't include any gas. And in almost all cases, that change dominates the reaction entropy. And since we're modeling oxygen as an ideal gas, the entropy of this gas is temperature independent. It's configuration entropy. It's the thing we've been talking about pretty much the whole semester, which is the entropy of randomly configured gas molecules. Good. Thank you. So now we have waved our hands, although we've done it scientifically. Scientifically waving of hands has happened, and we figured out that we can neglect the temperature dependence of the enthalpy and we can neglect the temperature dependence of the entropy, and this is going to make our lives even easier than they already were. We can solve for this thing. We can solve for that oxygen partial pressure. PO2 by atmosphere equals e the delta H0 over RT. e the minus delta S0 over R. This is the oxygen pressure at which a metal and its oxide coexist at equilibrium. All right, so you've got a metal. They've got it's oxide, and this is the oxygen partial pressure, so there's some oxygen here. This is the oxygen partial pressure at which this kind of situation is that equilibrium. What happens if the oxygen partial pressure is higher than the equilibrium value? What do you think happens spontaneously in this system? Think Le Chatelier's principle. I have the system in equilibrium and then I shove more oxygen gas in. I increase the oxygen partial pressure. What do you expect to happen? How will the system respond? AUDIENCE: Spontaneously oxidized? RAFAEL JARAMILLO: One more time. I'm sorry, I didn't quite hear that. AUDIENCE: Spontaneously oxidized? RAFAEL JARAMILLO: Yes, fantastic. Metal spontaneously. So you're going to have spontaneous conversion of more metal into oxide. We've already established that the oxygen can't dissolve in the metal and the oxygen can't dissolve in the oxide. They're pure components, so all we can do is convert metal to oxide, metal to oxide, metal to oxide. Excellent. Thank you. What about at lower? At lower oxygen pressure pressure? If we're below this value, what do we expect to happen spontaneously? AUDIENCE: We will see the reverse reaction happening. So more O2 gas will form? RAFAEL JARAMILLO: How does it form? AUDIENCE: Through the oxide decomposing back into a metal and O2 gas. RAFAEL JARAMILLO: Exactly, thank you. I'm going to use this term the oxide is reduced, but your expression of spontaneously decomposing is exactly the right thing. So if we pull down the oxygen pressure below the equilibrium value, the system will respond by converting oxide into metal and oxygen gas. And you can think of this in terms of Le Chatelier. If I add more oxygen, the system will try to counteract that by condensing the oxygen into more oxide, so it'll take more metal, convert it into oxide by pulling some of my extra oxygen out of the atmosphere. If I pull down the oxygen pressure, the system will try to counteract that by giving off more oxygen gas from the oxide. The oxide will decompose into more oxygen gas and more metal. That's right. So that's what happens. So for example, let's imagine Ti oxidizing. That's a nice reaction. At 298 Kelvin, I looked it up. The PO2 at this equilibrium is-- anybody have a guess? Forget the guess. It's completely naive. 10 to the minus 150 atmospheres. A completely silly, make believe number. It's 0. It's 0. It might as well be 0. What does that mean about titanium metal? AUDIENCE: It's very favorable for it to oxidize. RAFAEL JARAMILLO: Very favorable with oxidized. Titanium metal, if you see it in the air, will actually be titanium metal with a thin layer of its own native oxide on top of it. Because if you were to remove the oxide and wait a split second, the oxide would spontaneously reform. And so this spontaneous oxide formation process is really important for a lot of reasons. It's important for microelectronics, it's important for stainless steel. That's how stainless steel remains stainless. You have spontaneous formation of a passive rating, or you can think of it as a protecting oxide layer and the engineering of those materials continues. It's all interesting stuff. All right, so that is enough to take us to Richardson Ellingham diagrams. So we're going to do this. I can't really motivate this, because it sounds kind of random what we're going to do. So I'm just going to tell you what it is, and at the end tell you why it's useful. We're going to plot delta G0 as a function of temperature for metal oxidation reactions. As you know, it's like this. And so this is a pretty simple thing. This is a line with slope minus delta S0 and intercept delta H0. This is a really, really simple line. So here's is temperature. Let's see in Kelvin. Here's 0, here's 0. Here's delta G0, and we have line. We have a line, the slope, and an intercept. Slope, intercept. That is a Richardson Ellingham diagram. Now, why did we do that? A couple of reasons. Let's start by talking about the PO2 scale. What's that? The equilibrium constant is PO2 by atmosphere to the minus 1, which is equal to e to the minus G0 over RT. So I'm just going to rearrange that and write delta G0 equals RT log PO2 by atmosphere. This is a line on delta G0 by t plot with slope our log, PO2 and 0 intercepts. Let's draw that. All right, here is temperature in Kelvin. 0 is 0. Here's delta G0. I'll draw the oxidation of a given metal just to have something on there. And now, I'm going to draw these two contours, right? Delta G0 equals RT log PO2 over atmosphere. So I'm going to draw a series of lines all with 0 intercepts in varying slope. So there's a series of lines with 0 intercept and varying slope. And as I move in this direction, I'm increasing PO2. So that's what a Richardson Ellingham plot is. Now, I get to tell you why it's useful. All right, so if I rush through the explanation what it is, because it's just like, I have to tell you what it is. But now, I can show you why it's useful. So why is this useful? It's useful for materials processing. Let me illustrate that. Let's consider two metals, and I pick these just because they appear nicely separate on the plot. Let's consider tin in manganese. If I look up the Richard Ellingham diagram for tin and manganese, I could do that, but I'll just draw qualitatively what it looks like. It's temperature in Kelvin. It's delta G0. And I'm just here to tell you that the tin line sits well above the manganese line. So that's just a fact about these metals. This here is tin plus O2 going to tin O2. That's what that is, and this is 2 manganese plus O2 going to 2 manganese oxide. And here are my PO2 contours. What is this plot telling me? It's telling me that, at a given-- I have a typo on my scan lecture notes, I'm sorry. At a given temp at any given temperature, PO2 for the manganese oxidation reaction is lower than for the tin oxidation reaction. That's a useful fact. How do I see that? Let's say at a given temperature, let's see that temperature here. I have this point on the tin oxidation line. And this point on the manganese oxidation line. Well, the PO2 contour that runs through the manganese oxidation line intercepting of that temperature corresponds to a lower partial pressure than the PO2 contour that runs through the tin oxidation line for that given temperature. Again, at its given temperature, the PO2 partial pressure contour for manganese oxidation is a lower partial pressure than the contour for tin oxidation. What does that mean? Manganese metal will reduce tin oxide. So if you were to put tin oxide-- let's say an ore into a furnace in the presence of manganese metal and heat it up, the manganese will convert to manganese oxide and then tin oxide will convert to tin metal. Why is that? Manganese has higher affinity for oxygen than does tin. Or saying the same thing, the enthalpy of oxidation is more negative for manganese oxidation than for tin oxidation. So you can see that from the plots. The intercept of these plots is delta H. So you can see the intercepts for the manganese oxidation-- that's at this point down here-- is lower. Let me extend these lines. The intercept of the manganese oxidation reaction is lower than the intercept of the tin oxidation reaction. The enthalpy of manganese oxidation is more negative. Or if we are speaking a little bit colloquially, manganese metal pulls-- in quotation marks-- oxygen out of tin oxide. So everything I just wrote, these are all saying the same thing about metal oxygen bonds. All right, these are all equivalent statements. They're just ways of interpreting metal oxygen bonding and the implications of that for materials processing. If you're going to be processing manganese in the presence of tin, you need to know which one is going to have a higher PO2 at equilibrium for oxidation, because it's going to determine your process outcome. I want to share with you some pictures of Ellingham diagrams. So there's lots of Ellingham diagrams out there. Here is one representation that's a little busy, but you can see lots of things here. You can see many metals which are all represented. Let me grab my laser pointer. Sometimes the laser pointer is stubborn. So here's iron. Oxidation reaction is a nickel reaction copper. As you move down, you're going to metals that have higher affinity for oxygen. So copper is sometimes considered noble. So it doesn't readily form an oxide. You remember we talked about roof flashing. It takes a long time for it to turn green. As opposed to calcium, right? If you have calcium metal around, you should look out, because it's going to explosively oxidize. So down here, you have calcium, magnesium. Aluminum. Aluminum is a very energetic oxidation reaction, but we know it forms a passivating oxide. So you don't have exploding aluminum all over the place. You have almost instantaneous formation of aluminum oxide on the surface of aluminum. Titanium silicon manganese chromium. So this is a limited Richardson diagram, Ellingham diagram, because it only shows about a dozen or so metals. Of course, you can get very, very busy plots. I want to point out one more thing, which is the PO2 scale. PO2. And you see it's a series of numbers running from 1 atmosphere to 0 to the minus 200 atmospheres. And each of these numbers has a little tick mark, and the tick marks all point towards the origin. You see these tick marks, they're at different angles. Why is that? Because it's asking you to imagine a series of straight lines connecting the origin to those values. Those series of straight lines connecting the origin to-- I've never done this before. I draw the line. So here is-- oh, there we go. Look at that. So this would be a line of 10 to the minus 50 atmospheres of oxygen. So for example, you might say that zinc and its oxide are equilibrium at 400 degrees C and 10 to the minus 50 atmospheres of oxygen. That would be one way to read this plot. And you also see that the delta H of formation here is correlated to the electronegativity. So this idea that metal oxygen bond formation is energetic, well, we know that, but it's more energetic for less electronegative metals. So we expect noble metals like silver and copper to be fairly electronegative. They like their electrons, 1.9, 1.93. And we expect alkaline, alkaline Earth metals-- let's say alkaline earth-- alkali metals. Alkaline Earth metals like magnesium, right? They are happy to give up their electrons to oxygen. And that has a lower electronegativity calcium is down here at 1. So there's some solid chemistry here. There's one more thing I want to tell you about Ellingham diagrams, and that is the effect of phase transitions. So let's consider melting of metal or its oxide. Metal's metal-- oxide's also metal-- causes-- it causes discrete jumps in those values. So for example, let's consider solid metal plus O2 going to MO2 solid. Reaction one. This says standard entropy 1. It's negative. It's negative, because we're pulling oxygen gas out of the gas, right? Solids. Solids. Now, consider case where metal melts at lower temp than MO2. So metals have a lower melting point, which is true for most but all oxides. And raise the temp until then the metal solid becomes liquid metal. So that's what we're going to do. We're going to consider that, and now we're going to consider the oxidation again, except now we're oxidizing liquid metal. So the oxide is still solid but the metal is now liquid. And we'll call this reaction to have delta S02 Well, we know that the liquid metal has higher entropy than solid metal. We know that. So what that means is that delta S02 is going to be smaller than delta S01, and they're all less than 0. In other words, it's going to be even more negative. On a plot, it looks like as follows. There's temperature. This is delta G0, and we have a kink in the plot. The kink happens at the melting point. So here is metal melting at that temperature. The lower curve is a solid oxidation reaction-- solid metal oxidation. The upward curve is liquid metal oxidation. This intercept is delta H0 for solid metal oxidation, and this intercept is delta H0 for liquid metal oxidation. So let's go back to some real plots. So now we can understand why there are break points in these. There are break points in these curves, because at these transition points, the standard enthalpy and entropy of these reactions changes. So this melting point here of aluminum, it changes the slope very slightly. You can't really see it. You can't really see it, but there is a slope change. Then up here, the aluminum boils. So there's another break in the slope. Here, manganese melts. Here, the manganese melts. Here, zinc melts and zinc boils. So forth and so on. And if you want to really lose your mind, you can look at more complete Ellingham diagrams. So this is posted on the website. And this is a very, very, very thorough Ellingham diagram now with so many elements and melting points and boiling points and so forth. And you can start to look at these and learn what the melting points and the boiling points of the metals and the oxides are, and you can see how they change the slope of these curves. There are cases where the slope actually dips downward for a little while. That's typically a melting point of an oxide and so forth. And if you really like to explore this a little more, which I encourage you to do. It's a nice way to learn. Go to DoITPoMS page with Ellingham diagrams. For those of you who don't know DoITPoMS, you should know DoITPoMS. It's a really excellent resource for learning material science concepts. It's maintained by University of Cambridge, and it's a nice complement to things like OpenCourseWare and MIT. Let me share that just to show you where you might go to play with this a little more, get a feel for it. So here is the DoITPoMS page for Ellingham diagrams, and if you click on oxides, there's a whole bunch of oxides. And so let's go at random cobalt oxide. See Ellingham diagram. So it now shows you Ellingham diagrams for cobalt for its two different oxidation states. And it will give you data, you can mouse around, get the actual numbers. Change the temperature. It gives you free energy information, and so forth. And it's a nice learning tool. So now I will call it a day and stop recording.
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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: Hello, welcome again to 8701. In this short video, I'll talk about the books we are using and the literature we are using in this class. So let's dive right into. There's a sequence of textbooks I go back to when I prepare the material for the class. You know, the one which I use in order to derive the outline or the schedule for the class is Introduction to High Energy Physics by Perkins. But again, I use material from a sequence of textbooks and reading material for you guys as well. Nuclear physics is not covered in Perkins, so we have here Samuel Wong's book on Introductory Nuclear Physics. We spend about two weeks talking about nuclear physics towards the second part of the class. And a couple of basics, we talk about the introductory material really in the sequence. A book I like a lot is the Introduction to Elementary Particles by Griffiths. And you see me using examples out of that book a bit. Then on the nuclear physics side there is Kenneth Krane. It's an MIT book and a book which has been put together by MIT faculty and research scientists. And then there's Techniques for Nuclear and Particle Physics by Leo, which I like a lot. It's a little bit of an older book, but it goes into some of the technical details and material details which are important to understand how we build detectors. And then a more recent book is Modern Particle Physics by Mark Thomson. It dives right into particle physics, the energy frontier. And it's really nice to read. It's a modern book. And it's easy to read and comprehend. I recommend to have a look at the review articles by the particle data group. They are really concise articles which are for beginners or for introductory level maybe a little bit difficult. But as we go through the material in this class, you should be able to take those articles to review certain sections of this class. For example, QCD, or electroweak interactions, the Higgs mechanism. And while you do this, you also learn one of the latest results and measurement in this area. I'll be posting a set of papers as we go through the class. And you'll see in the course organization that I'll ask you to actually summarize some of those papers in our recitation section. So those are going to be important papers, for example, describing the experiment which was used to measure parity violation, or the paper on the Higgs discovery. That's it for literature. Please as always go ahead and ask me questions. You know this. You know if you Google particle physics or nuclear physics, you will find tons of literature available on as many good books. And you might find a different one from this listing which suits your appetite for reading and learning.
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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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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. PROFESSOR: --not of the vindictive sort. You skip class, you've skipped a lot of important stuff. But I'll get you on the quiz, that's all. Kidding aside, there were a number of things that I passed out. I think nobody needs a set of problem set number 13. That was the one that had something on symmetry constraints and working with second-rank tensors. If anybody missed that, you can see me during break. Some people, I think, missed problem set 14. And that's the one where you are invited to diagonalize some tensors, either using the method of successive approximations or the direct diagonalization- by-an-eigenvalue procedure. Anybody need one of that? And I'm sure that nobody has a copy of problem set number 15, which deals with piezoelectricity. And I know you don't have it, because I just put it together. So I'd like to hand it out and invite you to explore things that deal with third-rank tensors. And I hope that, even though doing the problem sets is optional, particularly at this juncture in the semester when things have come to a set of successive crunches. But if you don't know how to do it, for goodness' sakes, come see me. Or raise it in our next class. You know, some question like, I haven't the foggiest idea how to do problem number two. Could you say a little bit about that, please? And I'd be happy to oblige. All right. I will have for you next time the quizzes and also all the problem sets which will have been turned in up to that point. And what I spent my time doing instead is writing out notes for those people who missed the last lecture, and also notes covering what we're going to do today. Because it is very exquisitely intensive, algebraically. It's not hard, but there are a lot of variables with a lot of subscripts. So let me pass this around. I'll split it up into packs. These are notes on some basic relations in electromagnetism which you may or may not have forgotten. Take one off the top. And it's coming at you from either side, so you're going to pass it back. And the notes also cover everything that we're going to do on piezoelectricity. Most of it will take place today. And I can zip along a little more rapidly if you have notes to follow. I would like to ask you when you get a set of the notes-- I could really kick myself-- the introductory discussion reminds you of the definition of a dipole. And down in the middle of the page, on the cover sheet, two different types of polarizability are defined. And one of them involves the separation of charge on an individual atom. And that is called, quite appropriately, the electronic polarizability because it involves polarization of the electrons and protons on the individual atoms. And then there's another type of induced dipole moment that comes when the structure is ionic. And then an electric field will pull positive ions in one direction and negative ions in the opposite direction. And that is very often referred to as the ionic polarizability. And it's easy to keep them straight. One involves electrons, which all atoms have. The other involves ions. And not all structures and materials have ions. So the second one is unique to ionic structures. And then, these are sometimes also referred to as the dielectric polarizability. And I meant to purge that from the notes. And as I put this together to xerox it, I grabbed the uncorrected sheet. So please, just below you see the displaced positive and negative ion on the middle of the page, cross out "dielectric" polarizability. And change that to "ionic" polarizability. And I didn't catch that. There's one other little typo as we go partway through. In any case, we'll talk about third-rank tensor properties. We'll introduce some other ones other than piezoelectricity later on. But piezoelectricity is one of the primary examples of a third-rank tensor property. And it's one that has a lot of applications in devices-- pressure sensors, audio equipment, electronic devices. It's a very important property in terms of devices and present-day technology. But there are others. And we'll cover those in due course. Some of them are rather exotic. OK. But to review these basic concepts in electromagnetism, we remind you again of the definition of a dipole. We mentioned this last time. But to quickly review, a dipole is a pair of charges of opposite sign but equal magnitude, separated by a separation, d. And then one defines a dipole moment, which has a vector character, as the product of one of the charges of magnitude, q. And the vector that separates the two charges in the sense of the vector is defined as going from the negative charge and pointing towards the positive charge. It's purely a definition. But the reason it's convenient is that the vector sense and the product of charge and separation comes up again and again in all sorts of problems, among them definition of the piezoelectric effects. Also a dipole in an electric field is going to experience a torque because the electric field will pull on the positive charge in the same direction. It will pull on the negative charge in the opposite direction from the field. And that's going to create a torque on this little gizmo. OK. Now it's important, too-- and interesting to note-- that there are three different kinds of dipole moments. There are some molecules-- and water is the primary example. Water has seen an asymmetrical arrangement of hydrogen, relative to the oxygen ion to which they are connected. And that gives water a permanent dipole moment, which is what makes water such a darn good solvent. And its ability to dissolve primordial juices probably accounts for our fact, intelligent design notwithstanding, of why we are here today. On the other hand, dipoles, as I was just saying, can be induced when you impose an electric field on matter. And these are of two kinds. One is the dipole moment that is induced on an individual atom. And that results in displacement of the positive nucleus relative to the negative electron shell. It's found that the dipole moment is proportional to the magnitude of the electric field. And the proportionality constant, alpha, is called the polarizability. And for an individual atom, as I said a moment ago, it's defined as the electronic polarizability. We write it as a scalar quantity, but actually, by now you're probably sensitized to being a little bit skeptical when things that relate to vectors are described as a scalar. And in fact, the electronic polarizability is not a scalar, it's a tensor. And one should really write that the i-th component of the dipole moment is given by alpha i,j times the j-th component of the electric field. Second type of induced dipole moment involves, again as we said a moment ago, the effect of imposing an electric field on an ionic structure. And again, the field will pull the positive ions in one direction and negative ions in the other. And here quite clearly, if this pair of ions is in a structure, and that structure has some symmetry, we really have to consider this second origin to induce dipole moments as a tensor. And this is referred to as the ionic polarizability. So both of these types of dipole moments will be present in matter, in general. The relative importance of each depends on whether the electric field is a static field or an oscillatory electric field. And then the frequency dependence of these two polarizabilities has a consequence on the magnitude for fields of different frequencies. And not surprisingly, the ability of the ions in the structure to polarize is going to poop out with high-frequency electric fields a lot quicker than just the displacement of the light electrons about a positive nucleus. So there is a frequency dependence of the net polarizability. We won't go into that. But just keep in mind that at very high-frequency electric fields, the ionic polarizability will damp out. OK. Then we went through a rather simplified but amusing model for the electronic polarizability. And there's some rather severe assumptions that are made. But making those assumptions let's you get a rigorous result, which tells you something about the electronic polarizability. So we model the electron distribution on the atom as a uniform charge density in a distribution that goes up to some radius, r, and then quits. So there's a sharp cutoff to the distribution of electrons, which is obviously ridiculous. You know there are a collection of wave functions that give you charge probabilities that tail off slowly to large distances, getting progressively smaller and smaller. So this is not terribly realistic. And then the other thing we assume to make this model, something that we can solve exactly, is that the nucleus and the electron distribution displace as units. In other words, we start with a sphere of electrons, the center displaces, but it stays a sphere of uniformly distributed electrons. And then having made those assumptions and having lost any credibility for the model, if we carry through to see what the model predicts, it's rather interesting. We use a fact, again, known to freshman and sophomore students of electromagnetism, that a charge inside of a uniform distribution of charge experiences no force. And that's surprising, but it's something that you are very often invited to do on problem sets. So therefore, if the center of the electron distribution is displaced from the nucleus, then the restoring force between the electron sphere and the nucleus is simply the coulombic force between a nucleus of charge plus ze, and a fraction of the total number of electrons, namely that fraction of the electrons which are contained within a sphere that has a radius equal to the displacement. And that geometry and that algebra's carried out for you on the bottom of the first page. If you set that up and ask what the dipole moment will be, it comes out beautifully simple. It comes out to be equal to whatever proportionality constant you use in Coulomb's law. I use rationalized MKS units from force of habit. So there's a 4 pi epsilon 0 in there, then times the cube of the radius of the electron distribution, times the electric field. So the two items of note that come out of this simplified treatment is first of all, the induced dipole moment is proportional to the applied electric field, which is what we assumed. And so therefore, the polarizability, which is the quantity that relates the dipole moment to the magnitude of the field, is a constant. And it turns out to be equal to 4 pi epsilon 0, times the radius. So not only does this tell us that the electronic polarizability is something that relates dipole moment in direct proportion to the magnitude of the field. And secondly, the electronic polarizability involves the radius of the charge distribution, cubed. And even on a qualitative basis, this is interesting. It tells you that high-atomic-number, big, fat ions are going to have a very, very large polarizability. And things way down in the periodic table, like beryllium and lithium, and other low-z atoms, are going to be tough little nuts that don't display much polarization at all. And in point of fact, several individuals have tabulated empirical sets of electronic polarizabilities. One of the earliest ones are the so-called TKS values published a long time ago by Tessman, Kahn, and "Wild Bill" Shockley. And I give you the reference to those. There is another set of values that were assembled by a fellow at DuPont named Bob Shannon. And I'll give you a citation to those values. But in any case, if you look at these values, you find that the cation that has highest electronic polarizability is thallium, way down on the bottom of the periodic table, next to lead. And that's just a big, fat, flabby atom that can be deformed very, very easily. How do you get these polarizabilities? Well, the equations at the bottom of page two-- which we won't make any use of but, nevertheless, will tell you where they come from-- there's a relation between the square of the index of refraction and the sum of the polarizabilities of the individual species, times the number of those species per unit volume. And that is something that's called the Lorentz-Lorenz equation, equation. I can't resist saying everything twice, Lorenz and Lorentz. You can put this in another form that involves the molecular weight of a molecular structure and the polarizability per molecule. It's the same equation, but for organic compounds. It's a useful form. And it turns out that the dielectric constant of the material is directly related to the square of the index of refraction. So you can write those two equations in terms of either the dielectric constant or the index of refraction. And if you substitute dielectric constant in place of n squared in those equations, the equations get new names. And they're not called the Lorentz-Lorenz equation, equations. They're called the Clausius-Mossotti equations, which shows you that sometimes fame and immortality can be gained simply by a trivial substitution of variables. Nice to keep in mind if you can find something like that. Finally, we don't see dipole moments on individual atoms or molecules. We see evidence of polarization in bulk. And on page three is a little model that reminds you of what the polarization of a material is. And that's defined as simply the dipole moment per unit volume. And that's represented by a capital P rather than a small p. And this is also a vector quantity. And we can see how this is related to the individual dipole moments in the solid by dividing the solid up into individual cells. And I use the term of "cell" loosely. Is this a unit cell? Is this a box around each of the atoms? It really doesn't matter. Because all of these little dipoles on the individual unit cells are packed together back to front in the solid. So internally, the negative end of one induced dipole moment is always adjacent to the positive end of the neighboring dipole moment. And everything cancels out internally, except for the two surfaces of the solid. So macroscopically, if you impose an electric field and it induces dipole moments, what happens is you see that there is a charge on the surface of the piece of material. And that is a real charge. It's a bound charge. You can't draw that charge off as a current. Because the minute you reduce and eliminate the electric field, those dipoles disappear. And the charges all go back to where they came from. So this is the thing that keeps the exact model that we use for one of the cells in the solid [INAUDIBLE] of no importance whatsoever. You could say that the dipole moment on each atom is the charge times the separation of the positive and negative charge. And qi is the induced charge. The number of cells per unit volume, if we say that they're little cubes of the same edge length, delta, is unit volume one divided by the volume per cell, which will be delta cubed. And the dipole moment per unit volume is the number of cells per unit volume times the polarization on each. And you find that everything drops out. And the polarization indeed turns out to be equal to induced charge per unit area, whether that's induced charge per unit area of one of our cells or on a square centimeter of material. So polarization, dipole moment per unit volume is numerically equal to, and physically equivalent to, an induced charge per unit area. OK. So much for freshman electromagnetism in fast forward. And now I'd like to turn to something that involves tensors and materials. As I mentioned a moment ago, piezoelectricity, literally "pressure electricity," is a very technologically important property, useful property, but also provides a nice example of a tensor property that has to be defined in terms of a sensor of third rank. There are a number of different piezoelectric effects. The first one is the so-called direct piezoelectric effect. And it refers to the fact that if you subject a material to an applied stress, it will develop a surface charge. Remove the applied stress, the surface charge goes away. The surface charge can be described in terms of a polarization, a dipole moment per unit volume. And it will be in direct proportion to a stress tensor sigma j,k. And so what we'll assume-- and this is an assumption. And it's followed, except for very extreme conditions, that each component of the polarization, p sub i, is given by a linear combination of every one of the nine elements of stress, sigma i,j. Oops, not i,j. I have to use a different index. I'm used to writing i,j. So each component of the polarization, where i ranges from 1 to 3, is given by a linear combination of all nine of the elements of stress, sigma j,k. And the proportionality constants, di,j,k, are known as the piezoelectric moduli. And this is called the direct piezoelectric effect. OK. That's simply an assumption that the polarization is proportional to the applied stress. We know that stress, however, is a field tensor of second rank. We know that the polarization has the character of a vector, charge times the length. We know how second-rank tensors transform. We know how first-rank tensors transform. And therefore, knowing that, we can say that the coefficients, the array of coefficients, di,j,k, will transform like a third-rank tensor. And therefore they are a tensor. So we will have three components of polarization. And we will have nine components of stress. And so there will be 27 piezoelectric moduli, di,j,k. So things go up rapidly in terms of number of coefficients as the rank of a tensor increases. Now let's take a look at the nature of these equations. We'll have, for example, the x1 component of P being given by d1,1,1 times the element of stress sigma 1,1. The next term will be d1,2,2 times sigma 2,2. I'm putting down the tensor components first. But this is just an equation. I can write the terms in any order. Next will be a d1,3,3 times sigma 3,3. And the next one would be a d1,2,3 times a shear component of stress sigma 2,3 and other terms. I don't want to write out all nine of them. Because I think I have written enough to make my point. The point is that the value of i is always tied to the component of the resulting vector that we're defining. But the second pair of subscripts, j and k, always go together as an unseparable pair with the component of stress that they modify. So the 1,1 here goes with the 1,1. The 2,2 goes with the 2,2. The 3,3 goes with the 3,3. So why in the world, if they're always going to go together, do we have to use two indices to define the piezoelectric moduli? Why don't we use a single symbol? Well, there is a good reason for not doing it. But we'll issue that caveat later on. We could use an a, a b, and a c, or an alpha or a beta or a gamma, or some other esoteric symbol. But what makes sense since we're using Arabic numerals to represent subscripts, let's write the pair of indices in terms of an index that's related to the stress tensor, which is sigma 1,1; sigma 2,2; sigma-- whoops. Sigma 1,2; sigma 1,3. And then comes sigma 2,1; sigma 2,2; sigma 2,3; sigma 3,1; sigma 3,2; sigma 3,3. We know that this is a symmetric tensor. So these off-diagonal terms are always equal to one another. So we're numerically going to have to enter each of those twice. So to define a single index that represents the components of stress. And I'll describe it in this fashion so you can remember, as a mnemonic device, what makes this work. We'll go down the main diagonal of the stress tensor, this fashion. And then having reached the bottom, we'll go up along the right-hand side and then jog over to the left to pick up the last term. And we'll define a number associated with these indices that goes in the form 1 to 2, to 3, to 4, to 5, to 6. So we'll just use these three integers, ranging from 1 to 6, to represent pairs of integers in the stress tensor. So things tidy up quite nicely then. This would be P1 is d1,1 times sigma 1; plus d1,2 times sigma 2; plus d1,3 times sigma 3; plus d1,4 times sigma 4. And now, uh-oh, Houston. We've got a problem. Because there's another term in here that is d1,3,2 times sigma 3,2. And sigma 3,2 is required, because the stress tensor is symmetric, to be numerically equal to sigma 2,3. So this is really d1,2,3 times sigma 2,3. And then we've got a d1,3,2 times a sigma 3,2. I know that this is equal to this. But is there any reason d1,2,3 has to be equal to d1,3,2? I can't see any reason why. OK. Let me confide in you that, yes, they are equal. But we can't show it or claim it on the basis of what we've got before us right now. It's going to come later. But we can show that they are equal. OK. But equal or not, this means we'll have a term d1,4 sigma 4, and we're going to get it in there twice. So when the subscripts go up to 4, we're going to get a 2 out in front. And then we'll get for the term sigma 1,3 and 3,1 we'll have a d1,5 times a sigma 5. This would be sigma 1,3. And then we'd have another d1,5 for sigma 5 again. And this would be sigma 3,1. So what are we going to do? We're going to say, well, it's nice we're getting rid of an unneeded subscript. P1 is the d1,1 times sigma 1; plus d1,2 times sigma 2; plus d1,3 times sigma 3. Are we then going to say 2 d1,4 times sigma four, and say that the coefficient here is di,j when i is equal to 1, 2, 3. But the coefficient is 2 di,j when j is 4, 5, or 6? Hell of a matrix that would be. That's going to be a bother. Since it doesn't seem that we can measure these two coefficients, d1,2,3 and d1,3,2, independently anyhow, let's just write our relation in reduced subscripts with the two added together as the element d1,4. So we are going to define d1,4 equals d1,3,2 plus d1,2,3. So that's a definition. And d1,5 will be defined as the sum of d1,1,3 plus d1,3,1. And we're going to define d1,6, finally, as d1,1,2 plus d1,2,1. OK. So if we do that, then and only then are we entitled to write a nice, simple, compact little nugget that says that in our redefined form, Pi is equal to di,j times sigma of j. And i goes from 1 to 3, and j goes from 1 to 6. And this is kind of a neat compact, easily managed outcome. So the moral of this story is, I like to think, is you can have your cake if you eat its 2. Oh come on, this is a tough, tough crowd. You're just all worn out from the MRS meeting. That'll do it to anybody. So why do we worry about the fact that to the direct piezoelectric effect should be a third-rank tensor? The answer, my friends, is that this is no longer a tensor relationship. It's a matrix relationship. A tensor is a matrix, but it's a matrix with a difference. It's a matrix for which a law of transformation is defined, if you change axes. There's no such law defined for the di,j's. So they qualify as a matrix, but they are not a tensor. So to attempt to say-- as you might be inclined to do when we raise the issue of what the symmetry restrictions are on these tensors-- if you attempt to say, well, I'm going to find d2,1 prime. And that's going to be c2,i c1,j times di,j. Wrong. You go down in flames because that's just nonsense. That's just nonsense. So this is a matrix relation. And if you want to do exercises such as slice a piezoelectric wafer out of a piece of quartz, and then having done so, refer the properties to axes taken along those edges of the plate that you've cut, you've got to-- in order to find the piezoelectric moduli for that plate in that new coordinate system-- you've got to be prepared always to go back to the full tensor notation. If you want to derive symmetry restrictions, which we're going to do whether you want to or not-- but we won't do them exhaustively-- you've got to go from this matrix notation back to the full three-subscript tensor notation. OK? Yes, sir? AUDIENCE: Couldn't you extend c sub i,k's to be [? 3,6's ?] and extend most others [INAUDIBLE]? PROFESSOR: Oh, yeah. No, if we would write all these down, we'd have-- I don't know if that's what you're asking-- but we'd have P1 is equal to d1,j times sigma sub j; P2 is equal to d2,j sigma sub j; and P3 is equal to d3,j times sigma sub j. So I just did that for the line with i equal to 1, because I was too lazy to write down all three relations. Is that what you're asking? AUDIENCE: No, I'm saying you could have a transformation model if you were to extend a c sub i,j matrix to be 3,6. PROFESSOR: No. It just won't do it. I mean, think of what these indices are. They're 1, 2, and 3 standing for the x1 direction, the x2 direction, the x3 direction. What's the x5 direction? It's just not defined. Good try, but you just can't do it. AUDIENCE: I'll think of something next time. PROFESSOR: Oh, I'm sure you will. Whether it will work or not is an absolutely different question. OK. So beware. Do not try to transform tensors of third rank expressed with two subscripts to other coordinate systems. All right. What I would like to do now, then, is to examine the symmetry restrictions that are imposed on a third-rank tensor by crystallographic symmetry. And these are embedded in the notes that you have. But it's going to be bothersome to leaf through those. So I separated out the pages separately. And I'm going to look at a few transformations so that you can see how they work, and then point out some very, very curious consequences of this, which are non-intuitive. So first of all, how would we do it? If we want to get the form of di,j, the matrix for-- let's do a very, very simple one. Well, let's do one that we did last time, for those who were here, for inversion. For inversion, the direction cosine scheme is c1,1, 0; 0, 0-- I'm sorry. For 1 bar, the transformation of axes is c1,1, 0, 0; 0, c2,2, 0; 0, 0, c3,3. So we'll take di,j, change it to the full three-subscript notation, di,j,k. And the law for transformation of the third-rank tensor is di,j,k prime is equal to ci, capital I; cj, capital J; ck, capital K; times d, capital I, capital J, capital K. We've not done this much. But we've mentioned quite some time ago that this is the way a third-rank tensor would transform element by element. The only terms that are non-zero in this array, when the symmetry transformation is inversion-- where x1, x2, x3 goes to minus x1, minus x2, minus x3-- is whatever i is, minus 1; whatever j is-- only the diagonal terms are there, and they're all minus 1-- whatever k is, it's going to be minus 1, times di,j,k. So we find that for inversion, di,j,k prime is always, regardless of what the three indices are, is going to be minus di,j,k. And if inversion is a symmetry operation which the crystal possesses, we're demanding that the transform index be identical to the original index. But there's a minus sign in there. So we can say that every single element vanishes, has to vanish. So any crystal that has inversion in it is not going to be able to display the piezoelectric effect or any other third-rank tensor property. The electro-optic effect, piezoresistance, and there are a whole slew of them, which are examined for you in part on this sheet. One third-rank tensor property is the direct piezoelectric effect, which we've been discussing as our example. There is something called the converse piezoelectric effect, which describes the phenomenon where an applied electric field creates a strain. So the direct effect is you [? scush ?] your material, you develop charge. The converse piezoelectric effect says if you apply an electric field that's going to move the atoms around and induce polarizations, you are going to create a strain. And the absolutely mind-boggling thing is that the same array of 3 by 9 coefficients describe both the direct piezoelectric effect and the converse piezoelectric effect. So in one relation, we have that each component of polarization is di,j,k times the nine elements of stress. And that means we have to write three equations in nine variables, the elements of stress. And the converse piezoelectric effect, we're developing a strain. And there are, therefore, nine elements of strain. And we're applying a vector, e sub i, a field. So there are three components of that vector. 27 elements, 3 times 9. 27 elements, 9 times 3. So both of these are third-rank tensors. Strain transforms like a second-rank tensor. Field vectors transform like a first-rank tensor. Therefore, the coefficients are elements of a third-rank tensor. But what boggles the mind is that the same 27 numbers describe these two seemingly disparate phenomena. You don't prove this by symmetry. You don't prove this by the nature of stress and strain. This hinges on the thermodynamic argument, which I'm not going to go into. But what you do if you're willing to stretch tensor notation a little bit-- you can't have a 3 by 9 array when you've got a tensor of second rank on the left and a vector on the right. You have to write this in this fashion, that di,j,k times ei gives you the element of strain, epsilon j,k. And that's not proper tensor notation. But we're not going to quibble. Because if you allow this little departure from convention, then you can write both the converse piezoelectric effect and the direct piezoelectric effect in terms of the same matrix. But again, that is not intuitively obvious. It does not have to be the case. And it hinges on an argument in thermodynamics. Some other examples of piezoelectric relations that I've mentioned in the notes. If you apply a stress and you get a polarization, that stress produces a strain. So you also have to get a polarization if you're applying a stress, and write it in terms of the strain that's produced. Similarly, if you apply an electric field and it produces a strain, the crystal must be in a state of stress. So there must be relation between stress and applied electric field, which will also be a third-rank tensor. Those two relations are represented by coefficients given the symbol e. And again, the same array of 27 elements describes both effects. And these are not dignified with any special names. They're just tensor relations that have to be true because of the fact that stress and strain are coupled by elastic relations. Not surprisingly, then, the coefficients e must somehow involve the piezoelectric moduli di,j and the elastic properties of the material. Another effect which is a very interesting one is the electro-optic effect. If you apply an electric field to a material, you change the birefringence of the field. The birefringence is defined as the difference in index of refraction for light polarized in two orthogonal directions. Another tensor effect-- and I'll give you a note defining the terms next time we meet-- there is a piezoresistive effect, which you don't see talked about very much. But that's a property that Texas Instruments was interested in at one time. And I have a set of sheets defining those relations that the presenter of a paper at a meeting, one time, kindly gave to me. OK. So what other symmetry restrictions are there? Having shown that inversion will not permit any third-rank tensor property, we have gone from 32 point groups, lost interest in the 11 centrosymmetric point groups. And so there are only 21 piezoelectric point groups. And the way we would plod through all 21 of them would be to simply define, starting with a twofold axis. Let's say a twofold axis parallel to x3 would correspond to a change of axes ci,j that describes the new axes in terms of the original ones that consisted of minus 1, 0, 0; 0, minus 1, 0; 0, 0, 1. So if we look at an element in the piezoelectric matrix-- something like d1,6-- d1,6 actually corresponds in tensor notation to d1,3,2 plus d1,2,3. And d1,3,2 prime is going to be c1,i c3,j c2,k times all of the original tensor elements di,j,k. The only term of the form ci, something that is non-zero is c,1,1. And that has a value, minus 1. The only term of the form c3, something which is non-zero is c3,3. And that turns out to be plus 1. c2, something, the only form that's non-zero is c2,2. And that has value, minus 1. And this should be times d. And the only value of i that stayed was 1. The only value of j that stayed was 3. And then only value of k that stayed was 2. So this says that d1,3,2 should be equal to d1,3,2, which I don't like, unless it's supposed to be negative. I did something wrong here. Well, you see how easy it is, even if it didn't turn out right. And the form of the tensor for monoclinic crystal of symmetry 2, with a twofold access parallel to x3, has as shown in the lower left-hand corner of the handout on symmetry restrictions, it has eight non-zero terms. If you do the same thing for a mirror plane perpendicular to x3, you find that there are 10 non-zero terms. So we don't have the situation where all of the point groups that are able to show the property within a given crystal systems like monoclinic have exactly the same form of the property tensor. In fact, you'll notice a curious correspondence between the restrictions for symmetry 2 and symmetry m. All the terms that are 0 in symmetry 2 are non-zero in symmetry m, and vice versa. All the non-zero terms in symmetry 2 are 0 in symmetry m. And the reason for that is simply that this is the form of the direction cosine scheme for symmetry 2. The form of the direction cosine scheme for symmetry m, where the m is perpendicular to x3, would have the form 1, 0, 0; 0, 1, 0; 0, 0, minus 1. So ci,j, for a mirror plane perpendicular to x3, is exactly the negative of the direction cosine scheme for a twofold axis parallel to x3. And since the number of direction cosines is odd, this means that everything that has an equality between the di,j,k's for m would have the transformed element be the negative of the original one for 2, and vice versa. So that's why there's this complementary form of the tensors for the two monoclinic symmetries. Point out a couple of curious things in the tables. You really have to go through 21 of the symmetries independently. And you find that some of them do come out the same. Symmetry 4 bar 3m and symmetry 2:3 have restrictions of the same form. But for the most part-- Yes? AUDIENCE: I know why this is wrong. PROFESSOR: Why is that wrong? AUDIENCE: Because your notation of d1,6 is in fact not d1,3,2 but d1,2,1. Since you have two same indices, [? d3 ?] and minus [? d1. ?] PROFESSOR: OK. d1,6; d1,6. Ah, of course. Of course. d1,2,6 is the one up here. And that's 1, 2 and 2, 1 for the strains. OK. Thank you. So this is d1,1,2 plus d1,2,1. And so we would have 1 by 1,k and 2,k. 1, 1, and 2. And 1, 1, 1. 1,i; 1,j; and 2,j for this one. So we'd have c1,1 c1,1 c2,2. c c1,1 is minus 1; c1,1 is minus 1; c2,2 is minus 1. So d1,1,2 is minus d1,2,1, which means they have to be identically 0. Thank you. I'm sure nobody cares at this point. Very good. OK. Another curious thing that happens is that for cubic symmetry 4:3:2, it's acentric. But every single modulus is 0. And the reason, the explanation, is there's so many different transformations which have to leave the tensor invariant that the poor tensor just can't do it. It gives up, packs up, and goes home, leaving all the elements zero. Just no way you can get all the qualities to be satisfied. Another curious result that I point out for some of the hexagonal symmetries for symmetry 3:2, for symmetry 6:2:2, and for 3 over m and 6 bar 2m, all the elements of the form d3, something are identically 0. Which says you simply cannot create a polarization that has a component P3. You just cannot create a polarization perpendicular to the axis of high symmetry. Yeah. Got a question? AUDIENCE: So all of the cubic that has 4:3:2 symmetry or [? 4:1:0 ?], that means you can't get-- PROFESSOR: You just don't have any piezoelectric effect, even though the crystal is not centrosymmetric, requiring that all those transformations leave the tensor invariant; simultaneously, require that everything has to be 0. One final thing, and then I'm running a little bit over. But I'd like to go on to other aspects of piezoelectricity during the next hour. Remember something that we've shown and which I've asked you to look into again on one of the problem sets, and that is that the trace of the strain tensor gives you the change in volume. The first 3 by 3 blocks of this tensor that we've been looking at, were we to write the elements of strain, epsilon i,j, in terms of d-- epsilon j,k-- in terms of di,j,k times e sub k, it is this block in here which enters into the terms epsilon 1,1; epsilon 2,2; and epsilon 3,3. These terms give you the fractional change in volume. And the elements that are involved in the piezoelectric matrix are these terms in here. So if all of those terms are 0, it turns out that when you apply a stress and look at the electric field, or apply an electric field and look at the strain, if you apply a field and all of those nine terms are 0, you cannot create a strain. So there's no volume change. There can be a strain, but no volume change. The only deformation you can create is pure shear. So if you drive a piezoelectric oscillator with an electric field, for those property tensors for which that first 3 by 3 block are all zero, the thing can shear-- in the water, for example-- but it can't pulse. It can't have a volume change. And if you were to create a transducer for sonar applications, what you would want to do is to have piezoelectric device which expanded this way, to create a sound wave going through the water. If it just goes back and forth, it's going to slosh back and forth in the water and not create any sonic wave that could be used in sonar. Now that is true for elements being identically 0. It turns out it's also true for elements such as 4 bar, where two of those terms in the 3 by 3 block are the negative of one of the other. These will also have no volume change. So that's another very curious property of piezoelectric response that follows from these symmetry restrictions. OK. That's enough for our first session. When we come back, we'll look at the converse piezoelectric effect. And we'll also look at representation surfaces for specific piezoelectric devices and ask if it's possible for a third-rank tensor to have a representation surface that's analogous to the representation quadric for second-rank tensors. So I'm sure you'll all want to come back and hear the answer to that question.
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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 we will have weekly quiz tomorrow. There's been a lot of coverage, and so to focus you a bit, I'm going to confine the weekly quiz to glasses and chemical kinetics. So you don't have to worry about diffusion. We'll catch up on that, but I know there's so much material there. Let's keep it confined to glasses and chemical kinetics. And I'll be available today 4:30 to 5:30. If you can't be at that time, send me a note, and I'm sure we can figure out a time to get together. So what I want to do today is to start a new unit. We're going to start talking today about solutions, and do some solution chemistry, OK? Today we talk about solutions. And you might initially say, why are we talking about solutions? This is solid-state chemistry. I think up until now, you've seen that it's pretty rare that we use solids in their pure form. We usually have mixtures. So for example, when we studied glasses, we modified the glasses with an alkaline earth oxide. Well we, in fact, were making a solution of more than one component. So it's to get certain properties that we study solutions. Secondly, coming out of liquid phase is a way to make solids. So in terms of processing, we need to understand something about solutions. And then lastly, we're going to be studying, towards the end of the semester, a big unit on biochemistry. And biochemistry, why, you say, biochemistry? Why is he doing biochemistry? I thought this was solid-state chemistry. We are solid-state devices, but we're made of soft matter. At least the exoskeleton is soft matter. The endoskeleton is ceramic, right? Our bone structure is ceramic. Hydroxyapatite, calcium hydroxyapatite, and the outside is a polymer. So look at this. Confirmational changes in polymer. But the chemistry, the biochemistry, so much of it takes place in aqueous solution. So hence, we better know something about aqueous solution. So to this, we go almost right back to the first day. You remember, was the second lecture, and we showed this figure, and all the different categories of matter. And we started over here with elements, and we moved into some pure substance compounds, et cetera, et cetera. Now we're going to move over to here. So homogeneous mixture containing uniform composition and properties, as opposed to a heterogeneous mix. So we're now over here. We're working our way through the diagram. All right. So let's get a couple of basic definitions up. So the solution is really a mix of at least two constituents. One, the majority constituent is called the solvent, and then we can have one or more solutes. So the solvent, this is the majority constituent, and then the solutes are the minority constituents. And in some instances, it's pretty hard to tell which is which. And as far as types of solutions, I want to broaden this. You know, when I say solution, you're probably thinking about aqueous solution. But I want to take a minute and work through this chart. And this is all posted, so you don't have to write it all down. Just follow with me. So most of the general chemistry subjects will just stop at the end of the first line. They'll treat solutions as aqueous solutions. But in material science, we think of things broadly. So a good example of a simple aqueous solution is sodium chloride and water. Sodium chloride is the solute and water is the solvent. But you can have a liquid solute. So wine-- in case you've ever encountered this beverage, it's primarily water, but it can contain up to about 14% ethyl alcohol, and there are other constituents that give the color and the flavor and so on. And you can have a gas in a liquid, and that would be seltzer, where CO2 is dissolved. It's actually dissolved. If you take a look at a bottle of bubbly water on the shelf, you don't see the bubbles. The carbon dioxide is actually dissolved. You can have a gas as a solvent, and air would be an example of that, where the solvent is nitrogen. And then we have as solutes oxygen, argon, carbon dioxide, if you live next to a power plant, it's sulfur dioxide, if you live next to an aluminum smelter, it's tetrafluoromethane, and so on. So we have all kinds of solutes in the air. And then we can have solid solutions. And OK, as soon as you see the word solid, you know, that means you're 3.091. So what do we see for solids? Metal alloys. We saw carbon dissolved in iron. Well, that's a solution. It's homogeneous, single phase, just as the definition on that chart 1.11 said. Semiconductor, boron doping into silicon, the boron sits on a silicon lattice site. So this is a true solution. The boron is the solute, and silicon is the solvent. We can have the ceramic. We talked about the oxygen sensor. The oxygen sensor is zirconia, which has been stabilized with the addition of calcium oxide. As one example. In contemporary work, they might use another oxide. But what's the purpose of the calcium oxide? Well, I told you that it's to increase the vacancy population, give you a rapid response on your oxygen sensor. And as we're going to learn later, that's true, and there's a second value in putting in the calcium oxide. And it stabilizes, that's why they use the term stabilize, it stabilizes the cubic form of zirconia. Zirconia has a different crystallographic modifications. The cubic one is the one that gives us the best ability to transfer oxygen, and the addition of calcium oxide as a solute, the calcium ions actually sit on the zirconium lattice, that's a true solid solution. It also makes the cubic zirconia stable, and with Christmas coming, you know, it's the poor man's diamond, and that's the same material there. And to modify a glass, as I mentioned earlier, adding a alkaline earth oxide breaks the silicate network. That's a solution. Now here's an inverse one, where the solid is the solvent, and the liquid is the solute. So this was dentist's office practice going back to the time when I was your age. If you had a feeling, the dentist had silver and mercury in the dentist's office, and would add liquid mercury to silver and make up this amalgam, and then jam that into the tooth. Actually, that was the B part of the operation. There was this dentist here in the United States, and if I could find him, I would like to have a word with him. But he had this theory. You see, the amalgam is a metallic alloy. And so obviously, it has metallic bonding. And the tooth is-- it's a ceramic. So what kind of bonds are going to form between a metal and a ceramic? They're not very good. So this dentist had the theory that the way to increase the bonding capability of the amalgam to the tooth was to maximize contact areas. So when you went in with a tiny, tiny little cavity, the dentist would drill the tooth out, removing about 75% of the volume of the tooth, and then they would make this amalgam and shove that in. And you'd wear that for about 20 years, until one day you bite into a muffin that's got a little piece of walnut shell in it, and the amalgam flexes, and your thin-walled tooth goes bang! Like that. And now you get to go back to the dentist's office for yet some more medieval treatment. Actually, things have improved somewhat. Thanks to material science. But they're not using this amalgam anymore. But there are a number of us who are still walking around with silver and mercury in our mouths, and-- yeah. Enough about my dental problems. Now let's go on to intercalation. You can put a gas into a solid. And I told you about this one, where if you want to store hydrogen, and this is a big problem, if we're going to talk about hydrogen-powered vehicles. A problem as big as, how to get the fuel cell cheap enough, is where are you going to store the hydrogen on the car, and one of the materials that's been proposed is this alloy of lanthanum and nickel that can intercalate huge amounts of hydrogen. So that forms a solid solution. So those are examples of solutions, and we're going to go back and make sure that we cover the basic Gen Chem at the top here. But I want you know that solution chemistry is very broad. Now when you dissolve something, you actually have things down at the atomic level. So for example, in brine, you actually have below two nanometers as particle size. You actually have sodium ions and chloride ions dissolved completely, which means that the solution is clear, colorless, and transparent to visible light. Water is clear, colorless, transparent to visible light. If you had sodium chloride, it remains clear, colorless, transparent to visible light. You can't filter, and you can't wait for the sodium chloride to settle out, because it's bonded within the structure of the liquid water. And this has implications. There's so much water on the planet, but there's not so much fresh water on the planet. And desalination can't be accomplished by filtration. A filter that had a pore size small enough to trap sodium ions, the pore size would be so small, water molecules couldn't go through. So this is the concept. At the other end, you can get something called a suspension, where the particle size is greater than about 1,000 nanometers, and blood is a good example of that, where it's opaque. Visible light doesn't go through, but you can filter out some of the matter in blood. And if you let it sit for a while, it'll settle in it gravitational field. And in between, you have this whole zone of colloids. And so between colloids and suspensions, the only difference is particle size. And this whole thing we would just call a dispersion of another phase, either another solid phase, or another liquid phase. And it's kind of an interesting physical chemistry about the dispersion, what makes it work. For example, in milk, milk is a good example of a dispersion. You have a fatty phase, and you have an aqueous phase. The aqueous phase, where all the minerals are. Why? Because the minerals are-- what kind of compound? They're not metallic, they're not covalent. They're ionic. The ionic compounds dissolve in the aqueous phase, and then in the fatty phase, that's where you have the protein and so on. But the fatty phase is clear and colorless, and the aqueous phase is clear and colorless. And yet milk is white. It's, as the term implies, it's milky. Why? What's going on? You have a second phase here. This could be the fatty phase, and this is the aqueous phase. They're both clear and colorless, but they have different indices of refraction. And since this has n fatty phase, and this has n aqueous phase, this interface scatters the light. So if you want to start a business, if you want to try to tailor the index so that the index of the aqueous phase matches the index of refraction of the fatty phase, you'd have the two dissolve, one and the other, and it would be transparent, divisible light. So you would have milk that isn't milky. I mean, the public would be very confused. But anyways. So that's what you do. So why does this thing not settle? Because they do have a density difference, and again, going back to the days when I was not a college student but a youngster, there was still this form of milk called pasteurized milk. Well, all milk is pasteurized, but this milk was not homogenized. And what would happen is, there was this person that would deliver milk. This was a borosilicate glass bottle. And here would be the cream. The cream would rise to the top. We have all these expressions in our language. And then this would be the milk here, and you could skim this off for coffee or sugar, and then this would be a low-fat milk. But people wanted this all mixed, so then they went to homogenized milk. So what is homogenized milk? This is your red cap, now. All the milk is homogenized. This was a big thing. This was simply called pasteurized. I'm going to get to the physical chemistry here. It's very interesting. Because this is lower density, and yet in homogenized milk it doesn't rise. So let's take a look at what goes on in the physical chemistry of homogenized milk, because it's all about these various systems. So I'm going to take this particle here, and its sum insoluble cluster. And I'm not specifying the cluster size. It's probably greater than about two nanometers. So there are two forces acting on this. There's a settling force and there's a buoyancy force. Obviously, otherwise it wouldn't float. So there's some kind of a buoyancy force. Settling force and a buoyancy force. Well, the settling force, this is just the gravity. Right? This is the force of gravity. And we know the force of gravity. That goes with the mass. Come on, get that cell phone out of here. This force of gravity goes as the mass, and the mass, we know, goes as the volume. And the volume goes as the cube of the radius. I'm assuming this is a spherical particle, all right? Now, the buoyancy force. The buoyancy force is the interfacial force between the two. There's some binding across here. Maybe weak van der Waals, or if this fatty phase has molecules in it that are polar, then there could be dipole-dipole interaction. But in any case, there's some kind of an interfacial force here. And this is all chemical bonding between solute and solvent. But you see, the force is a weak force. If it were a really strong force, it would dissolve this thing. It won't quite dissolve it, but there is some kind of dipole-dipole weak force. And this one here is operating across the surface area. That's the contact. So this force goes as the area, and area goes as the square of the radius, whereas mass goes as the cube of the radius. So you know, from your math, that r cubed dominates r squared, but only at large r. It's not always the case, is it? Maybe before you got here, you thought that. But now that you've been at MIT a few months, you know that r squared can dominate r cubed at small r. Interfacial forces dominate. And that's exactly what happens in these dispersions, and that's why they don't settle out. Now, homogenized milk is simply milk that has been agitated in such a way as to reduce the fat globule size below a critical value so that these interfacial forces hold the fat globules in suspension. If you waited long enough, they would agglomerate and settle, but that time is probably longer than the shelf life of the milk. So the milk probably spoils before stuff settles out. So this is all very important to understand. The range that exists between insolubility and this sort of clustering, and suspension, and so on. By the way, a lot of pharmaceuticals are like this. A lot of pharmaceuticals. So when it says, shake well before using, they're not kidding. Because the active ingredient will settle. And you're drinking just the solvent, just the vehicle. And all of the potency is on the bottom of the bottle. Shake that thing up! I can't say what it does to the taste, but that's another problem. Actually, I threw in this slide here. We're not going to spend any time on it, but you can look at it at some point. This is a whole taxonomy of colloids. Solid-liquid emulsions, aerosols, they're all part of this magic zone between solubility and just brick, all right? There's this whole fine, pardon the pun, the whole fine structure. OK, So let's get to the chemistry. Obviously there's something to do with bonding here, right? So here's a simple experiment, and this is taken right from the reading. So I've just taken this episode that is written up in the reading. So we've got two beakers here, and in each beaker, we have a bilayer. We've poured in some carbon tetrachloride, liquid, and we've poured in some water. And these two are immiscible, because carbon tetrachloride is obviously a non-polar liquid, and water is a polar liquid with hydrogen bonding capability. And in one beaker, we introduce crystals of iodine. In the other beaker, we introduce crystals of potassium permanganate. And then we shake them up, and we wait. And eventually we see that on the left, the iodine dissolves in the carbon tetrachloride. And we're using the purple color as an indicator. And this is kind of cute, because both potassium permanganate and iodine will render things purple. So you're comparing purple to purple. They could have chosen something else, but this is kind of cute. All right. So here you end up with a solution of iodine and carbon tetrachloride, whereas on the right side, you end up with a solution of potassium permanganate and water, and nothing in the carbon tetrachloride. So what can we infer from this? Well, let's take a look at the possible interactions. So first of all let's categorize H2O. This is polar, it's a polar liquid, with hydrogen bonding capability. Carbon tetrachloride is non-polar. It's very toxic. When I was a child, we had this in the medicine cabinet. It's a non-polar solvent. It's fantastic for getting grease stains off. Every man had this. With a little handkerchief, you'd take a little grease stain off your tie. You can't buy this stuff. You can't even guy it for research anymore. It's too bad. It's great stuff. Highly toxic, though. But you know, there's a time and a place. All right. So then here's iodine. Iodine, we know, is non-polar. It's a homonuclear molecule, it has to be non-polar. And so what holds iodine together in the crystal? There's only one bond, and that's van der Waals, right? It's a van der Waals solid, whereas potassium permanganate is an ionic solid. Potassium permanganate is ionic, and it consists of potassium cations and permanganate anions. And what we find is that the non-polar solute dissolves in the non-polar solvent, and the ionic solute dissolves in the polar solvent with hydrogen bonding capability. So from this, we can infer the general rule, which is encapsulated in the language used in chemistry textbooks. Like dissolves like. And what they're really saying here, is that like solutes dissolve in like solvents. Solute-solvent is like dissolves like. OK. It's a good place to start, but it's an incomplete picture. So I want to show you that there's some sophistication here. This is taken from one of the other books. And it shows just the rules for ionic compounds in water. And I just showed you, from this example, that the ionic compound dissolved in water. And what you see here is that some ionic compounds dissolve in water, but there's a whole set or insoluble ionic compounds. So it's not straightforward. But we know from 3.091, we know that we can make sense of this on the basis of competition. Competition between what holds the compound in the solid state versus what will pull it into the liquid state. So for example, we can compare sodium chloride, which we know dissolves in water. So sodium chloride, as a crystalline solid, will dissolve to form sodium chloride aqueous solution. Whereas if you look at magnesium oxide, which is also an ionic crystal, it does not, to any standard imaginable, dissolve to form an aqueous solution of magnesium oxide. And what's the difference here? The difference here is, compare solvation energy, in other words, the energy that you got by pulling this into solution, and forming bonds between sodium and chlorine in water, with the crystallization energy. And what's that all about? Well, that's the Madelung constant, remember? Madelung constant. q1 q2 over 4 pi epsilon 0 r, where this is the cation anion. And you can see here that sodium chloride has sodium cations and chloride anions, and there's a certain binding energy in the crystal. Magnesium has 2 plus. Oxide is 2 minus. The binding energy between magnesium and oxygen is so great that there's no driving force to dissolve. Now, I don't expect you to be able to look at this and tell me whether something's going to dissolve or not. But if I were to say to you, explain why sodium chloride forms solutions with water, and magnesium oxide doesn't, that you could go through this rationalization. And here's a cartoon from the textbook that tries to illustrate this solvation. Here you can see a crystal of sodium chloride, in the inimitable fashion, as drawn by chemistry books, where the chloride ion is green, and the sodium ion is blue. And that's OK. I can live with the color-coding, as long as we agree amongst ourselves, these ions are clear and colorless, because they have octet stability in their electronic shell. And here's water, with the hydrogen shown in white, and the oxygen shown in red. And you can see that the oxide end, the oxygen end, the delta negative end of the water, is trying to wrest sodium cation out of the lattice, and ultimately surround it by a cage. And the same thing here. You see the hydrogen ends of the water trying to pull chloride out, and ultimately surround it with a water-like cage. So this is the competition I was talking about. In the case of sodium chloride, water wins. In the case of magnesium oxide, water loses. The binding energy between magnesium and oxygen and the crystal dominate. So you don't see that solvation. OK. Let's talk about metrics now. Let's look at some metrics of solubility. It's quantifiable. So we can express a measure of solubility in terms of a quantity known as molarity. So we can express moles of solute per liter of solvent. And this is called molarity, and the symbol is capital M. So we can say, for example, a 1 molar solution of sodium chloride in water, we'll write 1 molar NaCl, and then we'll write subscript aq, meaning aqueous. So it's an aqueous solution at a concentration of 1 mole of NaCl per liter. And the liter is named after a person, so that's capital L, per liter of solution. And remember, the solution is the sum of the water plus solute. For dilute solutions, there's very little difference between the total amount of water and the total amount of solution. But in certain instances, the presence of the solute actually has a volumetric change on here. So strictly speaking, it's per liter of solution And as I've shown you, there are degrees of solubility. So people represent the threshold of solubility as c of the solute. When c of the solute is less than about a million molar, 0.001 molar, we call this insoluble. So something is vanishingly soluble, we say that's the value. So we'll call this the threshold. And then, something that's quite soluble, we'd say that the concentration of the solute starts to exceed at about 0.1 molar. And then we would say, that qualifies as soluble. So now let's look at two extremes in solubility, operating off of this. So in the one case, we can have complete solubility. So examples of that, where things are completely miscible in one another. Complete solubility. By the way, some people will use the term miscibility. When something is miscible it means it's soluble. Same idea. If something is insoluble, some people might say it's immiscible. Same thing. OK? So complete solubility is ethyl alcohol and water. You can mix them in all proportions. Continuously variable. On the solid alloy is silver and gold. Silver and gold, you can make alloys of any composition between 100% gold and 100% silver. Now, that's the exception. Most cases are situations of limited solubility. So they go up to a maximum, which we can denote C-star or C saturation. This is the maximum solubility. So an example of something that's sparingly soluble in water is silver chloride. Let's look at silver chloride. So silver chloride, I'm going to start here, silver chloride as a crystal, and I'm going to dissolve that in water. So that gives me AgCl aq. So that's just simply the formation of an aqueous solution of silver chloride. And when the reaction moves from left to right, we call that dissolution. The silver chloride is dissolving, and when the system moves from right to left, we call that-- now here I'm going to nitpick with the book. The book calls the left reaction crystallization. And that's correct. It is crystallization in this case, because silver chloride is a crystal. But it is possible, in other systems, to have the solute come out of solution, and make a solid that is noncrystalline. And you know that all crystals are solids, but not all solids are crystals. You can have an amorphous solid. So what would happen if you were to bring out of solution an amorphous solid? It would be silly to call it crystallization. So I prefer to use the term precipitation. And there's another term that you can use, and I learned this one from reading the literature of geochemistry. What the geochemists call the reaction going from solution to make a solid, they say the system exsolves. This is dissolve, this is exsolve. So this is called exsolution. It's amazing what you can do when you know a little bit of Latin. So this exsolves. So that's the exsolution, or the crystallization reaction. Now I'm going to show you what won Arrhenius his Nobel prize. Arrhenius did not get the Nobel prize for his brilliant work on activation energy. He got his Nobel prize on the theory of electrolytic dissociation, which was, people knew that you could dissolve salts and water, but they didn't know how. And it was Arrhenius who said that the salts go into solution by dissociating and forming ions. So goes in as Ag plus Ag plus aq plus chloride ion-- thank you. And that, ladies and gentleman, was a Nobel prize for Arrhenius. And you can see that there's a relationship between the amount of silver chloride and the amount of ions. So there's a mass balance there, that the concentration of silver chloride dissolved, in fact, equals, in this case, by stoichiometry, the concentration of Ag plus, because of the nature of the dissociation reaction on a mass balance basis. And that also equals the concentration of the chloride ion, OK? So that's the way we can look at the system. And how do we know that this thing has limited solubility? Well, there's various ways of measuring it, and one of them involves conductivity. Here's the conductivity of pure water. And you know that water has very, very poor conductivity, and in fact, what's happening here, when we add silver chloride is, we're adding charge carriers, because the audience are charged species. So they can carry charge. And you can see that this is a measure of conductivity as a function of silver chloride concentration. And as you add silver chloride to higher and higher values, the conductivity goes up. And look at even the tiniest amount of silver chloride has a conductivity that's about, what, half an order of magnitude higher than the conductivity of pure water itself. So I would say that this is akin to doping, isn't it? So up here, when you've got 10 to the minus 6 Siemens per centimeter conductivity, that aqueous solution is demonstrating the extrinsic behavior. This is very similar to doping. And at some point, we get to this value here, around 10 to the minus 5 moles of silver chloride per liter, and then adding more silver chloride has no impact on conductivity. Which tells you that you've hit saturation. This is akin to adding more and more sugar to the cup of tea, until finally the sugar just falls to the bottom. You can stir all you want, but you won't get it to dissolve, because you've hit saturation. So that indicates the presence of a saturated solution. And so we can talk about that value, and we can say that for silver chloride, the concentration at saturation is equal to 1.3 times 10 to the minus 5 moles per liter. Moles of silver chloride per liter. And obviously, that's equal to, according to that, it's equal to the silver ion concentration. I'm going to use square brackets to indicate moles of silver, ion per liter of solution, which is also equal to-- pardon me-- it's also equal to the chloride ion concentration at saturation. Now I'm going to ask you a simple question. Suppose I've got a beaker here, and I know the maximum I can get here, the maximum is 1.3 times 10 to the minus 5. Now this is a really simple question. Suppose I am about to add silver chloride-- let's say this is 1 liter already. All right? I've got one liter. And you'd tell me, well, you can put in 1.3 times 10 to the minus 5 moles to get the saturation. Suppose instead of 1 liter of pure water, I had 1 liter of water already containing, say, 1 times 10 to the minus 5 molar silver chloride. Well, that's kind of obvious, isn't it? I'm only going to be able to put 0.3 moles in, because 0.3 times 10 to the minus 5, because I've already got silver chloride in there. That's easy. Now let's make the question a little more interesting. Suppose instead of a certain amount of silver chloride in there, I have no silver chloride in there. But I've got, say, 0.1 molar sodium chloride. It's a salt. It's a difference salt. So the question is, does the presence of a different salt have an impact on how much silver chloride I can put into this solution? And the answer is, yes. The answer is yes. So what we find is that the presence of the other salt, in this case, has an impact, because sodium chloride goes in as sodium plus, and chloride minus. So there's already a boatload of chloride ion in there, and that has an impact on this relationship. So how do we answer the question, what is the solubility of silver chloride in the presence of other chloride ions? And for that, we define something called the solubility product. And you need it in order to answer the question, how do you determine the solubility of a solute when there are other solutes present already? And it's denoted capital K, lowercase sp, subscript. Solubility product. And it's equal to simply the ion products of the constituents. So the solubility product of silver chloride is the product of the silver ion concentration, and the chloride ion concentration. And you know that in a solution of silver chloride alone, if nothing else, that the concentration of silver ion equals the concentration of chloride ion. That's the whole business of dissolving by itself. And so I can then just say that Ksp. will then equal the concentration of silver ion squared, which we also know is equal to the concentration of silver chloride aqueous. See, all of these are the same. So this solubility is the same. I can just put that in, which is just concentration of silver chloride. Square that. So if I square 1.3 times 10 to the minus 5, I end up with a solubility product of 1.8 times 10 to the minus 10. So now I can use this in order to determine how much solubility I get in the presence of another salt. So in this case, I'm going to put 0.1 molar sodium chloride. And these are strong salts, so we're going to get complete dissociation. Gives me 0.1 molar sodium ion, and 0.1 molar chloride ion, when it dissociates. Now you see the difference. Because the silver chloride, by itself, gives me 10 to the minus 5 molar chloride ions. When I add sodium chloride, I get 4 as a magnitude more chloride. So let's go back to the Ksp. So Ksp, this is for silver chloride. Ksp for silver chloride is going to be equal to the silver concentration and the chloride concentration. And in this case, the silver concentration is just equal to whatever that solubility is. Because there's only source of silver ion, and it's silver chloride. So I can write that as concentration of silver chloride. That's good. And now this one here is what? I've got two sources. I've got silver chloride, I I've got sodium chloride. So it's going to equal this thing here. 0.1 plus whatever I get from silver chloride. And it's vanishingly small, isn't it? The concentration of silver chloride, whatever it is. This is nothing, so I'm just going to neglect it. It's dominated now. The chlorine is flooded by the silver chloride. And this product is a constant. That's still equal to 10 to the minus 10. So I can turn this around and solve for the concentration of silver chloride, which is equal to the concentration of silver io. And that's equal to, what is it, 10 to of the minus 10 divided by 0.1, which then gives me-- what's the number here? Plug that in. And I end up with 10 to the 1.8 times 10 to the minus 9 molar. Right? So look. Look at what's happened by having the chloride present from sodium chloride in such a large amount. It's repressed. It has repressed the dissolution. The presence of chloride then has a negative impact on solubility, and instead of having 10 to the minus 5 molar, it's down to 10 to the the minus 9 molar. And this effect of repressing solubility by adding a second solute is called the common ion effect, OK? Solubility repression by second solute is known as the common ion effect. And this is used in processing. If I want to trigger the precipitation-- see, if I started with a solution containing 10 to the minus 5 molar of silver chloride, and I throw in some sodium chloride, it'll start precipitating out silver chloride. So if I wanted to make a fine precipitate of silver chloride, I make a pregnant solution. And I could drop the temperature, because you can imagine that solubility is a function of temperature, or I could keep it isothermal, throw in some sodium chloride, and out comes silver chloride. So since the common ion effect on it and its value in processing. OK. Good. Well, I think I'm going to hold it there. I"ll show you just one more thing. If you've got stoichiometry like this-- this is a difluoride of magnesium. If it goes into solution, you get magnesium cation plus 2 fluoride anions. And so if I wrote the Ksp for this reaction, it would be the product of the magnesium concentration and the fluoride ion concentration. Because there's the two here, this is squared. So this, too, will transfer up there, and then throw in some sodium fluoride, and cause the other thing to exsolve, and away we go. All right. I've got a couple of things to show you. We're talking a lot about Arrhenius. This is a book I have. It's an English translation of a book written by Arrhenius in the late 1800s, and it was printed in English around 1908. And Arrhenius was a genius. He wrote on all sorts of topics here. Biology, physics, you name it. And one of the things and he was interested in was the origins of the earth. So this chapter is called, Celestial Bodies as Abodes of Organisms. Already speculating on whether you could have life as we know it exist elsewhere in the universe. And one of the things he talks about is global warming. So this is Arrhenius on global warming. I'll read you the last paragraph of this chapter. There had been some major volcanic eruptions that had caused cooling when Krakatoa in 1883 and Martinique in 1902. Major plumes of soot that caused dramatic decreases in temperature. So now here comes the last paragraph. We often hear lamentations that the coal stored up in the earth is wasted by the present generation-- remember, this is written 100 years ago-- without any thought of the future. And we are terrified by the awful destruction of life and property which is followed the volcanic eruptions of our day. We may find a kind of consolation in the consideration that adheres in every other case. There is good mixed with the evil. By the influence of the increasing percentage of carbonic acid in the atmosphere-- that's CO2-- we may hope to enjoy ages with more equitable and better climates.-- Remember, he's a Swede; it's cold-- especially as regards the colder regions of the earth, ages when the earth will bring forth much more abundant crops than at present for the benefit of rapidly propagating mankind. So it's interesting to see the world-- it's a great book to read. And you can see people in the 1830s were already calculating heat transfer coefficients to how much the earth was changing. We're going to post these to the website. This was, last year, in the New York Times. Every Tuesday they have a science section. And this was about glass. And very, actually, with your 3.091 knowledge, you'll read this like a newspaper, and it'll be very easy. And they go through the structure. Here's the structure of a window glass. You can see the network, former network modifier, and so on. And they talk about how difficult it is from a first principle's standpoint to model the structure of glass. These oxide glasses are complex and not easy to model. So when you're trying to engineer the glasses, instead of trial and error, it's hard to do so by theory. And it talks about some efforts at theory, and so on. And the last thing I want to talk about is bulk metallic glasses. You recall that I showed you metallic glass that was made by melting gold silicon, and dripping it onto a water-cooled copper wheel that was zooming around to give us a cooling rate of about a million degrees a second. And those strips had to be very, very thin, the metal strips. Because you've got liquid dripping down, and we're pulling the solid away, and there's a finite thickness here. Let's use a Greek letter, xi. There's a finite thickness xi. Because what's happening is that this is the water-cooled copper wheel, and you're extracting heat here. But eventually, the thermal conductivity of the metal is the limiter. In other words, I don't care how cold. You can put this in liquid helium if you want. You can't get the heat through the metal fast enough. And what happens is, when you look down here, the lower part is amorphous and the upper part is crystalline. So you don't end up with metallic glass. You end up with some metallic glass, and the upper part is crystalline. So what happens when you end up with the limitation being the thermal conductivity of the metal? At that point, you're finished. And this was typically on the order of microns. And then they got up to, sort of, submillimeter. And that was it. So the glass that I showed you was foil. Now what happened was, with more research-- so here we are. This is Pol Duwez at Caltech, gold silicon. And this is the thickness in centimeters. So you were down here at some tens of microns, all right? Now, in the '60s, research at Harvard uncovered a set of palladium alloys that had better thermal conductivity and, remember the first day, about the analogy to the musical chairs? These things have slightly more complicated crystal structures. So for a given cooling rate, the metal has more difficulty finding the proper lattice site. And they were able to make metallic glasses that were on the order of 0.1 centimeters. That's still fairly thin. And then back to Caltech in the late '80s, early '90s, and a man by name of Bill Johnson was able to develop a set of alloys that can be made in bulk form. Totally amorphous metal. And these are known as bulk metallic glasses. And look at the complexity of the alloy designation. So now you see, well, why are they doing this? Because look, this is strength versus elastic limit. So you can either have things like polymers, that you can stretch very, very far, but they don't have much strength. Or you can have things like certain steels, that are very, very strong, but you can't bend them very far. And bulk metallic glass has put you right up here. You have strong alloys that have very, very high elastic limits. So they make great golf clubs and tennis rackets and so on, because they can flex way back, store enormous energy, and then spring. So here are some bulk metallic glasses, and here's a classic one. This is zirconium, titanium, copper, nickel, beryllium. All right, now how do we get that? Well, zirconium and titanium are body-centered cubic, we've got copper and nickel are face-centered cubic, and beryllium is hexagonal close-packed. So the idea here is the principle of confusion. So the alloy is fighting with itself. You know, am I face-centered cubic, am I body-centered cubic, am I hexagonal? And this confusion about what the crystal structure is to be leads to quenching and the disorder of a liquid state, and preventing the formation of grain boundaries, dislocations and so on. So I want to show you one example besides golf clubs. Dave, could we cut to the document camera here? Get this thing down. So this is a-- I'm not endorsing the product at all, but you know, this is one of the companies that makes, this is SanDisk, I think. And they make these flash memories. This just looks like a another piece of metal, and in fact, it's very disarming, because they call this model the Titanium. Well, it has about 13% titanium. The interesting thing here, what's so cool about this, this is bulk metallic glass. And what's attractive about the fact that it's bulk metallic glass is, that it can be shaped by casting, from the liquid state to very, very fine precision. So you see all of these slots and everything, and on the side, all of this kind of stuff, and on the end. All of this is done by casting from the liquid state, in one operation. And this is a clam shell. There's two pieces here. I don't know if you can see, but there's two pieces that have been sandwiched together. You can see on the edge there. And the impact that has on manufacturing costs is phenomenal, because normally you make the basic shells, then you have to drill, and you've got to auger out, and so on. This thing, one step operation, including the labeling. There's no subsequent processing. And this is done by a company out in Michigan called Liquid Metal. And they've licensed the technology and so on. And again, I'm not trying to make a commercial thing, but I'm just trying to show you that these ideas of changing properties for engineered materials, you know, are around us everywhere. And this is relatively recent. Bulk metallic glass. Fantastic example of structure property relations. OK.
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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 that we've calculated the change in potential energy between some initial and final heights for the gravitational problem, mg of y final minus y initial. For this conservative force of gravity, and we had our coordinate system like that, we were able to calculate the change in potential energy. And remember our theorem is that non-conservative work equals Delta K plus Delta U. Now what we'd like to do is establish the concept of a reference point for potential energy. And we'll do that as follows, see the change in potential energy only depended on say, our initial state and our final state. What we'd like to do is introduce the concept of a reference point. So let's identify some point Yp to be our reference point. And we'll define a potential Yp to be the reference potential. And if we want to talk about the potential energy in a final state, we'll always refer it to the difference between that and the reference point. So this is Delta U between the final state and the reference point. Likewise, for the initial state, we'll always refer that with respect to our reference point. So any state that we have, we can always refer the potential energy difference between some state and some reference point. Why do we do this? Because notice that if we look at U final minus U reference point, and we subtract from that U initial minus U reference point, then the difference-- the reference points here we have a minus, here we have a plus-- this is just equal to U final minus U initial, which is Delta U. So the change in potential energy between any two states is independent of how we choose our reference potential. But it can make calculations easier when we can identify what the potential energy is for a little state. Now let's look at our example. So for our gravitational problem we will choose y reference point to be 0. And we'll choose the potential energy at this reference point also to be 0. So our reference potential at the origin is 0, and I'll denote it like that. Then the potential energy at some initial state minus the reference point-- well, we can use our formula here, because this is between any two states. So this is mg y initial minus the reference point. But our reference point was 0, and so we see that U initial minus y reference point, which was also 0, is just equal to mg Yi. And so we have this statement that the potential energy difference between our initial state and the reference point is just mg where Yi is the height that the initial state is above the reference point. In a similar way, we have U final is mg y final. And so we see we recover what we expect. This is just mg y final minus y initial. Now this we can generalize just a little bit by saying that for any-- let's write that for us our mass, which is for any height y our potential energy function for the gravitational force U(y) is equal to mg y. That's a formula that many of you have seen before. But it's very important to note that with this formula, U of 0 equals 0, because that's our reference point. And that becomes our potential energy function for the gravitational force. Now our next step will be to do the same thing for spring forces and inverse square gravitational forces. And we'll also look at graphical analysis of this 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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So when we have a vector product of two vectors, A cross B equals C, let's compute that vector product in different coordinate systems. So let's begin by choosing two vectors. I hat, and J hat. And notice they're at a right angle, and because there is a unit vector, the area here is equal to and 1. And I want to define K hat to be equal to I hat cross J hat. Now, our angle theta here was 90 degrees. And so, if I use the right hand rule, then I hat, cross J hat is-- the right handed unit normal is out of the plane of the figure. And so I would write K hat like that, and notice K hat is out of plane of figure. And because I used my right hand rule, this is what we call a right handed coordinate system. Now, there's something very nice about this cross product definition. Notice that there's a cyclic order. IJK is a cyclic order. And if you interchange any two. For example, JIK, that's anti-cyclic. And the cross products satisfy this cyclic rule, in that, J cross K hat is I hat. And notice JKI, maintains that cyclic order, and K hat cross I hat is J hat. KIJ maintains that cyclic order, but because of the way we defined a cross-product in general A cross B is minus B cross A, because now you're using the opposite direction, so there's a minus sign. And therefore, any anti-cyclic permutation of these unit vectors, as an example, K hat cross J hat, has to be-- notice I've-- is minus I hat-- that's anti-communitive property of the cross-product. Similarly, I hat cross K hat is minus J hat. And lastly, J hat cross I hat is minus K hat. And so, in fact, you only need to know one. And this idea of cyclic, and anti-cyclic, to be able to write down all of the other 6. So we have one, two, three, four, five, six. Now, when you want to compute the cross products in Cartesian coordinates, for instance, it can be a little bit messy. There's going to be a lot of terms. If I write a vector A as A X I hat plus A Y J hat plus A Z K hat. And I write a vector B as BX I hat plus BY J hat plus BZ K hat. And now I want to compute the cross-product of these vectors to get the new one. Notice that there's going to be six terms, because I have I hat-- well, we actually should say one more thing. that I hat cross I hat-- the angle between these two vectors is zero. There's no perpendicular projection. So that's zero, as is J hat cross J hat, K hat cross K hat. So of these nine parts, when we take the cross-product, three of them will be 0, by this rule, and we'll we apply our cyclic or anti-cyclic rules for the others. And so what we have here-- let's just do it in-- so C equals A cross B. And now let's just go one by one. I hat cross I hat. That's 0. There's no perpendicular part. The area formed by these two vectors is 0. I hat cross J hat. That's cyclic. IJ is plus K hat. So our first non-zero term is AXBY K hat I hat cross J hat. And now let's do I hat cross K hat. Notice that's anti-cyclic. I K minus J. So our next term is minus AX BZ J hat. So there's the first two. And now let's just continue this process. J hat cross I hat. That's anti-cyclic. So we have minus AY BX K hat. AY J hat across BX I hat. J hat cross J hat. That's 0. So we have no contribution there. And J hat crossed K hat, that is cyclic. So that's plus AY BZ K hat. And now we have our last two terms K hat cross I hat. That cyclic. So that's plus AZ BX J hat. K hat cross J hat. That's anti-cyclic. So there's a minus I hat. So it's minus AZ BY I hat. And finally K hat cross K hat, well that's also 0. So we have six terms. And we can collect them equal to AX BY minus-- well let's check this one. Here we used the wrong symbol here. We have to be a little bit careful here. AY cross BZ is J hat cross K hat. That's plus I hat. So we have AXBY minus AY BX K hat. And now let's look at the I hat terms. I'll just check those off. We have AY BZ minus AZ BY I hat. And check those two terms off. And lastly, we have AZ BX minus AXPZ. So we have AZBX minus AX BZ and that's J hat. And that's how we calculate the cross-product in Cartesian coordinates.
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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: Great. So I will begin with phase shifts and do the introduction of how to make sure we can really-- so this is the important part of this. Just like when we added the reflected and transmitted wave we could find the solution I'm going to try to explain why with this things we can find solutions in general. So this is the subject of partial waves, and it's a nice subject, a little technical. There might seem to be a lot of formulas here, but the ideas are relatively simple once one keeps in mind the one dimensional analogies. The one dimensional analogies are very valuable here, and we will emphasize them a lot. So we will discuss partial waves and face shifts. So it's time to simplify this matters a little bit. And to do that I will assume from now on that the potential is central so v of r is equal to v of r. That will simplify the azimuthal dependence. There will be no azimuthal dependencies. You see, the thing is spherical is symmetric, but still you're coming from a particular direction, the z. So you can expect now that the scatter wave depends on the angle of the particle with respect to z because it's spherically symmetrical. But it shouldn't depend on five, the angle five, should just depend on theta. So expect f of theta. Now, a free particle is something we all know how to solve, e to the ikx. Why do we bother with the free particle in so many ways? Because free particle is very important. Part of the solution is free particles. To some degree far away it is free particles as well. And we need to understand free particles in spherical coordinates. So it's something we've done in 805 and sometimes in 804, and we look at the radial equation which is associated to spherical coordinates for a free particle. So we'll consider free particle and we'd say, well, that's very simple but it's not all that simple in spherical coordinates, and you'd say, OK, if it's not simple, it's spherical coordinates, why do we bother? We bother because scattering is happening in spherical coordinates. So we can't escape having to do the free particle in spherical coordinates. It is something you have to do. So what are solutions in spherical coordinates? We'll have solution SI of r. Remember the language with coordinates was a U of r divided by r and of Ylm of omega. That was a typical solution, a single solution of the showing our equation will-- the U only depends on l, the m disappears, so this is r. This r's are r's without the vector because you're already talking about the radial equations, and depend on the energy and depend on the value of the l quantum number. So what is the Schrodinger equation? The radial equation is minus h squared over 2m, the second the r squared plus. h squared over 2m l times l plus 1 over r squared. Remember the potential centrifugal barrier in the effective potential, then you would have v of r here, but it's free particle, so v of r is equal to 0. So if nothing else, U of El of little r is equal to the energy, which is h squared k squared over 2m UEl. And that's a parliamentary session of the energy in terms of the k squared, like that. Well, there's lots of h squared, k squared, and 2m's, so we can get rid of them. Cancel the h squared over 2m. You get minus d second dr squared plus l times l plus 1 over r. UEl is equal to k squared UEl. It's a nice equation. It's the equation of the free particle in spherical coordinates. Now, this is like the Schrodinger equation. And I think when you look at that you could get puzzled whether or not the value of k squared or the energy might end up being quantized. With the Schrodinger equation many times quantized is the energy, but here it shouldn't happen. This is a free particle. All values of k should be allowed, so there should be no quantization. This is an r squared here. You can see one reason, at least analytically, that there is no quantization is that you can define a new variable row equal kr and then this whole differential equation becomes minus the second the row squared plus l times l plus 1 over row squared. Well, I can put the other number in there as well, or should I not? No, it's not done here. UEl is equal to UEl, and the k squared disappeared completely. That tells you that the case will kind of get quantized. If there is a solution of this differential equation it holds for all values of k. And these are going to be like plane waves, and maybe that's another reason you can think that k doesn't get quantized because these solutions are not normalizable anyway, so it shouldn't get quantized. So with this equation in here we get the two main solutions. The solutions of this differential equation are vessel functions, spherical vessel functions. UEl is equal to a constant Al times row times the vessel function lowercase j of row. There's a row times that function. That's the way it shows up. It's kind of interesting. It's because in fact you have to divide U by r, so that would mean dividing U by row, and it means that the radial function is just the vessel function without anything else. And then there's the other vessel function, the n of l a row times of n of l of row. So those are spherical vessel functions. As you're familiar from the notation that j is the one that this healthy at row equals 0 doesn't diverge the n is the solution that diverges at the origin. And both of them behave nicely far away. So Jl of x goes like 1 over x sine of x minus l pi over 2, and ADA l of x behaves like minus 1 over x cosine of x minus l pi over 2. This is for x big, x much greater than 1, you have this behavior. So these are our solutions, and here is the thing that we have to do. We have to rewrite our solutions in terms of spherical waves because this was the spherical wave so we should even write this part as a spherical wave. And this is a very interesting and in some way strange representation of E to the ikz You have E to the ikz that you have an intuition for it as a plane wave in the z direction represent it as an infinite sum of incoming and outgoing spherical waves. That's what's going to happen. So this is the last thing we need do here. We have that e to the ikz is a plane wave solution, so it's a solution of a free particle, so I should be able to write the superpositions of the solutions that we have found. So it should be a superposition of solutions of this type. So it could be a sum of coefficients al times, well, alm you think of some a's times solutions. Remember, we're writing a full solution, so a full solution you divide by r. So you divide by this quantity. So you could have an alm Jl of row plus Blm ATA l of row times Ylm. So this should be a general solution, and that would be a sum over l's and m's of all those quantities. But that's a lot more than what you need. First, this does not diverge near r equals 0. It has no divergence anywhere and the ATAs or the n's, I think they're n such and not ATAs, the n's diverge for row equal to 0. So none of this are necessary, so I can erase those. l and m. But there is more. This function is invariant and there are some beautiful rotations. If you have your axis here, here's the z, and you have a point here and you rotate that the value of z doesn't change. It's independent of phi for a given theta, z just depends on r of cosine theta. So there's no phi dependence but all the Ylm's with m difference from 0 have phi dependent. So m cannot be here either. m must be 0. So you must be down to sum over l, al some coefficient, Jl of row, Yl0. And all of those would be perfectly good plane wave solutions. Whatever numbers you choose for the little al's, those are good solutions because we've build them by taking linear combinations of exact solutions of this equation. But to represent this quantity the al's must take particular values. So what is that formula? That formula is quite famous, and perhaps even you could discuss this in recitation. e to the ikz, which is e to the ikr cosine theta, is the sum 4 pi. Now you have to get all the constants right. Square root of 4 pi, sum from l equals 0 to infinity, square root of 2l plus 1. Coefficients are pretty funny. They get worse very fast. Now you have of i to the I, i to the l, Yl0 of theta doesn't depend on phi, Jl of kr. This is the expansion that we need. There's no way we can make problems with this problem unless we have this expansion. But now if Y the intuition that I was telling you of these waves coming in and out, well, you have e to the ikz, you sum an infinite sum over partial waves. A partial wave is a different value of l. These are partial waves. As I was saying, any solution is a sum of partial waves is a sum over l. And where are the waves? Well, the Jl of kr far away is a sine, and the sine of x is an exponential ix minus e to the minus ix over 2. So here you have exponentials of e to the ikr and exponentials of e to the minus ikr, which are waves that are here like outgoing waves and incoming waves. So the E to the ikz's are sum of ingoing and outgoing spherical waves. And that's an intuition that we will exploit very clearly to solve this problem. So we will do that next.
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https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/8.01sc-fall-2016.zip
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When a wheel is rolling without slipping, as we saw before, that the contact point here, the contact point is instantaneously at rest. Now, if the wheel is rolling on a surface with friction, then it's possible that we may have a static friction force. So there could be static friction may act. However, static fiction is always depends on the circumstances. Now let's just consider two cases. Suppose here's a wheel, Vcm. This is a horizontal surface. And in that case, the static friction, if the wheel is rolling with velocity V, in this case f static is 0. And because of that, the wheel will roll without any friction, which is an idealization. There's other types of friction called rolling resistance, air resistance, et cetera, which will slow the wheel down. But in a perfect hard wheel, idealized wheel in a vacuum, it will keep on rolling. However, if a wheel is going down an incline plane and it wants to maintain the rolling without slipping condition-- so here we have Vcm. And it has some angular speed-- we have to be careful because in this case if we differentiate Acm is our alpha. So what is making the wheel spin faster? It has a non-zero, angular acceleration. And if we looked at the forces acting on the wheel, then we have a normal force. We have gravity acting at the center of mass. And we also have static friction. And it's precisely the static friction that is producing a torque about the center of mass. And that torque about the center of mass will produce an angular acceleration. So in order for the wheel to continue rolling without slipping, it must have both a linear acceleration, which comes from gravitational force component going down an inclined plane minus the friction plus the alpha is coming from the torque. And so this side, if we wrote it as a vector equation, the torque about the center of mass would be the vector from-- let's right this is r cm to where the static friction is acting. So that's the vector r cm F static cross f static. So in this case, f static is non-zero. So there are many circumstances in which static friction can vary between a zero value, some possible value, and some maximum value. And here we see one more example. In order for the wheel to continue to accelerate down the inclined plane and roll without slipping that the static friction must produce a torque that causes the angular acceleration. And so we see that static friction depends on the other constraints in the problem.
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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 we've been trying to find the velocity of the cart and the velocity of the person in terms of the relative velocity of the person jumping in the reference frame of the moving cart, the mass of the person, and the mass of the cart. We've already solved this in the ground frame, now I would like to solve this in which my reference frame is the moving frame. So our pictures in the moving frame were actually much easier. Because the moving frame is traveling with the final speed of the cart, at the final interaction-- after the interaction is done-- in this moving frame, the cart is at rest. They're moving together. You're sitting in the cart. You're moving at the same speed. The person jumped off with speed u relative to the cart. The tricky part was to realize that, in the ground frame-- when the cart is at rest here in the ground frame-- an observer moving with speed Vc would see the person in the cart moving backwards in the moving train with a speed minus Vc. So now we can apply the momentum principle to these two states. In the moving frame, we have that momentum of the system initially equals the momentum of the system finally. Now, the initial momentum is just minus Mp plus M cart V cart-- notice the minus V cart. And the final momentum is equal to just M person u. And so we see that V cart equals minus M person u over M person plus Mc, which is exactly the same result that we got in the ground frame but it was actually much easier to solve in the moving frame.
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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: So last time, we started looking at the system of [? spins. ?] So there was a field S of x on the lattice. And the energy cost was proportional to differences of space on two neighboring sites, which if we go to the continuum, became something like gradient of the vector S. You have to integrate this, of course, over all space. We gave this a rate of K over 2. There was some energy costs [INAUDIBLE] of this [? form ?] so that the particular configuration was weighted by this factor. And to calculate the partition function, we had to integrate over all configurations of this over this field S. And the constraint that we had was that this was a unit vector so that this was an n component field whose magnitude was 1, OK? So this is what we want to calculate. Again, whenever we are writing an expression such as this, thinking that we started with some average system [INAUDIBLE] some kind of a coarse graining. There is a short distance [INAUDIBLE] [? replacing ?] all of these tiers. Now what we can do is imagine that this vector S in its ground state, let's say, is pointing in some particular direction throughout the system and that fluctuations around this ground state in the transverse direction are characterized by some vector of pi that is n minus 1 dimensional. And so this partition function can be written entirely [? rather ?] than fluctuations of the unit vector S in terms of the fluctuations of these transverse [? coordinates ?] [INAUDIBLE] pi. And we saw that the appropriate weight for this n minus 1 component vector of pi has within it a factor of something like square root of 1 minus pi squared. There's an overall factor of 2, but it doesn't matter. And essentially, this says that because you have a unit vector, this pi cannot get too big. You have to pay [? a cost ?] here, certainly not larger than 1. And then the expression for the energy costs can be written in terms of two parts [? where it ?] is the gradient of this vector pi. But then there's also the gradient in the other direction which has magnitude square root of 1 minus pi squared. So we have this gradient squared. And so basically, these are two ways of writing the same thing, OK? So we looked at this and we said that once we include all of these terms, what we have here is a non-linear [? activity ?] that includes, for example, interactions among the various modes. And one particular leading order term is if we expand this square root of 1 minus pi squared, if it would be something like pi square root of pi, so a particular term in this expansion as the from pi [? grad ?] pi multiplied with pi [? grad ?] pi. [INAUDIBLE], OK? So a particular way of dealing these kinds of theories is to regard all of these things as interactions and perturbations with respect to a Gaussian weight which we can compute easily. And then you can either do that perturbation straightforwardly or from the beginning to a perturbative [? origin, ?] which is the route that we chose. And this amount to changing the short distance cut off that we have here that is a to be b times a and averaging over all [? nodes ?] within that distance short wavelength between a and ba. And once we do that, we arrive at a new interaction. So the first step is to do a coarse graining between the range a and ba. But then steps two and three amount to a rescaling in position space so that the cut-off comes back to ba and the corresponding thing in the spin space so that we start with a partition function that describes unit vectors. And after this transformation, we end up with a new partition function that also describes unit vectors so that after all of these three procedures, we hope that we are back to exactly the form that we had at the beginning with the same cut-off, with the same unit vector constraint, but potentially with a new interaction parameter K And calculating what this new K is after we've scaled by a factor of b, the parts that correspond to 2 and 3 are immediately obvious. Because whenever I see x, I have to replace it wit bx prime. And so from integration, I then get a factor of b to the d from the two gradients, I would get a factor of minus 2. So the step that corresponds to this is trivial. The step that corresponds to replacing S with zeta S prime is also trivial. And it will give you zeta squared. Or do I have yet to tell you what is? We'll do that shortly. And finally, the first step, which was the coarse graining, we found that what it did was that it replaced K by a strongly-- sorry. I didn't expect this. All right-- the factor K, which is larger by a certain amount. And the mathematical justification that I gave for this is we look at this expression, and we see that in this expression, each one of these pi's can be a long wavelength fluctuation or a short wavelength fluctuation. Among the many possibilities is when these 2 pi's that are sitting out front correspond to the short wavelength fluctuations. These correspond to the long wavelength fluctuations. And you can see that averaging over these two will generate an interaction that looks like gradient of pi lesser squared. And that will change the coefficient over here by an amount that is clearly proportional to the average of pi greater squared. And that we can see in Fourier space is simply 1 over Kq squared over KK squared for modes that have wave number K. So if I, rather than write this in real space, I write it in Fourier space, this is what I would get for the average. And in real space, I have to integrate over this K appropriately [INAUDIBLE] within the wave numbers lambda over [? real ?] [? light. ?] And this is clearly something that is inversely proportional to K. And the result of this integration of 1 over K squared-- we simply gave a [? 9 ?], which was i sub d of b. Because it depends on the dimension. It depends on [? et ?] [? cetera. ?] AUDIENCE: Sir? PROFESSOR: Yes? AUDIENCE: Shouldn't there be an exponential inside the integral? PROFESSOR: Why should there be an exponential inside the integral? AUDIENCE: Oh, I thought we were Fourier transforming. PROFESSOR: OK, it is true, when we Fourier transform for each pi, we will have a factor of e to the IK. If we have 2 pi's, I will get e to the IK e to the IK prime. But the averaging [INAUDIBLE] set K and K prime can be opposite each other. So the exponentials disappear. So always remember the integral of any field squared in real space is the same thing as the integral of that field squared in Fourier space. This is one of the first theorems of Fourier transformation. OK, so this is a correction that goes like [INAUDIBLE]. And last time, to give you a kind of visual demonstration of what this factor is, I said that it is similar, but by no means identical to something like this, which is that a mode by itself has very low energy. But because we have coupling among different modes, here for the Goldstone modes of the surface, but here for the Goldstone modes of the spin, the presence of a certain amount of short range fluctuations will stiffen the modes that you have for longer wavelengths. Now, I'm not saying that these two problems are mathematically identical. All I'm showing you is that the coupling between the short and long wavelength modes can lead to a stiffening of the modes over long distances because they have to fight off the [? rails ?] that have been established by shorter wavelength modes. You have to try to undo them. And that's an additional cost, OK? Now, that stiffening over here is opposed by a factor of zeta over here. Essentially, we said that we have to ensure that what we are seeing after the three steps of RG is a description of a [? TOD ?] that has the same short distance cut-off and the same length so that the two partition functions can map on to each other. And again, another visual demonstration is that you can decompose the spin over here to a superposition of short and long wavelength modes. And we are averaging over these short wavelength modes. And because of that, we will see that the effective length once that averaging has been performed has been reduced. It has been reduced because I will write this as 1 minus pi squared over 2 to the lowest order. And pi squared has n minus 1 components. So this is n minus 1 over 2. And then I have to integrate over all of the modes pi alpha of K in this range. So I'm performing exactly the same integral as above. So the reduction is precisely the same integral as above, OK? So the three steps of RG performed for this model to lowest order in this inverse K, our [? temperature-like ?] variable is given by this one [INAUDIBLE] once I substitute the value of zeta over there. So you can see that the answer K prime of b is going to b to the b minus 2 zeta squared-- ah, that's right. For zeta squared, essentially the square of that, 1 minus n minus 1 over K, the 2 disappears once I squared it. Id of b divided by K-- that comes from zeta squared. And from here I can get the plus Id of be over K. And the whole thing gets multiplied by this K. And there is still terms at the order of temperature of 1 over K squared, OK? And finally, we are going to do the same choice that we were doing for our epsilon expansion. That is, choose a rescaling factor that is just slightly larger than 1. Yes? AUDIENCE: Sir, you're at n minus 1 over K times Id over K. PROFESSOR: Yeah, I gave this too much. Thank you. OK, thank You. And we will write K prime at Kb to be K plus delta l dK by dl. And we note that for calculating this Id of b, when b goes to 1 plus delta l, all I need to do is to evaluate this integrand essentially on the shell. So what I will get is lambda to the power of d minus 2. The surface area of a unit sphere divided by 2 pi to the d, which is the combination that we have been calling K sub d, OK? So once I do that, I will get that the dK by dl, OK? What do I get? I will get a d minus 2 here times K. And then I will get these two factors. There's n minus 1. And then there's 1 here. So that becomes n minus 2. I have a Idb, which is Kd lambda to the d minus 2. And then the 1 over K and K disappear. And so that's the expression that we have. Yes? AUDIENCE: Sorry, is the Kd [? some ?] angle factor again? PROFESSOR: OK, so you have to do this integration, which is written as the surface area inside an angle, K to the d minus 1 dK divided by 2 pi to the d. This is the combination that we have always called K sub d. AUDIENCE: OK. PROFESSOR: OK? Now, it actually makes more sense since we are making a low temperature expansion to define a T that is simply 1 over K. Its again, dimensionless. And then clearly dT by dl is going to be minus K squared dK by dl minus 1 over K squared. Minus 1 over K squared becomes minus T squared dK by dl. So I just have to multiply the expression that I have up here with minus T squared, recognizing that TK is 1. So I end up with the [? recursion ?] convention for T, which is minus d minus 2T. And then it becomes plus n minus 2 Kd lambda to the d minus 2 T squared. And presumably, there are high order terms that we have not bothered to calculate. So this is the [INAUDIBLE] we focused on. OK? So let's see whether this expression makes sense. So if I'm looking at dimensions that are less than 2, then the linear term in the expression is positive, which means that if I'm looking at the temperature axis and this is 0, and I start with a value that is slightly positive, because of this term, it will be pushed larger and larger values. So you think that you have a system at very low temperatures. You look at it at larger and larger scales, and you find that it becomes effectively something that has higher temperature and becomes more and more disordered. So basically, this is a manifestation of something that we had said before, Mermin-Wagner theorem, which is no [? long ?] range order in d less than 2, OK? Now, if I go to the other limit, d greater than 2, then something interesting happens, in that the linear term is negative. So if I start with a sufficiently small temperature or a large enough coupling, it will get stronger as we go towards an ordered phase, whereas the quadratic term for n greater than 2 has the opposite sign-- this is n greater than 2-- and pushes me towards disorder, which means that there should be a fixed point that separates the two behaviors. Any temperature lower than this will give me an ordered phase. Any temperature higher than this will give me a disordered phase. And suddenly, we see that we have potentially a way of figuring out what the phase transition is because this T star is a location that we can perturbatively access. Because we set this to 0, and we find that T star is equal to d minus 2 divided by n minus 2 Kd lambda to the d minus 2. So now in order to have a theory that makes sense in the sense of the perturbation that we have carried out, we have to make sure that this is small. So we can do that by assuming that this quantity d minus 2 is a small quantity in making an expansion in d minus 2, OK? So in particular, T star itself we expect to be related to transition temperature, not something that is universal. But exponents are universal. So what we do is we look at d by dl of delta T. Delta T is, let's say, T minus T star in one direction or the other. [INAUDIBLE] And for that, what I need to do it to linearize this expression. So I will get a minus epsilon from her. And from here, I will get 2 n minus 2 Kd lambda to the d minus 2 T star times delta T. I just took the derivative, evaluated the T star. And we can see that this combination is precisely the combination that I have to solve for T star. So this really becomes another factor of epsilon. I have minus epsilon plus [? 2 ?] epsilon. So this is epsilon delta T. So that tells me that my thermal eigenvalue is epsilon, a disorder clearly independent of n. Now, we've seen that in order to fully characterize the exponent, including things like magnetization, et cetera, it makes sense to also put a magnetic field direction and figure out how rapidly you go along the magnetic field direction. So for that, one way of doing this is to go and add to this term, which is h integral S of x. And you can see very easily that under these steps of the transformation, essentially the only thing that happens is that I will get h prime at scale d is h from the integration. I will get a factor of b to the d. From the replacement of s with s prime, I will get a factor of zeta. So this combination is simply my yh. And just bringing a little bit of manipulation will tell you that yh is d minus the part that comes from zeta, which is n minus 1 over 2 Id of b, which is lambda to the d minus 2 Kd. And then we have T star. And again, you substitute for lambda to the d minus 2 Kd T star on what we have over here. And you get this to be d minus n minus 1 over 2n minus 2 epsilon. And again, to be consistent to order of epsilon, this d you will have to replace with 2 plus epsilon. And a little bit of manipulation will give you yh, which is 2 minus n minus 3 divided by 2n minus 2 epsilon. I did this calculation of the two exponents rather rapidly. The reason for that is they are not particularly useful. That is, whereas we saw that coming from four dimensions the epsilon expansion was very useful to give us corrections to the [? mean ?] field values of 1/2, for example, for mu to order of 10% or so already by setting epsilon equals to 1. If I, for example, put here epsilon equal to 1 to [? access ?] 3 dimensions, I will conclude that mu, which is the inverse of yT is 1 in 3 dimensions independent of n. And let's say for super fluid it is closer to 2/3. And so essentially, this expansion in some sense is much further away from 3 dimensions than the 4 minus epsilon coming from 4 dimensions, although numerically, we would have said epsilon equaled 1 to both of them. So nobody really has taken much advantage of this 2 plus epsilon expansion. So why is it useful? The reason that this is useful is the third case that I have not explained, yet, which is, what happens if you sit exactly in 2 dimensions, OK? So if we sit exactly in 2 dimensions, this first term disappears. And you can see that the behavioral is determined by the second order and [? then ?] depends on the value of n. So if I look at n, let's say, that is less than 2, then what I will see is that along the temperature axis, the quadratic term-- the linear term is absent-- the quadratic term, let's say, for n equals 1 is negative. And you're being pushed quadratically very slowly towards 0. The one example that we know is indeed n equals 1, the Ising model. And we know that the Ising model in 2 dimensions has an ordered phase. It shouldn't really even be described by this because there are no Goldstone modes. But n greater than 2, like n equals 3-- the Heisenberg model, is interesting. And what we see is that here the second order term is positive. And it is pushing you towards high temperatures. So you can see a disordered behavior. And what this calculation tells you that is useful that you wouldn't have known otherwise is, what is the correlation [? length? ?] Because the recursion relation is now dT by dl is, let's say, n minus 2. And my d equals to 2. The lambda to the d minus 2, I can ignore. Kd is 2 pi from the [? solid ?] angle divided by 2 pi squared. So that's 1 over 2 pi times T squared, OK? So that's the recursion relation that we are dealing with. I can divide by 1 over T squared. And then this becomes d by dl of minus 1 over T equals n minus 2 divided by 2 pi. I can integrate this from, say, some initial value minus 1 over some initial temperature to some temperature where I'm at [? length ?] [? scale ?] l. What I would have on the right hand side would be n minus 2 over [? 2yl ?], OK? So I start very, very close to the origin T equals 0. So I have a very strong coupling at the beginning. So this factor is huge. T is [? more than ?] 1 over T is huge. I have a huge coupling. And then I rescale to a point where the coupling has become weak, let's say some number of order of 1, order of 1 or order of 0. In any case, it is overwhelmingly smaller than this. How far did I have to go? I had to rescale by a factor of l that is related to the temperature that I started with by this factor, except that I forgot the minus that I had in front of the whole thing. So the resulting l will be large and positive. And the correlation length-- the length scale at which we arrived at the coupling, which is of the order of 1 or 0, is whatever my initial length scale was times this factor that I have rescaled by, b, which is e to the l. And so this is a exponential of n minus 1 over 2 pi times 1 over T. The statement is that if you're having 2 dimensions, a system of, let's say, 3 component spins-- and that is something that has a lot of experimental realizations-- you find that as you go towards low temperature, the size of domains that are ordered diverges according to this nice universal form. And let's say around 1995 or so, when people had these high temperature superconductors which are effectively 2 dimensional layers of magnets-- they're actually antiferromagnets, but they are still described by this [INAUDIBLE] with n equals 3-- there were lots of x-ray studies of what happens to the ordering of these antiferromagnetic copper oxide layers as you go to low temperatures. And this form was very much used and confirmed. OK, so that's really one thing that one can get from this analysis that has been explicitly [? confirmed ?] [? for ?] experiments. And finally, there's one case in this that I have not mentioned so far, which is n equals 2. And when I am at n equals 2, what I have is that the first and second order terms in this series are both vanishing. And I really at this stage don't quite know what is happening. But we can think about it a little bit. And you can see that if you are n equals 2, then essentially, you have a 1 component angle. And if I write the theory in terms of the angle theta, let's say, between neighboring spins, then the expansions would simply be gradient of theta squared. And there isn't any other mode to couple with. You may worry a little bit about gradient of theta to the 4th and such things. But a little bit of thinking will convince you that all of those terms are irrelevant. So as far as we can show, there is reason that essentially this series for n equals 2 is 0 at all orders, which means that as far as this analysis is concerned, there is a kind of a line of fixed points. You start with any temperature, and you will stay at that temperature, OK? Still you would say that even if you have this gradient of theta squared type of theory, the fluctuations that you have are solutions of 1 over q squared. And the integral of 1 over 2 squared in 2 dimensions is logarithmically divergent. So the more correct statement of the Mermin-Wagner's theorem is that there should be no long range order in d less than or equal to 2. Because for d equals 2 also, you have these logarithmic divergence of fluctuations. So you may have thought that you are pointing along, say, the y direction. But you average more and more, and you see that the extent of the fluctuations in angle are growing logarithmically. You say that once that logarithm becomes of the order of pi, I have no idea where my angle is. There should be no true long range order. And I'm not going to try to interpret this too much. I just say that Mermin-Wagner's theorem says that there should be no true long range order in systems that have continuous symmetry in 2 dimensions and below, OK? And that statement is correct, except that around that same time, Stanley and Kaplan did low temperature series analysis-- actually, no, I'm incorrect-- did high temperatures series of these spin models in 2 dimensions. And what they found was, OK, let's [? re-plot ?] susceptibility as a function of temperature. We calculate our best estimate of susceptibility from the high temperature series. And what they do is, let's say, they look at the system that corresponds to n equals 3. And they see that the susceptibility diverges only when you get in the vicinity of 0 temperature, which is consistent with all of these statements that first of all, this is a [? direct ?] correlation [? when it ?] only diverges at 0 temperature. And divergence of susceptibility has to be coupled through that. And therefore, really, the only exciting thing is right at 0 temperature, there is no region where this is long range order, except that when they did the analysis for n equals 2, they kept getting signature that there is a phase transition at a finite temperature in d equals 2 for this xy model that described by just an [INAUDIBLE], OK? So there is lots of numerical evidence of phase transition for n equals 2 in d equals [? 2, ?] OK? So this is another one of those puzzles which [INAUDIBLE] if we interpret the existence of a diverging susceptibility in the way that we are used to, let's say the Ising model and all the models that we have discussed so far, in all cases that we have seen, essentially, the divergence of the susceptibility was an indicator of the onset of true long range order so that on the other side, you had something like a magnet. But that is rigorously ruled out by Mermin-Wagner. So the question is, can we have a phase transition in the absence of symmetry breaking? All right? And we already saw one example of that a couple of lectures back when we were doing the dual of the 3-dimensional Ising model. We saw that the 3-dimensional Ising model, its dual had a phase transition but was rigorously prevented from having true long range order. So there, how did we distinguish the different phases? We found some appropriate correlation function. And we showed that that correlation function had different behaviors at high and low temperature. And these two different behaviors could not be matched. And so the phase transition was an indicator of the switch-over in the behavior of the correlation functions. So here, let's examine the correlation functions of our model. And the simplest correlation that we can think of for a system that is described by unit spins is to look at the spin at some location and the spin at some far away location and ask how correlated they are to each other? And so basically, there is some kind of, let's say, underlying lattice. And we pick 2 points at 0 and at r. And we ask, what is the dot product of the spins that we have at these 2 locations? And clearly, this is invariant under the [? global ?] rotation. What I can do is I can pick some kind of axis and define angles with respect to some axis. Let's say with respect to the x direction, I define an angle theta. And then clearly, this is the expectation value of cosine of theta 0 minus theta r. Now, this quantity I can asymptotically calculate both at high temperatures and low temperatures and compare them. So let's do a high T expansion. For the high T expansion, I sort of go back to the discrete model and say that what I have here is a system that is characterized by a bunch of angles that I have to integrate in theta i. I have the cosine of theta 0 minus theta r. And I have a weight that wants to make near neighbours to be parallel. And so I will write it as product over nearest neighbors, p to the K cosine of theta i minus theta j. OK, so the dot product of 2 spins I have written as the cosine between [? nearest ?] neighbors. And of course, I have to then divide by [INAUDIBLE]. Now, if I'm doing the high temperature expansion, that means that this coupling constant K scaled by temperature is known. And I can expand this as 1 plus K cosine of theta i minus theta j [? plus ?] [? higher ?] orders in powers of K of course, OK? Now, this looks to have the same structure as we had for the Ising model. In the Ising model, I had something like sigma i sigma j. And if I had a sigma by itself and I summed over the possible values, I would get 0. Here, I have something like a cosine of an angle. And if I integrate, let's say, d theta 0 cosine of theta 0 minus something, just because theta 0 can be both positive as [? it just ?] goes over the entire angle, this will give me 0. So this cosine I better get rid of. And the way that I can do that is let's say I multiply cosine of theta 0 minus theta r with one of the terms that I would get in the expansion, such as, let's say, a factor of K cosine of theta 0 minus theta 1. So if I call the next one theta 1, I will have a term in the expansion that is cosine of theta 0 minus theta 1, OK? Then this will be non-zero because I can certainly change the origin. I can write this as integral d theta 0 minus theta 1-- I can call phi. This would be cosine of phi from here. This becomes cosine of theta 0 minus theta 1 minus phi. And this I can expand as cosine of a 0 minus theta 1 cosine of phi minus sine of theta 0 minus theta 1 sine of phi. Then cosine integrated against sine will give me 0. Cosine integrated against cosine will give me 1/2. So this becomes 1/2 cosine of theta 0 minus theta 1 theta r. What did I-- For theta 0, I am writing phi plus theta 1. So this becomes phi plus theta 1 minus theta r. This becomes cosine of theta 1 minus theta r. This becomes theta 1 minus theta r. This becomes theta 1 minus theta r. OK, so essentially, we had a term that was like a cosine of theta 0 minus theta r from here. Once we integrate over this bond, then I get a factor of 1/2, and it becomes like a connection between these two. And you can see that I can keep doing that and find the path that connects from 0 to r. For each one of the bonds along this path, I pick one of these factors. And this allows me to get a finite value. And what I find once I do this is that through the lowest order, I have to count the shortest number of paths that I have between the two, K. I will get a factor of K. And then from the averaging over the angles, I will get 1/2. So it would b K over 2 [INAUDIBLE]. By this we indicate the shortest path between the 2, OK? So the point is that K is a small number. If I go further and further away, this is going to be exponentially small in the distance between the 2 spaces, where [? c ?] can be expressed in something that has to do with K. So this is actually quite a general statement. We've already seen it for the Ising model. We've now seen it for the xy model. Quite generally, for systems at high temperatures, once can show that correlations decay exponentially in separation because the information about the state of one variable has to travel all the way to influence the other one. And the fidelity by which the information is transmitted is very small at high temperatures. So OK? So this is something that you should have known. We are getting the answer. But now what happens if I go and look at low temperatures? So for low temperatures, what I need to do is to evaluate something that has to do with the behavior of these angles when I go to low temperatures. And when I go to low temperatures, these angles tend to be very much aligned to each other. And these factors of cosine I can therefore start expanding around 1. So what I end up having to do is something like a product over i theta i cosine of theta 0 minus theta r. I have a product over neighbors of factors such as K over 2 theta i minus theta j squared, [? as ?] I expand the Gaussian, expand the cosine. And in the denominator I would have exactly the same thing without this. So essentially, we see that since the cosine is the real part of e to the i theta 0 minus theta r, what I need to do is to calculate the average of this assuming the Gaussian weight. So the theta is Gaussian distributed, OK? Now-- actually, this [INAUDIBLE] I can take the outside also. It doesn't matter. I have to calculate this expectation value. And this for any Gaussian expectation value of e to the i some Gaussian variable is minus 1/2 the average of whatever you have, weight. And again, in case you forgot this, just insert the K here. You can see that this is the characteristic function of the Gaussian distributed variable, which is this difference. And the characteristic function I can start expanding in terms of the cumulants. The first cumulant, the average is 0 by symmetry. So the first thing that will appear, which would be at the order of K squared, is going to be the variance, which is what we have over here. And since it's a Gaussian, all higher order terms in this series [? will. ?] Another way to do is to of course just complete the square. And this is what would come out, OK? So all I need to do is to calculate the expectation value of this quantity where the thetas are Gaussian distributed. And the best way to do so is to go to Fourier space. So I have integral. For each one of these factors of theta 0 minus theta r, I will do an integral d2 q 2 pi squared. I have 1 minus e to the iq.r, which is from theta 0 minus theta r. And then I have a theta tilde q. I have two of those factors. I have d2 q prime 2 pi squared. I have e to the minus iq prime dot r theta tilde q prime. And then this average simply becomes this average. And the different modes are independent of each other. So I will get a 2 pi to the d-- actually, 2 here, a delta function q plus q prime. And for each mode, I will get a factor of Kq squared because after all, thetas are very much like the pi's that I had written at the beginning, OK? So what I will have is that this quantity is integral over one q's. Putting these two factors together, realizing that q prime is minus q will give me 2 minus 2 cosine of q.r divided by Kq squared. So there is an overall scale that is set by 1 over K, by temperature. And then there's a function [? on ?] [? from ?], which is the Fourier transform of {} 1 over q squared, which, as usual, we call C. We anticipate this to be like a Coulomb potential. Because if I take a Laplacian of C, you can see that-- forget the q-- the [? Laplacian ?] of C from the cosine, I will get a minus q squared cancels that. I will have [INAUDIBLE] d2q 2 pi squared. Cosine itself will be left. The 1 over q squared disappears. This is e to the iqr you plus e to the minus iqr over 2. Each one of them gives a delta function. So this is just a delta function. So C is the potential that you have from a unit charge in 2 dimensions. And again, you can perform the usual Gaussian procedure to find that the gradient of C times 2 pi r is the net charge that is enclosed, which is [? unity ?]. So gradient of C, which points in the radial direction is going to be 1 over 2 pi r. And your C is going to be log of r divided by 2 pi. So this is 1 over K log of r divided by 2 pi. And I state that when essentially the 2 angles are as close as some short distance cut-off, fluctuations vanish. So that's how I set the 0 of my integration, OK? So again, you put that over here. We find that s0.s of r in the low temperature limit is the exponential of minus 1/2 of this. So I have log of r over a divided by 4 pi K. And I will get a over r to the power of 1 over 4 pi K. And I kind of want to check that I didn't lose a factor of 2 somewhere, which I seem to have. Yeah, I lost a factor of 2 right here. This should be 2 because of this 2, if I'm using this definition. So this should be 2. And this should be 2 pi K. OK. So what have we established? We have looked as a function of temperature what the behavior of this spin-spin correlation function is. We have established that in the higher temperature limit, the behavior is something that falls off exponentially with separation. We have also established that at low temperature, it falls off as a power law in separation, OK? So these two functional behaviors are different. There is no way that you can connect one to the other. So you pick two spins that are sufficiently far apart and then move the separation further and further away. And the functional form of the correlations is either a power-law decay, power-law decay, or an exponential decay. And in this form, you know you have a high temperature. In this form, you know you have a low temperature. So potentially, there could be a phase transition separating the distinct behaviors of the correlation function. And that could potentially be underlying what is observed over here, OK? Yes? AUDIENCE: So where could we make the assumption we're at a low temperature in the second expansion? PROFESSOR: When we expanded the cosines, right? So what I should really do is to look at the terms such as this. But then I said that I'm low enough temperature so that I look at near neighbors, and they're almost parallel. So the cosine of the angle difference between them is the square of that small angle. AUDIENCE: Thank you. PROFESSOR: Yeah. OK? So actually, you may have said that I could have done the same analysis for small angle expansions not only for n equals 2, but also for n equals 3, et cetera. That would be correct. Because I could have made a similar Gaussian analysis for n equals e also. And then I may have concluded the same thing, except that I cannot conclude the same thing because of this thing that we derived over here. What this shows is that the expansion around 0 temperature regarded as Gaussian is going to break down because of the non-linear coupling that we have between modes. So although I may be tempted to write something like this for n equals 3, I know why it is wrong. And I know the correlation length at which this kind of behavior will need to be replaced with this type of behavior because effectively, the expansion parameter became of the order of 1. But I cannot do that for the xy model. I don't have similar reason. So then the question becomes, well, how does this expansion then eventually break down so that I will have a phase transition to a phase where the correlations are decaying exponentially? And you may say, well, I mean, it's really something to do with having to go to higher and higher ordered terms in the expansion of the cosine. And it's going to be something which would be very difficult to figure out, except that it turns out that there is a much more elegant solution. And that was proposed by [? Kastelitz ?] and [? Thales. ?] And they said that what you have left out in the Gaussian analysis are topological defects, OK? That is, when I did the expansion of the cosine and we replaced the cosine with the difference of the angles squared, that's more or less fine, except that I should also realize that cosine maintains its value if the angle difference goes up by 2 pi. And you say, well, neighboring spins are never going to be 2 pi different or pi different because they're very strongly coupled. Does it make any difference? Turns out that, OK, for the neighboring spins, it doesn't make a difference. But what if you go far away? So let's imagine that this is, let's say, our system of spins. And what I do is I look at a configuration such as this. Essentially, I have spins [? radiating ?] out from a center such as this, OK? There is, of course, a lot of energy costs I have put over here. But when I go very much further out, let's say, very far away from this plus sign that I have indicated over here, and I follow what the behavior of the spins are, you see that as I go along this circuit, the spins start by pointing this way, they go point this way, this way, et cetera. And by the time I carry a circuit such as this, I find that the angle theta has also rotated by 2 pi, OK? Now, this is clearly a configuration that is going to be costly. We'll calculate its cost. But the point is that there is no continuous deformation that you can make that will map this into what we were expanding around here with all of the cosines being parallel to each other. So this is a topologically distinct contribution from the Gaussian one, the Gaussian term that we've calculated. Since the direction of the rotation of the spins is the same as the direction of the circuit in this case, this is called a plus topological defect. There is a corresponding minus defect which is something like this. OK, and for this, you can convince yourself that as you make a circuit such as this that the direction of the arrow actually goes in the opposite direction, OK? This is called a negative sign topological defect. Now, I said, well, let's figure out what the energy cost of one of these things is. If I'm away from the center of one of these defects, then the change in angle is small because the change in angle if I go all the way around a circle of radius r should come back to be 2 pi. 2 pi is the uncertainty that I have from the cosines. So what I have is that the gradient of theta [? between ?] neighboring angles times 2 pi r, which is this radius, is 2 pi. And this thing, here it is plus 1. Here it is minus 1. And in general, you can imagine possibilities where this is some integer that is like 2 or minus 2 or something else that is allowed by this degeneracy of this cosine, OK? So you can see that when you are far away from the center of whatever this defect is, the gradient of theta has magnitude that is n over r, OK? And as you go further and further, it becomes smaller and smaller. And the energy cost out here you can obtain by essentially expanding the cosine is going to be proportional to the change in angle squared. So the cost of the defect is an integral of 2 pi rdr times this quantity n over r squared multiplied by the coefficient of the expansion of the cosine, or K over 2. And this integration I have to go all the way to ends up of my system. Let's call it l. And then I can bring it down, not necessary to the scale of the lattice spacing, but maybe to scale of 5 lattice spacing or something like this, where the approximations that I have used of treating this as a continuum are still valid. So I will pick some kind of a short distance cut-off a. And then whatever energy is at scales that are below a, I will add to a core energy that depends on whatever this a is. I don't know what that is. So basically, there is some core energy depending on where I stop this. And the reason that this is more important-- because here I have an integral of 1 over r. And an integral of 1 over r is something that is logarithmically divergent. So I will get K. I have 2 cancels the 2. I have pi en squared log of l over a, OK? So you can see that creating one of these defects is hugely expensive. And energy that as your system becomes bigger and bigger and we're thinking about infinite sized systems is logarithmically large. So you would say these things will never occur because they cost an infinite amount of energy. Well, the thing is that entropy is also important. So if I were to calculate the partition function that I would assign to one of these defects, part of it would be exponential of this energy. So I would have this e to the minus this core energy. I would have this exponential of K pi n squared log of l over a. So that's the [INAUDIBLE] weight for this. But then I realize that I can put this anywhere on the lattice so there is an entropy [? gain ?] factor. And since I have assigned this to have some kind of a bulk to it as some characteristic size a, the number of distinct places that I can put it is over the order of l over a squared. So basically, I take my huge lattice and I partition it into sizes of a's. And I say I can put it in any one of these configurations. You can see that the whole thing is going to be e to the minus this core energy. And then I have l over a to the power of 2 minus pi K n squared, OK? So the logarithmic energy cost is the same form as the logarithmic entropy gain that you have over here. And this precise balance will give you a value of K such that if K is larger than 2 over pi n squared, this is going to be an exponentially large cost. There is a huge negative power of l here that says, no, you don't want to create this. But if K becomes weak such that the 2, the entropy factor, [? wins ?], then you will start creating [INAUDIBLE]. So you can see that over here, suddenly we have a mechanism along our picture over here, maybe something like K, which is 2 over pi, such that on one side you would say, I will not have topological defect and I can use the Gaussian model. And on the other side, you say that I will spontaneously create these topological defects. And then the Gaussian description is no longer valid because I have to now really take care of the angular nature of these variables, OK? So this is a nice picture, which is only a zeroed order picture. And to zeroed order, it is correct. But this is not fully correct. Because even at low temperatures, you can certainly create hairs of plus minus defects, OK? And whereas the field for one of them will fall off at large distances, the gradient of theta is 1 over r, if you superimpose what is happening for two of these, what you will convince yourself that if you have a pair of defects of opposite sign at the distance d that the distortion that they generate at large distances falls off not the as 1 over r, which is, if you like, a monopole field, but as d over r squared, which is a dipole field. So whenever you have a dipole, you will have to multiply by the separation of the charges in the dipole. And that is compensated by a factor of 1 over r in the denominator. There is some angular dependence, but we are not so interested in that. Now, if I were to integrate this square, we can see that it is something that is convergent at large distances. And so this is going to be finite. It is not going to diverge as the size of the system, which means that whereas individual defects there was no way that I could create in my system, I can always create pairs of these defects. So the correct picture that we should have is not that at low temperatures you don't have defects, at high temperatures you have these defects spontaneously appearing. The correct picture is that at low temperatures what you have is lots and lots of dipoles that are pretty much bound to each other. And when you go to high temperatures, what happens is that you will have these pluses and minuses unbound from each other. So if you like, the transition is between molecules to a plasma as temperature is changed. Or if you like, it is between an insulator and a conductor. And how to mathematically describe this phase transition in 2 dimensions, which we can rigorously, will be what we will do in the next couple of lectures.
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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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CATHERINE DRENNAN: The idea was in a big class of like 300 students, most of the students are not going to have an opportunity really to meet the professors. They may go to office hours, but even then, the office hours, you can't schedule your office hours at a time where all 300 people in the class are available. So you'll just get to know maybe a few. But with the pizza forums, they're every few weeks during the semester, and we encourage at least one person from each recitation to come. And it's just a way for the students to get to know the faculty and vice versa. And sometimes we discuss things about chemistry, you know, especially if people have suggestions of what they like and don't like. But a lot of times, they're just talking about their experiences at MIT. And there have been some really useful things, I think, for the course that have come out of it, like learning, for example, that we had our problems sets due at the same time as pretty much all of their other classes. So we moved it to a different day because that made no sense. And also, kind of some of the educational ways-- what clickers they're using, how they like them. We heard some horrible things about clickers that had been used in physics. And so then they started to change them. And when we looked at clickers, we learned from the physics experience. But it really came through knowing that from the students, rather than talking to the other faculty. We had sort of heard what worked and didn't work. So having those kind of direct conversations. And they feel like they're representing the student body in saying, oh we like this, or we don't like that, or have you thought about doing something this way. And it's just also a lot of fun. And everyone goes around and says where they're from. I love to ask them, what is one thing about MIT that is exactly what you expected and what's one thing that really surprised you when you got here? And I just find it really interesting to see what people said. And I often get, I expected everyone to be smart, everybody is smart. I expected everyone to be just more serious, but people are fun and nice and silly and they're kind of like normal people and that sort of surprised me. So that is always a lot of fun to hear that and get to know them.
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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 ocw.mit.edu. PROFESSOR: Anything lingering and disturbing or bewildering? No? Nothing? All right. OK, so the story so far is basically three postulates. The first is that the configuration of a particle is given by, or described by, a wave function psi of x. Yeah? So in particular, just to flesh this out a little more, if we were in 3D, for example-- which we're not. We're currently in our one dimensional tripped out tricycles. In 3D, the wave function would be a function of all three positions x, y and z. If we had two particles, our wave function would be a function of the position of each particle. x1, x2, and so on. So we'll go through lots of details and examples later on. But for the most part, we're going to be sticking with single particle in one dimension for the next few weeks. Now again, I want to emphasize this is our first pass through our definition of quantum mechanics. Once we use the language and the machinery a little bit, we're going to develop a more general, more coherent set of rules or definition of quantum mechanics. But this is our first pass. Two, the meaning of the wave function is that the norm squared psi of x, norm squared, it's complex, dx is the probability of finding the particle- There's an n in their. Finding the particle-- in the region between x and x plus dx. So psi squared itself, norm squared, is the probability density. OK? And third, the superposition principle. If there are two possible configurations the system can be in, which in quantum mechanics means two different wave functions that could describe the system given psi 1 and psi 2, two wave functions that could describe two different configurations of the system. For example, the particles here or the particles over here. It's also possible to find the system in a superposition of those two psi is equal to some arbitrary linear combination alpha psi 1 plus beta psi 2 of x. OK? So some things to note-- so questions about those before we move on? No questions? Nothing? Really? You're going to make he threaten you with something. I know there are questions. This is not trivial stuff. OK. So some things to note. The first is we want to normalize. We will generally normalize and require that the integral over all possible positions of the probability density psi of x norm squared is equal to 1. This is just saying that the total probability that we find the particle somewhere had better be one. This is like saying if I know a particle is in one of two boxes, because I've put a particle in one of the boxes. I just don't remember which one. Then the probability that it's in the first box plus probability that it's in the second box must be 100% or one. If it's less, then the particle has simply disappeared. And basic rule, things don't just disappear. So probability should be normalized. And this is our prescription. So a second thing to note is that all reasonable, or non stupid, functions psi of x are equally reasonable as wave functions. OK? So this is a very reasonable function. It's nice and smooth. It converges to 0 infinity. It's got all the nice properties you might want. This is also a reasonable function. It's a little annoying, but there it is. And they're both perfectly reasonable as wave functions. This on the other hand, not so much. So for two reasons. First off, it's discontinuous. And as you're going to show in your problem set, discontinuities are very bad for wave functions. So we need our wave functions to be continuous. The second is over some domain it's multi valued. There are two different values of the function. That's also bad, because what's the probability? It's the norm squared, but if it two values, two values for the probability, that doesn't make any sense. What's the probability that I'm going to fall over in 10 seconds? Well, it's small, but it's not actually equal to 1% or 3%. It's one of those. Hopefully is much lower than that. So all reasonable functions are equally reasonable as wave functions. And in particular, what that means is all states corresponding to reasonable wave functions are equally reasonable as physical states. There's no primacy in wave functions or in states. However, with that said, some wave functions are more equal than others. OK? And this is important, and coming up with a good definition of this is going to be an important challenge for us in the next couple of lectures. So in particular, this wave function has a nice simple interpretation. If I tell you this is psi of x, then what can you tell me about the particle whose wave function is the psi of x? What can you tell me about it? What do you know? AUDIENCE: [INAUDIBLE]. PROFESSOR: It's here, right? It's not over here. Probability is basically 0. Probability is large. It's pretty much here with this great confidence. What about this guy? Less informative, right? It's less obvious what this wave function is telling me. So some wave functions are more equal in the sense that they have-- i.e. they have simple interpretations. So for example, this wave function continuing on infinitely, this wave function doesn't tell me where the particle is, but what does it tell me? AUDIENCE: Momentum. PROFESSOR: The momentum, exactly. So this is giving me information about the momentum of the particle because it has a well defined wavelength. So this one, I would also say is more equal than this one. They're both perfectly physical, but this one has a simple interpretation. And that's going to be important for us. Related to that is that any reasonable function psi of x can be expressed as a superposition of more equal wave functions, or more precisely easily interpretable wave functions. We saw this last time in the Fourier theorem. The Fourier theorem said look, take any wave function-- take any function, but I'm going to interpret in the language of quantum mechanics. Take any wave function which is given by some complex valued function, and it can be expressed as a superposition of plane waves. 1 over 2pi in our normalization integral dk psi tilde of k, but this is a set of coefficients. e to the ikx. So what are we doing here? We're saying pick a value of k. There's a number associated with it, which is going to be an a magnitude and a phase. And that's the magnitude and phase of a plane wave, e to the ikx. Now remember that e to the ikx is equal to cos kx plus i sin kx. Which you should all know, but just to remind you. This is a periodic function. These are periodic functions. So this is a plane wave with a definite wavelength, 2pi upon k. So this is a more equal wave function in the sense that it has a definite wavelength. We know what its momentum is. Its momentum is h bar k. Any function, we're saying, can be expressed as a superposition by summing over all possible values of k, all possible different wavelengths. Any function can be expressed as a superposition of wave functions with a definite momentum. That make sense? Fourier didn't think about it that way, but from quantum mechanics, this is the way we want to think about it. It's just a true statement. It's a mathematical fact. Questions about that? Similarly, I claim that I can expand the very same function, psi of x, as a superposition of states, not with definite momentum, but of states with definite position. So what's a state with a definite position? AUDIENCE: Delta. PROFESSOR: A delta function, exactly. So I claim that any function psi of x can be expanded a sum over all states with a definite position. So delta of-- well, what's a state with a definite position? x0. Delta of x minus x0. OK? This goes bing when x0 is equal to x. But I want a sum over all possible delta functions. That means all possible positions. That means all possible values of x0, dx0. And I need some coefficient function here. Well, the coefficient function I'm going to call psi of x0. So is this true? Is it true that I can take any function and expand it in a superposition of delta functions? Absolutely. Because look at what this equation does. Remember, delta function is your friend. It's a map from integrals to numbers or functions. So this integral, is an integral over x0. Here we have a delta of x minus x0. So this basically says the value of this integral is what you get by taking the integrand and replacing x by x0. Set x equals x0, that's when delta equals 0. So this is equal to the argument evaluated at x0 is equal to x. That's your psi of x. OK? Any arbitrarily ugly function can be expressed either as a superposition of states with definite momentum or a superposition of states with definite position. OK? And this is going to be true. We're going to find this is a general statement that any state can be expressed as a superposition of states with a well defined observable quantity for any observable quantity you want. So let me give you just a quick little bit of intuition. In 2D, this is a perfectly good vector, right? Now here's a question I want to ask you. Is that a superposition? Yeah. I mean every vector can be written as the sum of other vectors, right? And it can be done in an infinite number of ways, right? So there's no such thing as a state which is not a superposition. Every vector is a superposition of other vectors. It's a sum of other vector. So in particular we often find it useful to pick a basis and say look, I know what I mean by the vector y, y hat is a unit vector in this direction. I know what I mean by the vector x hat. It's a unit vector in this direction. And now I can ask, given that these are my natural guys, the guys I want to attend to, is this a superposition of x and y? Or is it just x or y? Well, that's a superposition. Whereas x hat itself is not. So this somehow is about finding convenient choice of basis. But any given vector can be expressed as a superposition of some pair of basis vectors or a different pair of basis vectors. There's nothing hallowed about your choice of basis. There's no God given basis for the universe. We look out in the universe in the Hubble deep field, and you don't see somewhere in the Hubble deep field an arrow going x, right? So there's no natural choice of basis, but it's sometimes convenient to pick a basis. This is the direction of the surface of the earth. This is the direction perpendicular to it. So sometimes particular basis sets have particular meanings to us. That's true in vectors. This is along the earth. This is perpendicular to it. This would be slightly strange. Maybe if you're leaning. And similarly, this is an expansion of a function as a sum, as a superposition of other functions. And you could have done this in any good space of functions. We'll talk about that more. These are particularly natural ones. They're more equal. These are ones with different definite values of position, different definite values of momentum. Everyone cool? Quickly what's the momentum associated to the plane wave e to the ikx? AUDIENCE: [INAUDIBLE]. PROFESSOR: h bar k. Good. So now I want to just quickly run over some concept questions for you. So whip out your clickers. OK, we'll do this verbally. All right, let's try this again. So how would you interpret this wave function? AUDIENCE: e. PROFESSOR: Solid. How do you know whether the particle is big or small by looking at the wave function? AUDIENCE: [INAUDIBLE]. PROFESSOR: All right. Two particles described by a plane wave of the form e to the ikx. Particle one is a smaller wavelength than particle two. Which particle has a larger momentum? Think about it, but don't say it out loud. And this sort of defeats the purpose of the clicker thing, because now I'm supposed to be able to know without you guys saying anything. So instead of saying it out loud, here's what I'd like you to do. Talk to the person next to you and discuss which one has the larger AUDIENCE: [CHATTER]. All right. Cool, so which one has the larger momentum? AUDIENCE: A. PROFESSOR: How come? [INTERPOSING VOICES] PROFESSOR: RIght, smaller wavelength. P equals h bar k. k equals 2pi over lambda. Solid? Smaller wavelength, higher momentum. If it has higher momentum, what do you just intuitively expect to know about its energy? It's probably higher. Are you positive about that? No, you need to know how the energy depends on the momentum, but it's probably higher. So this is an important little lesson that you probably all know from optics and maybe from core mechanics. Shorter wavelength thing, higher energy. Higher momentum for sure. Usually higher energy as well. Very useful rule of thumb to keep in mind. Indeed, it's particle one. OK next one. Compared to the wave function psi of x, it's Fourier transform, psi tilde of x contains more information, or less, or the same, or something. Don't say it out loud. OK, so how many people know the answer? Awesome. And how many people are not sure. OK, good. So talk to the person next to you and convince them briefly. All right. So let's vote. A, more information. B, less information. C, same. OK, good you got it. So these are not hard ones. This function, which is a sine wave of length l, 0 outside of that region. Which is closer to true? f has a single well defined wavelength for the most part? It's closer to true. This doesn't have to be exact. f has a single well defined wavelengths. Or f is made up of a wide range of wavelengths? Think it to yourself. Ponder that one for a minute. OK, now before we get talking about it. Hold on, hold on, hold on. Since we don't have clickers, but I want to pull off the same effect, and we can do this, because it's binary here. I want everyone close your eyes. Just close your eyes, just for a moment. Yeah. Or close the eyes of the person next to you. That's fine. And now and I want you to vote. A is f has a single well defined wavelength. B is f has a wide range of wavelengths. So how many people think A, a single wavelength? OK. Lower your hands, good. And how many people think B, a wide range of wavelengths? Awesome. So this is exactly what happens when we actually use clickers. It's 50/50. So now you guys need to talk to the person next to you and convince each other of the truth. AUDIENCE: [CHATTER]. All right, so the volume sort of tones down as people, I think, come to resolution. Close your eyes again. Once more into the breach, my friends. So close your eyes, and now let's vote again. f of x has a single, well defined wavelength. And now f of x is made up of a range of wavelengths? OK. There's a dramatic shift in the field to B, it has a wide range of wavelengths, not a single wavelength. And that is, in fact, the correct answer. OK, so learning happens. That was an empirical test. So does anyone want to defend this view that f is made of a wide range of wavelengths? Sure, bring it. AUDIENCE: So, the sine wave is an infinite, and it cancels out past minus l over 2 and positive l over 2, which means you need to add a bunch of wavelengths to actually cancel it out there. PROFESSOR: Awesome, exactly. Exactly. If you only had the thing of a single wavelength, it would continue with a single wavelength all the way out. In fact, there's a nice way to say this. When you have a sine wave, what can you say about it's-- we know that a sine wave is continuous, and it's continuous everywhere, right? It's also differentiable everywhere. Its derivative is continuous and differentiable everywhere, because it's a cosine, right? So if yo you take a superposition of sines and cosines, do you ever get a discontinuity? No. Do you ever get something whose derivative is discontinuous? No. So how would you ever reproduce a thing with a discontinuity using sines and cosines? Well, you'd need some infinite sum of sines and cosines where there's some technicality about the infinite limit being singular, because you can't do it a finite number of sines and cosines. That function is continuous, but its derivative is discontinuous. Yeah? So it's going to take an infinite number of sines and cosines to reproduce that little kink at the edge. Yeah? AUDIENCE: So a finite number of sines and cosines doesn't mean finding-- or an infinite number of sines and cosines doesn't mean infinite [? regular ?] sines and cosines, right? Because over a finite region [INAUDIBLE]. PROFESSOR: That's true, but you need arbitrarily-- so let's talk about that. That's an excellent question. That's a very good question. The question here is look, there's two different things you can be talking about. One is arbitrarily large and arbitrarily short wavelengths, so an arbitrary range of wavelengths. And the other is an infinite number. But an infinite number is silly, because there's a continuous variable here k. You got an infinite number of wavelengths between one and 1.2, right? It's continuous. So which one do you mean? So let's go back to this connection that we got a minute ago from short distance and high momentum. That thing looks like it has one particular wavelength. But I claim, in order to reproduce that as a superposition of states with definite momentum, I need arbitrarily high wavelength. And why do I need arbitrarily high wavelength modes? Why do we need to arbitrarily high momentum modes? Well, it's because of this. We have a kink. And this feature, what's the length scale of that feature? It's infinitesimally small, which means I'm going to have to-- in order to reproduce that, in order to probe it, I'm going to need a momentum that's arbitrarily large. So it's really about the range, not just the number. But you need arbitrarily large momentum. To construct or detect an arbitrarily small feature you need arbitrarily large momentum modes. Yeah? AUDIENCE: Why do you [INAUDIBLE]? Why don't you just say, oh you need an arbitrary small wavelength? Why wouldn't you just phrase that [INAUDIBLE]? PROFESSOR: I chose to phrase it that way because I want an emphasize and encourage-- I emphasize you to think and encourage you to conflate short distance and large momentum. I want the connection between momentum and the length scale to be something that becomes intuitive to you. So when I talk about something with short features, I'm going to talk about it as something with large momentum. And that's because in a quantum mechanical system, something with short wavelength is something that carries large momentum. That cool? Great. Good question. AUDIENCE: So earlier you said that any reasonable wave function, a possible wave function, does that mean they're not supposed to be Fourier transformable? PROFESSOR: That's usually a condition. Yeah, exactly. We don't quite phrase it that way. And in fact, there's a problem on your problem set that will walk you through what we will mean. What should be true of the Fourier transform in order for this to reasonably function. And among other things-- and your intuition here is exactly right-- among other things, being able to have a Fourier transform where you don't have arbitrarily high momentum modes is going to be an important condition. That's going to turn to be related to the derivative being continuous. That's a very good question. So that's the optional problem 8 on problem set 2. Other questions? PROFESSOR: Cool, so that's it for the clicker questions. Sorry for the technology fail. So I'm just going to turn this off in disgust. That's really irritating. So today what I want to start on is pick up on the discussion of the uncertainty principle that we sort of outlined previously. The fact that when we have a wave function with reasonably well defined position corresponding to a particle with reasonably well defined position, it didn't have a reasonably well defined momentum and vice versa. The certainty of the momentum seems to imply lack of knowledge about the position and vice versa. So in order to do that, we need to define uncertainty. So I need to define for you delta x and delta p. So first I just want to run through what should be totally remedial probability, but it's always useful to just remember how these basic things work. So consider a set of people in a room, and I want to plot the number of people with a particular age as a function of the age of possible ages. So let's say we have 16 people, and at 14 we have one, and at 15 we have 1, and at 16 we have 3. And that's 16. And at 20 we have 2. And at 21 we have 4. And at 22 we have 5. And that's it. OK. So 1, 1, 3, 2, 4, 5. OK, so what's the probability that any given person in this group of 16 has a particular age? I'll call it a. So how do we compute the probability that they have age a? Well this is easy. It's the number that have age a over the total number. So note an important thing, an important side note, which is that the sum over all possible ages of the probability that you have age a is equal to 1, because it's just going to be the sum of the number with a particular age over the total number, which is just the sum of the number with any given age. So here's some questions. So what's the most likely age? If you grabbed one of these people from the room with a giant Erector set, and pull out a person, and let them dangle, and ask them what their age is, what's the most likely they'll have? AUDIENCE: 22. PROFESSOR: 22. On the other hand, what's the average age? Well, just by eyeball roughly what do you think it is? So around 19 or 20. It turns out to be 19.2 for this. OK. But if everyone had a little sticker on their lapel that says I'm 14, 15, 16, 20, 21 or 22, how many people have the age 19.2? None, right? So a useful thing is that the average need not be an observable value. This is going to come back to haunt us. Oops, 19.4. That's what I got. So in particular how did I get the average? I'm going to define some notation. This notation is going to stick with us for the rest of quantum mechanics. The average age, how do I compute it? So we all know this, but let me just be explicit about it. It's the sum over all possible ages of the number of the number of people with that age times the age divided by the total number of people. OK? So in this case, I'd go 14,14, 16, 16, 16, 20, 20, 21, 21, 21 21, 22, 22, 22, 22, 22. And so that's all I've written here. But notice that I can write this in a nice way. This is equal to the sum over all possible ages of a times the ratio of Na to N with a ratio of Na to n total. That's just the probability that any given person has a probability a. a times probability of a. So the expected value is the sum over all possible values of the value times the probability to get that value. Yeah? This is the same equation, but I'm going to box it. It's a very useful relation. And so, again, does the average have to be measurable? No, it certainly doesn't. And it usually isn't. So let's ask the same thing for the square of ages. What is the average of a squared? Square the ages. You might say, well, why would I ever care about that? But let's just be explicit about it. So following the same logic here, the average of a squared, the average value of the square of the ages is, well, I'm going to do exactly the same thing. It's just a squared, right? 14 squared, 15 squared, 16 square, 16 squared, 16 squared. So this is going to give me exactly the same expression. So over a of a squared probability of measuring a. And more generally, the expected value, or the average value of some function of a is equal-- and this is something you don't usually do-- is equal to the sum over a of f of a, the value of f given a particular value of a, times the probability that you measure that value of a in the first place. It's exactly the same logic as averages. Right, cool. So here's a quick question. Is a squared equal to the expected value of a squared? AUDIENCE: No. PROFESSOR: Right, in general no, not necessarily. So for example, the average value-- suppose we have a Gaussian centered at the origin. So here's a. Now a isn't age, but it's something-- I don't know. You include infants or whatever. It's not age. Its happiness on a given day. So what's the average value? Meh. Right? Sort of vaguely neutral, right? But on the other hand, if you take a squared, very few people have a squared as zero. Most people have a squared as not a 0 value. And most people are sort of in the middle. Most people are sort of hazy on what the day is. So in this case, the expected value of a, or the average value of a is 0. The average value of a squared is not equal to 0. Yeah? And that's because the squared has everything positive. So how do we characterize-- this gives us a useful tool for characterizing the width of a distribution. So here we have a distribution where its average value is 0, but its width is non-zero. And then the expectation value of a squared, the expected value of a squared, is non-zero. So how do we define the width of a distribution? This is going to be like our uncertainty. How happy are you today? Well, I'm not sure. How unsure are you? Well, that should give us a precise measure. So let me define three things. First the deviation. So the deviation is going to be a minus the average value of a. So this is just take the actual value of a and subtract off the average value of a. So we always get something that's centered at 0. I'm going to write it like this. Note, by the way, just a convenient thing to note. The average value of a minus it's average value. Well, what's the average value of 7? AUDIENCE: 7. PROFESSOR: OK, good. So that first term is the average value of a. And that second term is the average value of this number, which is just this number minus a. So this is 0. Yeah? The average value of a number is 0. The average value of this variable is the average value of that variable, but that's 0. So deviation is not a terribly good thing on average, because on average the deviation is always 0. That's what it means to say this is the average. So the derivation is saying how far is any particular instance from the average. And if you average those deviations, they always give you 0. So this is not a very good measure of the actual width of the system. But we can get a nice measure by getting the deviation squared. And let's take the mean of the derivation squared. So the mean of the derivation squared, mean of a minus the average value of a squared. This is what I'm going to call the standard deviation. Which is a little odd, because really you'd want to call it the standard deviation squared. But whatever. We use funny words. So now what does it mean if the average value of a is 0? It means it's centered at 0, but what does it mean if the standard deviation of a is 0? So if the standard deviation is 0, one then the distribution has no width, right? Because if there was any amplitude away from the average value, then that would give a non-zero strictly positive contribution to this average expectation, and this wouldn't be 0 anymore. So standard deviation is 0, as long as there's no width, which is why the standard deviation is a good useful measure of width or uncertainty. And just as a note, taking this seriously and taking the square, so standard deviation squared, this is equal to the average value of a squared minus twice a times the average value of a plus average value of a quantity squared. But if you do this out, this is going to be equal to a squared minus 2 average value of a average value of a. That's just minus twice the average value of a quantity squared. And then plus average value of a squared. So this is an alternate way of writing the standard deviation. OK? So we can either write it in this fashion or this fashion. And the notation for this is delta a squared. OK? So when I talk about an uncertainty, what I mean is, given my distribution, I compute the standard deviation. And the uncertainty is going to be the square root of the standard deviations squared. OK? So delta a, the words I'm going to use for this is the uncertainty in a given some probability distribution. Different probability distributions are going to give me different delta a's. So one thing that's sort of annoying is that when you write delta a, there's nothing in the notation that says which distribution you were talking about. When you have multiple distributions, or multiple possible probability distributions, sometimes it's useful to just put given the probability distribution p of a. This is not very often used, but sometimes it's very helpful when you're doing calculations just to keep track. Everyone cool with that? Yeah, questions? AUDIENCE: [INAUDIBLE] delta a squared, right? PROFESSOR: Yeah, exactly. Of delta a squared. Yeah. Other questions? Yeah? AUDIENCE: So really it should be parentheses [INAUDIBLE]. PROFESSOR: Yeah, it's just this is notation that's used typically, so I didn't put the parentheses around precisely to alert you to the stupidities of this notation. So any other questions? Good. OK, so let's just do the same thing for continuous variables. Now for continuous variables. I'm just going to write the expressions and just get them out of the way. So the average value of some x, given a probability distribution on x where x is a continuous variable, is going to be equal to the integral. Let's just say x is defined from minus infinity to infinity, which is pretty useful, or pretty typical. dx probability distribution of x times x. I shouldn't use curvy. I should just use x. And similarly for x squared, or more generally, for f of x, the average value of f of x, or the expected value of f of x given this probability distribution, is going to be equal to the integral dx minus infinity to infinity. The probability distribution of x times f of x. In direct analogy to what we had before. So this is all just mathematics. And we define the uncertainty in x is equal to the expectation value of x squared minus the expected value of x quantity squared. And this is delta x squared. If you see me dropping an exponent or a factor of 2, please, please, please tell me. So thank you for that. All of that is just straight up classical probability theory. And I just want to write this in the notation of quantum mechanics. Given that the system is in a state described by the wave function psi of x, the average value, the expected value of x, the typical value if you just observe the particle at some moment, is equal to the integral over all possible values of x. The probability distribution, psi of x norm squared x. And similarly, for any function of x, the expected value is going to be equal to the integral dx. The probability distribution, which is given by the norm squared of the wave function times f of x minus infinity to infinity. And same definition for uncertainty. And again, this notation is really dangerous, because the expected value of x depends on the probability distribution. In a physical system, the expected value of x depends on what the state of the system is, what the wave function is, and this notation doesn't indicate that. So there are a couple of ways to improve this notation. One of which is-- so this is, again, a sort of side note. One way to improve this notation x is to write the expected value of x in the state psi, so you write psi as a subscript. Another notation that will come back-- you'll see why this is a useful notation later in the semester-- is this notation, psi. And we will give meaning to this notation later, but I just want to alert you that it's used throughout books, and it means the same thing as what we're talking about the expected value of x given a particular state psi. OK? Yeah? AUDIENCE: To calculate the expected value of momentum do you need to transform the-- PROFESSOR: Excellent question. Excellent, excellent question. OK, so the question is, how do we do the same thing for momentum? If you want to compute the expected value of momentum, what do you have to do? Do you have to do some Fourier transform to the wave function? So this is a question that you're going to answer on the problem set and that we made a guess for last time. But quickly, let's just think about what it's going to be purely formally. Formally, if we want to know the likely value of the momentum, the likely value the momentum, it's a continuous variable. Just like any other observable variable, we can write as the integral over all possible values of momentum from, let's say, it could be minus infinity to infinity. The probability of having that momentum times momentum, right? Everyone cool with that? This is a tautology, right? This is what you mean by probability. But we need to know if we have a quantum mechanical system described by state psi of x, how do we can get the probability that you measure p? Do I want to do this now? Yeah, OK I do. And we need a guess. Question mark. We made a guess at the end of last lecture that, in quantum mechanics, this should be dp minus infinity to infinity of the Fourier transform. Psi tilde of p up to an h bar factor. Psi tilde of p, the Fourier transform p norm squared. OK, so we're guessing that the Fourier transform norm squared is equal to the probability of measuring the associated momentum. So that's a guess. That's a guess. And so on your problem set you're going to prove it. OK? So exactly the same logic goes through. It's a very good question, thanks. Other questions? Yeah? AUDIENCE: Is that p the momentum itself? Or is that the probability? PROFESSOR: So this is the probability of measuring momentum p. And that's the value p. We're summing over all p's. This is the probability, and that's actually p. So the Fourier transform is a function of the momentum in the same way that the wave function is a function of the position, right? So this is a function of the momentum. It's norm squared defines the probability. And then the p on the right is this p, because we're computing the expected value of p, or the average value of p. That make sense? Cool. Yeah? AUDIENCE: Are we then multiplying by p squared if we're doing all p's? Because we have the dp times p for each [INAUDIBLE]. PROFESSOR: No. So that's a very good question. So let's go back. Very good question. Let me phrase it in terms of position, because the same question comes up. Thank you for asking that. Look at this. This is weird. I'm going to phrase this as a dimensional analysis question. Tell me if this is the same question as you're asking. This is a thing with dimensions of what? Length, right? But over on the right hand side, we have a length and a probability, which is a number, and then another length. That looks like x squared, right? So why are we getting something with dimensions of length, not something with dimensions of length squared? And the answer is this is not a probability. It is a probability density. So it's got units of probability per unit length. So this has dimensions of one over length. So this quantity, p of x dx, tells me the probability, which is a pure number, no dimensions. The probability to find the particle between x and x plus dx. Cool? So that was our second postulate. Psi of x dx squared is the probability of finding it in this domain. And so what we're doing is we're summing over all such domains the probability times the value. Cool? So this is the difference between discrete, where we didn't have these probability densities, we just had numbers, pure numbers and pure probabilities. Now we have probability densities per unit whatever. Yeah? AUDIENCE: How do you pronounce the last notation that you wrote? PROFESSOR: How do you pronounce? Good, that's a good question. The question is, how do we pronounce these things. So this is called the expected value of x, or the average value of x, or most typically in quantum mechanics, the expectation value of x. So you can call it anything you want. This is the same thing. The psi is just to denote that this is in the state psi. And it can be pronounced in two ways. You can either say the expectation value of x, or the expectation of x in the state psi. And this would be pronounced one of two ways. The expectation value of x in the state psi, or psi x psi. Yeah. That's a very good question. But they mean the same thing. Now, I should emphasize that you can have two ways of describing something that mean the same thing, but they carry different connotations, right? Like have a friend who's a really nice guy. He's a mensch. He's a good guy. And so I could see he's a nice guy, I could say he's [? carinoso ?], and they mean different things in different languages. It's the same idea, but they have different flavors, right? So whatever your native language is, you've got some analog of this. This means something in a particular mathematical language for talking about quantum mechanics. And this has a different flavor. It carries different implications, and we'll see what that is later. We haven't got there yet. Yeah? AUDIENCE: Why is there a double notation of psi? PROFESSOR: Why is there a double notation of psi? Yeah, we'll see later. Roughly speaking, it's because in computing this expectation value, there's a psi squared. And so this is to remind you of that. Other questions? Terminology is one of the most annoying features of quantum mechanics. Yeah? AUDIENCE: So it seems like this [INAUDIBLE] variance is a really convenient way of doing it. How is it the Heisenberg uncertainty works exactly as it does for this definition of variance. PROFESSOR: That's a very good question. In order to answer that question, we need to actually work out the Heisenberg uncertainty relation. So the question is, look, this is some choice of uncertainty. You could have chosen some other definition of uncertainly. We could have considered the expectation value of x to the fourth minus x to the fourth and taken the fourth root of that. So why this one? And one answer is, indeed, the uncertainty relation works out quite nicely. But then I think important to say here is that there are many ways you could construct quantities. This is a convenient one, and we will discover that it has nice properties that we like. There is no God given reason why this had to be the right thing. I can say more, but I don't want to take the time to do it, so ask in office hours. OK, good. The second part of your question was why does the Heisenberg relation work out nicely in terms of these guys, and we will study that in extraordinary detail. We'll see that. So we're going to derive it twice soon and then later. The later version is better. So let me work out some examples. Or actually, I'm going to skip the examples in the interest of time. They're in the notes, and so they'll be posted on the web page. By the way, the first 18 lectures of notes are posted. I had a busy night last night. So let's come back to computing expectation values for momentum. So I want to go back to this and ask a silly-- I want to make some progress towards deriving this relation. So I want to start over on the definition of the expected value of momentum. And I'd like to do it directly in terms of the wave function. So how would we do this? So one way of saying this is what's the average value of p. Well, I can phrase this in terms of the wave function the following way. I'm going to sum over all positions dx. Expectation value of x squared from minus infinity to infinity. And then the momentum associated to the value x. So it's tempting to write something like this down to think maybe there's some p of x. This is a tempting thing to write down. Can we? Are we ever in a position to say intelligently that a particle-- that an electron is both hard and white? AUDIENCE: No. PROFESSOR: No, because being hard is a superposition of being black and white, right? Are we ever in a position to say that our particle has a definite position x and correspondingly a definite momentum p. It's not that we don't get too. It's that it doesn't make sense to do so. In general, being in a definite position means being in a superposition of having different values for momentum. And if you want a sharp way of saying this, look at these relations. They claim that any function can be expressed as a superposition of states with definite momentum, right? Well, among other things a state with definite position, x0, can be written as a superposition, 1 over 2pi integral dk. I'll call this delta tilde of k. e to the ikx. If you haven't played with delta functions before and you haven't seen this, then you will on the problem set, because we have a problem that works through a great many details. But in particular, it's clear that this is not-- this quantity can't be a delta function of k, because, if it were, this would be just e to the ikx. And that's definitely not a delta function. Meanwhile, what can you say about the continuity structure of a delta function. Is it continuous? No. Its derivative isn't continuous. Its second derivative. None of its derivatives are in any way continuous. They're all absolutely horrible, OK? So how many momentum modes am I going to need to superimpose in order to reproduce a function that has this sort of structure? An infinite number. And it turns out it's going to be an infinite number with the same amplitude, slightly different phase, OK? So you can never say that you're in a state with definite position and definite momentum. Being in a state with definite position means being in a superposition of being in a superposition. In fact, I'm just going right down the answer here. e to the ikx0. Being in a state with definite position means being in a superposition of states with arbitrary momentum and vice versa. You cannot be in a state with definite position, definite momentum. So this doesn't work. So what we want is we want some good definition. So this does not work. We want some good definition of p given that we're working with a wave function which is a function of x. What is that good definition of the momentum? We have a couple of hints. So hint the first. So this is what we're after. Hint the first is that a wave-- we know that given a wave with wave number k, which is equal 2pi over lambda, is associated, according to de Broglie and according to Davisson-Germer experiments, to a particle-- so having a particle-- a wave, with wave number k or wavelength lambda associated particle with momentum p is equal to h bar k. Yeah? But in particular, what is a plane with wavelength lambda or wave number k look like? That's e to the iks. And if I have a wave, a plane wave e to the iks, how do I get h bar k out of it? Note the following, the derivative with respect to x. Actually let me do this down here. Note that the derivative with respect to x of e to the ikx is equal to ik e to the ikx. There's nothing up my sleeves. So in particular, if I want to get h bar k, I can multiply by h bar and divide by i. Multiply by h bar, divide by i, derivative with respect to x e to the ikx. And this is equal to h bar k e to the ikx. That's suggestive. And I can write this as p e to the ikx. So let's quickly check the units. So first off, what are the units of h bar? Here's the super easy to remember the units of-- or dimensions of h bar are. Delta x delta p is h bar. OK? If you're ever in doubt, if you just remember, h bar has units of momentum times length. It's just the easiest way to remember it. You'll never forget it that way. So if h bar has units of momentum times length, what are the units of k? 1 over length. So does this dimensionally make sense? Yeah. Momentum times length divided by length number momentum. Good. So dimensionally we haven't lied yet. So this makes it tempting to say something like, well, hell h bar upon i derivative with respect to x is equal in some-- question mark, quotation mark-- p. Right? So at this point it's just tempting to say, look, trust me, p is h bar upon idx. But I don't know about you, but I find that deeply, deeply unsatisfying. So let me ask the question slightly differently. We've followed the de Broglie relations, and we've been led to the idea that using wave functions that there's some relationship between the momentum, the observable quantity that you measure with sticks, and meters, and stuff, and this operator, this differential operator, h bar upon on i derivative with respect to x. By the way, my notation for dx is the partial derivative with respect to x. Just notation. So if this is supposed to be true in some sense, what is momentum have to do with a derivative? Momentum is about velocities, which is like derivatives with respect to time, right? Times mass. Mass times derivative with respect to time, velocity. So what does it have to do with the derivative with respect to position? And this ties into the most beautiful theorem in classical mechanics, which is the Noether's theorem, named after the mathematician who discovered it, Emmy Noether. And just out of curiosity, how many people have seen Noether's theorem in class. Oh that's so sad. That's a sin. OK, so here's a statement of Noether's theorem, and it underlies an enormous amount of classical mechanics, but also of quantum mechanics. Noether, incidentally, was a mathematician. There's a whole wonderful story about Emmy Noether. Ville went to her and was like, look, I'm trying to understand the notion of energy. And this guy down the hall, Einstein, he has a theory called general relativity about curved space times and how that has something to do with gravity. But it doesn't make a lot of sense to me, because I don't even know how to define the energy. So how do you define momentum and energy in this guy's crazy theory? And so Noether, who was a mathematician, did all sorts of beautiful stuff in algebra, looked at the problem and was like I don't even know what it means in classical mechanics. So what is a mean in classical mechanics? So she went back to classical mechanics and, from first principles, came up with a good definition of momentum, which turns out to underlie the modern idea of conserved quantities and symmetries. And it's had enormous far reaching impact, and say her name would praise. So Noether tells us the following statement, to every symmetry-- and I should say continuous symmetry-- to every symmetry is associated a conserved quantity. OK? So in particular, what do I mean by symmetry? Well, for example, translations. x goes to x plus some length l. This could be done for arbitrary length l. So for example, translation by this much or translation by that much. These are translations. To every symmetry is associated a conserved quantity. What symmetry is associated to translations? Conservation of momentum, p dot. Time translations, t goes to t plus capital T. What's a conserved quantity associated with time translational symmetry? Energy, which is time independent. And rotations. Rotational symmetries. x, as a vector, goes to some rotation times x. What's conserved by virtue of rotational symmetry? AUDIENCE: Angular momentum. PROFESSOR: Angular momentum. Rock on. OK So quickly, I'm not going to prove to you Noether's theorem. It's one of the most beautiful and important theorems in physics, and you should all study it. But let me just convince you quickly that it's true in classical mechanics. And this was observed long before Noether pointed out why it was true in general. What does it mean to have transitional symmetry? It means that, if I do an experiment here and I do it here, I get exactly the same results. I translate the system and nothing changes. Cool? That's what I mean by saying I have a symmetry. You do this thing, and nothing changes. OK, so imagine I have a particle, a classical particle, and it's moving in some potential. This is u of x, right? And we know what the equations of motion are in classical mechanics from f equals ma p dot is equal to the force, which is minus the gradient of u. Minus the gradient of u. Right? That's f equals ma in terms of the potential. Now is the gradient of u 0? No. In this case, there's a force. So if I do an experiment here, do I get the same thing as doing my experiment here? AUDIENCE: No. PROFESSOR: Certainly not. The [? system ?] is not translationally invariant. The potential breaks that translational symmetry. What potential has translational symmetry? AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, constant. The only potential that has full translational symmetry in one dimension is translation invariant, i.e. constant. OK? What's the force? AUDIENCE: 0. PROFESSOR: 0. 0 gradient. So what's p dot? Yep. Noether's theorem. Solid. OK. Less trivial is conservation of energy. I claim and she claims-- and she's right-- that if the system has the same dynamics at one moment and a few moments later and, indeed, any amount of time later, if the laws of physics don't change in time, then there must be a conserved quantity called energy. There must be a conserved quantity. And that's Noether's theorem. So this is the first step, but this still doesn't tell us what momentum exactly has to do with a derivative with respect to space. We see that there's a relationship between translations and momentum conservation, but what's the relationship? So let's do this. I'm going to define an operation called translate by L. And what translate by L does is it takes f of x and it maps it to f of x minus L. So this is a thing that affects the translation. And why do I say that's a translation by L rather than minus L. Well, the point-- if you have some function like this, and it has a peak at 0, then after the translation, the peak is when x is equal to L. OK? So just to get the signs straight. So define this operation, which takes a function of x and translates it by L, but leaves it otherwise identical. So let's consider how translations behave on functions. And this is really cute. f of x minus L can be written as a Taylor expansion around the point x-- around the point L equals 0. So let's do Taylor expansion for small L. So this is equal to f of x minus L derivative with respect to x of f of x plus L squared over 2 derivative squared, two derivatives of x, f of x plus dot, dot, dot. Right? I'm just Taylor expanding. Nothing sneaky. Let's add the next term, actually. Let me do this on a whole new board. All right, so we have translate by L on f of x is equal to f of x minus L is equal to f of x. Now Taylor expanding minus L derivative with respect to x of f plus L squared over 2-- I'm not giving myself enough space. I'm sorry. f of x minus L is equal to f of x minus L with respect to x of f of x plus L squared over 2 to derivatives of x f of x minus L cubed over 6-- we're just Taylor expanding-- cubed with respect to x of f of x and so on. Yeah? But I'm going to write this in the following suggestive way. This is equal to 1 times f of x minus L derivative with respect to x f of x plus L squared over 2 derivative with respect to x squared times f of x minus L cubed over 6 derivative cubed with respect to x plus dot, dot, dot. Everybody good with that? But this is a series that you should recognize, a particular Taylor series for a particular function. It's a Taylor expansion for the AUDIENCE: Exponential. PROFESSOR: Exponential. e to the minus L derivative with respect to x f of x. Which is kind of awesome. So let's just check to make sure that this makes sense from dimensional grounds. So that's a derivative with respect to x as units of 1 over length. That's a length, so this is dimensionless, so we can exponentiate it. Now you might look at me and say, look, this is silly. You've taken an operation like derivative and exponentiated it. What does that mean? And that is what it means? [LAUGHTER] OK? So we're going to do this all the time in quantum mechanics. We're going to do things like exponentiate operations. We'll talk about it in more detail, but we're always going to define it in this fashion as a formal power series. Questions? AUDIENCE: Can you transform operators from one space to another? PROFESSOR: Oh, you totally can. But we'll come back to that. We're going to talk about operators next time. OK, so here's where we are. So from this what is a derivative with respect to x mean? What does a derivative with respect to x do? Well a derivative with respect to x is something that generates translations with respect to x through a Taylor expansion. If we have L be arbitrarily small, right? L is arbitrarily small. What is the translation by an arbitrarily small amount of f of x? Well, if L is arbitrarily small, we can drop all the higher order terms, and the change is just Ldx. So the derivative with respect to x is telling us about infinitesimal translations. Cool? The derivative with respect to a position is something that tells you, or controls, or generates infinitesimal translations. And if you exponentiate it, you do it many, many, many times in a particular way, you get a macroscopic finite translation. Cool? So this gives us three things. Translations in x are generated by derivative with respect to x. But through Noether's theorem translations, in x are associated to conservation of momentum. So you shouldn't be so shocked-- it's really not totally shocking-- that in quantum mechanics, where we're very interested in the action of things on functions, not just in positions, but on functions of position, it shouldn't be totally shocking that in quantum mechanics, the derivative with respect to x is related to the momentum in some particular way. Similarly, translations in t are going to be generated by what operation? Derivative with respect to time. So derivative with respect to time from Noether's theorem is associated with conservation of energy. That seems plausible. Derivative with respect to, I don't know, an angle, a rotation. That's going to be associated with what? Angular momentum? But angular momentum around the axis for whom this is the angle, so I'll call that z for the moment. And we're going to see these pop up over and over again. But here's the thing. We started out with these three principles today, and we've let ourselves to some sort of association between the momentum and the derivative like this. OK? And I've given you some reason to believe that this isn't totally insane. Translations are deeply connected with conservation of momentum. Transitional symmetry is deeply connected with conservation momentum. And an infinitesimal translation is nothing but a derivative with respect to position. Those are deeply linked concepts. But I didn't derive anything. I gave you no derivation whatsoever of the relationship between d dx and the momentum. Instead, I'm simply going to declare it. I'm going to declare that, in quantum mechanics-- you cannot stop me-- in quantum mechanics, p is represented by an operator, it's represented by the specific operator h bar upon I derivative with respect to x. And this is a declaration. OK? It is simply a fact. And when they say it's a fact, I mean two things by that. The first is it is a fact that, in quantum mechanics, momentum is represented by derivative with respect to x times h bar upon i. Secondly, it is a fact that, if you take this expression and you work with the rest of the postulates of quantum mechanics, including what's coming next lecture about operators and time evolution, you reproduce the physics of the real world. You reproduce it beautifully. You reproduce it so well that no other models have even ever vaguely come close to the explanatory power of quantum mechanics. OK? It is a fact. It is not true in some epistemic sense. You can't sit back and say, ah a priori starting with the integers we derive that p is equal to-- no, it's a model. But that's what physics does. Physics doesn't tell you what's true. Physics doesn't tell you what a priori did the world have to look like. Physics tells you this is a good model, and it works really well, and it fits the data. And to the degree that it doesn't fit the data, it's wrong. OK? This isn't something we derive. This is something we declare. We call it our model, and then we use it to calculate stuff, and we see if it fits the real world. Out, please, please leave. Thank you. [LAUGHTER] I love MIT. I really do. So let me close off at this point with the following observation. [LAUGHTER] We live in a world governed by probabilities. There's a finite probability that, at any given moment, that two pirates might walk into a room, OK? [LAUGHTER] You just never know. [APPLAUSE] But those probabilities can be computed in quantum mechanics. And they're computed in the following ways. They're computed the following ways as we'll study in great detail. If I take a state, psi of x, which is equal to e to the ikx, this is a state that has definite momentum h bar k. Right? We claimed this. This was de Broglie and Davisson-Germer. Note the following, take this operator and act on this wave function with this operator. What do you get? Well, we already know, because we constructed it to have this property. P hat on psi of x-- and I'm going to call this psi sub k of x, because it has a definite k-- is equal to h bar k psi k of x. A state with a definite momentum has the property that, when you hit it with the operation associated with momentum, you get back the same function times a constant, and that constant is exactly the momentum we ascribe to that plane wave. Is that cool? Yeah? AUDIENCE: Question. Just with notation, what does the hat above the p [INAUDIBLE]? PROFESSOR: Good. Excellent. So the hat above the P is to remind you that P is on a number. It's an operation. It's a rule for acting on functions. We'll talk about that in great detail next time. But here's what I want to emphasize. This is a state which is equal to all others in the sense that it's a perfectly reasonable wave function, but it's more equal because it has a simple interpretation. Right? The probability that I measure the momentum to be h bar k is one, and the probability that I measure it to be anything else is 0, correct? But I can always consider a state which is a superposition. Psi is equal to alpha, let's just do 1 over 2 e to the ikx. k1 x plus 1 over root 2 e to the minus ikx. Is this state a state with definite momentum? If I act on this state-- I'll call this i sub s-- if I act on this state with the momentum operator, do I get back this state times a constant? No. That's interesting. And so it seems to be that if we have a state with definite momentum and we act on it with momentum operator, we get back its momentum. If we have a state that's a superposition of different momentum and we act on it with a momentum operator, this gives us h bar k 1, this gives us h bar k2. So it changes which superposition we're talking about. We don't get back our same state. So the action of this operator on a state is going to tell us something about whether the state has definite value of the momentum. And these coefficients are going to turn out to contain all the information about the probability of the system. This is the probability when norm squared that will measure the system to have momentum k1. And this coefficient norm squared is going to tell us the probability that we have momentum k2. So I think the current wave function is something like a superposition of 1/10 psi pirates plus 1 minus is 1/100 square root. To normalize it properly psi no pirates. And I'll leave you with pondering this probability. See you guys next time. [APPLAUSE] CHRISTOPHER SMITH: We've come for Prof. Allan Adams. PROFESSOR: It is I. CHRISTOPHER SMITH: When in the chronicles of wasted time, I see descriptions of fairest rights, and I see lovely shows of lovely dames. And descriptions of ladies dead and lovely nights. Then in the bosom of fair loves depths. Of eyes, of foot, of eye, of brow. I see the antique pens do but express the beauty that you master now. So are all their praises but prophecies of this, our time. All you prefiguring. But though they had but diving eyes-- PROFESSOR: I was wrong about the probabilities. [LAUGHTER] CHRISTOPHER SMITH: But though they had but diving eyes, they had not skill enough you're worth to sing. For we which now behold these present days have eyes to behold. [LAUGHTER] But not tongues to praise. [APPLAUSE] It's not over. You wait. ARSHIA SURTI: Not marbled with gilded monuments of princes shall outlive this powerful rhyme. But you shall shine more bright in these contents that unswept stone besmear its sluttish tide. When wasteful war shall statues overturn and broils root out the work of masonry. Nor Mars his sword. Nor war's quick fire shall burn the living record of your memory. Gainst death and all oblivious enmity shall you pace forth. Your praise shall still find room, even in the eyes of all posterity. So no judgment arise till you yourself judgment arise. You live in this and dwell in lover's eyes. [APPLAUSE] CHRISTOPHER SMITH: Verily happy Valentine's day upon you. May your day be filled with love and poetry. Whatever state you're in, we will always love you. [LAUGHTER] [APPLAUSE] Signed, Jack Florian, James [INAUDIBLE]. [LAUGHTER] PROFESSOR: Thank you, sir. Thank you. CHRISTOPHER SMITH: Now we go. [APPLAUSE]
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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. I am going to start now by telling you a little bit more about Lewis theory. Last time we went through aspects of the cube theory of Lewis and showed how that could account for single bonds and double bonds, but not triple bonds. And that when that was superseded by the electron pair theory, the idea being that these electron pairs would, in fact, be oriented at the vertices of a tetrahedron. And I showed you that tetrahedron, as described in a cube last time, that you could then account for triple bonds. And, if you looked at the notes, you also saw that certain aspects of the dynamic behavior of certain bond types, rotation around single bonds, hindered restricted rotation around double bonds, and so forth. Those were accounted for by this electron pair theory in which the four pairs of electrons were oriented at these vertices of a tetrahedron. So, that was another quite important triumph of that part of Lewis's theory. And the problem was, although that electron pair theory initially put forward by Lewis was so successful at accounting for the properties of many different kinds of molecules and was a good description of their electronic structure, there was one class of molecules, a very important class of molecules, namely those known as aromatic. And the most important member of the class of aromatic compounds is, in fact, the benzene molecule. Benzene has the formula C six H six. And we can easily calculate the number of valence electrons in benzene if we just say 4 for carbon times 6, plus 1 times 6 for the number of valance electrons from each of the six hydrogens. And that is 30 electrons. So, in trying to understand how these 30 electrons in a valence shell of benzene hold this molecule together, it is known to be a planar molecule, how do those electrons not only hold it together, -- -- but how do they account for the structure of benzene, and how do they account for its amazing stability? And this picture that I have drawn here is a representation of benzene, I will explain to you in a moment, but it comes from a person by the name of Ernest C. Crocker. Ernest C. Crocker was an MIT Bachelor of Science degree holder, who earned that degree in 1917. He was an MIT undergraduate like you. And he published a paper. Let me write down the reference for that paper. This is the Journal of the American Chemical Society, 1922, Volume 44, Page 1618. That paper was the first paper in which the Lewis electron pair theory was applied to an understanding of the electronic structure of aromatic molecules, in particular, benzene. And the amazing thing about that paper is -- Well, there are many amazing things, but one of the amazing things about that paper is that there is only one author on that paper, Ernest C. Crocker. And Ernest, who was a very bright individual, nonetheless did not go on in graduate school to earn a Ph.D. in chemistry, had many various chemistry-related interests. He explained the chemistry of many different kinds of fragrances and odors. I think he was referred to as "the man with the million dollar nose." So, this person had quite a variety of interests. He worked in what was then the Applied Chemistry Laboratory at MIT after his graduation with his Bachelor of Science degree. And one of the things he was thinking about was how to use modern electronic structure descriptions, such as Lewis theory, to explain molecules like benzene. And so, if you go and read this paper, you are going to find a very lucid discussion of how the Lewis electron pair theory could represent benzene according to this formula that I have drawn here. And in this formula, there is considered to be an electron pair between each carbon and each of the hydrogen nuclei. There is an electron pair along each carbon-carbon axis, as shown here, and here, and so on, all the way around the ring. And then, finally, you have six more electrons to come up to the number of 30, which is the number of electrons in benzene. And so, the question really is what to do with this remaining six electrons. And I have shown them around the outside of the ring, here, which is where Ernest arranged them in his work. And I just want to point out that he put forward the idea that these six electrons were circulating around the plane of the ring and involved a net one-half bond between each pair of adjacent carbon atoms in the benzene ring. So, in effect, when we draw benzene this way, with a circle in the middle, we know that that circle represents the circulating six electrons in what we are going to call the pi system. But Ernest C. Crocker was the man who put the circle into the middle of benzene. And he was an MIT undergrad. And this was a sole-authored paper. Aromaticity has a vast history in chemistry, and it is still a very active and unfolding history because of the problems, to our understanding, posed by electrons that seem to be circulating around a whole molecule rather than localized between pairs of nuclei. So, Ernest Crocker, Bachelor of Science, 1917, MIT, had a huge hand in that. I thought you might find that interesting. And, having looked at benzene rings like that, I will now draw them perhaps another way that is also useful, which comes from KekulÈ. Because I want to continue our discussion of Lewis acid-base theory. And I am going to draw two molecules here that are going to be pretty similar. When I write a molecule, as I have done here on the left, I haven't explicitly indicated each of the hydrogens that are present on the periphery of this substituted benzene ring. But you should understand from a formula like this that this is the molecule boron C 18 H 15. And over here I am drawing explicitly, at each of these peripheral positions on the substituted benzene rings, fluorine atoms in place of the hydrogens. So, this is a different molecule with formula B C 18 F These are both Lewis acids. And based on our discussion, last time, if you were to add ammonia to one of these molecules, where would the lone pair of electrons on the ammonia bind? The boron. So, yes, you have here a trigonal planer boron center. And, if you were to add an ammonia molecule, the lone pair of electrons would come in and stick to the boron -- -- because the boron is electron deficient. Just like the aluminum we discussed last time, it has only six electrons around it, and it wants eight. But what if we were to add one ammonia molecule to a flask containing both of those Lewis acids that would then be competing for the ammonia molecule? What I am asking you to do is something I will ask you to do throughout this semester, and that is to analyze a molecule's properties based on its structure and its composition. Exactly. She said that it would preferentially stick to the one with the fluorines because these very electronegative fluorines are drawing electron density away from the boron. This is one of our most electronegative elements. So you have a whole bunch of fluorines in that molecule. The whole thing, we call it perfluorinated. It is a perfluorinated triaryl boron reagent. These, in fact, are great Lewis acids, really powerful Lewis acids. And they are modern Lewis acids whose implementation in chemical processes has come about really in the last 10, 15 years. And, in fact we were talking a little bit about Professor Schrock last time. In some of his research, he has used that perfluorinated Lewis acid as an activator in catalysis to get catalytic polymerization reactions to work. And that is a very popular approach these days in Lewis acid chemistry. The design of new kinds of Lewis acids with interesting molecular architectures is something that is very much a current topic of interest in research in chemistry, because you can make lots of chemical processes happen when you have something that can tug on electron pairs. And this one tugs a lot harder than that one because this one has very electronegative fluorines to pull electron density away from that boron and to adjust the distribution of the electron density in the molecule. And we will be talking more about electron density and distribution in a few minutes in connection with what I am going to show you now. And that is -- That has to do with this molecule, which is the SO two molecule. Anyone know where SO two comes from in nature? Or in the environment, I should say? Volcanoes, absolutely. And also coal-burning power plants. Coal is a very dirty fuel, and it contains a lot of sulfur. And, when you burn that coal without controlling the way you burn it, you emit SO two into the atmosphere. So, that can be a big problem. And we will try to understand why. One of the things that SO two can do when it gets into the atmosphere is it can react with dioxygen. And that can give you -- -- SO three. And if SO two and SO three are present in the atmosphere and if there is also water present in the atmosphere, acid rain, that is exactly the type of process that we will be talking about here. SO two and SO three are examples of what we call anhydrides. Anhydride is a word that means without water. And so you should not be surprised that SO two can react with water. And when it reacts with water, it takes up water. And the product of that reaction will look like this. I draw it like that. Okay. So, H two O plus SO two going to H two SO three. The name of this molecule is sulfurous acid. And, alternatively, when SO three reacts with water -- I have one extra electron pair that should not have been there. So, both SO two and SO three, as they react with water, are going from a situation in which they are electron deficient to a situation in which the sulfur attains an octet. Okay? And you should verify that the number of electrons that I have drawn up here actually is consistent with the elements that I am using with the stated charge that I am using. But when SO three reacts with H two O to give H two SO four, we have now got sulfuric acid. Okay. And, as I did over there, I have two acids that I want here now to compare in terms of their relative strengths. Here I have a Lewis acid SO three and a Lewis acid SO two engaging in a hydration reaction, which produces sulfurous acid and sulfuric acid. And the type of acids that these are on the bottom is Bronsted acids, -- -- distinguished from Lewis acids in that the way that they behave as acids is through ionization that produces a proton. They are also Lewis acids because the Lewis definition of acidity is far more general, in saying that acids are simply entities that can accept a pair of electrons. Protons can accept a pair of electrons, so they are Lewis acids. But if you are talking about Bronsted acids, you are talking exclusively about protons that are produced by ionization of some kind of a Bronsted acid. Which one of these is the stronger acid? Sulfuric acid. And why? Yes, down here. You got the other one right. Exactly. When this ionizes, you get SO four minus. I will draw is a slightly different way that is quicker. You get HSO four minus. There is your ionization. And the idea now is that this O minus that you have, the negative charge can actually be shared among a greater number of electronegative oxygens here, namely four, as compared to here, where we have only three electronegative oxygens. It is a consideration of the very electronegative elements in your molecule that will help you understand the properties that these molecules will have in terms of acid-base chemistry. Now, how many electrons do we have in the valance shell of SO three? And, that being the case, what molecule from last time does that remind you of? Maybe seeing a picture of it will help refresh your memory. AlCl three. Exactly. I am going to show it to you anyway. And this is going to be faster than the last time, if I set this up right. Just to remind you. And if we could have the lights down just a little bit for a moment. I want to refresh your memory of the electron density distribution here in AlCl three. This is an electron density isosurface of AlCl three. And what you are noticing is that the electron density drops to a low value in between the central aluminum and the radially disposed chlorides, the three chlorides that surround that central aluminum ion. And the coloring in this is such that the blue regions represent regions in space where there is a high probability of finding paired electrons. So, basically you have three Cl minus's that are packed tightly around an Al three plus. This is a very ionic compound. See the empty region in space between aluminum and chloride, and the polarization of that otherwise spherical cloud of electrons around the chloride in the direction of that positively charged aluminum? There is your electron density distribution for that. And now I want you to keep that in mind, and we will compare the isoelectronic SO three molecule to it. SO three. Here is another case where we have 3 times 8, 24 valance electrons in a system. But the character of this molecule in terms of electron density distribution is very different. While on the blackboard, I am not able to really tell you very much about the difference between SO three and AlCl three, here I think you can see that indeed they are quite different. This is an electron density isosurface at the same contour level as what we were looking at for AlCl three. Now, to explain this and to understand just what is going on here, you need to remember that the electronegativity difference between the central sulfur and the peripheral oxygens is not very great compared to the electronegativity difference between aluminum, which is a very electropositive and metallic element, and chlorine, which is a very electronegative halogen. And so, what that results in, as shown here, is a much more equal sharing of the electrons between that central sulfur and these peripheral oxygens. So, even though these things are both Lewis acids and they both have 24 valance electrons, the electron density distribution in three-dimensional space for these molecules and the covalent verses ionic character of these molecules, is really quite different. Our section that is going to be devoted to bonding has not really kicked into gear yet, but the nice thing is that Lewis theory applies both to acid-base chemistry and to bonding, so we are able to talk a little bit about that. In a few moments, I will tell you a little more about an issue that is very important in chemistry as regards bonding. And it has to do with what happens when acids, like sulfuric acid, ionize in water. When Bronsted acids ionize in water, we get this ion produced. H three O plus, which is the hydronium ion. That is to say that if you ionize in water some Bronsted acid, the protons that are produced through that ionization are not floating around freely, naturally, because they are positively charged and they are attracted to negatively charged electrons. So, they look around in solution and find the next source of an electron that they can. And you know that if we draw out a molecule like water, according to the Lewis dot structure, it has two extra pairs of electrons, in addition to those two pairs of electrons it is using in making bonds to the two hydrogens that are on the oxygen of the water molecule. So, H plus is not just isolated around by itself in solution. It perches on an oxygen lone pair. So, H three O plus is what you get when Bronsted acids ionize in water. And furthermore, when you put these things in solution you find that you organize the water molecules that are close to the hydronium ion. Let's draw here a neighboring water molecule. And another one. And I think you can imagine that throughout a solution, we might have many of the kinds of interactions that I have drawn here as dotted peach-colored lines. Those lines represent what we call hydrogen bonds. And hydrogen bonds are enormously important in chemistry. Later, when we talk about the structure of proteins in DNA, in particular, you may be aware that the DNA double helix is held together by a network of hydrogen bonds between complimentary base pairs. So hydrogen bonds are not only restricted to the hydronium ion in aqueous solution. There are many other types of molecules that can form what we call hydrogen bonds. Another really interesting thing is that in water, the hydronium ion can move around really rapidly, much more rapidly than molecules normally diffuse through aqueous solution. And the reason for that is if you look at the arrangement of electrons and nuclei here, all I have to do is, without even moving the nuclei much at all, reorganize the hydrogen bonding network as such. And now you can see that through just a slight set of motions, our hydronium ion has moved from the left-hand side of this hydrogen bonded network, where you can see that it is indicated with the positive charge and the three solid lines drawn to the oxygen, over to the right-hand side. But it did that not by coming off and moving over, but rather through just redistribution of the electron density, so that the positively charged part ends up down on the other side. And so this way of propagating hydronium ions in aqueous solution is one of the really special aspects of Bronsted acid chemistry that takes place in water. And I think I will also show you what a hydrogen bond looks like from the standpoint of electron density. First, I am just going to show you the position of the nuclei in a very simple hydrogen bonded system. Here what you can see, the oxygens are drawn in red as spheres and the hydrogens are drawn in white as spheres. You can see that the geometry around the oxygen atoms is slightly pyramidal. And that is due, of course, to the presence of that extra lone pair here, up above one oxygen and here, up above the other oxygen. And what we have now is a hydrogen serving in a bridging fashion. And the number of electrons in this system is exactly 2 times 8, because we have two water molecules and we have an H plus. So this is a positively charged ion in which a hydronium ion, and you can pick either side, actually, is interacting with one of the lone pairs of the other water molecule. And you could imagine lots of different types of water clusters like this that are singly positively charged. And people have done a lot of work to study such clusters in solution. What you should remember, though, is that the size of the spheres that I have drawn there to represent those oxygens and hydrogens is somewhat arbitrary. But what is not arbitrary is the way that the electron density represents a molecule like this. So, we will show that to you next. Here it is. And if we could have the lights down just a little bit, please, since this one is a little harder to see. What you should see here is that we have the same structure now surrounding that representation of the water molecule hydrogen bonded to the hydronium ion that I drew a moment ago. We now have this sort of mesh, which is exactly what we have been looking at with these other molecules, namely, an electron density isosurface. And what you can see is that the electron density is falling to a pretty small value in the middle, here, where we have the proton that is the connecting glue binding together these two water molecules in this 16 valance electron system. And after one more representation of that, we will be onto our next topic. And this one is a solid display of the electron density isosurface associated with this hydrogen bonded cluster. And it is, once again, color mapped with this function that tells us about the probability of finding electrons paired up together in space. There is H three O plus hydrogen bonded to H two O. The blue color represents those regions in space where you are most likely to find pairs of electrons. And you can see that the two OH bonds over here are nicely colored blue. The lone pair of electrons up here is nicely colored blue. And then we have an interesting situation where there is some blue in between that bridging H plus and the two lone pairs that are pointed at it that produces, in fact, our hydrogen bond. So, there is a picture of hydrogen bonding in terms of electron density. And it is a type of bonding that compliments the straight, covalent, and ionic bonding that I was talking about in terms of SO three and AlCl three. So, we have added this third type of hydrogen bonding to our list of bonding interests. And now, we will talk more about what we can do when we consider Bronsted acids ionizing. Here is a generic representation of the formula of a Bronsted acid, HA. A might be, for example, the HSO four minus ion that we showed over there. When we put a Bronsted acid in aqueous solution, as I said before, we can get ionization into H plus and A minus. But we know that it is not just H plus. It is actually H three O plus **H3O^+**. And H three O plus is further hydrogen bonded in networks in the water system. But, for simplicity, I will just write it as H plus right here. Recently, in your crash review of thermodynamics, you were talking about equilibria and equilibrium constants. And we are going to make use of some of that right here because we are going to talk about the acidity constant, Ka. And that is going to be defined as equal to the hydrogen or hydronium ion concentration times the concentration of the conjugate base, A minus -- When a Bronsted acid ionizes it produces what we call the conjugate base of the acid. Here is conjugate base -- -- divided by the concentration of the acid. And this is at equilibrium. And let me just emphasize something so that you don't forget. This is an important piece of nomenclature. These square brackets here refer to concentration, usually in molarity. Okay? So that is what we are talking about. And concentration is something that can be measured. You may be familiar, for example, with the pH meter invented by Arnold O. Beckman and its utility in measuring the concentration of hydrogen ions in solution. Well, we can make use of information like that to talk about the properties of our Bronsted acids. How can we do that? Well, let's say we are going to take a particular acid such as this one, which is acetic acid. You know the smell of acetic acid if you have ever been in an establishment where they were making barbecued chicken wings. That is the smell of acetic acid. A beautiful smell. Anyway, what you do here is you are trying to figure out what is going on. You have some concentration of the acid HA, which is, we are going to talk about, acetic acid. And in solution there may also be a hydrogen ion or a hydronium ion. And then there may also be A minus, which in the case of acetic acid would be acetate, -- -- where we have two electronegative oxygens, among which the negative charge can be shared and the acetate ion, which is the conjugate base of acidic acid. So we make a table. We need to have some initial concentration. That is to say let's consider, for example, tenth molar acetic acid. We are just choosing tenth molar as a concentration for acetic acid solution. What that means is you have pure acetic acid. And then you dissolve it in water and bring it up to a total volume such that the concentration was 0.1 molar, assuming that none of it had been ionized yet. And so that means we have an initial concentration of acetic acid of 0.1 molar. And initially, before the ionization, we have zero H plus or hydronium and zero A minus. And then, the concentration changes. And it changes because the HA ionizes to some particular extent, depending on the KA value for the acetic acid. And what is going to happen is that some of the HA ionizes. And the amount of the HA that is undergoing ionization is x, so we are going to lose x. And then, for every HA that ionizes, we get that same amount of H plus produced and that same amount of A minus. And so then, after the system reaches equilibrium, we will finally have 0.1 minus x as our concentration of HA and x and x will be our concentrations, respectively, of H plus and A minus. And so let me put this board all the way up. Therefore, we can write the following, that Ka is equal to x squared over 0.1 over x by substituting into the expression for the acidity constant. Ka is our acidity constant. And we can go to a table and look up the acidity constant for acetic acid because it is a known quantity. And it turns out that that is 1.8x10^-5. And now that we have this equation for the acidity constant and we know what the acidity constant is, we can solve this for x. Of course, this is a cubic equation. We are going to get two roots. You will see that you get a positive root and a negative root. The negative root is meaningless because concentration cannot be negative, so you pick the positive root. And when you have done that, you can then go ahead and answer questions, like, what is the pH of the solution? And what is the percent ionization? And we can talk about that. So next time, at the beginning of class, we will do that calculation. We will find what the pH of a tenth molar solution of acetic acid would be. We will also go on and talk about pH and the pKa scale, and also a general equation for discussing titrations and buffers.
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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: --that Brown and Goldstein carried out, which in conjunction with many other experiments and experiments by other investigators have led to the model that you see here. And so we'll just briefly go through this model, which, again, was the basis for thinking about the function of PCSK9 that you learned about recitation last week, as well as providing the foundation for thinking about the recitation. This week, we really care how you sense cholesterol levels in membranes, which is not an easy thing to do given that it's lipophilic and so are many other things. OK. So the LDL receptor-- that was their model, that there is a receptor-- is generated in the endoplasmic reticulum. If you looked at the handout, you'll see that it has a single transmembrane-spanning region, which means it's inserted into a membrane. And the membrane where it functions, at least at the start of its life, is in the plasma membrane. So somehow, it has to get from the ER to the plasma membrane. And this happens by forming coated vesicles. We'll see a little bit of that, but we're not going to talk about this methodology in any detail. But Schekman's lab won the Nobel Prize for this work, either last year or the year before, of how do you take proteins that are not very soluble and get them to the right membrane. And they do this through coated vesicles that, then, move through the Golgi stacks that we talked about at the very beginning. And then, eventually, they arrive at the plasma membrane and become inserted. So these little flags are the LDL receptor. OK. So that's the first thing that has to happen. And I just know that this whole process is extremely complex. And patient mutants are observed in almost every step in this overall process. It's not limited to the one set of types of experiments, where something binds and doesn't bind to LDL receptor that we talked about last time. So the next thing that has to happen-- again, and we haven't talked about the data for this at all, but not only do these receptors have to arrive at the surface, but they, in some way, need to cluster. And it's only when they cluster that they form the right kind of a structure that, then, can be recognized by the LDL particles that we've talked about. And so they bind in some way. And that's the first step in the overall process. And then, this receptor, bound to its cargo, its nutrients-- and, again, this is going to be a generic way of bringing any kinds of nutrients into cells. It's not limited to cholesterol-- undergoes what's now been called receptor-mediated endocytosis. And so when the LDL binds to the receptor, again, there's a complex sequence of events that leads to coding of the part that's going to bud off, by a protein called clathrin. Again, this is a universal process. We know quite a bit about that. And it buds off. And it gives you a vesicle. And these little lines along the outside are the clathrin coat. I'll show you a picture. I'm not going to talk about it in any detail, but I'll show you a picture of it. So the LDL binding, we talked about. We talked about binding in internalization. Those are the experiments we talked about last time in class that led, in part, to this working hypothesis. And so we have clathrin-coated pits. And it turns out that there's a zip code. And we'll see zip codes throughout-- we'll see zip codes again, in a few minutes, but we'll see zip codes which are simply short sequences of amino acids that signal to some protein that they're going to bind. So how do you target clathrin to form these coated pits? How do you form a pit, anyhow, in a circle? And how does it bud off? And where do you get the curvature from? Many people study these processes. All of these are interesting machines that we're not going to cover in class. So you form this coated pit, and then it's removed. So once it's formed, and you've got a little vesicle, it's removed. And then it can go on and do another step. And another step that it does is that it fuses with another organelle called an endosome, which is acidic pH. How it does that, how it's recognized, why does it go to the endosome and not directly to the lysosome-- all of these things, questions, that should be raised in your mind if you're thinking about the details of how this thing works, none of which we're going to discuss. But it gets into the endosome, and then what you want to do is separate the receptor from its cargo, the LDL. And we know quite a bit about that. If you read-- I'm not going to talk about that either, but if you read the end of the PowerPoint presentation, there's a model for actually how this can happen. And you can separate the receptor from the cargo. And the receptors bud off, and they are recycled in little vesicles to the surface, where they can be reused. The LDL particles can also, then-- and what's left here can then fuse with the lysosome. And that's, again-- we've talked about this-- it's a bag of proteases and a bag of esterases, hydrolysis, lipids. That's what we have in the LDL particle-- hydrolysis. We talked about ApoB being degraded with iodinated tyrosine, last time. That's where this happens and gives you amino acids and gives you cholesterol. OK. And then, again, depending on what's going on in the environment of the cell, the cholesterol would then be shuttled, somehow, to the appropriate membranes. OK. So you can see the complexity of all of this. If the cholesterol is present, and we don't need anymore in the membranes, then it can become esterified with long-chain fatty acids. Those become really insoluble, and they form these little globules inside the cell. And then the process can repeat itself. And the question we're going to focus on in lectures 4 and 5, really, are how do you control all of this. OK. So this is the model. And so I think what's interesting about it is people have studied this in a lot of detail. It was the first example of receptor-mediated endocytosis. So we know something about the lifetime of the receptor. We know it can make round trip from surface inside, back to the surface in 10 minutes. We also know it doesn't even have to be loaded to make that round trip. It could be one of the ones that isn't the clustering of the receptors, which is required for clathrin-coated vesicles to form. And so you can tell how many trips it makes in its lifetime. And so the question, then, what controls all of this? But before we go on and do that, I just want to briefly talk about, again, mutations that have been found in the LDL receptor processing. And they're really, basically, at every step in the pathway. So the initial ones we found, that we talked about, we'll come to in a minute. But we had some patients with no LDL receptor express at all. So somehow, it never makes it to the surface. OK? There are other examples-- and these have all been studied by many people over the decades-- that it takes a long time to go through this processing. And it gets stuck somewhere in the processing. That may or may not be surprising, in that you have transmembrane insoluble regions. And if the processing goes a little astray or some mutation changes, then you might be in trouble. So we talked about this last time. We talked about that they had just looked at 22 patients. Some of the patients had no binding of LDL to the surface of the fibroblast that they were using as a model, at all. Some have defective binding. So if they compared it to a normal, they had a range of dissociation constants. And we'll talk quite a bit about dissociation constants, not this week but next week, in recitation. It's not so easy to measure dissociation constants when things bind tightly. And thinking about how to measure them correctly, I think, is really important. And I would say, probably, I could pull out 10 papers out of current journals, really good journals, where people haven't measured dissociation constant correctly, when you have tight binding. So this is something that we put in because I think it's important that people need to know how to think about this problem. So anyhow, let's assume that Brown and Goldstein did these experiments correctly, which I'm sure they did. And they got a range of binding. And we also saw that the patient we looked at, JD, had normal binding. That indicates he was the same as normal patients, but something else was problematic. And that something else wasn't that it failed to form coated pits, but that it failed to bring this into the cell. So it failed to internalize the LDL. That was JD's defect. We also, in recitation last week-- hopefully, you've had time, now, to go back and look at the paper a little bit. But LDL, in the model we were just looking at, gets recycled. It goes in and gets back to the surface. But what happens if, on occasion, instead of budding off into vesicles and returning to the surface, it, with the LDL cargo, goes to the lysosome and gets degraded? Well, that was the working hypothesis for what PCKS did. It targeted to the wrong place and degraded it. And the phenotypes of those patients were interesting, and that's why it was pursued. So there are many, many defects. And despite the fact that we have these statins, people are still spending a large amount of time thinking about this because of the prevalence of coronary disease. So I'm not going to talk about this, but I'm just going to show you two slides. And you can go back and think about this yourself. But this is the LDL receptor. We know quite a bit about it now. And one of the questions you can ask yourself, which is an interesting question we're not going to describe-- but you have LDL particles that are different sizes. How do you recognize all these different sizes? And how does the clustering do that? And so that's done up here. And there's calcium binding. We know quite a bit about that, but I don't think we really understand the details. You have a single transmembrane helix in the plasma membrane. And this is the part-- this part up here-- that actually binds the LDL particle. And the last thing I just want to briefly say, because we're going to see this again but without going through any details, remember that eventually we form what are called clathrin-coated pits. That's a picture of what the clathrin-coated pits look like. And the key thing-- and I just wanted to mention this briefly because we're going to see this again, over and over-- is the LDL receptor, itself, has a little zip code. And that's enough-- it's at the tail. That's enough for it to attract this green protein called to AP-2, which is key to starting clathrin binding, and formation of the curvature, and eventually being able to bud off these vesicles surrounded by clathrin. And when you do that, you start budding. And then, somehow, it turns out there's a little machine, a GTPase-- we've seen GTPases all over the place-- that's involved-- this is the name of it-- that allows you to bud off. And you use ATP energy to do all of this. We've seen this over and over again. And so the point I wanted to make here is we've seen this with these seminal experiments, by Brown and Goldstein. But in fact, we now know that this is sort of a generic mechanism for taking nutrients into the cell. So it's not limited to LDL receptor and LDL. And in fact, we're going to see, we're going to talk about, in module 7, Epidermal Growth Factor Receptor. And we're going to talk, in module 6, the receptor that takes iron into the cell, both of which do this kind of signaling. So this is a generic mechanism to do that. All of these things are interesting. We know quite a bit about it. And if you want to study that, you could have spent another weeks worth of lectures studying this. So the idea, then, is that we have nutrient sensing. And this is a general way to try to get nutrients into the cell, that is, you have a receptor, and it's undergoing receptor-mediated endocytosis. So that's the end of lecture 3. I think I'm one lecture behind, but that's not too bad. So what I'm going to do now is-- let's make sure I get this right-- I'm going to start on lecture 4. And now we're sort of into the question of how do we sense cholesterol. OK. So what I've done in the original handout, I had lecture 4 and 5 in the single PowerPoint. They're still in a single PowerPoint, but I've just split them into two. So I'll tell you how I've split them. So lecture 4 is going to be focused on sensing and transcriptional regulation. And lecture 5 will be focused on sensing and post-transcriptional regulation by a protein-mediated degradation. So I'm going to split that in two parts. And so today's lecture will be mostly focused on transcriptional regulation. And the key issue is how do we sense cholesterol-- what is the mechanisms by which we sense cholesterol. And the outline for the lecture is that the transcriptional regulation involves a sterol-responsive element. So this is sterol-responsive element. This is a DNA sequence of about 10 base pairs. And it also involves a transcriptional factor, so TF. This is a transcriptional factor-- transcription factor. And this is called SRE-BP. So this is Sterol-Responsive Element Binding Protein. So BP is Binding Protein. OK. So the first thing I'm going to talk about, then, is the discovery of SRE-BP. So that'll be the first section. And then what we're going to do is we want to know what are the players that allow us to understand how this transcription factor works. What we'll see that's sort of amazing-- it was amazing at the time, but now it's been found in a number of systems-- is where would you expect a transcription factor to be located? AUDIENCE: In the nucleus. JOANNE STUBBE: In the nucleus. OK. And what they found from their studies that it's located in the ER membrane. So this was a major discovery. So this protein is located in the ER membrane. They didn't know it at the time. But now, you're faced with the issue, transcription factors do work in the nucleus. So somehow, we have to get it from the ER membrane into the nucleus. And so to do that, what we need are players for SRE-BP to go from the ER to the nucleus. And we're going to see that these players are called SCAP, and they're called INSIG. And we'll come back, and we're going to talk about those in some detail. And then the last thing we'll focus on is-- we'll see it throughout. I'm going to give you-- what I usually do when we're talking about some complex mechanism, I give you the model upfront so you sort of see where you're going. Hopefully, you've all had time now-- we've been in this module for a long time-- to read the review articles. But we want a model for transcriptional regulation. So that's where we're going. And so what I want to do, before we get into the model, is come back where we started to try to keep you grounded on what we're doing. And what we're doing here is our cartoon of the cell that I showed you in the very beginning. We know that metabolism of hydrocarbons, fatty acids, and cholesterol all focus on a central player. And the central player is acetyl CoA. Acetyl CoA can be obtained from fatty acids in the diet. We've talked about the distribution of fatty acids using lipoproteins, including LDL. And we get to acetyl CoA-- this all happens in the mitochondria. But acetyl CoA cannot get across membranes. And that's true. There are a number of things that can't get across membranes. And so carriers in the mitochondrial membrane are key to metabolism. And I think once you look at it and think about metabolism overall, it's not so confusing. But you might not have chosen those. If you were the designer, you might not have chosen these to be the carriers to move in between organelles. So I think this happens quite frequently, so you need to pay attention to it. And so what happens in this case is acetyl CoA combines with oxaloacetic acid to form citrate. Citrate is an intermediate in the Krebs cycle. The TCA cycle is part of all of central metabolism. We're going to see citrate again. It plays a central role in iron homeostasis as well. And citrate-- there is a transporter that gets this into the cytoplasm. So here's the cytoplasm. There's an enzyme citrate, lyase that uses ATP to generate acetyl CoA. OK. So acetyl CoA is a central player. And really, what we're thinking about now, in general-- I'm going back through this-- is what do we expect sterol-responsive element-binding protein to regulate. And I'm going to show you it doesn't just regulate cholesterol homeostasis. There's a big picture [AUDIO OUT] all of this. So you can make-- you talked about this as a prelude to the polyketide synthases, the natural products Liz introduced you to. Anyhow, you can make fatty acids. Fatty acids can do a number of things. If you have a ton of them, then you can react them with glycerol to form triacylglycerol. And they're insoluble messes. If you look at the structures, they form little globules. So we have all these little insoluble globules inside the cell. And people are actually quite interested in studying these things. Now, we don't know that much about whether they are proteins or metabolic enzymes that could be sitting on the surface of these globules. A lot of people are trying to figure that out. But also, fatty acids are required in the presence of glycerol 3-phosphate, which comes from the glycolysis pathway, the other pathway that everybody learns about in an introductory course, to form phospholipids, which are the key component of all of your membranes. Alternatively, acetyl CoA, depending on the regulation of all of this-- that's the key-- gets converted to hydroxymethylglutaryl-CoA and mevalonic acid. Mevalonic acid-- that reduction between these two is a target of statins-- then ends up making cholesterol. And where does cholesterol have to go? So cholesterol is made, and a lot of it's happening in the membranes. A lot of it is associated with the ER, but only a small amount of the total cholesterol is in the ER membrane. Somehow, it's got to be transferred to all these other membranes. So that's a problem we haven't talked about. That's a big problem. Most of the cholesterol is in the plasma membrane. If you have excessive of cholesterol, you can esterify it, and, again, form little droplets of fats, which have fatty acids and cholesterol. So that's the big picture. And so this is the picture of the regulatory network. So I'll say this is a PowerPoint for the regulatory network. And it's governed by-- it turns out there are three SRE-BPs. They have a slightly-- and they're structurally homologous to each other, and they work in ways that they interact with other protein factors and control this whole homeostatic process between fatty acids and cholesterol biosynthesis. So I think there are two things that you need to think about. So we want to control basically its lipid metabolism. And I should say at the outset, we're focusing on SRE-BP, but some of you, in maybe a more advanced biology course, know that there are other transcription factors involved in regulating cholesterol homeostasis. This is a major one, and that's all we're going to talk about in this class. But what else do you need to make molecules, if you're going to make fatty acids, if you were going to make cholesterol? What you need is NADPH. So that's the other thing that you need to think about when you're looking at the regulatory network. So we need to control-- how do we make lipids? Where did they come from? They come from acetyl CoA. And the second thing we need to think about is a source of energy to actually form the molecules. We're after the long-chain fatty acids. Go back and look at that-- or cholesterol, if you go back and look at the pathway we talked about in the first couple lectures. So NADPH is at the center. And I forgot to point out before and probably many of you have heard of but never really thought about malic enzyme in the cytosol. You can go back and think about that, but that's a major source of NADPH. What is another source of NADPH in the cytosol. Anybody know? Where do you get most of your NADPH from? It's key to biosynthesis of any kind of anabolic pathways. Does anybody know? AUDIENCE: [INAUDIBLE] JOANNE STUBBE: No. OK. Did you ever hear of the pentose phosphate pathway? Well, hopefully, you've heard of it. Reproducing it might be challenging, but the pentose phosphate pathway is central to providing us with NADPH. It's central for controlling reactive oxygen species, which is going to be module 7. It's central for providing NADPH for nucleotide metabolism. So the pentose phosphate pathway and malic enzyme are the key sources of NADPH. And if you're becoming biochemists, I think, now, all of these pathways, these central pathways that we talked about in 5.07, should just-- you don't need to know all the details, but you need to know how things go in and out. And it's central to thinking about anything. And if you ever do any genetic studies, you can never figure out anything unless you know how all these things are connected. So knowing these central pathways and how things go in and out and connect is really critical in thinking about many, many kinds of reactions you might be doing in the lab. Because you might see something over here, but it might be way over here that you had the effects. And knowing these connections, I think, is why I spent another-- whatever-- five minutes describing the regulation. OK. So if we look at this, what we see here-- and this is an old slide, so this might have changed. But all of the enzymes in italics are all regulated by SRE-BP. So here's acetyl CoA. What do we see in this path, where we're making cholesterol? So many of the enzymes-- we're not going to talk about them-- that we talked about when we went through the pathway are all regulated by SRE-BP and is predominantly-- again, there's overlap of the regulation between the different forms of the sterol-responsive element-binding protein. But you can see, we have HMG CoA reductase, which is the rate-limiting step. So that might be expected. But many of the other enzymes that are also controlled by this transcription factor. And the one that turns out, I think, to be quite interesting for most recent studies is-- remember, we briefly talked about how you get from a linear chain, and then we had to use a monooxygenase to make the epoxide. That enzyme is a key regulatory enzyme, people now think. It wasn't thought to be so not all that long ago. So anyhow, all of these enzymes that we've talked about are regulated in some way by SRE-BP. But it doesn't stop there. If you go over here, you sort of have a partitioning between acetyl CoA also going into lipids and forming phospholipids or triacylglycerols, depending on whether you store or whether you're dividing and need more membranes. So all of this, again, it's about regulation. And if you look at this, you can see that many of the enzymes in this pathway, for formation of monoacylglycerol and triacylglycerols are also involved. OK. So that gives you the big picture that I want you to think about. So when you wonder where you're going, you should go back and take a look at the first few slides. So what I want to do now is really focus on the first thing. The first factor was how did they identify. So this is identification of SRE-BP. And so probably most people wouldn't talk about this, but I think it's sort of amazing. So I'm going to just show you what had to be done. And this is not an easy set of experiments. First of all, transcription factors, in general, aren't present in very large amounts. To get them out, they also stick to DNA. So that poses a problem. Unlike using his tags and all this stuff, none of that stuff works to isolate transcription factors. And this was all done before the-- a long time ago. And so this was this is quite a feat. And the key to this feat was that Brown and Goldstein recognized that in the front of the gene for HMGR-- Hydroxymethylglutaryl-CoA reductase-- in the LDL receptor, they found a 10-- I'm not going to write out the sequence-- base-pair sequence that was the same. So that suggested to them that there's a little piece of nucleic acid with 10 base pairs that might be recognized by a protein, which could be the transcription factor. So this was the key, this 10 base-pair sequence. And I'll just say, see PowerPoint. And this is the SRE, before the genes, again. And this has now been found in front of many genes. I just showed you that many, many genes are regulated, in some way, by these proteins. But this was an observation they made a long time ago. OK. So where would you expect-- we just went through this. Where would you expect SRE-BP, the transcription factor, to be located? You'd expect it to be in the nucleus. OK. That's a reasonable expectation. And so what step might you do, in the very beginning, to try to help you purify this protein? And let me just tell you at the outset that the protein had to be purified 38,000-fold. OK. Now, you guys, none of you have ever experienced, really, protein purification, starting with kilograms of anything. I have done that and spent three months purifying a microgram of protein. And I would argue that some people still need to do that, because when you do recombinant expression, lots of times, you miss a lot of stuff. So somewhere along the way, somebody needs to really know what the endogenous protein is like, and not the recombinant protein. So we're going to have to do a 38,000-fold purification. And I would say that's not uncommon. I've done 20 liter by 20 liter gradients that take three weeks to get through the gradients and looking for your proteins. So if your protein is not stable, even if you're in the cold room, what happens? Or if there are proteases, it gets degraded. So I'm just saying, transcription factors are not easy to deal with. And this was sort of an amazing feat. Anyhow, they started with-- over here-- 100 liters of tissue culture cells. So most of you have probably seen tissue culture plates. And that's what you work with. They started with 100 liter, and that's why they're using HeLa cells, because you can grow them on this scale. You can probably grow a lot of things on this scale, now. We have much better ways than-- this was a long time ago. So their approach was-- so the first thing-- I got sidetracked again. But the first thing is that if it's in the nucleus, what would you do to try to enrich in the transcription factor? What would be the first thing you might do after you've isolated the cells? AUDIENCE: [INAUDIBLE] JOANNE STUBBE: I can't hear you. AUDIENCE: Maybe, something involving nuclear-binding proteins that transport things into [INAUDIBLE]---- that have transported things into the-- JOANNE STUBBE: OK. So I still can't hear you. You're going to have to speak louder. I'm going deaf. And I will get a hearing aid, but I don't have one now. So you have to speak loud, and you have to articulate. Yeah? AUDIENCE: Wait, so just the absolute first step? JOANNE STUBBE: Yeah. AUDIENCE: How we're just lysing cells and pelleting them? AUDIENCE: Yeah. JOANNE STUBBE: But is there a certain way you would pellet them? AUDIENCE: You would have to do a sucrose gradient. JOANNE STUBBE: You would do some kind of gradient to try to separate the-- well, you have to pellet the cells first. But then, what you want to do is separate the nucleus from all the organelles. The issue is-- we already told you this-- most of the protein is not found in the nucleus. And that was part of this. They didn't know that at all, but that's what they did. They did some kind of a gradient to separate nuclei from the rest of it, because they were trying to enrich, which was a totally reasonable thing for them to have done. OK. So I'm not going to write that down, but that's the first thing they did. The second thing they did is they made an affinity column all out of the SRE. So this is a nucleotide affinity column. And they ended up using that a couple of times. And they ended up using a couple of other kinds of columns and eventually got protein out after a lot of effort. After a lot of effort, they got protein out. And the size of the protein-- so they went through this column. And they went through additional columns. I'm not going to go through the-- and they ended up with proteins that were actually smaller than the SRE-BP, but they still bound to the affinity column. So they ended up with proteins-- I don't remember. And again, the details of this really aren't so important. But they ended up with smaller proteins. Somewhere, I have the size written down. 59 to 68 kilodaltons. So either protein had been degraded, or we will see the protein has been processed, or was being processed during all this workup. And there are many things that could have happened to this process. But what this allowed them to do-- and this was the key to allowing them to do this better-- was they could generate antibodies. So they took this protein that they isolated, and they generated antibodies. And we're going to be talking about antibodies this week. But we're going to be, also, talking about use of antibodies with fluorescent probes, the last recitation, as well. So what did this allow them to do? The antibodies, then, allowed them to go back into the cells and look for expression of SRE-BP. And instead of finding it in the nucleus, what they found was that most of it was localized in the ER membrane. So these antibodies revealed that SRE-BP is predominantly in ER membrane. And again, this question of antibodies-- which Liz brought up-- and the question of specificity, and, moreover, the question of sensitivity is really key. Because now, when you're looking at eukaryotic cells, we know things move around. They move around all over the place, and they move around dependent on the environment. So you could easily miss location. This might be the predominant one under the conditions you looked, but it could be somewhere else. And I think they didn't realize so much about that back in these days, but we now know that a lot. So anyhow, that was a surprise. And then, that provided the basis for them going back and thinking much more about this system. And so what I'm going to show you is the model that's resulted. And if some of you have started working on problem set 7 that's due this week-- the problem deals with some of the experiments-- then I'm going to tell you what the answer is. And you're supposedly looking at the primary data from where this model came-- a small amount of the primary data from where this model came. OK. So this is the model. And I'll write this down in minute. But the model is at low sterol concentrations. So at low sterol concentrations, what do we want to do? We want to-- this transcription factor-- I should write this down somewhere. But the transcription factor activates transcription. It could repress transcription, but it activates. So if you have low sterols, what do you want to do? You want to turn on the transcription factor. So it needs to somehow move from this location in the membrane to the nucleus. So that's where this model is coming from. And we'll walk through it step by step. So what you'll see-- these are cartoons for the factors we're going to be looking at. So this SRE-BP has two transmembrane regions. We'll come back to that. This little ball here, which turns out to be at the N terminus, is a helix-loop-helix, which is a DNA-binding motif. We'll come back to this in a minute. I'm just giving you an overview, and then we'll come back. There's a second protein. And this is the key sensor that we're going to see of cholesterol levels, called SCAP. And it also resides in the ER membrane. And it has a little domain on it that recognizes and interacts with part of SRE-BP. And so this is located in the lumen. And these guys, especially this guy, is located in the cytosol. And we don't want it inside the lumen, we want it on the outside so it can go into the nucleus eventually. So what happens is somehow, when you have low sterols-- and we're going to look at the model for how this happens-- both of these proteins, SCAP and SRE-BP, are transferred by coated vesicles-- we'll come back to this in a minute-- into the Golgi. So they go together into the Golgi. And I would say that, right now, a lot of people are asking the question, once you do the processing to get SRE-BP into the nucleus, what happens to SCAP. And there are lots of papers, now, that are focusing on the fact the SCAP can recycle from the Golgi back to the ER. So it's never as simple. This thing's continually going on that not that much is wasted. So this can actually recycle. And I'm not going to talk about that. And then, in the Golgi apparatus, there are two proteins, called S1P and S2P. And they're both proteases. We'll come back to this in a second. So what's unusual is that we want to get this guy into the nucleus. And one of the proteases cuts here. So then we get this piece. And then the second protease cuts here, and then we get a little soluble piece that can move into the nucleus. Now, this is also revolutionary, in that nobody had ever known there were proteases that are actually found in membranes. Now, we know there are lots of proteases found in membranes. And any of you work in Matt's lab? What is the factor that is regulated just like SRE-BP? Do you know? OK. So go look up the AFT4. Anyhow, so to me, what is common is once we found this, we've now discovered this in many other systems. So this system is a paradigm for many things that people have discovered since the original discovery. But of course, the thing that's amazing-- first of all, this was amazing. The fact that this thing is in the membrane and gets to the nucleus is amazing. And at low cholesterol, what you want to do is activate transcription. And you saw all the genes that could be activated in the previous slide. And it's complicated. There are many factors involved. And so the key question, then, is how do you sense this movement from one place to the other and what do we know about that. So what I'm going to do is look a little bit at the model. So the model will start with-- and the players. So this is part 2-- the players. And the players are-- so if you look at the ER membrane, what we have is two domains. And whenever you see a line through the membrane, that means a single trans helix membrane spanning region. We see that a lot. So I'm not going to write that out. But this is really sort of a single transmembrane helix. And the key thing is at the N terminus, you have the helix-loop-helix. And this binds to DNA. So this is a DNA-binding motif. And so this is the protein SRE-BP. And so the second protein-- and this is the protein you're focused on for your problem set-- has a SCAP. 2, 3, 4, 5, 6, 7, 8. So it has eight transmembrane helices. And they've studied all of this using some of the methods that you're going to be looking at in your problem set. And to me, there's a couple of things that we're going to be talking about in detail, but your problem sets are focused on-- all right. So I haven't really shown you where the loops are, but there are a couple of loops, loop one and loop six, which is what the problem set is focused on. And how do you know these are interesting and important. And we'll come back to this in a little bit. So now, at low sterols-- so we want to turn on the machinery to make more cholesterol-- so that low sterols. And one of the key questions is what is the structure of the sterol. Can more than one do that? We'll see different sterols turn on different domains. And we'll see that there's a domain within SCAP-- so this protein here is called SCAP. And we'll see that SCAP has a sterol-sensor domain, as does another protein called INSIG, as does HMG-CoA reductase. So somehow, you have these transmembrane regions that can bind some kind of sterol, that then changes the conformations, that is going to allow all of this chemistry to happen. So here, for example, we're not going to talk about this now. We're going to talk about that in the last lecture. But here's SCAP with its sterol-sensing domain. So what happens, then, is this has to move. And as I said before, this can return. This moves to the Golgi. So this is the Golgi. And the Golgi are complicated. And so I haven't defined where within the Golgi this is. And these are transferred by COPII vesicles. OK. And so what you then have, again, is your 1, 2, 3, 4, 5, 6, 7, 8. And you have your sterol-responsive element-binding protein. And now what you see-- and so nothing happens in terms of processing, until you get into the Golgi. And then, there's one protein, S1P, which is a protease. And I'm not going to go into the details of it, but if you look over here, what's unusual about this protease? If I gave you this cartoon, what would you say about that protease? Is it unusual compared to, say, trypsin or chymotrypsin. Can you see it? You can pull out your handouts. What are the catalytic groups? AUDIENCE: [INAUDIBLE] JOANNE STUBBE: Huh? Where have you seen those before? AUDIENCE: [INAUDIBLE] JOANNE STUBBE: Yeah, so they're aspartic acid, histidine, and serine. You see these over and over and over again. There are 150 serine-type proteases. OK. But what's unusual about this? Huge-- huge. OK. And then, the other thing that's unusual about it is that you have a transmembrane region. So it's completely different from serine proteases, so there's got to be some little domain that's actually doing all of this. So I just want to note that it's huge. But you could still pick up D, H, S and know that that's the protease domain. And you could study that. You could mutate serine to alanine or something. And then you have S2 domains. So we've gotten here. And this protease ends up clipping. so within the membrane-- so somehow, these things got to come together. And the active side of this protease needs to clip SRE-BP. So it does that. And when does that, what you end up with-- I'm not drawing the whole thing out, but what you end up with, then, is your helix-loop-helix. So this part is still embedded in the membrane. And then you have your second protease. I don't know. I probably have the wrong numbers. So this is S2P. And if you look at S2P, what's unusual about it and what people picked up on is that it has another little sequence motif. And this is what you see over and over again, in enzymology. Once you sort of know something in detail, you know, even though there's no homology between the proteins at all, you can pick up little motifs, just like you can pick out little motifs that are zip codes that move things around inside the cell. This little motif is the key player that tells you that this is probably a zinc-dependent metalloprotease. So this turns out to be a zinc metalloprotease. And this, then, does cleavage. But now, we actually-- it's pretty close to the membrane. OK. It does cleavage. And now what you've done is you've released this thing. It pulls itself out of the membrane. And what you can do, then-- I'll just put this in here for a second. But what you can do now is we now move to the nucleus. And we have our pieces of DNA. And we have our SRE. And now we have this helix-loop-helix that activates transcription. OK. So this is really sort of what I just told you in the other cartoon. And I just want to repoint out again that we now believe that these SCAP proteins can recycle back into the ER and be used again. And so controlling the levels of all these things-- we're going to see at the very end-- is also related to protein-mediated degradation that we're just now beginning to appreciate. OK. So here's the model. This now sets the stage for you to solve problem set 7 that's due. Because the key question you want to ask yourself is how do we know about the structure of SCAP. And so problem set-- sorry, I'm over again. But problem set 7 is focused on how do you know that this little loop here, this little loop here, and this little zip code plays a key role in this whole process of moving from the ER into the Golgi. OK. And we'll come back and talk about this briefly. We're not going to talk in detail about the experiments. And then we're going to move on and look at the post-transcriptional regulation of cholesterol homeostasis.
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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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JANET RANKIN: Another active learning technique that can be very, very effective in supporting student learning is the jigsaw. And it gets its name from the activity involved when you do a jigsaw puzzle. And usually, most people when they do a jigsaw will look for the-- they'll find all the edge pieces, or they'll find all the blue pieces, or they'll find all the pieces that have a certain image on them. You do the homogeneous aspects of the exercise. So you get all the edge pieces, the blue pieces, the particular image pizzas. But eventually, you have to take those pieces and put them together. You have to link them up with pieces that don't look quite like those pieces, so that's the heterogeneous component of it. And so what you do in a jigsaw is you ask students to become experts, if you will, homogeneous experts, in a particular area, on a particular topic. And you can do this by giving them a reading in class, or you can do this by asking them to reflect on something that you've addressed in a previous class or that they've addressed in a previous homework. But they sit together, or they group together in this, what I call, homogeneous group, where everyone is talking about the same particular topic. And they just discuss that topic and they become experts in that homogeneous topic. And then you can have three or four different topics, whatever, depending on the size of the class. And you have these local groups of experts, like the edge pieces, and the blue pieces, and the picture pieces. And then after giving them a period of time, maybe five minutes, 10 minutes, depending on the class, you break them up into heterogeneous groups. So you get one with one representative from Group A, goes with one representative from Group B, and one representative from Group C, and one representative from group D. And now you have a new heterogeneous group that has an A, a B, a C, and a D in it. Or if we use the puzzle analogy, you have a group that has some person that's the edge expert, some person that's the picture expert, and some person that's the blue piece expert. And they're all in a group. And now they share what they know individually with each other and then they combine that expertise, they synthesize that expertise, to make something bigger to come up with a more global understanding of the problem. And then each group does that in their own way. So the heterogeneous groups then form a more holistic, or a more heterogeneous, solution to the problem, as if they're putting together the different kinds of puzzle pieces. And then again, as with the other activities, we ask them to then report out. Before that I usually will circulate in the room the same way I do with the pair share to hear what each of the groups are saying, to see if any groups are having particular issues, or if they have some particular breakthroughs that I want to make sure they share with the class as a whole. There are some logistical issues. One, you want to make sure that the students really can become these homogeneous experts. So that if you have a group of four people, that they've all really come up to speed on the topic. So ideally, you would give a pre-class assignment and you'd say, yes. I need everybody to read this article and to be able to explain this point, this point, and this point, and you would assign that to that particular group of students. And then you'd do the same with a few other topics. You're assuming then that the students have read the paper, or have read the article, whatever it is you want them to read, and that they understand it to the level that you hope they understood it. Sometimes you don't have time for that or it just doesn't work out, so you give them something to read within class. And again, that can be tricky. Students may not have the same level of understanding, so groups may be a little bit spotty. The other thing that can happen with the jigsaw, which is a really silly problem but sometimes it just trips you up, is if you have, depending on the number of students, you have to kind of think on your feet about how many students are in the homogeneous group versus how many students are in the heterogeneous group. If it's a perfect square, it works fine. So if you've got nine students, or 16 students, or 25 students, you know you have five groups in one situation and then five groups in the other, that means you have to have the same number of groups as heterogeneous topics. So if I went A, B, C, D, I have to then have four groups because I had to have four A's, four B's, four C's, and four D's. So sometimes just the number of topics and the number of students that you have in your class makes it kind of hard to make groups that don't have extra people or aren't short a particular expert. And that's just something you want to plan beforehand. And if somebody doesn't come to class, if you know you have 25 students in your class and someone doesn't show up, then you have to kind of deal with it on the fly. But that can be the trickiest thing with the jigsaw, is just getting the number of students right.
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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: All right, so this is where we got to last time. So we were on the road to deriving the leading order SCET Lagrangian. We said that there were some things that we had to deal with, in particular expanding it. And we started out, as a first step, getting rid of two of the components of the field that we said are not going to be ones that we project onto in the high energy limit. And that led us to this Lagrangian right here, but we haven't yet expanded it. We haven't yet separated the ultra soft and collinear momenta or the ultra soft and collinear gauge fields. And so in order to do that, I argued that you have to make multipole expansion. And I talked a little bit about what the multipole expansion would look like in position space. And then I said, it's better for us to do it in momentum space. And we started to do that. And we're going to continue to do that today. So first of all, let's just go to momentum space. So I've called the field here-- since it's not the final field that we want, I introduced a crummy notation for it, Cn hat. And once we get to the final field, it won't have a hat. And that'll be the one that we keep using from then on. So let's take this hatted guy, which is not the final guy, Fourier transform him and talk about doing things in momentum space. In momentum space, I'm going to make an analogy with what we did in HQET. In HQET, we split the momentum into two pieces. We had a piece that was large and a piece that was small. And then we could expand. I'm going to do the same thing here, split it into a large piece and a small piece where I'll call the large pieces the one in the lambda components. And then the small piece will just be all the pieces that are of order lambda squared. So if you draw a picture for this, it looks as follows. How should I think about what I'm doing here? Here's how I want you to think about it. So let's draw a two-dimensional picture in minus perp space. In plus space, there's no split because there's no plus here. So it's really in the minus and the perp space that we have to make a split. Yeah, let's see. All right, so I'm going to draw a grid. And the way I want to make this split precise is I'm going to think-- OK, so I should have maybe made my axes in a different color to distinguish them from the grid. Let's make them orange. So the way I want to think about these two types of momenta is I'm going to think about these Pl's as discrete and the Pr's as continuous. So let's think about the Pl's as labeling these points at the center of each of these boxes in the grid. This one was supposed to be-- that's fine. My squares are not the same size, but they roughly should be. And so then we can think about, if we specify some momentum in the space-- let's say we specify momentum that points to this point right here. The way that I can point to that point is I take one vector that points to the center. And then I take another small vector that points to that point. The large vector is the Pl. The small vector is the Pr. OK, so this guy here is Pl. Maybe I should make the other one a different color, Pr. And so the way that we should think about this is in the following sense. We should think about the average spacing in between these grid points here is something that gets a power counting that makes it of order Q lambda. And the average spacing between grid points in this direction is something that makes it order 1. But if I ask about distances inside a box, if I ask about a distance like this one, that's of order lambda squared. So to get between boxes, you need to make a big jump to the larger momentum. But to move around in a box, that's only small momenta. AUDIENCE: So why can't you have order lambda basically inside the box in the P minus dimension? IAIN STEWART: Yeah. So you could. So you can ask, where is in order lambda momentum in the P minus? AUDIENCE: Yeah. IAIN STEWART: And it's in a nether land. Either way, you can think about it as being part of this one or that one. It doesn't matter. Because technically, if I just talked about minus, I'd only be interested in separating a 1 and a lambda squared. And whether I call this lambda squared or I call it lambda or I call it eta-- let eta equal lambda squared-- I'm only trying to separate two things. So if you talk about some momentum that's in between, you could ask the same question actually about, what about a lambda to the 3/2 momentum in the perp space? And that's the same question you just asked about the minus space, right? And that's something that we don't really care about because there's not going to be momenta that are scaling like that. So we don't really care about separating it. OK, so Pl are discrete grid points. And Pr-- continuous. And this picture also makes clear that, if you take any P, it uniquely is given by some Pl and some Pr. If I point at any place in this grid, you can tell me exactly which Pl I should pick and then what Pr I should pick. This is, in some ways, just going to be a tool. And in the end of the day, we can kind of dispense with this picture. But it'll be a useful way of thinking about what's going on and getting to sort of the final results that allow us to dispense with this picture. So what about integration? So if I integrate over all P and P is collinear, what does that mean? So it corresponds to summing over all the grid points and then integrating over all the residuals except for one location. And that's the special box here that has label equal 0 because we know that the power counting of the collinear momentum is such that it always has to have a non-0 Pl. And so the one box where Pl is equal to 0, 0, that particular box doesn't have a collinear momentum. And actually, that box is exactly what you think of as the ultra soft momentum. So if you have an integral d4P and P is ultra soft, there's no sum. And you just have interval d4Pr. And that's like integrating over the 0, 0 location in the grid. OK, so you can see that together we cover all possible places for an integration. And we actually don't do any double counting with this setup. What am I going to do with my fields? Well, because of this split into two different pieces, I'm going to say that my fields actually can be labeled by Pl and carry Pr also as an index or as a function of Pr. So in order to make this explicit, we're going to write our Fourier transform guy with a label Pl and an argument Pr like that. So the name label comes from the fact that we do this. And there will be some analogies with why we called this a label in HQET. We'll see to what extent that analogy carries through in a minute. Another thing you should note from this grid picture is that you have separate momentum conservation in the label and the residual. So if you think about some integral d4x of something that you split between label pieces and residual pieces, then the right way of thinking about this is that you have a discrete [INAUDIBLE] delta for the label and then a continuous Dirac delta for the continuous variables. And there's some 2 pis. And so this is what would be meant by-- if you have two momenta that are equal, you know that they're in the same grid square. And then they have the same little green arrow as well. That's what that's saying. So when we talked about what happens with the multipole expansion in position space, we said that effectively you have a non-conservation of momenta. OK. We had a field at 0. We worked through what that meant. And it meant, basically, that the momenta was conserved on the other line, but not on the one that you were doing the expansion on. The nice thing about this type of setup is that we have conservation of momenta, but we actually have two conservations of momenta. So we don't give up momentum conservation. It's just that we go over to a picture where we have two of them. So let me show you how that works. Let's say we have an ultra soft particle, and it's hitting a collinear quark. The collinear quark, because it's collinear, carries two types of momentum. It carries a label P and a residual. But the ultra soft guy, it's only residual. So let's just call that k residual. There's no label to that ultra soft guy. So if I just look then what this guy's momentum is, it's P label for the large piece because that is not affected by the ultra soft guy. And then it's P residual plus k residual. OK, so we have a conservation in the residual space, which is the standard one. And we have a conservation in the label space, but the ultra soft particles just don't contribute to that. So we have two conservation laws. And we just know that certain fields can carry these label momentum and other fields can't. All right, well, since Pr here and kr, the residual components, is just standard quantum field theory-- the only thing that's special is the Pl. For the Pr, what we're going to do for the residual momenta is we're going to transform back to position space. And that's kind of what we do in HQET as well, right? We treat this guy in momentum space as a label and this guy as a position space for the residual. And the reason we really want to do that is related to locality. And I'll say some more words about that. AUDIENCE: [INAUDIBLE] IAIN STEWART: Yeah. AUDIENCE: I have a question. IAIN STEWART: Sure. AUDIENCE: So this top equation-- IAIN STEWART: Yeah. AUDIENCE: So are you bounding the d4Pr by lambda squared, or is there some reason why the integral should fall off before you get there or something? IAIN STEWART: Yeah, we're going to talk about that. So, so far, this is just notation. If you think about it from this grid picture, you'd think some hard walls put in. That's not quite the right way of thinking about it. It's actually not right for several reasons. And we will come back and talk much more about what the right way of thinking about it is. But the physical picture that you're in this box is the right way of thinking about it. The tricky thing is that these boxes are infinite. Each of these boxes is infinite. And we'll have to see why that is. In what I've told you so far, you'd think that the d4Pr would really be a box with some fixed edges. And it's going to turn out that the d4Pr-- we're going to use dimensional regularization-- it's going to be an infinite box. But we're still going to be able to do exactly what I'm telling you. And it's going to make sense. We're going to want infinite boxes to avoid breaking symmetries, like gauge symmetry and Lorentz symmetry. And we'll come back and see how that works. So in some sense, what I've told you is a Wilsonian story. And we're going to have to make it into a more continuum friendly story. But the moral of the story, the way you think about where these momentum modes live, what kind of momentum they carry, that is just true. So that part of the story, which you can think of so far as Wilsonian, is true. And it's OK to think like that. So we'll see actually what we do in practice a little later on. Where this will really come in is when we start wanting to think about doing loops, but we won't get there for a little while yet. So the conservation law in the residual just corresponds to locality. We learned all about locality yesterday in [INAUDIBLE] seminar. And so we can make things local by just going back to the x base for the residual momenta. OK, so we go back to the x base, which is the Fourier transform conjugate variable to the residual momenta. So the fields we're actually going to use are these ones, where we Fourier transform back in the residual, these guys here. So it's kind of like HQET, where you think about this part as momentum space and that part in position space. OK, so if you talk about ultra soft interactions-- you can see it from over there-- they conserve the labels. So there's label conservation of a field for ultra soft interactions. The label of the field here is the same as the label of the field there, so that ultra soft guys do not change labels. So this is Pl. This is Pl. And if this is ultra soft, it's still labeled by Pl. And the thing that can change is the residual momenta. And that's now encoded in locality in saying these fields all sit at the same point x. On the other hand, the collinear gluons cause us to hop between boxes. So the collinear gluons are label changing. And that's what's different than HQET. In HQET, we don't really have interactions in the theory in the Lagrangian that do that, but here we do. So if we have a collinear gluon come in, the collinear gluon, again, carries both labels and residuals if I just draw the labels. And this is some Ql coming in. And this will be Pl plus Ql. And so the label of the outgoing field has changed. There's one more label. And that's this n, right? We also label these guys by a direction. And that is unchanged also by the collinear interactions. So it's preserved by both ultra soft and collinear interactions. It can only be changed by hard interactions where you would have some hard production that could produce, say, two guys in two different directions. And those type of interactions are typically only going to occur once in your quantum field theory, kind of like a decay, a weak decay where it occurs once. And then you do perturbation theory around that. Hard interactions are that way, too. So as far as dynamics is concerned, this label n is something that you can think of as being conserved for the field. It's not going to be changed by the Lagrangian, for example, all right? OK, so, now, we have kind of what we want. And now, I want to write the action that we had before in terms of this guy and see what it looks like. This split of these guys into a Pl and Pr or Pl and x allows us to expand very easily in momenta. It's allowing us to do the multipole expansion because we have two things to talk about. And we just can compare them. So let's put the pieces together. So Cn hat of x is the full Fourier transform of this Cn twiddle. And that can be split for a collinear guy. We can split the measure, and we can split the phase. And we can write the field in this way. And then we decided that we wanted to use this field. Let me see, [INAUDIBLE] two pis here. So that means we take this measure, we take this phase, we take this guy, and we write it back in terms of that field. So that leaves, if we want to relate the field that their Lagrangian was in terms of in terms of this field that I told you we're going to use, it's this equation, OK? All right-- a little bit more formalism. So we're going to actually want to have two different types of derivatives, derivatives that can talk to the momentum conjugate to x and derivatives that give this Pl. So we define two types of derivative operators. One is just, if you like, a standard derivative in the x base of this guy, acts only on that coordinate. And that derivative scales like lambda squared. So that's residual, picking out talking to residual momenta. Then we need something that can give us the labels. So we define an operator that does that, which we'll call curly p. It doesn't care about x. And it's just defined to give the Pl mu of the field it acts on. So there's no plus component to this. There's an order 1 minus component. And then perp is order lambda. I want to say how big that guy is. So then we have a power counting that tells us how big the p is and how big the partial is. And we can compare them, and we can expand in them, OK? So the bottom line is that we now can just say that, like we had a comparison for the gauge field, we have a comparison for derivatives. And so this is exactly analogous to what we said before where we had M bar dot A ultra soft is much less than N bar dot A collinear and A perp ultra soft being much less than A perp collinear. And in fact, the way we're setting things up is going to make the gauge symmetry also easy. Because by gauge symmetry, these derivatives and these fields should go together. They are the same size in the power counting. And these guys should go with those guys. So we're kind of doing this same thing we thought of for fields we're doing for momenta here. So when we talk about gauge symmetry, this will make gauge symmetry easier, this homogeneity between the derivatives and the fields. So that's one of the powers of thinking like this. OK, so let's see that this notation is actually kind of friendly. So let me try to convince you of that. It looks kind of daunting, but maybe it's not so daunting. So let's look at this guy, which was equal to this. So what I can do is I can take this and replace it by e to the minus i label operator dot x, right? But the label operator then doesn't care about the sum. So I can push it through. So as an operator, it goes through the sum. I have that. And for a lot of purposes, in a sense of suppressing indices, we can just use a simplified notation for this guy where we suppress the sum. So we were thinking about this as a discrete sum. And we can always write out that sum if we get confused. But we could also adopt a notation where we suppress the sum. And for many manipulations, just like you might suppress some indices in a quantum field theory, that's perfectly fine. And having this label operator makes it easy to do that. So if you have, for example, field products, say I had two fermions, we try to make the notation work for us as nice as possible. Well, if you have a sum of momenta and then it's exponentiated, if you have a product and it's a product of exponentials, then it just becomes the sum in the exponential. So this is true if we do it for two fields, right? I'd have the label operator acting on this guy and that guy. And if I were to write out this twice with two sums, then because the product of those two exponentials can be just written as a single exponential, this is true. So this guy acts on both fields. So in this sense, the notation is friendly. And what this phase factor here is really doing for you, if you think about it, if you think about it in a whole product of fields, it's just when you Fourier transform giving the label conservation, OK? So there's going to be this kind of overall e to the minus i p dot x. And it's just sitting there to conserve momentum in the label space. So that's its purpose. OK, so there's one more step that we have to do before we start talking about the Lagrangian. And that's this little caveat that I had right at the beginning where I said, let's consider only quarks. So far, we've only considered the quark part, not the antiquark part. So let's see what we want to do about antiquarks. Technically, the way that the field theory wants to work is that the quarks and the antiquarks kind of want to be different fields, but that's very cumbersome. And so we're not going to do that. And we're going to find, again, sneaky notation to treat them as one field even though they want to be two. The reason that they want to be two is because, if you look at the phase factor for the quarks in the antiquarks in the kind of standard mode decomposition, it has the opposite sign. So look at the mode expansion in QCD. I can write it in a covariant type notation or with a d4p p if I put in the proper projection. And then we have the quark part with the e to the minus ip dot x. And then we have antiquark part, our b daggers, and the e to the plus ip dot x, OK? So let me think of that as a psi plus plus a psi minus. And we've so far only talked about psi plus. So for psi plus of x, we have a sum over Pl not equal to 0 e to the minus i Pl dot x of sort of a Cn Pl plus of x. So this is the case we did. If we were to repeat everything we did for this case, then what we would get for psi minus-- and this is why I say that they want to kind of look like separate fields-- is we'd get a plus i Pl dot x. And in this notation, if you think about the P0, the P0 is the sum of the P minus and the P plus. So saying that, in this mode decomposition, P0 is greater than 0 for these two equations and when we go over to sort of a collinear scaling, it becomes that Pl minus is greater than 0 in these equations. OK, so that's the sense in which these are different. With a positive Pl minus, we have an opposite phase factor for these two guys. And in both of these cases, it's true that just as kind of a way of remark, we still have this. That didn't change, remember. OK, but talking about minus and plus fields would just make our life really difficult. Actually, when we first derived SCET, sometimes you're bone-headed. And we went through everything in a much more detailed way that I'm going to present it to you now, wrote out everything in terms of plus fields and minus fields because of this little issue. Then we noticed the following thing that I'm going to tell you, that we could just put everything back together with a simple definition. And that's because really the only difference is this sign. So if we want a kind of convenient notation, we can do the following. So far, if you like, we have this guy with Pl's not equal to 0, but the Pl minus should be greater than 0. So we can make use of kind of negative labels by the following trick. Just define a composite field that looks like this where I put the minuses and the pluses back together. If I have a positive Pl minus here, then that's giving the pluses. If I have a negative Pl minus here, then that's giving the minuses because of what I just told you. We don't just think about this guy as having plus Pl minus. We allow it to have negative. And we take the negative Pl minuses as a way of encoding what's going on with this guy. So we have different possibilities for M bar dot Pl. And if you like, what the plus guy is doing is destroying particles. And so let's say this is for the-- if I think about Cn of-- let me make my chart like this. If I think about what Cn of Pl is doing, it's destroying particles and the antiparticles are created. But the antiparticles are going to be created for the opposite sign of the Pl. And then if I think about what the barred field would be doing with this notation, if the M bar dot Pl is greater than 0, then the particles are created. So this guy is really just the field we were talking about before if Pl is greater than 0. And if Pl is less than 0, it just repeats the whole story for the antiparticles. OK, so I could write everything out in terms of this plus and minus field. But if I just make this simple definition, I could put everything back together and just work in terms of one field, which kind of does what we're used to. And it's just a sign convention to remember for this Pl. And that sign convention is also easy to remember because the Pl is really following the fermion flow. If you think about the fermion, then if you think about the physical momentum of that fermion, then the physical momentum follows the fermion in the case of the fermion. And it's opposite to the fermion in the case of the antifermion. So here's a fermion coming into some vertex. There's a antiparticle coming into the vertex. And the Pl goes this way. It follows the fermion flow, whereas the physical momentum is going this way. And that's the sign we're just keeping track of. And so the P labels are always following the arrows of the fermion lines. They're always along the fermion number with the conventions we've set up. OK, so with that little piece of notation, we take into account the antiparticles as well. There's no real additional complications since any way you draw a picture it's completely clear what's an antiparticle and what's a particle. And it just follows your intuition. And furthermore, because the only difference in the mode decomposition was that sign and we've now accounted for it by just making the labels negative, we have this equation, too, for both particles and antiparticles. So we get the opposite phase here because this guy has the opposite sign. So there's really no additional complications from adding the antiparticles once we set up this notation. OK, any questions so far? What about collinear gluons? So the gluon field is Hermitian, the original one. And what does that translate into? Well, it translates into the B's and the A's, right? They're both appearing in the decomposition of the field. There's no B's. There are just A's. And if you work out, using the type of body decomposition that we're doing and using the type of notation that we're doing, what that means, it means that the star of this guy just flips the sign. So that's what Hermiticity of the original field becomes for this notation. And again, because of the theta functions that show up in the mode decomposition, you can think of associating ql minus greater than 0 as sort of destruction and then ql minus less than 0 as creation. If you want, although it's not absolutely necessary, you could adopt this convention. And then that would just be putting a particular direction to your gluons. And everything that we said before goes through. And in particular, if we think of decomposing sort of the hatted version of the fields, then this-- just like before. And we can just also, again as before, if we like, adopt the notation where we suppress the labels if we don't need them. So the gluons really, in some sense, it's just simpler because we don't have to worry about the particles and the antiparticles because of their Hermiticity. It's sort of they're all the same thing. And for most purposes, there's no sense in even keeping track of this convention. So that's why I went through it quickly because the convention is always kind of obvious from the context. If you're really talking about production of a real gluon or annihilation of a real gluon, then you can keep track of the signs that way. The fermions are a little different. So because any starred field is just the negative of this guy, we could always write everything in terms of fields that have no daggers for the gluons. But for the fermions, some of the fields are barred or daggered. OK, so what would be the form of having a general label operator act on some combination of daggered fields and some combination of undaggered fields where the gluons are always undaggered? Well, it just gives the sum of these momenta. And the way it does it-- so sum of the label of momenta. So I'm thinking of all these denominators as label momenta. I'll maybe not put the l. And the way that the convention works is that we get a minus sign for the daggered guys. That's our sign convention. And the reason for that sign convention is effectively that, if you think about e to the minus i curly p dot x on sum 5 p1 and then you take the dagger of that, then it becomes 5 p1 dagger of e top the plus i kind of p dagger dot x. And basically, I'm saying that I want to be able to write that like this. And that convention is useful. So if I want to act forward on this field, which is a daggered field, then I'm going to take the opposite side. And if I have a daggered p, then it would take the plus sign. And that convention is the useful one for keeping track of momentum conservation because then every field has the same convention. We just have e to the minus i p dot x for every single field. All right, so if we started with some partial derivative-- I mean some regular derivative. And it was acting on one of our hatted fields. So if I write it out in terms of the labels-- put a Pl on it if you like-- we'd have that. And so there's two contributions from this derivative. It either hits the phase or it hits the field. And so we can write it like-- if we go over to the label operator, we can write it that this partial derivative is the sum of two derivatives, which then we can take the sum over Pl inside and write it as-- so every derivative for the purposes of this exercise is to convince you of the sanity of the notation. Every regular derivative acting on one of these hatted fields where we hadn't yet decomposed momenta just becomes a sum of these two derivatives. And that's just like the fact that we split the full momentum into the sum of the label in the residual piece. OK, this is the field version of P mu equals Pl mu plus P residual mu. OK, but the advantage of this notation is that we can then drop the partials if they're smaller. That's the whole reason for this setup. So now, we can expand because we've split up both the fields and the momentum. So we're able to expand. So let's go back to the thing we wanted to expand. So we had this guy. And we wanted to expand it. And now, we know how to expand it. We've set up a nice way of doing the multipole expansion by introducing some additional notation. So because of what I just said and because of what I said last night for the gauge field, we have a very simple decomposition for the covariant derivative. It's the sum of a collinear kind of piece and an ultra soft kind of piece. So this P mu plus i partial mu is just this. And the split of the gauge field into these two pieces is what we talked about last time. And that split of the gauge field actually led to some things that we didn't talk about that we'll talk about later. So if we now look at the components that are appearing in here, we can figure out what to keep in each case. And it's pretty simple. Well, for the plus, there's nothing to drop. We just keep everything. And there's also only one type of derivative, so we just keep this. This is exact. All terms here are order lambda squared. For the perp, we just keep the collinear parts. So if I write out the other part, just for completeness we can drop those terms relative to these terms. Because these terms here are lambda, and these terms here are order lambda squared. So we drop these guys from the leading order and then, similarly, for the minus. So here, this guy is order lambda squared. And this guy is order lambda 0. So we drop this guy. OK. So now, just plug those things into our Lagrangian. And then we get the leading order Lagrangian. All of these phase factors can be encoded in an overall phase factor. We get the full n dot D derivative. But for the collinear parts, we only get these piece, this piece, and the piece over there. And so you can think of that as a collinear covariant derivative. No. So sort of in the obvious notation, we take the two pieces that are collinear and put them into a collinear D and likewise in the n bar direction. OK, so this Lagrangian here is the leading order Lagrangian with this additional notation. And if you like, we have this kind of D that has both pieces in it for that n dot direction. And then for the other two, we just have the collinear pieces. And that's the leading order action. Where we've carried out the multipole expansion, after we introduced notation, it became something trivial. So the power of the notation is that, once you've swallowed it, everything else is simple. If I want to go to higher order, I just work out what these terms are. Then I should also worry about these terms. And we'll do that later on. OK, so what can we note about this? Well, one thing that we worked out earlier was that this should be of order lambda. And that actually implies that, our new field, we were just pulling out phases. And that doesn't change the overall power counting. So our new field is of order lambda as well. And likewise, before we sort of had a power counting for the measure, which you can think of in our new notation as a power counting for this thing, and that just scales like lambda to the minus 4 as before. So all the power counting that we were doing even before we sort of figured out what the right field to use, that all carries through. And that's why I did it earlier. OK, so what's going on here, which I didn't really spell out, but I can spell out now, is since this phase is order 1, the power counting from momenta induces a power counting on the x. And it's kind of the larger momentum that dominates. So x minus scales like 1 over p plus x plus-- remember in the dot product, it's always plus times minus, a confusing thing about light cone coordinates. And so if you count up the powers of lambda, you get minus 4 of them because you get 2 here, none here, and then 2 there. So that's all going through as before. And then if you ask, what's the order of the Lagrangian, the Lagrangian density here is order lambda to the 4. Because we have two fields, and then we have a lambda squared here and then a lambda here, a lambda 0 here, and a lambda there. And our power counting made that explicit. So the whole thing is lambda to the 4. This lambda to the 4 of the Lagrange density cancels the lambda to the minus 4 of the measure. And you get that, when you integrate, this is lambda to the 0. And that's what we wanted. That was our convention for the leading order term, OK? So the power counting, nothing really changes. We had a power counting for momenta. We just now have a power counting for all these objects that we're talking about. What about locality? Well, all the fields in this Lagrangian are at x. So it is local in residual momenta. We said that already when we were drawing Feynman diagrams. And that's just true of the action. And so what that means is the action is local at the smallest momentum scale in our problem. And we would have been surprised if we'd encountered anything different than that. The action is actually also local at the scale lambda. Even though we adopted this label convention just to separate the momenta, there's no inverse perp derivatives in this action. And so the perp derivatives appear in the numerator. And so it's actually local at that scale, too. And so if you like this, we've just written something that's in momentum space. And it's the momentum space version of locality at the scale lambda. And we've hidden all the momentum space so it even looks local. But if we were to write out the indices again, we would see that it's the momentum space version of locality. And so the only scale that it's non-local at is actually scale that we want to integrate out, the hard scale. And that's due to sort of the presence of this kind of 1 over M bar dot Pl type factor. So that we already sort of saw earlier when we were talking about building up the Wilson line, right? We said, well, there's some off-shell fields. Let's start integrating them out. And we got inverse powers of this M bar dot Pl, but that was at the hard scale. So we had no choice but to integrate them out. And you can think about this factor here. If it bothers you, you can think about it in the same way that it's not any worse than what we already did. OK, let's see if this Lagrangian does what we want. Let's look at propagators. If we have a collinear propagator with a collinear gluon, say, coming in, let's look at this diagram. These are fields in the n direction. This is some guy that's Pl Pr. And let me just take it to be Pr plus for convenience. Then if you look at this guy and you ask what that propagator is, it comes out exactly how we want it to come out. So there's nothing that gets dropped in the denominator in particular in the sense that all components of the momentum are showing up. We're still dropping residuals. But if we're just talking about collinear particles, then the residuals are kind of redundant. The purpose of the residual's momenta is when we're talking about both collinear and ultra soft particles. So the other case that we need to treat is when we have an ultra soft particle. We could, of course, have both of them at the same time, but let's treat this case. So we have Pl, Pl plus some k, which is purely residual. And now, this guy, because of the derivatives, these derivatives don't see the k. None of these derivatives see the k. So in the n bar in the perp, we don't see the k. It would look like that. If you like, if you want to make it kind of look more like the Lagrangian, convince yourself that this is the inverse, just divide by this factor. Divide by this factor, then it puts it under here. And that's exactly what this is doing if you ignore the Dirac structure. It's not playing a role really for the propagator. And same here, if you take this factor, divide it out, then you got the plus piece which is this derivative. And then you get perp squared over minus. And that's what happens if you put this down stairs, OK? For linear, all the components show up because we didn't drop anything. For ultra soft, only the plus shows up because it only changes the plus momentum of the collinear field. So the Lagrangian is smart. It knows how to deal with the fact that the propagator should be different for these two situations. And that was one of the things that we wanted, and we achieved it. So even though the Lagrangian doesn't know what's going to happen, it's smart enough to deal with any possible thing that could happen and give the right propagator. So we don't have to expand any further. The Lagrangian knows how to give the leading order propagator for any situation that we might be interested in at leading order, which are these two situations. What about Feynman rules for interactions? So that's also interesting. So if we have an ultra soft gluon, and let's say it has an index mu, there's only one term that comes in because it's only the n dot D that had an ultra soft field. All the other ultra soft fields got dropped. So there's only n dot A ultra soft in the Lcc0. And so there's only one term. Things are a little more complicated for the collinear. And this is kind of the price that we pay for setting up the notation that we did the way we did. So this guy here has all terms. Just by pulling a gluon field out of any one of the terms it can couple to any component of the gluon field just like in QCD, but our Feynman rule is a little more complicated because of the way we set things up. And in some ways, what's happened is that we've made the propagator simpler for the collinear fields at the expense of complicating the interaction. OK, so we couple to all components-- the perp component, the n component, and the n bar component if this is a collinear field. So these are all collinear fields. So one way of thinking about why that kind of had to happen is the following. What if we didn't have ultra softs? Imagine that we were going through this whole story, but we just had collinears. Then you wouldn't really need to go through this multipole expansion because there would never be two momentum of different size. And you could actually, just for the collinears, use QCD. The purpose of having this multipole expansion is really because we want to talk about both of these things at the same time. In fact, if you only had collinears, you just boost to the frame where they're not having this boosted scaling. It's really relative scalings that the effective theory is trying to encode, relative scaling between collinear and ultra soft or between, say, two collinears going off back to back. That's where the power of the effective theory is, in relative thing. So it's only when you have two things at the same time and you want to describe both of them that you get into the kind of complications we're talking about. If you just have one thing, then things are simple. You could just use QCD. But if you just had QCD for the collinear sector, you'd know that you have a very simple Feynman rule, igTA gamma mu, and a more complicated propagator because you'd have a P slash. Well, here we don't have any P slash. The propagator, if you write out the spin, is simply n slash over 2. But the P slash has to come back somewhere. And it's coming back because it gets encoded in the way we set things up in this vertex. So these factors of P slash here are just reproducing P slashes. Then in the full QCD context, it would be in the propagator. You can actually work out, if you start writing down Feynman diagrams, that that's exactly what's going on, OK? So that's kind of the price that we paid for sort of making a Lagrangian that dealt with both things at the same time and made this part really simple. That part became slightly more complicated. But once we know what I just told you, we can oftentimes, you know, carry out calculations here by making correspondences with the full QCD Lagrangian. Sometimes that's done anyway. They couple to all four components of the field. Actually, once you start learning how to deal with numerators like this, it's different than what you're used to, but it's just, in some sense, as complicated using this Lagrangian. It's just that you're writing things out in components, which you might not want to do, right? Why decompose things into components, which we're doing here? Why do that if you don't have sort of something that's projecting onto a component? That's, in some sense, the only complication. OK. And there's one other thing, of course, which is that the Lagrangian also had terms like this. So it's not simply just one gluon interaction. We could have, for example, two covariant derivatives, two of these perp covariant derivatives, for example. And there's a Feynman rule for that, too, which comes out of the action that we wrote down. OK, questions? AUDIENCE: So I guess this, I mean, the effective theory is reproducing the collinear [INAUDIBLE] divergences of QCD? IAIN STEWART: It is, yeah. AUDIENCE: So in [INAUDIBLE] QCD, the higher divergence cancel with interference [INAUDIBLE]?? IAIN STEWART: And that'll happen here, too. AUDIENCE: And you have to loops? Or it's happening like a leading order [INAUDIBLE]?? IAIN STEWART: No, you have to look at loops, too. So, yeah, there can be cancellations between real and virtual graphs. And there actually can be such cancellations separately for ultra soft gluons and for collinear gluons. So we'll talk about what you're talking about a little later on. But there is an important point that I'll emphasize since you ask the question. If you ask about the divergences, the infrared divergences of an ultra soft gluon, so the ultra soft gluon, if we go back to sort of where it was living, it was littered here. Collinear gluon was living here. And if you ask about divergences, what the divergences are coming from are momenta going to 0 or momentum going collinear. And this guy here that lives kind of here can go both collinear, and it can go soft. So the names collinear and soft, which correspond to the names collinear and soft divergences, are not a one-to-one correspondence. An ultra soft particle can actually have both collinear and soft divergences. The collinear particle can also actually have both types of divergences. And so the names are associated to where they live in momentum space, not the type of divergences that they're encoding. If I wanted to do something along the lines of setting up the field theory to just exactly correspond to the divergences, well, first of all, no one's ever figured out how to do that. But you could imagine how it would be different than what I'm doing. Because if I wanted to do that, then somehow my collinear field would sort of be sort of sensitive to angles. It should be like an angular variable because the divergences are coming from angles going to 0. And it's not an angular variable. It's just a regular field. So setting things up in this way is kind of how we usually think of an effective field theory. We're just talking about momenta small or big. And that's what we're doing, but it comes in the sense that you don't have a direct correspondence with the divergences. AUDIENCE: So you have more divergences [INAUDIBLE]?? IAIN STEWART: The way of saying it is that you have more divergences in a way, but they're the same divergences-- yeah. Yeah. So you don't have more divergences, but really what's happening is-- so if you look at this kind of picture, if you look at the kind of infrared divergences that you could have, so let's regulate them with an off-shellness. You could have infrared divergences that would be like that. But when you do some diagrams, there's kind of infrared divergences that could look like this, OK? I'm just telling you something that you have to trust me. And this is coming from an ultra soft scale. P squared over P minus, that's this hyperbola. That's the scale of this hyperbola. And the P squared here is the scale of this hyperbola. What the effective theory does is, any ultra soft diagram, that'll give all these logs. Any collinear diagram will give all these logs. You could say, well, P minus is a hard scale. So really the only the IR divergences are just P squared, but the effective theory is kind of dividing it up into these two types of logs. And that's actually going to be useful for us. We'll see that that allows us to do some things that are not so obvious how to do in the full theory. So, yes, it is kind of splitting and IR divergence in these two different types of terms, but that's actually going to be something that's useful. Because in some sense, there's two physical scales in the infrared. There's a sort of collinear hyperbola and the ultra soft hyperbola. And the effective theory is making that explicit. Other questions? All questions have been good so far. Keep asking them. AUDIENCE: Could you try using two different lambdas for the ultra soft and the collinear? IAIN STEWART: Yeah. So you could if there was a physical reason why you wanted to. And the thing is that, if the lambdas order each other, then there's no point. If there's a hierarchy, then what that would change would be that the n dot D should get expanded, too. But if the n dot D gets expanded, then kind of what is happening is you're really decoupling everything. AUDIENCE: [INAUDIBLE] IAIN STEWART: Yeah. And so you're kind of losing the dynamical connection that we have so far in this theory. So I think it's not so interesting, but it's worth thinking about. OK. We're going to erase this. OK. So now, that we have a leading order Lagrangian, it does everything that we want it to do. We can use it to calculate Feynman diagrams. It defines the leading order collinear sector in the effective theory. There's one more thing I want to do. And that's come back to the fact that I told you before, but now that we can actually make more explicit. And that fact was, when we were talking about currents and integrating that off-shell particles, we saw that you start with these guys. You start attaching them. And you end up with a Wilson line. I'd now like to make it obvious that that kind of manipulation always is true, that we can always get rid of this field and put it in terms of a Wilson line. And the notations that we've introduced are going to allow us to do that. So let's have a little aside on that fact. So if you are in momentum space, the Wilson line has an equation of motion, or a defining equation if you like, that's this. This defines the direction, that the Wilson line is long n bar dot n bar direction. And the x and the minus infinity are telling you what the end points are. So this is the defining equation. You can look back at Peskin if you want to see how that all works. And that's the defining equation in position space. But we've just advocated for a kind of momentum space notation. So what's the corresponding equation to momentum space? It's pretty simple. If we go over to our momentum space Wilson line, we just go over to our momentum space covariant derivative. And the defining equation is this. So the notation, in some sense, is pretty simple. We just have to switch objects. If we write out what that derivative is, then it's that, OK? So now, consider the following thing. Consider this combination, but let me think about it as an operator equation. So let me think about it as if there could be something to the right. If there's something to the right, than the P bar can act on that thing, too. So I get 0 if the P bar acts on this thing. But if the P bar acts through, then I get just P bar on the operator. So as an operator equation, I can trace the O. This is true for any O. And we have this operator equation. But I know something else I know that the Wilson line is at e to the i something. And actually, it's unitary. So W dagger W is 1. And again, in momentum space notation, that's kind of the obvious thing. W dagger W is 1. And that means I can multiply this equation here on the left or the right by W daggers. And so if I do that, let me multiply it by W dagger. And I can write i n bar dot Dn is Wn P bar Wn dagger. So the gauge field, g n bar dot An, because of gauge symmetry, it's always going to show up in the covariant derivative. So it's always going to show up in this combination. And this equation is telling me I can switch that for Wilson lines in this P bar if I want to. I could have also multiplied things the other way around and written it P bar. So sort of by way of completeness, I could have written that as well. It's also easy to convince yourself of the following. 1 over P bar is Wn dagger 1 over i n bar dot D W. And 1 over i n bar dot D-- I'll check that I got the daggers in the right place. But these two equations are true, kind of the inverse of those equations. So how do we check that? Well, just multiply this times this. You should get 1, right? And if I multiply, say, on the right here, then the W dagger here kills the W dagger. Then the P bar kills the P bar. And then the W kills the W dagger. So those are true. And so we can put this into the Lagrangian if we want. And we can do what I said, that we can actually, if we want to, get rid of the n bar dot A field for the collinears in terms of Wilson lines. And we'll use this from time to time when it's convenient. If we use the D Lagrangian, the Lagrangian becomes this thing. So that 1 over i n bar dot D becomes this thing. And in some sense, you might think that this is easier to expand if you know the expansion of the Wilson line. And we just have to let a 1 over P bar act on this full combination of fields. And it just gives the total n bar dot P momentum of that full combination of fields. I claimed this was true. And now, I show you in a kind of non-trivial case how, with these operator manipulations, we could just quickly go back and forth. And it's related to the fact that it appears in this combination. We'll talk more about gauge symmetry later on. I'm alluding to it from time to time, but we'll talk about it more precisely later on. We have to do this story for the collinear gluon Lagrangian. And much of this is just repeating what we've said for the quarks for the gluons. And once we know how to write down the covariant derivatives, then it's actually not so complicated. And there's just one little complication that we should be careful about, which I want to emphasize to you. And so we'll go through it. Let me write the Lagrangian in QCD in terms of traces. And let me use a general covariant gauge. So this is a general covariant gauge fixing term. That should be called the gauge parameter tau. Then we have some ghost fields as well. And of course, g mu nu is the usual g mu nu A TA. And that's my sign convention commutator of two derivatives with the pi over g. So we do the same steps to get SCET. And so really, if you think about writing all the g's in terms of D's, we already know how to split the D's in terms of ultra soft and collinear parts. And so that's what we do. Let me write out a D, which is kind of a leading order D. And we'll just set up some notation. And then I can just basically write down the answer. So let's set up a curly D that has all the sort of pieces that will survive at leading order. So all the pieces that survived were the collinear pieces and perp in n bar and then, in the n direction, both pieces. And effectively, I'm just changing all the regular Roman D's in my gluon action into these curly D's. And that's almost the entire story. So from a kind of power counting perspective, once you take into account that, when you dot things, n's are always dotting with the n bars, so it's this times this. So this is lambda squared. This is order lambda 0. This guy dots into himself. So you get lambda squared there as well. So for power counting purposes, when you start squaring things, they're all going to be homogeneous with this setup. So basically, that's what I'm saying here. And then there's one complication, and that's this partial derivative. Well, there's two partial derivatives. There's one there, and there's one there. So this is in the gauge fixing term. These terms are both related to the gauge fixing. And then those terms, actually, I want to make them covariant. So let me define one more thing. I want to make them covariant, but I want to make them only covariant under the ultra soft part of the action. So here is just pulling out the pieces that would have the ultra soft field making a derivative that I'll call curly D ultra soft. So one way of thinking about that is that it would be like background field gauge. In background field gauge, you'd make this partial view into the D mu of the background involving the background field. And that would be a background field gauge. So from the point of view of the ultra soft field being a background field, it's very natural to have a replacement that looks like this. And that's one way of arguing. Really what's going on in the effective theory is that this is general covariant gauge for A n. That's what we want it to be for A n mu. This should be the gauge fixing term for the collinear gluons. And it shouldn't be breaking any gauge symmetry of ultra soft gluons. So we'd actually like the Lagrangian to both have gauge symmetry for collinear gluons and a separate gauge symmetry for the ultra soft gluons. Since this guy is supposed to be gauge fixing for the collinear gluons, we don't want it to be doing anything to the ultra soft gluons. And that's accomplished if we make this replacement that i partial goes to i curly D, all right? So we could write out what Lcg0 is, but it's just these two replacements. Since we're out of time, I'll just say that and, you know, sort of write out the fields and the notation that we developed, OK? So we'll talk more next time. What we'll do next time is we'll talk about symmetries. We'll finally talk about gauge symmetry. And we'll talk about other symmetries. And in particular, symmetries are going to be important for understanding how much of this story that I've told you really carries through once you start doing loops. Symmetries protect you from certain loop corrections. Gauge symmetry, for example, connects terms to all orders in perturbation theory. And we'll see how those things kind of work next time. Really, the story I've told you so far is all tree level. We haven't done any loops. One way of handling what loops can do is by looking at symmetries and saying, once we impose those symmetries, what's the most general possible things that loops could do. And we'll do that next time.
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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: A quick review of what we talked about last time. So the first thing we did last time was to discuss the age of universe, considering so far only at this point, flat matter-dominated universes where the scale factor goes like t to the 2/3. And we were easily able to see that the age of such a universe was 2/3 times h inverse. And we did discuss what happens if you plug-in numbers into that formula using the best current value of h, the value obtained by the Planck team last March. The age turns out to be about 9.5 to 9.9 billion years. And that can't be the real age of the universe, we think, because there are stars that are older than that. The age of the stars seems to indicate that the university should be at least, according to one paper I cited last time, 11.2 billion years old, and this is younger. So the conclusion is that our universe is not a flat, moderate, matter-dominated universe. We do in fact have good evidence that the universe is very nearly flat, so it's the matter-dominated part that has to fail, and it does fail. We also have good evidence that the universe is dominated today by dark energy, which we'll be talking about later. But one of the pieces of evidence for this dark energy is this age calculation. The age calculation just does not work unless you assume that the universe has a significant component of this dark energy, which we'll be discussing later. We then talked about the big bang singularity, which is an important part of understanding, when you talk about the age, what exactly you mean by the age, age since when. And the point that I tried to make there is that the big bang singularity, which gives us mathematically the statement that the scale factor at some time which we call zero was equal to zero, and if you put back that formula into other formulas, you discover that the mass density, for example, was infinite at this magical time that we call zero. That singularity is certainly part of our mathematical model and doesn't go away even when we make changes in the mathematical model. But we don't really know if it's an actual feature of our universe, because there's certainly no reason to trust this mathematical model all the way back to t equals zero, where the densities become infinite. We know a lot about matter and we think we can predict how matter will behave, even the temperatures and energy densities somewhat beyond what we measure in laboratory. But we don't think we can't necessarily extrapolate all the way to infinite matter density. So these equations do break down when you get very close to t equals zero and nobody really knows exactly what should be said about t equals zero. When we talk about the age, we're really talking about the age since the extrapolated time at which a would have been zero in this model, but we don't really know that it ever actually was zero. We next discussed the concept of the horizon distance. If the universe, at least the universe as we know it, we want to be agnostic about what happened before t equals zero, but certainly the universe as we know it, really began at t equals zero in the sense that that's when structure and complicated things started to develop. So since t equals zero there's been only a finite amount of time elapsed. And since light travels at a finite speed, that means that light could only have traveled some finite distance since the big bang, since t equals zero. And that means that there's some object which is the furthest possible object that we could see, and any object further than that would be in a situation where light from that object would not yet have had time to reach us. And that leads to this notion of our horizon distance, where the definition of the horizon distance is that it is defined to be the present distance of the most distant objects which we are capable of seeing, limited only by the speed of light. And once able to calculate that, and in particular for the model that we so far understood at this point, the matter-dominated flat universe, the horizon distance turned out to be three times c t. Tt Now remember, if the universe were just static and appeared at time t ago, then the horizon distance would just be c t, the distance that light could travel in time t. What makes it larger is the fact that the universe is expanding, and that means that everything was closer together in the early time and light could make more progress at the early time, and then these objects have since moved out to much larger distances. So that allows the horizon distance to be larger than c t, and in this particular model, it's three times c t. Next we began a calculation, which is what we're going to pick up to continue on now. We were calculating how to extend our understanding of a of t, the behavior of the scale factor, away from the flat case, to ultimately discuss the two other cases, the open and closed universe. And we decided on a flip of a coin-- I promise you I flipped a coin at some point-- to start with the closed universe. We could have done either one first. The equation that we start with is basically the same, it's the sine of k that makes the difference. This is the so-called Friedmann equation, and for a closed universe k is positive. This is the evolution equation, but it has to be coupled with an equation that describes how rho behaves with time. And for a matter-dominated universe, rho is just representing non-relativistic matter, which is nothing but spread as the universe expands. And the spreading gives a factor of one over a cubed as the volume grows as a cubed. And that means that rho times a cubed is a constant, and that expresses everything that there is to know about how rho behaves with time. OK, then after writing these equations, we said that things will simplify a little bit, not a lot, but a little bit if we redefine variables basically to incorporate all the constants that appear in these equations into one overall constant. And we decided, or I claimed, that a good way to do that, an economical way to do that, is to define things so that the variables all have units which are easily understood. And in this case the units of length can describe everything that we need, so we chose to express everything in terms of variables that have units of length. So the scale factor itself is units of meters per notch, and that's not a length. And notches we'd like to get rid of because we know they're un-physical. That is, there's no standard for what the notch should be. So if we divide a of t by the square root of k, the notches disappear, and we get something which just has the units of meters, or units of length, and I call that a twiddle of t. Similarly, but more obviously, t can be turned into a length by just multiplying by c, the speed of light. So I defined a variable t tilde which is just c times t. So both of these new variables, with the tildes, have units of length. And then the Friedmann equation can be rewritten, just reshuffling things according to these new definitions, in this way where all the constants are lumped into this variable alpha, where alpha is this complicated expression, which absorbs all the constants from the earlier equation. Alpha also has units of length, and even though it has a rho times a tilde cubed in it, and both rho and a tilde each depend on time, the product of rho times a tilde cubed is a constant because rho times a cubed is a constant and a tilde differs only by the square root of k, which is also a constant. So alpha is a constant even though it has time dependent factors. The time dependence of those two factors cancel each other out to give something which is time independent and can be evaluated any old time. So this now is our equation, and we proceeded to manipulate it. So the first thing we did was to take it's square root, and to rearrange it so that d t tilde appeared on one side, and everything else was on the other side. And everything else depends on a tilde but not explicitly on time. So this completely separates everything that depends on t tilde on the left, and everything that depends on a tilde on the right. And now we can just integrate both sides of that equation, and and they point to that, that there are basically two ways of proceeding here, one of which we already did when we did the flat case. When we did the flat case, we integrated both sides as indefinite integrals. And when you carry out an indefinite integration, you get a constant of integration, which then becomes a constant in your solution. And in that case we discovered that the constant really just shifted the origin of time. And since we had not said anything previously that in any way determined the origin of time, we used that constant to arrange the origin of time so that a of zero would equal zero, and that eliminated the constant. Just for variety, I am going to do it another way this time. Instead of doing an indefinite integral, I will do a definite integral. And if you do a definite integral, you have to make sure you're integrating both sides over the same range, or at least corresponding ranges. We have different names for the variables, but the range of integration of the two sides has to match in order to maintain the equality between the two sides. So we're going to integrate the left side from zero to some final time. And I'm going to call the final time t tilde variable sub f, for f to stand for final. And there's no real final time here. It could be any time, it's just the final time for the integration, and in the end discover then how things behave at time t tilde sub f. And then once we've figured that out we can drop the effort. It will be a formula that will be valid for any time. So the left hand side is integrated from zero to time t tilde f, the right hand side has to be integrated over the corresponding time interval. And we would like to define a tilde and the origin of time so that a tilde equals zero at time zero, the same convention we used for the other case. The standard convention in cosmology, t equals zero, is the instance of the Big Bang. And the instance of the Big Bang is the time at which the scale factor vanished. So that means when we have a lower limit of integration of t tilde equals zero on the left hand side, we should have a lower limit of integration of a tilde equals zero on the right hand side. And similarly the upper limits of integration should correspond to each other. So the upper limit of integration on the left hand side was t tilde sub f, really just an arbitrary time that we designated by the subscript f. So the right hand side the limit should be the value of a tilde at that time. And I'll call that a tilde sub f. So a tilde sub f is just defined to be the value of a tilde at the time t tilde sub f. And in this way the limits of integration on the two sides correspond and now we can integrate them and we don't need any new integration constants. These definite integrals have definite values, which is why they're called definite integrals, I suppose. OK any questions about that, because that's where we're ready to take off and start doing new material. Yes. AUDIENCE: I have a general question. So regarding the issue of how we're not really sure how to extrapolate up to t equals zero, this is just kind of a general question. I was wondering how we're kind of assuming throughout that time kind of flows uniformly, and so I've heard about something like gravitational time dilation. So at the beginning especially when there's such a high density of matter or radiation, then wouldn't that affect how, I guess, time flows? PROFESSOR: OK good question, good question. The question was, does things like general realistic time dilation affect how time flows, and are we perhaps being overly simplistic and assuming that time just flows smoothly from time zero onward. The answer to that is that general relativity does predict an extra time dilation, which in fact is built into the Doppler shift calculation that we already did. And there are other instances where similar things happen. If a photon travels from the floor of this room to the ceiling of this room, there's a small Doppler shift, a small shift in the timing. And you could see it in principle with clocks as well. If you had a clock on the floor, and a clock on the ceiling, they would not run at quite the same rate. But to talk about time dilation, you always have to have two clocks to compare. In the case of the universe, we have this thing that we call cosmic time, which can be measured on any clock. The homogeneity assumption implies that all clocks will do the same thing, so so the issue of time dilation really does not arise. Our definition of cosmic time defines time that is the time variable that we will use. And when we say it flows uniformly from zero up to the present time, that word uniformly sounds sensible. But if you think about it, I don't know what it means. So I don't even know how to ask if it's really uniform or not. It's certainly true that our time variable evolves from zero up to some final time, but smoothly or uniformly is not really a question we can ask until one has some other clock that one can compare it to. PROFESSOR: Yes. AUDIENCE: So you said at the beginning that something like an infinite density is just an effect of our equation. Aren't a lot of theories like inflation come out of assuming that the universe had gotten some infinite, dense, small area? So what effect does this assumption that may or may not be correct have on theories? PROFESSOR: OK, the question is, if we are not sure we should believe the t equals zero singularity, how does that affect other theories like inflation, which are based on extrapolating backwards to very early times. And there is an answer to that. The answer may or may not sound sensible, and it may or may not be sensible. It's hard to know for sure. But one can be more detailed and ask how far back do we think we can trust our equations? And nobody really knows the answer to that of course, that's part of the uncertainty here. But a plausible answer which is kind of the working hypothesis for many of us, is that the only obstacle to extrapolating backwards is our lack of knowledge of the quantum effects of gravity, and therefore the quantum effects of what space time looks like. And we can estimate where that sets in. And it's at a time called the Planck time, which is about 10 to the minus 43 seconds, and inflation, which we'll see later as sort of a natural timescale of about probably 10 to the minus 37 seconds. So although that's incredibly small, it's actually incredibly large compared to 10 to the minus 43 seconds. So we think there is at least a basis for believing that things like our discussions of inflation which we'll be talking about later, are valid even though we don't think we can extrapolate back to t equals zero. Yes. AUDIENCE: I have a question about use of the definite integrals. PROFESSOR: Yes. AUDIENCE: So we have a twiddle defined as a over square root of k. And we noticed in our equation that a goes to zero as t goes to zero, so a twiddle also goes to zero. How do we know then that that definite integral is convergent, because then we have zero over zero competing case. [INAUDIBLE] PROFESSOR: Let me think. Yeah, we'll see, I think is the answer. How do we know it's convergent? Well, we're going to actually do the integral, and then the integral does converge. You are right. The integrand does become zero over zero, but that in fact means the integrand is some finite number actually. Both numerator and denominator go to zero, I mean let me think about this. I guess a tilde squared becomes negligible, so the denominator goes like one over the square root of a tilde. So in fact the numerator divided by the denominator goes like the square root of a tilde as time goes to zero. Because you have a square root in the denominator, the a tilde squared becomes negligible, so it's manifestly convergent. The integrand does not even blow up, even though a tilde does go to zero. It's certainly worth looking at, you're right. One should always check to make sure integrals are convergent. But since we will actually be explicitly doing the interval, if it was divergent we would get a divergent answer, and we will not as you'll see in a couple minutes. Any other questions? OK, so in that case, to the blackboard. OK, so I will write on the blackboard the same equation we have up there, so we can continue. t twiddle f is equal to the integral from zero to a twiddle sub f, a twiddle d a twiddle over the square root of two alpha a twiddle minus a twiddle squared. OK, so what we'd like to do now is to carry out the integral on the right hand side. Ideally what we'd like to do is to carry out the integral on the right hand side and get some function of a tilde f and then we'd like to convert that function to be able to write a tilde f as a function of time. We actually won't quite be able to do that. We'll end up with what's called the parametric solution, and you'll see how that arises and what that means. I don't need to try to describe exactly in advance what that means. Doing the integral can be done by some tricks, some substitutions. And the first substitution is based on completing the square in the denominator, and that motivates the substitution that we will make. So we can rewrite this just by doing some algebra on the denominator, which is called completing the square for reasons that you'll see when I write down what it is. The numerator will stay a tilde d a tilde. And the denominator can be written as alpha squared minus a tilde minus alpha quantity squared. So completing the square just means to put the a tilde inside a perfect square. And the nice thing about this is that now we can shift our variable of integration, and turn this into just a single variable instead of the sum of the two. And then you have an expression, which is clearly simpler looking than this one, which had a tildes in both places. Now the variable integration will appear only there. And substitution which does that obviously enough, is we're going to let something, we can choose whatever we want, and then once I call it x, we're going to let x equal a tilde minus alpha. And we're just going to rewrite that integral in terms of x. So what we get when we do that is t sub f tilde is equal to the rewriting of that integral, and just substituting the a tilde becomes x plus alpha. So we get x plus alpha where a tilde was, and d a tilde becomes just d x. And the denominator, which was our motivation for making the substitution in the first place, becomes just alpha squared minus x squared. So this is perfectly straight forward. The next step which is important, is to get the limits of integration right, because with this definite integral method we really have to make sure that our limits of integration are correct. So straightforward to do that, the lower limit of integration was a tilde equals zero, and if a tilde equals zero, x is equal to minus alpha. So the lower limit of integration expressed as a value of x is minus alpha. And the upper limit expressed as x was a tilde f, and that means x is equal to a tilde f minus alpha. So the limit here is a tilde sub f minus alpha for the upper limit of integration. OK, now this integral is still not easy, but it can be made easy by one more substitution. And the one more substitution is a trigonometric substitution to simplify the denominator. We can let x equal minus alpha cosine of theta, where theta is our new variable of integration. And then alpha squared minus x squared becomes alpha squared minus alpha squared cosine squared theta. And the alphas factor out and you have the square root of 1 minus cosine squared theta. 1 minus cosine squared theta is sine squared theta, which is a convenient thing to take the square root of, you just get sine theta. And everything else also simplifies. And the bottom line, which I will just give you, is that now we find that t sub s tilde is just the integral of 1 minus cosine theta d theta. That's it. Everything simplifies to that which is easy to integrate. We also have to keep track of our limits of integration. The limit of integration, the lower limit of integration, is where x equals minus alpha. And if x equals minus alpha, that means cosine theta equals 1. And cosine theta equals 1 means theta equals 0. So the limit of integration on theta is easy, it starts at 0. And the final value is really just a value that corresponds to the final value of x which is a twiddle f minus alpha. For now I'm just going to call it theta sub f. Things that we have just determined is the value of theta that goes with the upper limit there. We'll figure out in a second how to write it more explicitly. Let me first do the integral to just get that out of the way. Doing the integral, you get alpha times 1 minus cosine theta sub f. So this in fact becomes half of our solution. It expresses t sub f in terms of theta sub f. And now that we're done with the whole problem, I'm going to drop the subscript f. There's just some time that we're interested in which is called t and the value of a at that time will be called a, and the value of theta at that time will be called theta. So I'm just going to drop the subscript everywhere because we are now in a situation were the subscript is everywhere, so dropping everywhere does not lose any information. So one of our equations is going to be simply t is equal to alpha times 1 minus cosine theta. And another equation will come from figuring out what theta sub f really is, which I said comes from making sure that the upper limit of integration here corresponds to the upper limit of integration in the previous integral. So theta sub f, I had to determine this on another blackboard and then put the final equations together. AUDIENCE: [INAUDIBLE] PROFESSOR: Yes, that's right I don't want to drop twiddles. T twiddle, thank you, is equal to alpha times 1 minus cosine theta, and and then we have the final value of x. Xx x is equal to a tilde minus alpha. So the final value of x is equal to the final value of a tilde minus alpha. And the final value of x is also related to theta by this equation, which is what we're trying to get, how theta is related to the other variables. So this is equal to minus alpha cosine of theta sub f. And now we might want to, for example, solve this for a tilde sub f, which just involves looking at the right hand half of the equation, bringing this alpha to the other side making a plus alpha. So this implies that a tilde sub f is equal to alpha times 1 minus cosine theta sub f. So this equation now just says that theta sub f means what it should mean to give us a final limit of integration that corresponds to the final limit of integration on our original integral, which we called a tilde sub f. Yes. AUDIENCE: Sorry, perhaps I missed it. But when you are doing the integral of 1 minus cosine theta, how come-- PROFESSOR: Oh, I got it wrong. Good point. The integral of cosine theta d theta is sine theta, not cosine theta. AUDIENCE: [INAUDIBLE] PROFESSOR: Wait a minute. Hold on. Do I still have it wrong? AUDIENCE: [INAUDIBLE] PROFESSOR: OK hold on. There was an alpha missing here. That's part of the problem, coming from the original equation here. Let's see if I have this right now. This is copying from a long line altogether. So I got both factors wrong. And then wait a minute. I think this is right now. There's still a wrong [INAUDIBLE]? AUDIENCE: [INAUDIBLE] PROFESSOR: If I differentiate sine, I get cosine. So if I differentiate minus sine, I get minus cosine. So differentiating this, I get this. That should mean that integrating this I get that. I think I have it right. Sometimes I get things right. It's a surprise, but-- AUDIENCE: [INAUDIBLE] PROFESSOR: What? AUDIENCE: The last equation you wrote. PROFESSOR: The last equation needs to be changed, right. This was copied from that, right. Thanks for reminding me. So t tilde is equal to alpha times theta minus sine theta. OK, and now I was working out the relationship between theta and a tilde. And let me just remind you that all I did was make sure that the upper limits of integration correspond to each other. So I'm just basically rewriting the upper limit of integration in terms of the new variable each time when we change variables starting from a tilde going to a tilde to x, and from x to theta. And then I use the equality of these two to write a tilde in terms of theta. And these equations now hold with the subscript f present. When subscript f's appear everywhere we can just drop them and say that the time that we called t sub f now we're just going to call t. And now we could put together our two final results which are maybe right over here. We have ct. I'll eliminate my tildes altogether now. ct, which is t tilde, is equal to alpha times theta minus sine theta. And a divided by the square root of k, which is a tilde and its sub f, but we're dropping the sub f, is equal to alpha times 1 minus cosine theta. And this is as good as it gets. Ideally, it would be nice if we could solve the top equation for theta as a function of t, and plug that into the bottom equation, and then we would get a as a function of t. That's what ideally we would love to have. But there's no way to do that analytically. In principle of course, this equation can be inverted. You could do it numerically. For any particular value of t you could figure out what value of theta makes this work, and then plug that value of theta into here. But there's no analytic way that you can write theta as a function of t. It's not a soluble equation. So this is called a parametric solution in the sense that theta is a parameter. And as theta varies, both t and a vary in just the right way so that a is always related to t in the correct way to solve our original differential equation. That's what's meant by a parametric solution. We can also see from this equation, or maybe from thinking back about how things are defined in the first place, how theta varies over the lifetime of our model universe. Theta we discovered starts at 0. We discovered that when we wrote our integral over there. And we could also probably see it from here. We start our universe at a equals 0. And at a equals 0, we want 1 minus cosine theta to be 0 and theta equals 0 does that. So theta starts at 0, which corresponds to a equals 0, and it also corresponds putting theta equals 0 here to t equals 0. So we have arranged things the way we intended so that a equals 0 happens the same time t equals 0 happens. And then theta starts to grow. As theta grows, a increases, the universe gets bigger until theta reaches pi. And when theta reaches pi, cosine of pi is minus 1, you get a factor of 2 here, 2 alpha. That's as big as our universe gets. And then as theta continues beyond that, 1 minus cosine theta starts getting smaller again. So our universe reaches a maximum size when theta equals pi, and then starts to contract. And then by the time theta equals 2 pi, you are back to where you started from, a is again equal to 0. We have universe which starts at 0 size, goes to a maximum size, goes back to 0 size, giving a Big Crunch at the end. And that's the way this closed universe behaves. It turns out that these equations actually correspond to some simple geometry. It corresponds to a cycloid. And you may or may not remember what a cycloid is. There are the equations written on the screen, which are hopefully the same equations that I have on the blackboard. Can't always count on these things it turns out. But yeah, they are the same equations, that's healthy. And I have a diagram here which explains this cycloid correspondence. A cycloid is defined as what happens in this picture. Let me explain the picture. We have a disk shown in the upper left in its original position with a dot on the disk, which is initially at the bottom. And initially we're going to put this disk at the origin of our coordinate system to make things as simple as possible. And then we're going to imagine that this disk rolls without slipping to the right. And the path that this dot traces out, which is shown along that line, is a cycloid by definition. That's what defines a cycloid. It's the path that a point on a rolling disk evolved through. So what I would like to do is convince you that this geometric picture corresponds to those equations. Let me put the equations higher to make sure everybody can see them. And it's actually not so complicated once you figure out how to parse the pictures. So what I've drawn in the upper right is a blow up of this disk after it's rolled through some angle theta. And I even made it the same angle theta in the two cases. But this is just a bigger version of what's in the corner there showing the disk after it rolled a little bit. So after it's rolled through an angle theta, what we want to verify is that the horizontal and vertical coordinates here correspond to these two equations. And if they do, it means that we're tracing out the behavior of those two equations. So look first at the horizontal component. The horizontal component is the ct axis. So that should correspond to alpha times theta minus alpha times sine theta. And you can see that in the picture. We're talking about the horizontal component of the coordinates of this point P. And the point is that you can get to P starting at the origin. Now remember we're only looking at horizontal motion so we could start anywhere on that line. We could start at the origin, go to the right by alpha theta, and to the left by alpha times sine theta, and we get there. And that's exactly what this formula says. So if we just understand the alpha theta and the alpha sine theta of those two lines we have it made. So let's look first at what happens where this first arrow comes from, the alpha theta line. That's just the total distance that the point of contact has moved during the rolling process. And the claim is that as something rolls, really the definition of rolling without slipping, is that the arc length that is swept out by the rolling is the same as the length along the surface on which it's rolling. You could imagine, if you like, that as it rolls there's a tape measure that's wrapped around it that gets left on the ground as it rolls. And if you could picture that happening, the existence of that movie really guarantees that the length on the ground is the same as what the length was when the tape measure was wrapped around the cylinder. So the length that the point of contact has moved is just alpha times the angle through which the disk is rolled. So that explains the alpha theta label on that line. To get the alpha sine theta on the line above, that's the distance between the point P and a vertical line that goes through the center of the disk. And that's just simple trigonometry on this triangle. The radius of our circle is alpha. And then by simple trigonometry, this length is alpha times sine theta, which is what the label says. So to summarize what we've got here we can find the x-coordinate of the horizontal according to the point P by going to the right a distance alpha theta, and then back to the left the distance of minus alpha theta. And that gives us an x-coordinate, which is exactly the formula that appears in the ct formula. So ct works. The horizontal component of that dot is just where it should be to trace out the equations that describe the evolution of a closed universe. Similarly we can now look at the vertical components of that dot. Again it's most easily seen as the difference of two contributions. We're trying to reproduce this formula that says that's alpha minus alpha cosine theta. So if we start at vertical coordinate 0, which means on the x-axis, we can begin by going up to the center of the disk. And the disk has radius alpha, so that's going up the distance alpha. And then we go down the distance of this piece of the triangle, going from the center of the disk to the point which is parallel to the point P. And that again is just trigonometry on that triangle, and it's alpha cosine theta by simple trigonometry. So we can get to the elevation of point P by going up by alpha and down by alpha times cosine theta. And that's exactly that formula. So it works. The x and y components, the horizontal and vertical components of that dot, are exactly the two formulas here. So one of them can be thought of as the x-axis, and one of them can be thought of as the y-axis. And the rolling of the disk just traces out the evolution of our closed universe. So closed universes evolve like a cycloid. Any questions about that? OK, great. OK, Let me just mention that this angle theta is sometimes given a name. It's called the development angle of the universe or of the solution. And that is just intended to have the connotation that theta describes how developed the universe is and theta has a fixed scale. It always goes from 0 to 2 pi over the lifetime of this closed universe, no matter how big the closed universe might be. That brings me to my next question I want to mention. How many parameters do we have in this solution? The way we've written it, it looks like it can depend on both alpha and k because both of them appear in the answer. And k is positive for our closed universes. These formulas will not make sense if k were negative, square root of k appears there. We don't want anything to be imaginary. But k can have any value in principle and these equations would still be valid. So on the surface it appears like there's a two parameter class of closed universe solutions. But that's actually not true. Can somebody tell me why it's not true? Yes. AUDIENCE: [INAUDIBLE] PROFESSOR: Exactly. Yes, since k has units of 1 over notch squared, you can change k to anything you want by changing your definition of a notch. And there's nothing fixed about the definition of a notch. So k is an irrelevant parameter. If we change k we're just rescaling the same solution, and not actually creating a new solution. So there's really only one parameter class of solutions. One could, for example, fix k to always be 1, and then we'd have a one parameter class of solutions indicated by alpha. Alpha, unlike k, really does have a clear, physical meaning related to the behavior of the universe. And we could see what that is if we ask, what is the total lifetime of this universe from beginning to end, from Big Bang to Big Crunch. We can answer that by just looking at the ct equation. From Big Bang to Big Crunch, we know that theta evolves from 0 to pi back to 2 pi, which is the same as 0. So theta goes through one cycle of 2 pi during a lifetime of our model universe. As theta goes from 0 to 2 pi, sine theta starts at 0 and eventually comes back to 0 when theta equals 2 pi. But theta increases from 0 to 2 pi over one cycle. So over one cycle of our universe, ct increases by alpha times 2 pi. So that tells us what the total lifetime of the universe is, I'll call it t total. And we get it by just writing c times t total equals 2 pi times alpha. And I think I can do this one without making a mistake. t total is then 2 pi alpha divided by c. And we can even check our units there. Alpha has units of length, so length divided by c becomes a time, c being a velocity. So it has the right units. And that's the total lifetime of the universe. It's just determined by alpha. So alpha can be viewed as just the measure of the total lifetime of the universe, which can be anything for different sized closed universes. Alpha is also related to the maximum value of a over root k. And a has no fixed meaning, this is meters per notch. But a over root k does have units of meters. We haven't yet really seen what it means physically, because that's related to the geometry of the closed universe which we'll be discussing later. But in any case as a mathematical fact, we could always say that the maximum value of a over root k is determined by alpha. So a max over root k is equal to the maximum value of this expression. And this expression has a maximum value when theta equals pi, which gives cosine theta equals minus 1, which gives a 2 here. And that's as big as it ever gets, so that's just equal to 2 alpha. So alpha is related in a very clear way to the total lifetime of our universe, and is also related to a over root k, although we haven't really given a physical meaning to a over root k yet. But we know it has dimensions of meters. OK, the next calculation I want to do is to calculate the age of the universe as a function of measurable things. We learned for the flat matter dominated case that there was a simple answer to that. The age was just 2/3 times the inverse Hubble parameter. So what you do now is get the analogous formula here, it follows in principle immediately from our description of the evolution. But we have to do a fair number of substitutions before we can really see how to express the age in terms of things that we're interested in. The formula here tells us directly how to express the age of the universe. t is the age of the universe as a function of alpha and theta. But if you tell an astronomer to go out and measure alpha and theta so I could calculate the age, he says what in the world are alpha and theta. So what we'd like to do is to express the age in terms of things that astronomers know about. And the characterizations of the universe like this that an astronomer would know about would be the Hubble expansion rate, and some notion of the mass density. And the easiest way to talk about the mass density is in terms of omega, the fraction of the critical density that the actual mass density has. So our goal is going to be to manipulate these equations. All the information is already there. But our goal will be to manipulate these equations to be able to express the age t in terms of h and omega. OK, so first we need to remind ourselves what omega is. The critical density is defined as that density which makes the universe flat. And we've calculated that the critical density is equal to 3 h squared over 8 pi times newtons constant, capital G. We can then write the mass density rho as omega times the critical density, which is just the definition of omega. Omega is rho divided by rho c, the actual mass density divided by the critical density. And putting what rho c is, we can express rho as 3h squared omega divided by 8 pi G. And being very pedantic, I'm just going to rewrite that in the form that we're actually going to use it by multiplying through. 8 pi over 3 G rho, taking these factors and bringing them to the other side, becomes just equal to h squared times omega. And you might recognize this particular combination as what appears in the Friedmann equation. The Friedmann equation told us that a dot over a squared is equal to 8 pi over 3 G rho, minus kc squared over a squared. And in order to get the substitutions that I want, I'm going to just rewrite this putting h squared for a dot over a squared. 8 pi over 3 G rho we said we could write as 8 squared times omega. And then we have minus kc squared over a squared. Note that we really have here a tilde squared. This is a squared divided by k if I put them together. So this term can be written as minus c squared over a tilde squared. And this accomplishes one of our goals. It allows us to express a tilde in terms of the quantities that we want in our answers, h and omega. And if we can do the same for theta, we have everything we need to express the age. So the implication here is that a tilde squared is equal to c squared divided by h squared times omega minus 1. To take the square root of that equation, to find out what a tilde is, we need to think a little bit about signs and things like that. A tilde is always positive. This is the scale factor divided by the square root of k, square root of k is positive, scale factors are always positive the way we defined it. So we can take the square root of that taking the positive square root of the right hand side. Omega is bigger than 1 for our case, so omega minus 1 is a positive number, 8 squared is a positive number. So taking the square root there offers no real problem. We can write a tilde is equal to c over, I guess this point I might not notice until later. h h can be positive or negative over the course of our calculation. We're going to talk about an expanding phase and a contracting phase. So when we take the square root of h squared, we want the positive number to give us a positive a tilde. So it's the positive square root that we want, which is the magnitude of h, not necessarily h. When h is positive, the magnitude of h is h. h could be negative though, and the magnitude of h is still positive, and then the square root of omega minus 1, which is always positive. So that's our formula for a tilde in terms of h and omega. OK, now we want to evaluate alpha, and I guess I did not keep the formula for alpha quite as long as we needed it. When we defined alpha in the first place, let me remind you how it was defined, 4 pi over 3 G rho a tilde cubed over c squared. And that can be evaluated using our formula for rho and we'll put that in for rho, and what we get is c over 2 times the magnitude of h using this formula for a tilde as well. And then omega over omega minus 1 to the 3/2 power. Just using this formula, and we know how to express rho from the right hand side, and we know how to express a tilde from this formula here. So everything's straightforward, and this is what we get. And now I want to use these to rewrite this equation, a over the square root of k is equal to alpha times 1 minus cosine theta. I'm going to replace a over root k by this formula. I'm going to replace alpha by that formula. So this implies, rewriting it, that c over the magnitude of h times the square root of omega minus 1 is equal to c over twice the magnitude of h times omega minus 1 to the 3/2 power times 1 minus cosine theta. And now we've had to survive some boring algebra. But notice that now most things cancel away here. We get a very simple relationship between theta and h and omega, actually just omega . In particular, when we solve that, we get simply that cosine theta is equal to two minus omega over omega. So theta is directly linked to omega. If you know omega, you know theta, If you know theta, you know omega, by that formula. And we can rewrite this the other way around by solving for omega if we want. Omega is equal to 2 over 1 plus cosine theta. Now we can look at this qualitatively to understand how omega is going to behave. At the very beginning cosine theta is equal to 1, theta is equal to 0, so omega is 2 over 1 plus 1, which is 1. So at very early times omega is driven to 1 even in a closed universe. As theta gets larger, cosine theta gets less than 1. This then becomes more than 1. So omega starts to grow as the universe starts to evolve. At the turning point, when the universe has reached its maximum size, theta is pi, cosine theta is minus 1, omega is infinite at the turning point. That may or may not be a surprise. But it you think about it, it's obvious. At the turning point h is 0, therefore the critical density is 0, but the actual density is not 0. And the only way the actual density can be 0 while the critical density is 0 is for omega to be infinite, so we should have expected that. And then the return trip, the collapsing phases, a mirror image of the expanding phase, omega goes from infinity at the turning point back to 1 at the moment of the Big Crunch. Yes. AUDIENCE: I'm confused with what the universe would look like when it gets to infinity. PROFESSOR: It would look static. It's temporarily static. It's reached a maximum size and is about to turn around and collapse, so h is 0. AUDIENCE: OK, but like with the density. PROFESSOR: Well we could calculate the density. It's some number which depends on alpha. AUDIENCE: OK, but that doesn't diverge or anything. PROFESSOR: It doesn't diverge or anything, no. It's just some finite density. At the turning point, it's just a finite density that can be expressed in terms of alpha. Sounds like a good homework problem. I think maybe I'll do that. OK, so this now allows us to express theta as a function of omega, which is what we wanted. If we express theta as a function of omega, and alpha as a function of omega and h, we have our answer. We have t expressed as a function of h and theta. So there are choices about how exactly to express omega in terms of theta that will involve inverse trigonometric functions. And anything that can be expressed in terms of an inverse cosine, can also be expressed as an inverse sine by doing a little bit of manipulations. We have our choice here about what we want to do. But in any case, the answer already has a factor of sine theta in it. So it's most useful to express theta as the inverse sine of what we get from that formula, to express the answer in what at least to me is the simplest form. So I'm going to manipulate this a little bit to find out what sine theta is in terms of omega, and then invert that to express theta in terms of omega. So sine theta is of course plus or minus the square root of 1 minus cosine squared theta. And cosine theta we know in terms of omega, so we can express this in terms of omega. And if you do that it's straightforward enough algebra. It's plus or minus, depending on which sign of the square root is relevant, and we'll talk about that in a minute. It's plus or minus the square root of 2 times omega minus 1 over omega. So we can express theta as the inverse sine of this expression. And now what I want to do is to make use of this to put into this expression using the value for alpha that we've already calculated and wrote over there. So we get our final answer, which is that t is equal to omega over twice the magnitude of the Hubble expansion rate times omega minus 1 to the 3/2 power times the arc sine or inverse sine of twice the square root of omega minus 1 over omega. That's just the theta that appears in this formula written in terms of sine theta or the inverse sine of the quantity that we determined was the sine of theta. And then I see here I should have a plus or minus because we haven't figured out our signs yet. Either is actually possible, depending on where we are in the evolution. And then minus or plus twice the square root of omega minus 1 over omega. That's a plus or minus, this is a minus or plus. And the reason I wrote one upstairs and one downstairs is that we don't yet really know how to evaluate theta, but the sign of this term is always going to be the negative of the sign of that term, this or that minus sign. So theta can be the inverse sine of this expression with either sign of the sine. Sorry for the puns. But whichever it is, it's the same on there as it is here but with a minus sign in front, that minus sign. OK, so this is our final formula for the age. But we still need to think a little bit about the s-i-g-n signs of these inverse functions that appear here, that and that, and straightforward if you just take it case by case. And in the notes we have a table which I'll put shortly on the screen. But let's start by just talking about the earliest phase where the universe is shortly after the Big Bang, so that the development angle is a small angle. We know that theta is going to go from 0 to 2 pi over the lifetime of our universe. So I now want to think of theta being small. And small means small compared to any number you think about. So when theta is small the sine theta is nearly theta. Both are positive. And in that case the sine theta being positive means this is the positive root in this equation, and therefore the positive root in that equation. So for early times this would be the plus sign, and that would mean that this would be the minus sign. Again the minus sign just coming from there and theta being positive. So for early times it's a plus and minus. And the arc sine is itself ambiguous. For early times the angle we know is going to be just a little bit bigger than 0. So that's the evaluation that we make of the arc sine function. Pi plus that would also give us the same sine. It would be another possible value for the arc sine. And of course 2 pi plus that would also be another possible root. So you have to know which root to take to know the right answer here because as an angle, 0 is equivalent to 2 pi, but as a time, 0 is not at all equivalent to 2 pi. So you do have to know the right one to take. We'll continue doing that on a case by case basis. Those are the equations, that's the formula for the age, and that's the formula for the age with a description of which roots to take for each case, which just comes out by following the evolution, we know that theta is going from 0 to 2 pi, and this last column, the inverse sine of the expression which means the expression that appears here. For the smallest angle is 0 to pi over 2. We can think of this actually as defining our columns. Theta starts at 0 so the time lengths between 0 and pi over 2, a time length between pi over 2, a time length between pi and 3 pi over 2, and a final time length between 3 pi over 2 and 2 pi. And the first two correspond to the expanding phase, second two correspond to the contracting phase. We can easily see what values of omega are relevant in those cases. Omega we said starts at 1 and gets larger, the borderline where the angle is pi over 2 one could just plug into these formulas and see amounts to omega equals 2. So that is a division line between these first two quadrants just calculated from the value of theta. Omega then goes to infinity as we said, comes backwards and back to 1 in the end. And in this column we just figure out which sign choice corresponds to getting the right value for omega and the angle that appears in the arc sine of our formula for the age, the formula there. So any one of these I claim is very obvious. Seeing the whole picture takes time because I think you really have to look at each case one at a time to make sure you understand it in detail. But if you understand the initial expanding phase that's what corresponds to our universe if our universe were closed. And the others are just as easy. You just have to take them one at a time I think. OK, we're going to end there. We will continue on Thursday.
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